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1300 | https://www.cs.cornell.edu/courses/cs422/2008sp/A6/Ellipse.pdf | Using the Ellipse to Fit and Enclose Data Points A First Look at Scientific Computing and Numerical Optimization Charles F. Van Loan Department of Computer Science Cornell University Problem 1: Ellipse Enclosing Suppose P a convex polygon with n vertices P1, . . . , Pn. P1, . . . , Pn. Find the smallest ellipse E that encloses P1, . . . , Pn.
What do we mean by “smallest”?
Problem 2: Ellipse Fitting Suppose P a convex polygon with n vertices P1, . . . , Pn. Find an ellipse E that is as close as possible to P1, . . . , Pn.
What do we mean by “close”?
Subtext Talking about the these problems is a nice way to introduce the field of scienific computing.
Problems 1 and 2 pose research issues, but we will keep it simple.
We can go quite far towards solving these problems using 1-D minimization and simple linear least squares.
Outline • Representation We consider three possibilities with the ellipse.
• Approximation We can measure the size of an ellipse by area or perimeter. The latter is more complicated and requires approximation.
• Dimension We use heuristics to reduce search space dimension in the enclosure problem. Sometimes it is better to be approximate and fast than fool-proof and slow.
• Distance We consider two ways to measure the distance between a point set and an ellipse, leading to a pair of radically different best-fit algorithms.
Part I. →Representation Approximation Dimension Distance Representation In computing, choosing the right representation can simplify your algorithmic life.
We have several choices when working with the ellipse: 1. The Conic Way 2. The Parametric Way 3. The Foci/String Way The Conic Way The set of points (x, y) that satisfy Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 defines an ellipse if B2 −4AC < 0.
This rules out hyperbolas like xy + 1 = 0 and 3x2 −2y2 + 1 = 0.
This rules out parabolas like x2 + y = 0 and −3y2 + x + 2 = 0.
To avoid degenerate ellipses like 3x2 + 4y2 + 1 = 0 we also require D2 4A + E2 4C −F > 0 The Conic Way (Cont’d) Without loss of generality we may assume that A = 1.
An ellipse is the set of points (x, y) that satisfy x2 + Bxy + Cy2 + Dx + Ey + F = 0 subject to the constraints B2 −4C < 0 and D2 4 + E2 4C −F > 0 An ellipse has five parameters.
The Parametric Way This is an ellipse with center (h, k) and semiaxes a and b: ⎛ ⎜ ⎝x −h a ⎞ ⎟ ⎠ 2 + ⎛ ⎜ ⎝y −k b ⎞ ⎟ ⎠ 2 = 1 i.e., the set of points (x(t), y(t)) where x(t) = h + a cos(t) y(t) = k + b sin(t) and 0 ≤t ≤2π.
Where is the fifth parameter?
The Parametric Way (Cont’d) The tilt of the ellipse is the fifth parameter.
Rotate the ellipse counter-clockwise by τ radians: x(t) = h + cos(τ) [ a cos(t) ] −sin(τ) [ b sin(t) ] y(t) = k + sin(τ) [ a cos(t) ] + cos(τ) [ b sin(t) ] In matrix-vector notation: ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ x(t) y(t) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦= ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ h k ⎤ ⎥ ⎥ ⎥ ⎥ ⎦+ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos(τ) −sin(τ) sin(τ) cos(τ) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ a cos(t) b sin(t) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ Example 1: No tilt and centered at (0,0) ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ x(t) y(t) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦= ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 5 cos(t) 3 sin(t) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ −8 −6 −4 −2 0 2 4 6 8 −6 −4 −2 0 2 4 6 a = 5, b = 3, (h,k) = (0,0), tau = 0 degrees focii Example 2. 30o tilt and centered at (0,0) ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ x(t) y(t) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦= ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos(30o) −sin(30o) sin(30o) cos(30o) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 5 cos(t) 3 sin(t) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ −8 −6 −4 −2 0 2 4 6 8 −6 −4 −2 0 2 4 6 a = 5, b = 3, (h,k) = (0,0), tau = 30 degrees focii Example 3. 30o tilt and centered at (2,1) ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ x(t) y(t) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦= ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 2 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦+ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos(30o) −sin(30o) sin(30o) cos(30o) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ 5 cos(t) 3 sin(t) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ −8 −6 −4 −2 0 2 4 6 8 −6 −4 −2 0 2 4 6 a = 5, b = 3, (h,k) = (2,1), tau = 30 degrees focii The Foci/String Way Suppose points F1 = (x1, y1) and F2 = (x2, y2) are given and that s is a positive number greater than the distance between them.
The set of points (x, y) that satisfy 5 (x −x1)2 + (y −y1)2 + 5 (x −x2)2 + (y −y2)2 = s defines an ellipse.
The points F1 and F2 are the foci of the ellipse.
The sum of the distances to the foci is a constant designated by s and from the “construction” point of view can be thought of as the “string length.” The Foci/String Way (Cont’d) s = 8.00 F1 F2 5.32 2.68 Ellipse Construction: Anchor a piece of string with length s at the two foci. With your pencil circumnavigate the foci always pushing out against the string.
Conversions: Conic − →Parametric If Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 specifies an ellipse and we define the matrices M0 = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ F D/2 E/2 D/2 A B/2 E/2 B/2 C ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ M = ⎡ ⎢ ⎢ ⎢ ⎣ A B/2 B/2 C ⎤ ⎥ ⎥ ⎥ ⎦ then a = 5 −det(M0)/(det(M)λ1) b = 5 −det(M0)/(det(M)λ2) h = (BE −2CD)/(4AC −B2) k = (BD −2AE)/(4AC −B2) τ = arccot((A −C)/B)/2 where 0 < λ1 ≤λ2 are the eigenvalues of M.
Conversions: Parametric − →Conic If c = cos(τ) and s = sin(τ), then the ellipse x(t) = h + c [ a cos(t) ] −s [ b sin(t) ] y(t) = k + c [ a cos(t) ] + c [ b sin(t) ] is also specified by Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 where A = (bc)2 + (as)2 B = −2cs(a2 −b2) C = (bs)2 + (ac)2 D = −2Ah −kB E = −2Ck −hB F = −(ab)2 + Ah2 + Bhk + Ck2 Conversions: Parametric − →Foci/String Let E be the ellipse x(t) = h + cos(τ) [ a cos(t) ] −sin(τ) [ b sin(t) ] y(t) = k + sin(τ) [ a cos(t) ] + cos(τ) [ b sin(t) ] If c = √ a2 −b2 then E has foci F1 = (h −cos(τ)c, k −sin(τ)c) F2 = (h + cos(τ)c, k + sin(τ)c) and string length s = 2a Conversions: Foci/String − →Parametric If F1 = (α1, β1), F2 = (α2, β2), and s defines an ellipse, then a = s/2 b = 5 (s/2)2 −((α1 −α2)2 + (β1 −β2)2)) h = (α1 + α2)/2 k = (β1 + β2)/2 τ = arctan((β2 −β1)/(α2 −α1)) Summary of the Representations Parametric: h , k, a, b, and τ.
E = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩(x, y) e e e e e e e e e e ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ x y ⎤ ⎥ ⎥ ⎥ ⎥ ⎦= ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ h k ⎤ ⎥ ⎥ ⎥ ⎥ ⎦+ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ cos(τ) −sin(τ) sin(τ) cos(τ) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ a cos(t) b sin(t) ⎤ ⎥ ⎥ ⎥ ⎥ ⎦, 0 ≤t ≤2π ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭ Conic: B, C, D, E, and F E = F (x, y) | x2 + Bxy + Cy2 + Dx + Ey + F = 0 k Foci/String: α1, β1, α2, β2, and s.
E = l (x, y) e e e e e 5 (x −α1)2 + (y −β1)2 + 5 (x −α2)2 + (y −β2)2 = s.
M Representation Part II. →Approximation Dimension Distance The Size of an Ellipse How big is an ellipse E with semiaxes a and b?
Two reasonable metrics: Area(E) = πab Perimeter(E) = $2π 0 5 (a sin(t))2 + (b cos(t))2) dt There is no simple closed-form expression for the perimeter of an ellipse.
To compute perimeter we must resort to approximation.
Some Formulas for Perimeter Approximation Perimeter(E) ≈ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ π (a + b) π(a + b) · 3 − √ 1 −h 2 π(a + b) · (1 + h/8)2 π(a + b) · (3 −√4 −h) π(a + b) · 64 −3h2 64 −16h π(a + b) · (1 + 3h 10 + √ 4 −3h) π(a + b) · 256 −48h −21h2 256 −112h + 3h2 π 5 2(a2 + b2) π > :2(a2 + b2) −(a −b)2 2 h = ⎛ ⎜ ⎝a −b a + b ⎞ ⎟ ⎠ 2 (1) (2) (3) (4) (5) (6) (7) (8) (9) Relative Error as a Function of Eccentricity 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10 −16 10 −14 10 −12 10 −10 10 −8 10 −6 10 −4 10 −2 10 0 e = Eccentricity Relative Error (1) (8) (3) (6) (5) e = > :1 − ⎛ ⎜ ⎝b a ⎞ ⎟ ⎠ 2 e = 0 ⇒circle, e = .99 ⇒cigarlike Perimeter via Quadrature Apply a quadrature rule to Perimeter(E) = 8 2π 0 5 (a sin(t))2 + (b cos(t))2) dt For example: function P = Perimeter(a,b,N) % Rectangle rule with N rectangles t = linspace(0,2pi,N+1); h = 2pi/N; P = hsum(sqrt((acos(t)).^2 + (bsin(t)).^2)); How do you chose N?
What is the error?
Efficiency and Accuracy Compared to a formula like π(a+b), the function Perimeter(a,b,N) is much more expensive to evaluate.
The relative error of Perimeter(a,b,N) is about O(1/N2).
Can we devise an approximation with an easily computed rigorous error bound?
Computable Error Bounds For given n, define inner and outer polygons by the points (a cos(kδ), b sin(kδ) for k = 0:n −1 and δ = 2π/n.
⎛ ⎜ ⎜ ⎝Inner Polygon Perimeter ⎞ ⎟ ⎟ ⎠≤Perimeter(E) ≤ ⎛ ⎜ ⎜ ⎝Outer Polygon Perimeter ⎞ ⎟ ⎟ ⎠ Relative Error as a Function of n e n = 10 n = 102 n = 103 n = 104 n = 105 n = 106 0.00 1.590 · 10−1 1.551 · 10−3 1.550 · 10−5 1.550 · 10−7 1.555 · 10−9 1.114 · 10−11 0.50 1.486 · 10−1 1.449 · 10−3 1.448 · 10−5 1.448 · 10−7 1.448 · 10−9 1.461 · 10−11 0.90 1.164 · 10−1 1.157 · 10−3 1.156 · 10−5 1.156 · 10−7 1.156 · 10−9 1.163 · 10−11 0.99 6.755 · 10−2 1.015 · 10−3 1.015 · 10−5 1.015 · 10−7 1.015 · 10−9 1.003 · 10−11 e = eccentrity = > :1 − ⎛ ⎜ ⎝b a ⎞ ⎟ ⎠ 2 Error mildly decreases with eccentricity.
Summary of the Area vs. Perimeter Issue The “inverse” of the enclosing ellipse problem is the problem of inscribing the largest possible polygon in an ellipse.
Max Area Vertices Max Perimeter Vertices The choice of objective function, Area(E) or Perimeter(E), matters. For the enclosing ellipse problem we will have to make a choice.
Representation Approximation Part III. →Dimension Distance Enclosing Data with an Ellipse Given a point set P = { (x1, y1), . . . , (xn, yn)}, minimize Area(E) subject to the constraint that E encloses P.
Everything that follows could be adapted if we used Perimeter(E).
Simplification: Convex Hull The ConvHull(P) is a subset of P which when connected in the right order define a convex polyon that encloses P.
The minimum enclosing ellipse for ConvHull(P) is the same as the minimum enclosing ellipse for P. This greatly reduces the “size” of the problem.
Checking Enclosure Is the point (x, y) inside the ellipse E?
Compute the distances to the foci F1 = (α1, β1) and F2 = (α2, β2) and compare the sum with the string length s.
In other words, if 5 (x −β1)2 + (y −β1)2 + 5 (x −α2)2 + (y −β2)2 ≤s then (x, y) is inside E.
The “Best” E given Foci F1 and F2 If F1 = (α1, β1) and F2 = (α2, β2) are fixed, then area is a function of the string length s. In particular, if d = 5 (α1 −α2)2 + (β1 −β2)2 then it can be shown that Area(E) = π s 4 √ s2 −d2 If E is to enclose P = { (x1, y1), . . . , (xn, yn)}, and have minimal area, we want the smallest possible s, i.e., s(F1, F2) = max 1 ≤i ≤n 5 (xi −α1)2 + (yi −β1)2 + 5 (xi −α2)2 + (yi −β2)2 The “Best” E given Foci F1 and F2 (Cont’d) Locate the point whose distance sum to the two foci is maximal. This deter-mines sopt and Eopt Area(Eopt) = π 4sopt > : ⎛ ⎝sopt 2 ⎞ ⎠ 2 − ⎛ ⎜ ⎝d 2 ⎞ ⎟ ⎠ 2 The “Best” E given Center (h, k) and Tilt τ Foci location depends on the space between them d: F1(d) = ⎛ ⎝h −d 2 cos(τ) , k −d 2 sin(τ) ⎞ ⎠ F2(d) = ⎛ ⎝h + d 2 cos(τ) , k + d 2 sin(τ) ⎞ ⎠ Optimum string length s(F1(d), F2(d)) is also a function of d The minimum area enclosing ellipse with center (h, k) and tilt τ, is defined by setting d = d∗where d∗minimizes fh,k,τ(d) = π s 4 √ s2 −d2 s = s(F1(d), F2(d)) The “Best” E given Center (h, k) and Tilt τ (Cont’d) For each choice of separation d along the tilt line, we get a different minimum enclosing ellipse. Finding the best d is a Golden Section Search problem.
The “Best” E given Center (h, k) and Tilt τ (Cont’d) Typical plot of the function fh,k,τ(d) = π s 4 √ s2 −d2 s = s(F1(d), F2(d)) for d ranging from 0 to max { 5 (x(i) −h)2 + (y(i) −k)2}: 0 0.5 1 1.5 6 6.5 7 7.5 8 8.5 Heuristic Choice for Center(h, k) and Tilt τ Assume that (xp, yp) and (xq, yq) are the two points in P that are furthest apart. (They specify the diameter of P.) Instead of looking for the optimum h, k, and τ we can set h = xp + xq 2 k = yp + yq 2 τ = arctan ⎛ ⎝yq −yp xq −xp ⎞ ⎠ and then complete the specification of an approximately optimum E by deter-mining d∗and s(F1(d∗), F2(d∗)) as above.
We’ll call this the hkτ-heuristic approach. Idea: the major axis tends to be along the line where the points are dispersed the most.
The “Best” E given Center (h, k) The optimizing d∗for the function fh,k,d(d) depends on the tilt parameter τ.
Denote this dependence by d∗(τ) .
The minimum-area enclosing ellipse with center (h, k) is defined by setting τ = τ∗where τ∗minimizes ˜ fh,k(τ) = π s 4 5 s2 −d∗(τ)2 s = s(F1(d∗(τ)), F2(d∗(τ))) The “Best” E given Center (h, k) (Cont’d) Typical plot of the function ˜ fh,k(τ) = π s 4 5 s2 −d∗(τ)2 s = s(F1(d∗(τ)), F2(d∗(τ))) across the interval 0 ≤τ ≤360o: 0 50 100 150 200 250 300 350 0 0.5 1 1.5 2 2.5 3 3.5 4 tau Area The “Best” E The optimizing τ∗for the function ˜ fh,k(τ) depends on the center coordinates h and k.
Denote this dependence by τ∗(h, k)) .
The minimum-area enclosing ellipse is defined by setting (h, k) = (h∗, k∗) where h∗and k∗minimize F(h, k) = ˜ fh,k(τ∗(h, k)) = π s 4 5 s2 −d∗(τ∗(h, k))2 where s = s(F1(d∗(τ∗(h, k))), F2(d∗(τ∗(h, k)))) The “Best” E Through these devices we have reduced to two the dimension of the search for the minimum area enclosing ellipse.
Representation Approximation Dimension Part IV. →Distance Approximating Data with an Ellipse We need to define the distance from a point set P = {P1, . . . , Pn} to an ellipse E.
Goodness-of-Fit: Conic Residual The Point set P: { (α1, β1), . . . , (αn, βn) } The Ellipse E: x2 + Bxy + Cy2 + Dx + Ey + F = 0 The distance from P to E: dist(P, E) = n 3 i=1 w α2 i + αiβiB + β2 i C + αiD + βiE + F W2 Sum the squares of what’s “left over” when you plug each (αi, βi) into the ellipse equation.
A Linear Least Squares Problem with Five Unknowns dist(P, E) = E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ α1β1 β2 1 α1 β1 1 α2β2 β2 2 α2 β2 1 α3β3 β2 3 α3 β3 1 .
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αn−1βn−1 β2 n−1 αn−1 βn−1 1 αnβn β2 n αn βn 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ B C D E F ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ + ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ α1 α2 α3 .
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αn−1 αn ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E E 2 2 Goodness-of-Fit: Point Proximity The Point set P: { (α1, β1), . . . , (αn, βn) } The Ellipse E: x(t) = h + cos(τ) [ a cos(t) ] −sin(τ) [ b sin(t) ] y(t) = k + sin(τ) [ a cos(t) ] + cos(τ) [ b sin(t) ] The distance from P to E: dist(P, E) = n 3 i=1 ((x(ti) −αi)2 + (y(ti) −βi)2 where (x(ti), y(ti)) is the closest point on E to (αi, βi), i = 1:n.
Distance from a Point to an Ellipse Let E be the ellipse x(t) = h + cos(τ) [ a cos(t) ] −sin(τ) [ b sin(t) ] y(t) = k + sin(τ) [ a cos(t) ] + cos(τ) [ b sin(t) ] To find the distance from point P = (α, β) to E define d(t) = 5 (α −x(t))2 + (β −y(t))2 and set dist(P, E) = min 0 ≤t ≤2π d(t) Note that d(t) is a function of a single variable.
“Drop” Perpendiculars −4 −3 −2 −1 0 1 2 3 4 5 6 −2 −1 0 1 2 3 4 5 6 The nearest point on the ellipse E is in the same “ellipse quadrant”.
Comparison: Conic Residual vs Point Proximity The two methods render different best-fitting ellipses.
Conic Residual method is much faster but it may render a hyperbola if the data is “bad”.
Conic Residual Method w/o Constraints 0 2 4 6 8 10 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3 There are fast ways to solve the conic residual least squares problem with the constraint C > B2/4 which forces the solution x2 + Bxy + Cy2 + Dx + Ey + F = 0 to define an ellipse.
Overall Conclusions • Representation The Parametric Representation if not “friendly” when you want to check if a point is inside an ellipse. The Conic representation led to a very simple algorithm for the best-fit problem.
• Approximation There are many ways to approximate the perimeter of an ellipse. Al-though we defined the size of an ellipse in terms of its easily-computed area, it would also be possible to work with perimeter.
• Dimension We use heuristics to reduce search space dimension in the enclosure problem.
• Distance We consider two ways to measure the distance between a point set and an ellipse, leading to a pair of radically different best-fit algorithms. |
1301 | https://courses.lumenlearning.com/calculus2/chapter/center-of-mass-and-moments/ | Center of Mass and Moments | Calculus II
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Calculus II
Module 2: Applications of Integration
Search for:
Center of Mass and Moments
Learning Outcomes
Find the center of mass of objects distributed along a line.
Locate the center of mass of a thin plate.
Use symmetry to help locate the centroid of a thin plate.
Center of Mass and Moments
Let’s begin by looking at the center of mass in a one-dimensional context. Consider a long, thin wire or rod of negligible mass resting on a fulcrum, as shown in Figure 1(a). Now suppose we place objects having masses m 1 m 1 and m 2 m 2 at distances d 1 d 1 and d 2 d 2 from the fulcrum, respectively, as shown in Figure 1(b).
Figure 1. (a) A thin rod rests on a fulcrum. (b) Masses are placed on the rod.
The most common real-life example of a system like this is a playground seesaw, or teeter-totter, with children of different weights sitting at different distances from the center. On a seesaw, if one child sits at each end, the heavier child sinks down and the lighter child is lifted into the air. If the heavier child slides in toward the center, though, the seesaw balances. Applying this concept to the masses on the rod, we note that the masses balance each other if and only if m 1 d 1=m 2 d 2.m 1 d 1=m 2 d 2.
In the seesaw example, we balanced the system by moving the masses (children) with respect to the fulcrum. However, we are really interested in systems in which the masses are not allowed to move, and instead we balance the system by moving the fulcrum. Suppose we have two point masses, m 1 m 1 and m 2,m 2, located on a number line at points x 1 x 1 and x 2,x 2, respectively (Figure 2). The center of mass, ¯¯¯x,x¯, is the point where the fulcrum should be placed to make the system balance.
Figure 2. The center of mass ¯¯¯x x¯ is the balance point of the system.
Thus, we have
m 1|x 1−¯¯¯x|=m 2|x 2−¯¯¯x|m 1(¯¯¯x−x 1)=m 2(x 2−¯¯¯x)m 1¯¯¯x−m 1 x 1=m 2 x 2−m 2¯¯¯x¯¯¯x(m 1+m 2)=m 1 x 1+m 2 x 2¯¯¯x=m 1 x 1+m 2 x 2 m 1+m 2.m 1|x 1−x¯|=m 2|x 2−x¯|m 1(x¯−x 1)=m 2(x 2−x¯)m 1 x¯−m 1 x 1=m 2 x 2−m 2 x¯x¯(m 1+m 2)=m 1 x 1+m 2 x 2 x¯=m 1 x 1+m 2 x 2 m 1+m 2.
The expression in the numerator, m 1 x 1+m 2 x 2,m 1 x 1+m 2 x 2, is called the first moment of the system with respect to the origin. If the context is clear, we often drop the word first and just refer to this expression as the moment of the system. The expression in the denominator, m 1+m 2,m 1+m 2, is the total mass of the system. Thus, the center of mass of the system is the point at which the total mass of the system could be concentrated without changing the moment.
This idea is not limited just to two point masses. In general, if n n masses, m 1,m 2,…,m n,m 1,m 2,…,m n, are placed on a number line at points x 1,x 2,…,x n,x 1,x 2,…,x n, respectively, then the center of mass of the system is given by
¯¯¯x=n∑i=1 m i x i n∑i=1 m i.x¯=∑n i=1 m i x i∑n i=1 m i.
Center of Mass of Objects on a Line
Let m 1,m 2,…,m n m 1,m 2,…,m n be point masses placed on a number line at points x 1,x 2,…,x n,x 1,x 2,…,x n, respectively, and let m=n∑i=1 m i m=∑n i=1 m i denote the total mass of the system. Then, the moment of the system with respect to the origin is given by
M=n∑i=1 m i x i M=∑i=1 n m i x i
and the center of mass of the system is given by
¯¯¯x=M m.x¯=M m.
We apply this theorem in the following example.
Example: Finding the Center of Mass of Objects along a Line
Suppose four point masses are placed on a number line as follows:
m 1=30 kg,placed a t x 1=−2 m m 2=5 kg,placed at x 2=3 m m 3=10 kg,placed at x 3=6 m m 4=15 kg,placed at x 4=−3 m.m 1=30 kg,placed a t x 1=−2 m m 2=5 kg,placed at x 2=3 m m 3=10 kg,placed at x 3=6 m m 4=15 kg,placed at x 4=−3 m.
Find the moment of the system with respect to the origin and find the center of mass of the system.
Show Solution
First, we need to calculate the moment of the system:
M=4∑i=1 m i x i=−60+15+60−45=−30.M=∑i=1 4 m i x i=−60+15+60−45=−30.
Now, to find the center of mass, we need the total mass of the system:
m=4∑i=1 m i=30+5+10+15=60 kg.m=∑i=1 4 m i=30+5+10+15=60 kg.
Then we have
¯¯¯x=M m=−30 60=−1 2.x¯=M m=−30 60=−1 2.
The center of mass is located 1/2 m to the left of the origin.
Try It
Suppose four point masses are placed on a number line as follows:
m 1=12 kg,placed at x 1=−4 m m 2=12 kg,placed at x 2=4 m m 3=30 kg,placed at x 3=2 m m 4=6 kg,placed at x 4=−6 m.m 1=12 kg,placed at x 1=−4 m m 2=12 kg,placed at x 2=4 m m 3=30 kg,placed at x 3=2 m m 4=6 kg,placed at x 4=−6 m.
Find the moment of the system with respect to the origin and find the center of mass of the system.
Hint
Use the process from the previous example.
Show Solution
M=24,¯¯¯x=2 5 m M=24,x¯=2 5 m
We can generalize this concept to find the center of mass of a system of point masses in a plane. Let m 1 m 1 be a point mass located at point (x 1,y 1)(x 1,y 1) in the plane. Then the moment M x M x of the mass with respect to the x x-axis is given by M x=m 1 y 1.M x=m 1 y 1. Similarly, the moment M y M y with respect to the y y-axis is given by M y=m 1 x 1.M y=m 1 x 1. Notice that the x x-coordinate of the point is used to calculate the moment with respect to the y y-axis, and vice versa. The reason is that the x x-coordinate gives the distance from the point mass to the y y-axis, and the y y-coordinate gives the distance to the x x-axis (see the following figure).
Figure 3. Point mass m 1 m 1 is located at point (x 1,y 1)(x 1,y 1) in the plane.
If we have several point masses in the xy-plane, we can use the moments with respect to the x x– and y y-axes to calculate the x x– and y y-coordinates of the center of mass of the system.
Center of Mass of Objects in a Plane
Let m 1,m 2,…,m n m 1,m 2,…,m n be point masses located in the xy-plane at points (x 1,y 1),(x 2,y 2),…,(x n,y n),(x 1,y 1),(x 2,y 2),…,(x n,y n), respectively, and let m=n∑i=1 m i m=∑n i=1 m i denote the total mass of the system. Then the moments M x M x and M y M y of the system with respect to the x x– and y y-axes, respectively, are given by
M x=n∑i=1 m i y i and M y=n∑i=1 m i x i.M x=∑n i=1 m i y i and M y=∑n i=1 m i x i.
Also, the coordinates of the center of mass (¯¯¯x,¯¯¯y)(x¯,y¯) of the system are
¯¯¯x=M y m and¯¯¯y=M x m.x¯=M y m and y¯=M x m.
The next example demonstrates how to apply this theorem.
Example: Finding the Center of Mass of Objects in a Plane
Suppose three point masses are placed in the xy-plane as follows (assume coordinates are given in meters):
m 1=2 kg, placed at(−1,3),m 2=6 kg, placed at(1,1),m 3=4 kg, placed at(2,−2).m 1=2 kg, placed at(−1,3),m 2=6 kg, placed at(1,1),m 3=4 kg, placed at(2,−2).
Find the center of mass of the system.
Show Solution
First we calculate the total mass of the system:
m=∑3 i=1 m i=2+6+4=12 kg.m=∑i=1 3 m i=2+6+4=12 kg.
Next we find the moments with respect to the x x– and y y-axes:
M y=∑3 i=1 m i x i=−2+6+8=12,M x=∑3 i=1 m i y i=6+6−8=4.M y=∑i=1 3 m i x i=−2+6+8=12,M x=∑i=1 3 m i y i=6+6−8=4.
Then we have
¯¯¯x=M y m=12 12=1 and¯¯¯y=M x m=4 12=1 3.x¯=M y m=12 12=1 and y¯=M x m=4 12=1 3.
The center of mass of the system is (1,1/3),(1,1/3), in meters.
Try It
Suppose three point masses are placed on a number line as follows (assume coordinates are given in meters):
m 1=5 kg, placed at(−2,−3),m 2=3 kg, placed at(2,3),m 3=2 kg, placed at(−3,−2).m 1=5 kg, placed at(−2,−3),m 2=3 kg, placed at(2,3),m 3=2 kg, placed at(−3,−2).
Find the center of mass of the system.
Hint
Use the process from the previous example.
Show Solution
(−1,−1)(−1,−1) m
Watch the following video to see the worked solution to the above Try It.
Closed Captioning and Transcript Information for Video
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “2.6 Moments and Centers of Mass” here (opens in new window).
Center of Mass of Thin Plates
So far we have looked at systems of point masses on a line and in a plane. Now, instead of having the mass of a system concentrated at discrete points, we want to look at systems in which the mass of the system is distributed continuously across a thin sheet of material. For our purposes, we assume the sheet is thin enough that it can be treated as if it is two-dimensional. Such a sheet is called a lamina. Next we develop techniques to find the center of mass of a lamina. In this section, we also assume the density of the lamina is constant.
Laminas are often represented by a two-dimensional region in a plane. The geometric center of such a region is called itscentroid. Since we have assumed the density of the lamina is constant, the center of mass of the lamina depends only on the shape of the corresponding region in the plane; it does not depend on the density. In this case, the center of mass of the lamina corresponds to the centroid of the delineated region in the plane. As with systems of point masses, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the x x– and y y-axes.
We first consider a lamina in the shape of a rectangle. Recall that the center of mass of a lamina is the point where the lamina balances. For a rectangle, that point is both the horizontal and vertical center of the rectangle. Based on this understanding, it is clear that the center of mass of a rectangular lamina is the point where the diagonals intersect, which is a result of thesymmetry principle, and it is stated here without proof.
The Symmetry Principle
If a region R is symmetric about a line l l, then the centroid of R lies on l l.
Let’s turn to more general laminas. Suppose we have a lamina bounded above by the graph of a continuous function f(x),f(x), below by the x x-axis, and on the left and right by the lines x=a x=a and x=b,x=b, respectively, as shown in the following figure.
Figure 4. A region in the plane representing a lamina.
As with systems of point masses, to find the center of mass of the lamina, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the x x– and y y-axes. As we have done many times before, we approximate these quantities by partitioning the interval [a,b][a,b] and constructing rectangles.
For i=0,1,2,…,n,i=0,1,2,…,n, let P={x i}P={x i} be a regular partition of [a,b].[a,b]. Recall that we can choose any point within the interval [x i−1,x i][x i−1,x i] as our x∗i.x i∗. In this case, we want x∗i x i∗ to be the x x-coordinate of the centroid of our rectangles. Thus, for i=1,2,…,n,i=1,2,…,n, we select x∗i∈[x i−1,x i]x i∗∈[x i−1,x i] such that x∗i x i∗ is the midpoint of the interval. That is, x∗i=(x i−1+x i)/2.x i∗=(x i−1+x i)/2. Now, for i=1,2,…,n,i=1,2,…,n, construct a rectangle of height f(x∗i)f(x i∗) on [x i−1,x i].[x i−1,x i]. The center of mass of this rectangle is (x∗i,(f(x∗i))/2),(x i∗,(f(x i∗))/2), as shown in the following figure.
Figure 5. A representative rectangle of the lamina.
Next, we need to find the total mass of the rectangle. Let ρ ρ represent the density of the lamina (note that ρ ρ is a constant). In this case, ρ ρ is expressed in terms of mass per unit area. Thus, to find the total mass of the rectangle, we multiply the area of the rectangle by ρ.ρ. Then, the mass of the rectangle is given by ρ f(x∗i)Δ x.ρ f(x i∗)Δ x.
To get the approximate mass of the lamina, we add the masses of all the rectangles to get
m≈n∑i=1 ρ f(x∗i)Δ x m≈∑n i=1 ρ f(x i∗)Δ x
This is a Riemann sum. Taking the limit as n→∞n→∞ gives the exact mass of the lamina:
m=lim n→∞n∑i=1 ρ f(x∗i)Δ x=ρ∫b a f(x)d x m=lim n→∞∑n i=1 ρ f(x i∗)Δ x=ρ∫a b f(x)d x
Next, we calculate the moment of the lamina with respect to the x x-axis. Returning to the representative rectangle, recall its center of mass is (x∗i,(f(x∗i))/2).(x i∗,(f(x i∗))/2). Recall also that treating the rectangle as if it is a point mass located at the center of mass does not change the moment. Thus, the moment of the rectangle with respect to the x x-axis is given by the mass of the rectangle, ρ f(x∗i)Δ x,ρ f(x i∗)Δ x, multiplied by the distance from the center of mass to the x x-axis: (f(x∗i))/2.(f(x i∗))/2. Therefore, the moment with respect to the x x-axis of the rectangle is ρ([f(x∗i)]2/2)Δ x.ρ([f(x i∗)]2/2)Δ x. Adding the moments of the rectangles and taking the limit of the resulting Riemann sum, we see that the moment of the lamina with respect to the x x-axis is
M x=lim n→∞n∑i=1 ρ[f(x∗i)]2 2 Δ x=ρ∫b a[f(x)]2 2 d x.M x=lim n→∞∑n i=1 ρ[f(x i∗)]2 2 Δ x=ρ∫a b[f(x)]2 2 d x.
We derive the moment with respect to the y y-axis similarly, noting that the distance from the center of mass of the rectangle to the y y-axis is x∗i.x i∗. Then the moment of the lamina with respect to the y y-axis is given by
M y=lim n→∞n∑i=1 ρ x∗i f(x∗i)Δ x=ρ∫b a x f(x)d x M y=lim n→∞∑n i=1 ρ x i∗f(x i∗)Δ x=ρ∫a b x f(x)d x
We find the coordinates of the center of mass by dividing the moments by the total mass to give ¯¯¯x=M y/m and¯¯¯y=M x/m.x¯=M y/m and y¯=M x/m. If we look closely at the expressions for M x,M y,and m,M x,M y,and m, we notice that the constant ρ ρ cancels out when ¯¯¯x x¯ and ¯¯¯y y¯ are calculated.
We summarize these findings in the following theorem.
Center of Mass of a Thin Plate in the xy-Plane
Let R denote a region bounded above by the graph of a continuous function f(x),f(x), below by the x x-axis, and on the left and right by the lines x=a x=a and x=b,x=b, respectively. Let ρ ρ denote the density of the associated lamina. Then we can make the following statements:
The mass of the lamina is m=ρ∫b a f(x)d x.m=ρ∫a b f(x)d x.
The moments M x M x and M y M y of the lamina with respect to the x x– and y y-axes, respectively, are M x=ρ∫b a[f(x)]2 2 d x and M y=ρ∫b a x f(x)d x.M x=ρ∫a b[f(x)]2 2 d x and M y=ρ∫a b x f(x)d x.
The coordinates of the center of mass (¯¯¯x,¯¯¯y)(x¯,y¯) are ¯¯¯x=M y m and¯¯¯y=M x m.x¯=M y m and y¯=M x m.
In the next example, we use this theorem to find the center of mass of a lamina.
Example: Finding the Center of Mass of a Lamina
Let R be the region bounded above by the graph of the function f(x)=√x f(x)=x and below by the x x-axis over the interval [0,4].[0,4]. Find the centroid of the region.
Show Solution
The region is depicted in the following figure.
Figure 6. Finding the center of mass of a lamina.
Since we are only asked for the centroid of the region, rather than the mass or moments of the associated lamina, we know the density constant ρ ρ cancels out of the calculations eventually. Therefore, for the sake of convenience, let’s assume ρ=1.ρ=1.
First, we need to calculate the total mass:
m=ρ∫b a f(x)d x=∫4 0√x d x=2 3 x 3/2|4 0=2 3[8−0]=16 3.m=ρ∫a b f(x)d x=∫0 4 x d x=2 3 x 3/2|0 4=2 3[8−0]=16 3.
Next, we compute the moments:
M x=ρ∫b a[f(x)]2 2 d x=∫4 0 x 2 d x=1 4 x 2|4 0=4 M x=ρ∫a b[f(x)]2 2 d x=∫0 4 x 2 d x=1 4 x 2|0 4=4
and
M y=ρ∫b a x f(x)d x=∫4 0 x√x d x=∫4 0 x 3/2 d x=2 5 x 5/2|4 0=2 5[32−0]=64 5.M y=ρ∫a b x f(x)d x=∫0 4 x x d x=∫0 4 x 3/2 d x=2 5 x 5/2|0 4=2 5[32−0]=64 5.
Thus, we have
¯¯¯x=M y m=64/5 16/3=64 5⋅3 16=12 5 and¯¯¯y=M x y=4 16/3=4⋅3 16=3 4.x¯=M y m=64/5 16/3=64 5·3 16=12 5 and y¯=M x y=4 16/3=4·3 16=3 4.
The centroid of the region is (12/5,3/4).(12/5,3/4).
Try It
Let R R be the region bounded above by the graph of the function f(x)=x 2 f(x)=x 2 and below by the x x-axis over the interval [0,2].[0,2]. Find the centroid of the region.
Hint
Use the process from the previous example.
Show Solution
The centroid of the region is (3 2,6 5).(3 2,6 5).
Watch the following video to see the worked solution to the above Try It.
Closed Captioning and Transcript Information for Video
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “2.6 Moments and Centers of Mass” here (opens in new window).
We can adapt this approach to find centroids of more complex regions as well. Suppose our region is bounded above by the graph of a continuous function f(x),f(x), as before, but now, instead of having the lower bound for the region be the x x-axis, suppose the region is bounded below by the graph of a second continuous function, g(x),g(x), as shown in the following figure.
Figure 7. A region between two functions.
Again, we partition the interval [a,b][a,b] and construct rectangles. A representative rectangle is shown in the following figure.
Figure 8. A representative rectangle of the region between two functions.
Note that the centroid of this rectangle is (x∗i,(f(x∗i)+g(x∗i))/2).(x i∗,(f(x i∗)+g(x i∗))/2). We won’t go through all the details of the Riemann sum development, but let’s look at some of the key steps. In the development of the formulas for the mass of the lamina and the moment with respect to the y y-axis, the height of each rectangle is given by f(x∗i)−g(x∗i),f(x i∗)−g(x i∗), which leads to the expression f(x)−g(x)f(x)−g(x) in the integrands.
In the development of the formula for the moment with respect to the x x-axis, the moment of each rectangle is found by multiplying the area of the rectangle, ρ[f(x∗i)−g(x∗i)]Δ x,ρ[f(x i∗)−g(x i∗)]Δ x, by the distance of the centroid from the x x-axis, (f(x∗i)+g(x∗i))/2,(f(x i∗)+g(x i∗))/2, which gives ρ(1/2){[f(x∗i)]2−[g(x∗i)]2}Δ x.ρ(1/2){[f(x i∗)]2−[g(x i∗)]2}Δ x. Summarizing these findings, we arrive at the following theorem.
Center of Mass of a Lamina Bounded by Two Functions
Let R denote a region bounded above by the graph of a continuous function f(x),f(x), below by the graph of the continuous function g(x),g(x), and on the left and right by the lines x=a x=a and x=b,x=b, respectively. Let ρ ρ denote the density of the associated lamina. Then we can make the following statements:
The mass of the lamina is m=ρ∫b a[f(x)−g(x)]d x.m=ρ∫a b[f(x)−g(x)]d x.
The moments M x M x and M y M y of the lamina with respect to the x x– and y y-axes, respectively, are M x=ρ∫b a 1 2([f(x)]2−[g(x)]2)d x and M y=ρ∫b a x[f(x)−g(x)]d x.M x=ρ∫a b 1 2([f(x)]2−[g(x)]2)d x and M y=ρ∫a b x[f(x)−g(x)]d x.
The coordinates of the center of mass (¯¯¯x,¯¯¯y)(x¯,y¯) are ¯¯¯x=M y m and¯¯¯y=M x m.x¯=M y m and y¯=M x m.
We illustrate this theorem in the following example.
Example: Finding the Centroid of a Region Bounded by Two Functions
Let R be the region bounded above by the graph of the function f(x)=1−x 2 f(x)=1−x 2 and below by the graph of the function g(x)=x−1.g(x)=x−1. Find the centroid of the region.
Show Solution
The region is depicted in the following figure.
Figure 9. Finding the centroid of a region between two curves.
The graphs of the functions intersect at (−2,−3)(−2,−3) and (1,0),(1,0), so we integrate from −2 to 1. Once again, for the sake of convenience, assume ρ=1.ρ=1.
First, we need to calculate the total mass:
m=ρ∫b a[f(x)−g(x)]d x=∫1−2[1−x 2−(x−1)]d x=∫1−2(2−x 2−x)d x=[2 x−1 3 x 3−1 2 x 2]|1−2=[2−1 3−1 2]−[−4+8 3−2]=9 2.m=ρ∫a b[f(x)−g(x)]d x=∫−2 1[1−x 2−(x−1)]d x=∫−2 1(2−x 2−x)d x=[2 x−1 3 x 3−1 2 x 2]|−2 1=[2−1 3−1 2]−[−4+8 3−2]=9 2.
Next, we compute the moments:
M x=ρ∫b a 1 2([f(x)]2−[g(x)]2)d x=1 2∫1−2((1−x 2)2−(x−1)2)d x=1 2∫1−2(x 4−3 x 2+2 x)d x=1 2[x 5 5−x 3+x 2]|1−2=−27 10 M x=ρ∫a b 1 2([f(x)]2−[g(x)]2)d x=1 2∫−2 1((1−x 2)2−(x−1)2)d x=1 2∫−2 1(x 4−3 x 2+2 x)d x=1 2[x 5 5−x 3+x 2]|−2 1=−27 10
and
M y=ρ∫b a x[f(x)−g(x)]d x=∫1−2 x[(1−x 2)−(x−1)]d x=∫1−2 x[2−x 2−x]d x=∫1−2(2 x−x 4−x 2)d x=[x 2−x 5 5−x 3 3]|1−2=−9 4.M y=ρ∫a b x[f(x)−g(x)]d x=∫−2 1 x[(1−x 2)−(x−1)]d x=∫−2 1 x[2−x 2−x]d x=∫−2 1(2 x−x 4−x 2)d x=[x 2−x 5 5−x 3 3]|−2 1=−9 4.
Therefore, we have
¯¯¯x=M y m=−9 4⋅2 9=−1 2 and¯¯¯y=M x y=−27 10⋅2 9=−3 5.x¯=M y m=−9 4·2 9=−1 2 and y¯=M x y=−27 10·2 9=−3 5.
The centroid of the region is (−(1/2),−(3/5)).(−(1/2),−(3/5)).
Try It
Let R be the region bounded above by the graph of the function f(x)=6−x 2 f(x)=6−x 2 and below by the graph of the function g(x)=3−2 x.g(x)=3−2 x. Find the centroid of the region.
Hint
Use the process from the previous example.
Show Solution
The centroid of the region is (1,13 5).(1,13 5).
Watch the following video to see the worked solution to the above Try It.
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For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “2.6 Moments and Centers of Mass” here (opens in new window).
Try It
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Inequality
The subject of mathematical inequalities is tied closely with optimization methods. While most of the subject of inequalities is often left out of the ordinary educational track, they are common in mathematics Olympiads.
Contents
[hide]
1 Overview
2 Solving Inequalities
2.1 Linear Inequalities
2.2 Polynomial Inequalities
2.3 Rational Inequalities
3 Complete Inequalities
4 List of Theorems
4.1 Introductory
4.2 Advanced
5 Problems
5.1 Introductory
5.2 Intermediate
5.3 Olympiad
6 Resources
6.1 Books
6.1.1 Intermediate
6.1.2 Olympiad
6.2 Articles
6.2.1 Olympiad
6.3 Classes
6.3.1 Olympiad
7 See also
Overview
Inequalities are arguably a branch of elementary algebra, and relate slightly to number theory. They deal with relations of variables denoted by four signs: .
For two numbers and :
if is greater than , that is, is positive.
if is smaller than , that is, is negative.
if is greater than or equal to , that is, is nonnegative.
if is less than or equal to , that is, is nonpositive.
Note that if and only if , , and vice versa. The same applies to the latter two signs: if and only if , , and vice versa.
Some properties of inequalities are:
If , then , where .
If , then , where .
If , then , where .
Solving Inequalities
In general, when solving inequalities, same quantities can be added or subtracted without changing the inequality sign, much like equations. However, when multiplying, dividing, or square rooting, we have to watch the sign. In particular, notice that although , we must have . In particular, when multiplying or dividing by negative quantities, we have to flip the sign. Complications can arise when the value multiplied can have varying signs depending on the variable.
We also have to be careful about the boundaries of the solutions. In the example , the value does not satisfy the inequality because the inequality is strict. However, in the example , the value satisfies the inequality because the inequality is nonstrict.
Solutions can be written in interval notation. Closed bounds use square brackets, while open bounds (and bounds at infinity) use parentheses. For instance, ![Image 49: $x \in 3,6)$ means .
Linear Inequalities
Linear inequalities can be solved much like linear equations to get implicit restrictions upon a variable. However, when multiplying/dividing both sides by negative numbers, we have to flip the sign.
Polynomial Inequalities
The first part of solving polynomial inequalities is much like solving polynomial equations -- bringing all the terms to one side and finding the roots.
Afterward, we have to consider bounds. We're comparing the sign of the polynomial with different inputs, so we could imagine a rough graph of the polynomial and how it passes through zeroes (since passing through zeroes could change the sign). Then we can find the appropriate bounds of the inequality.
Rational Inequalities
A more complex example is .
Here is a common mistake: The problem here is that we multiplied by as one of the last steps. We also kept the inequality sign in the same direction. However, we don't know if the quantity is negative or not; we can't assume that it is positive for all real . Thus, we may have to reverse the direction of the inequality sign if we are multiplying by a negative number. But, we don't know if the quantity is negative either.
A correct solution would be to move everything to the left side of the inequality, and form a common denominator. Then, it will be simple to find the solutions to the inequality by considering the sign (negativeness or positiveness) of the fraction as varies. We will start with an intuitive solution, and then a rule can be built for solving general fractional inequalities. To make things easier, we test positive integers. makes a good starting point, but does not solve the inequality. Nor does . Therefore, these two aren't solutions. Then we begin to test numbers such as , , and so on. All of these work. In fact, it's not difficult to see that the fraction will remain positive as gets larger and larger. But just where does , which causes a negative fraction at and , begin to cause a positive fraction? We can't just assume that is the switching point; this solution is not simply limited to integers. The numerator and denominator are big hints. Specifically, we examine that when (the numerator), then the fraction is , and begins to be positive for all higher values of . Solving the equation reveals that is the turning point. After more of this type of work, we realize that brings about division by , so it certainly isn't a solution. However, it also tells us that any value of that is less than brings about a fraction that has a negative numerator and denominator, resulting in a positive fraction and thus satisfying the inequality. No value between and (except itself) seems to be a solution. Therefore, we conclude that the solutions are the intervals ![Image 78: $(-\infty,-5)\cup\frac{3}{2},+\infty)$.
For the sake of better notation, define the "x-intercept" of a fractional inequality to be those values of that cause the numerator and/or the denominator to be .To develop a method for quicker solutions of fractional inequalities, we can simply consider the "x-intercepts" of the numerator and denominator. We graph them on the number line. Then, in every region of the number line, we test one point to see if the whole region is part of the solution. For example, in the example problem above, we see that we only had to test one value such as in the region , as well as one value in the region ![Image 83: $(-\infty,-5]$]( and ![Image 84: $\frac{3}{2},+\infty)$; then we see which regions are part of the solution set. This does indeed give the complete solution set.
One must be careful about the boundaries of the solutions. In the example problem, the value was a solution only because the inequality was nonstrict. Also, the value was not a solution because it would bring about division by . Similarly, any "x-intercept" of the numerator is a solution if and only if the inequality is nonstrict, and every "x-intercept" of the denominator is never a solution because we cannot divide by .
Complete Inequalities
A inequality that is true for all real numbers or for all positive numbers (or even for all complex numbers) is sometimes called a complete inequality. An example for real numbers is the so-called Trivial Inequality, which states that for any real , . Most inequalities of this type are only for positive numbers, and this type of inequality often has extremely clever problems and applications.
List of Theorems
Here are some of the more useful inequality theorems, as well as general inequality topics.
Introductory
Arithmetic Mean-Geometric Mean Inequality
Cauchy-Schwarz Inequality
Titu's Lemma
Chebyshev's Inequality
Geometric inequalities
Jensen's Inequality
Nesbitt's Inequality
Rearrangement Inequality
Power mean inequality
Triangle Inequality
Trivial inequality
Schur's Inequality
Advanced
Aczel's Inequality
Callebaut's Inequality
Carleman's Inequality
Hölder's inequality
Radon's Inequality
Homogenization
Isoperimetric inequalities
Maclaurin's Inequality
Muirhead's Inequality
Minkowski Inequality
Newton's Inequality
Ptolemy's Inequality
Can someone fix that Ptolemy's is in Advanced?
Problems
Introductory
Practice Problems on Alcumus
Inequalities (Prealgebra)
Solving Linear Inequalities (Algebra)
Quadratic Inequalities (Algebra)
Basic Rational Function Equations and Inequalities (Intermediate Algebra)
A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than . What's the largest number of matches she could've won before the weekend began? (1992 AIME Problems/Problem 3)
Intermediate
Practice Problems on Alcumus
Quadratic Inequalities (Algebra)
Advanced Rational Function Equations and Inequalities (Intermediate Algebra)
General Inequality Skills (Intermediate Algebra)
Advanced Inequalities (Intermediate Algebra)
Given that , and show that . (weblog_entry.php?t=172070 Source)
Olympiad
See also Category:Olympiad Inequality Problems
Let be positive real numbers. Prove that (2001 IMO Problems/Problem 2)
Resources
Books
Intermediate
Introduction to Inequalities
Geometric Inequalities
Olympiad
Advanced Olympiad Inequalities: Algebraic & Geometric Olympiad Inequalities by Alijadallah Belabess.
The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele.
Problem Solving Strategies by Arthur Engel contains significant material on inequalities.
Inequalities by G. H. Hardy, J. E. Littlewood, G. Pólya.
Articles
Olympiad
Inequalities by MIT Professor Kiran Kedlaya.
Inequalities by IMO gold medalist Thomas Mildorf.
Classes
Olympiad
The Worldwide Online Olympiad Training Program is designed to help students learn to tackle mathematical Olympiad problems in topics such as inequalities.
See also
Mathematics competitions
Math books
Retrieved from "
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1303 | https://math.stackexchange.com/questions/3400800/understanding-the-proof-of-excercise-10b-from-chapter-5-of-spivak-s-calculus | limits - Understanding the proof of Excercise 10b from chapter 5 of Spivak’s Calculus. - Mathematics Stack Exchange
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Understanding the proof of Excercise 10b from chapter 5 of Spivak’s Calculus.
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The question is “Prove that the limit as x approaches zero of f(x)=the limit as x approaches a of f(x-a).
I have taken a few stabs at the problem and was told that the solution was the image I attached.
The given proof largely makes sense to me, only it seems to be a proof that the limit as x approaches a of f(x) equals the limit as y approaches zero of g(y-a). I don’t see how this is a proof of what was asked, and I also don’t see how to proceed in the other direction. I think my issue is that I’m struggling to connect my intuitive/verbal understanding with epsilon-delta limit notation, and changing the variable from x to y also confuses me. Can anyone help make sense of this?
calculus
limits
proof-explanation
epsilon-delta
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asked Oct 19, 2019 at 22:56
SamuelSamuel
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lim x→0 f(x)=L⟺lim x→0 f(x)=L⟺
∀e>0∃d>0∀x(0<|x|0∃d>0∀x(0<|x|<d⟹|f(x)−L|<e)⟺
∀e>0∃d>0∀x′(0<|x′−a|0∃d>0∀x′(0<|x′−a|<d⟹|f((x′−a))−L|<e)⟺
lim x′→a f(x′−a)=L⟺lim x′→a f(x′−a)=L⟺
lim x→a f(x−a)=L.lim x→a f(x−a)=L.
Regardless of whether or not a a is a 0 0 of f.f.
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edited Mar 26, 2020 at 5:26
Martin R
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answered Oct 20, 2019 at 0:39
DanielWainfleetDanielWainfleet
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Ah, thank you. So my confusion was due to the fact that the proof was more general than I was expecting. Also, I’m new here; how do I post with nice notation like you?Samuel –Samuel 2019-10-20 01:45:48 +00:00 Commented Oct 20, 2019 at 1:45
On the top bar, right, click on Help. Choose the Help Center. Click on "How Do I Format Mathematics Here?" Read it and its linked Quick Tutorial.DanielWainfleet –DanielWainfleet 2019-10-20 07:24:07 +00:00 Commented Oct 20, 2019 at 7:24
You can also look at MathJax.Org. This site has MathJax in it. E.g. typing a dollar sign before and after \lim_{x\to 0}f(x)=L\iff gives lim x→0 f(x)=L⟺lim x→0 f(x)=L⟺.DanielWainfleet –DanielWainfleet 2019-10-20 07:33:43 +00:00 Commented Oct 20, 2019 at 7:33
As a general rule of thumb, you should not 'answer' with a question. This would be more suited for a comment.MathsofData –MathsofData 2020-03-26 05:23:01 +00:00 Commented Mar 26, 2020 at 5:23
@MathsofData: This was posted as an answer last year, but replaced by an unrelated question by a new user. Unfortunately, two users approved this edit suggestion.Martin R –Martin R 2020-03-26 07:46:22 +00:00 Commented Mar 26, 2020 at 7:46
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1304 | http://content.nroc.org/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U16_L2_T3_text_final.html | Multiplication of Multiple Term Radicals
Multiplication of Multiple Term Radicals
Learning Objective(s)
·Multiply and simplify radical expressions that contain more than one term.
Introduction
When multiplying multiple term radical expressions it is important to follow the Distributive Property of Multiplication, as when you are multiplying regular, non-radical expressions.
Radicals follow the same mathematical rules that other real numbers do. So, although the expression may look different than , you can treat them the same way.
Using the Distributive Property
Let’s have a look at how to apply the Distributive Property. First let’s do a problem with the variable a, and then solve the same problem replacing a with .
Example
ProblemSimplify.
Use the Distributive Property of Multiplication over Subtraction.
Answer
Example
ProblemSimplify.
Use the Distributive Property of Multiplication over Subtraction.
Apply the rules of multiplying radicals: to multiply .
AnswerBe sure to simplify radicals when you can: , so .
The answers to the previous two problems should look similar to you. The only difference is that in the second problem, has replaced the variable a(and so has replaced a 2). The process of multiplying is very much the same in both problems.
In these next two problems, each term contains a radical.
Example
ProblemSimplify.
Use the Distributive Property of Multiplication over Addition to multiply each term within parentheses by.
Apply the rules of multiplying radicals.
, so can be pulled out of the radical.
Answer
Example
ProblemSimplify.
Use the Distributive Property.
Apply the rules of multiplying radicals.
Identify cubes in each of the radicals.
Answer
In all of these examples, multiplication of radicals has been shown following the pattern . Then, only after multiplying, some radicals have been simplified—like in the last problem. After you have worked with radical expressions a bit more, you may feel more comfortable identifying quantities such as without going through the intermediate step of finding that . In the rest of the examples that follow, though, each step is shown.
Multiply and simplify.
A)
B)
C)
D)
Show/Hide Answer
A)
Correct. Multiplying by and , you find it is equal to , or .
B)
Incorrect. , but what does simplify to? The correct answer is .
C)
Incorrect. You subtracted , and then multiplied by . Remember that you cannot subtract radicals unless the indices and the radicands are the same. The correct answer is .
D)
Incorrect. This is the correct product, but it is not in simplest form. Look for squares that exist within and . The correct answer is .
Multiplying Radical Expressions as Binomials
Sometimes, radical expressions appear in binomials as well. In these cases, you still follow the rules of binomial multiplication—but it is very important that you be precise and structured when you are multiplying the different terms.
As a refresher, here is the process for multiplying two binomials. If you like using the expression “FOIL” (First, Outside, Inside, Last) to help you figure out the order in which the terms should be multiplied, you can use it here, too.
Example
ProblemMultiply.
First:
Outside:
Inside:
Last: Use the Distributive Property.
Record the terms, and then combine like terms.
Answer
Here is the same problem, with replacing the variable x.
Example
ProblemMultiply.
First:
Outside:
Inside:
Last: Use the Distributive Property to multiply. Simplify using .
Record the terms, and then combine like terms.
Answer
The multiplication works the same way in both problems; you just have to pay attention to the index of the radical (that is, whether the roots are square roots, cube roots, etc.) when multiplying radical expressions.
Multiplying Binomial Radical Expressions
To multiply radical expressions, use the same method as used to multiply polynomials.
·Use the Distributive Property (or, if you prefer, the shortcut FOIL method);
·Remember that ; and
·Combine like terms.
Example
ProblemMultiply.
First:
Outside:
Inside:
Last: Use FOIL to multiply.
Record the terms, and then combine like terms (if possible). Here, there are no like terms to combine.
Answer
Multiply and simplify.
A)
B)
C)
D)
Show/Hide Answer
A)
Incorrect. Using the FOIL method, the product is , which simplifies to . It looks like you forgot the variables and added the coefficients 8 + 2 – 3 in order to arrive at 7. The correct answer is .
B)
Correct. Using the FOIL method, you find that the product of the binomials is , which simplifies to .
C)
Incorrect. Remember that , not . The correct answer is .
D)
Incorrect. You have multiplied the variable terms correctly, but you have the incorrect coefficients. Using the FOIL method, the product is , which simplifies to . The correct answer is .
Summary
To multiply radical expressions that contain more than one term, use the same method that you use to multiply polynomials. First, use the Distributive Property (or, if you prefer, the shortcut FOIL method) to multiply the terms. Then, apply the rules , and to multiply and simplify. Finally, combine like terms. |
1305 | https://www.wikihow.com/Use-the-3-4-5-Rule-to-Build-Square-Corners | Home
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How to Make Sure a Corner is Square with the 3-4-5 Rule
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Line up any project perfectly with this simple formula
Co-authored by
Mark Spelman
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Last Updated: July 18, 2025
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Squaring a Corner with the 3-4-5 Method
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This article was co-authored by Mark Spelman and by wikiHow staff writer, Luke Smith, MFA. Mark Spelman is a General Contractor based in Austin, Texas. With over 30 years of construction experience, Mark specializes in constructing interiors, project management, and project estimation. He has been a construction professional since 1987.
This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources.
This article has been viewed 1,238,030 times.
The 3-4-5 rule is a handy and foolproof way to determine if a corner is perfectly square when doing carpentry or other DIY projects, from building a deck to laying tile. And if your corners aren’t square when laying a foundation or building a frame for a deck, your later measurements will be off, too. We’ll show you how to use the 3-4-5 rule to quickly and accurately measure your corners, then we’ll show you how and why it works, so that you can add a handy new tool to your belt.
Things You Should Know
Measure and mark 3 units from the corner along one side of your project, then measure and mark 4 units from the corner along the other side.
Line up a tape measure between your marks and check to see if the length measures 5 units. If it does, the corner is square.
The 3-4-5 rule uses the Pythagorean Theorem (A2 + B2 = C2) to ensure your corner forms a right triangle with a 90º angle.
Steps
Section 1 of 2:
Squaring a Corner with the 3-4-5 Method
Download Article
1
Measure 3 units from the corner along one side. Use a tape measure to measure 3 units out from the corner of your project, marking the measurement with a pencil.
X
Research source Use feet, meters, or any other unit to make your measurements, so long as you use the same unit each time.
2. 2
Measure 4 units from the corner along the other side. Using the same unit of measurement, measure along the second side and make a mark at 4 units along its length.
X
Research source
Multiply each number in the 3-4-5 rule by the same amount to scale it up for a larger project (to get a more accurate measurement), or to convert it to different units. For example, try 30-40-50 cm if you’re using the metric system. For a large room, use 6-8-10 or 9-12-15 feet or meters.
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3. 3
Measure the distance between your marks—if it's 5, your corner is square. Line up your tape measure between the 2 marks. If the distance is 5 units, your corner is square
X
Research source If the distance is less than 5 units, your corner is less than 90º. If the distance is over 5 units, your corner has a measurement of more than 90º. Depending on your project, you may need to adjust your measurements or plans, or rebuild the corner to achieve a more precise angle. Use a framing square as a guide when you do this to avoid further mistakes.
Once you’ve got a square corner, check the project’s other 3 corners to ensure they’re the same.
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Section 2 of 2:
Why does the 3-4-5 rule work?
Download Article
The 3-4-5 rule uses the Pythagorean Theorem to make a right angle. If a triangle has sides measuring 3, 4, and 5 units long, it must be a right triangle with a 90º angle between the short sides.
X
Research source If you can "find" this triangle in your corner, you know the corner is square. This is based on the Pythagorean Theorem from geometry: A2 + B2 = C2 for a right triangle. C is the longest side (hypotenuse) and A and B are the two shorter "legs."
X
Research source
3-4-5 is a very convenient measurement to check because of the low, whole numbers. The math checks out: 32 + 42 = 9 + 16 = 25 = 52.
The 3-4-5 rule works with any numbers of the same proportions. For example, finding a triangle with sides that are 6-8-10 or 9-12-15 units long will also give you a 90º angle.
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Joseph Meyer
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Joseph Meyer is a High School Math Teacher based in Pittsburgh, Pennsylvania. He is an educator at City Charter High School, where he has been teaching for over 7 years. Joseph is also the founder of Sandbox Math, an online learning community dedicated to helping students succeed in Algebra. His site is set apart by its focus on fostering genuine comprehension through step-by-step understanding (instead of just getting the correct final answer), enabling learners to identify and overcome misunderstandings and confidently take on any test they face. He received his MA in Physics from Case Western Reserve University and his BA in Physics from Baldwin Wallace University.
Joseph Meyer
Math Teacher
Use this visual trick to understand the Pythagorean Theorem. Imagine a right triangle with squares constructed on each leg and the hypotenuse. by rearranging the smaller squares within the larger square, the areas of the smaller squares (a² and b²) will add up visually to the area of the larger square (c²).
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So, I should measure 3 ft on one side, 4 ft on the other side, and 5 ft for the center to get a square corner?
Mark Spelman
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Mark Spelman is a General Contractor based in Austin, Texas. With over 30 years of construction experience, Mark specializes in constructing interiors, project management, and project estimation. He has been a construction professional since 1987.
Mark Spelman
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Yes. Your longest side (hypotenuse) should measure 5 feet from point to point.
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What is the square footage of a 30 x 40 building?
Mark Spelman
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Mark Spelman is a General Contractor based in Austin, Texas. With over 30 years of construction experience, Mark specializes in constructing interiors, project management, and project estimation. He has been a construction professional since 1987.
Mark Spelman
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The square footage of a 30 foot x 40 foot building is 1,200 square feet.
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This method can be more accurate than using a carpenter's (framing) square, which may be too small to get precise measurements over greater lengths.
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About This Article
Co-authored by:
Mark Spelman
Construction Professional
This article was co-authored by Mark Spelman and by wikiHow staff writer, Luke Smith, MFA. Mark Spelman is a General Contractor based in Austin, Texas. With over 30 years of construction experience, Mark specializes in constructing interiors, project management, and project estimation. He has been a construction professional since 1987. This article has been viewed 1,238,030 times.
27 votes - 96%
Co-authors: 23
Updated: July 18, 2025
Views: 1,238,030
Categories: Walls and Ceilings
Article SummaryX
To build square corners with the 3-4-5 rule, first measure 3 units from the corner on 1 side. Turn in a perpendicular direction from the first line and measure 4 units. Then, measure the diagonal between the ends of your 2 lines. If it measures 5 units, your corner is square. If it's less than 5 units, your corner is less than 90 degrees; if it's greater than 5 units, your corner angle is too large. For more tips and an explanation of the math behind the 3-4-5 rulel, read on!
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Co-authored by:
Mark Spelman
Construction Professional
Co-authors: 23
Updated: July 18, 2025
Views: 1,238,030
96% of readers found this article helpful.
27 votes - 96%
Click a star to add your vote
% of people told us that this article helped them.
Mick Fowles
Jun 20, 2016
"I had to build a brick wall in a trench at right angles to an existing wall, and this system helped me obtain a right angle and build it with confidence. "..." more
Ross Grenz
Mar 20, 2019
"I was always confused by the 3-4-5 rule, but your site made it easy to understand. Thank you, I will use it on an upcoming project."..." more
Ian Burnby
Apr 27, 2016
"Building a patio in the middle of my lawn and want it to be square. I couldn't remember if it was 2/3/4 or 3/4/5. Thanks."..." more
Manny Garcia
Mar 28, 2017
"Very helpful. I heard about this, but not sure how or if it was an actual working thing. It works."
Jon B.
Aug 14, 2018
"I was installing floor tiling and needed to make sure my rows looked sweet. Thanks again!"
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Determining the Characteristics of a Quadratic Function Using Various Methods
Determine the following characteristics of the quadratic function y=−2x2+4x+6:
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From the graph of the parabola, determine the: • vertex• axis of symmetry• y-intercept• x-intercepts• domain• range• minimum/maximum value
Identifying Characteristics of Quadratic function in General Form: y=ax2+bx+c y=2x2−12x+10 is a quadratic function in general form. i) Determine: • y-intercept • x-intercepts • vertex ii) Sketch the graph.
Identifying Characteristics of Quadratic Functions in Vertex Form: y=a(x−p)2+q y=2(x−3)2−8 is a quadratic function in vertex form. i) Determine: • y-intercept • x-intercepts • vertex ii) Sketch the graph.
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Characteristics of quadratic functions
Jump to:NotesPrerequisitesRelated
Notes
Three properties that are universal to all quadratic functions: 1) The graph of a quadratic function is always a parabola that either opens upward or downward (end behavior); 2) The domain of a quadratic function is all real numbers; and 3) The vertex is the lowest point when the parabola opens upwards; while the vertex is the highest point when the parabola opens downward.
Prerequisites
Factoring trinomials
Solving quadratic equations using the quadratic formula
Completing the square
Shortcut: Vertex formula
Related
Even and odd functions
What is a polynomial function?
Characteristics of polynomial graphs
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1308 | https://openoregon.pressbooks.pub/socialprovisioning3/chapter/aggregate-demand-in-the-keynesian-model/ | Skip to content
12.4 – Aggregate Demand in the Keynesian Cross Model
Learning Objectives
By the end of this section, you will be able to:
Explain the components of the Keynesian cross diagram
The fundamental ideas of Keynesian economics were developed before the AD/AS model was popularized. From the 1930s until the 1970s, Keynesian economics was usually explained with a different model, known as the expenditure-output approach. This approach is strongly rooted in the fundamental assumptions of Keynesian economics: it focuses on the total amount of spending in the economy, with no explicit mention of aggregate supply or of the price level (although as you will see, it is possible to draw some inferences about aggregate supply and price levels based on the diagram).
The Axes of the Keynesian Cross Diagram
The expenditure-output model, sometimes also called the Keynesian cross diagram, determines the equilibrium level of real GDP by the point where the total or aggregate expenditures in the economy are equal to the amount of output produced. The axes of the Keynesian cross diagram presented in Figure 1 show real GDP on the horizontal axis as a measure of output and aggregate expenditures on the vertical axis as a measure of spending.
Remember that GDP can be thought of in several equivalent ways: it measures both the value of spending on final goods and also the value of the production of final goods. All sales of the final goods and services that make up GDP will eventually end up as income for workers, for managers, and for investors and owners of firms. The sum of all the income received for contributing resources to GDP is called national income (Y). At some points in the discussion that follows, it will be useful to refer to real GDP as “national income.” Both axes are measured in real (inflation-adjusted) terms.
The Potential GDP Line and the 45-degree Line
The Keynesian cross diagram contains two lines that serve as conceptual guideposts to orient the discussion. The first is a vertical line showing the level of potential GDP. Potential GDP means the same thing here that it means in the AD/AS diagrams: it refers to the quantity of output that the economy can produce with full employment of its labor and physical capital.
The second conceptual line on the Keynesian cross diagram is the 45-degree line, which starts at the origin and reaches up and to the right. A line that stretches up at a 45-degree angle represents the set of points (1, 1), (2, 2), (3, 3) and so on, where the measurement on the vertical axis is equal to the measurement on the horizontal axis. In this diagram, the 45-degree line shows the set of points where the level of aggregate expenditure in the economy, measured on the vertical axis, is equal to the level of output or national income in the economy, measured by GDP on the horizontal axis.
When the macroeconomy is in equilibrium, it must be true that the aggregate expenditures in the economy are equal to the real GDP—because by definition, GDP is the measure of what is spent on final sales of goods and services in the economy. Thus, the equilibrium calculated with a Keynesian cross diagram will always end up where aggregate expenditure and output are equal—which will only occur along the 45-degree line.
The Aggregate Expenditure Schedule
The final ingredient of the Keynesian cross or expenditure-output diagram is the aggregate expenditure schedule, which will show the total expenditures in the economy for each level of real GDP. The intersection of the aggregate expenditure line with the 45-degree line—at point E0 in Figure 1—will show the equilibrium for the economy, because it is the point where aggregate expenditure is equal to output or real GDP. After developing an understanding of what the aggregate expenditure schedule means, we will return to this equilibrium and how to interpret it.
Building the Aggregate Expenditure Schedule
Aggregate expenditure is the key to the expenditure-income model. The aggregate expenditure schedule shows, either in the form of a table or a graph, how aggregate expenditures in the economy rise as real GDP or national income rises. Thus, in thinking about the components of the aggregate expenditure line—consumption, investment, government spending, exports and imports—the key question is how expenditures in each category will adjust as national income rises.
Consumption as a Function of National Income
How do consumption expenditures increase as national income rises? People can do two things with their income: consume it or save it (for the moment, let’s ignore the need to pay taxes with some of it). Each person who receives an additional dollar faces this choice. The marginal propensity to consume (MPC), is the share of the additional dollar of income a person decides to devote to consumption expenditures. The the share of the additional dollar of income a person decides to save is the marginal propensity to save (MPS). It must always hold true that:
MPC + MPS = 1
For example, if the marginal propensity to consume out of the marginal amount of income earned is 0.9, then the marginal propensity to save is 0.1.
With this relationship in mind, consider the relationship among income, consumption, and savings shown in Figure 2. (Note that we use “Aggregate Expenditure” on the vertical axis because all consumption expenditures are parts of aggregate expenditures.)
An assumption commonly made in this model is that even if income were zero, people would have to consume something. In this example, consumption would be $600 even if income were zero. We’ll call this autonomous consumption (Ca), where ‘autonomous’ just means independent of the level of income. Then, the MPC is 0.8 and the MPS is 0.2. Thus, when income increases by $1,000, consumption rises by $800 and savings rises by $200. At an income of $4,000, total consumption will be the $600 that would be consumed even without any income, plus $4,000 multiplied by the marginal propensity to consume of 0.8, or $ 3,200, for a total of $ 3,800. The total amount of consumption and saving must always add up to the total amount of income. (Exactly how a situation of zero income and negative savings would work in practice is not important, because even low-income societies are not literally at zero income, so the point is hypothetical.) This relationship between income and consumption, illustrated in Figure 2 and Table 1, is called the consumption function. It can be written in equation form as:
C = Ca + (mpc)(Y)
Similarly, the savings function (not shown graphically) can be written as:
S = -Ca + (mps)(Y)
The pattern of consumption shown in Table 1 is plotted in Figure 2. To calculate consumption, multiply the income level by 0.8, for the marginal propensity to consume, and add $600, for the amount that would be consumed even if income was zero. Consumption plus savings must be equal to income.
Table 1. The Consumption Function
| Income | Consumption | Savings |
| $0 | $600 | –$600 |
| $1,000 | $1,400 | –$400 |
| $2,000 | $2,200 | –$200 |
| $3,000 | $3,000 | $0 |
| $4,000 | $3,800 | $200 |
| $5,000 | $4,600 | $400 |
| $6,000 | $5,400 | $600 |
| $7,000 | $6,200 | $800 |
| $8,000 | $7,000 | $1,000 |
| $9,000 | $7,800 | $1,200 |
However, a number of factors other than income can also cause the entire consumption function to shift. These factors were summarized in earlier int he chapter. When the consumption function moves, it can shift in two ways: either the entire consumption function can move up or down in a parallel manner, or the slope of the consumption function can shift so that it becomes steeper or flatter. For example, if a tax cut leads consumers to spend more, but does not affect their marginal propensity to consume, it would cause an upward shift to a new consumption function that is parallel to the original one. However, a change in household preferences for saving that reduced the marginal propensity to save would cause the slope of the consumption function to become steeper: that is, if the savings rate is lower, then every increase in income leads to a larger rise in consumption.
Investment as a Function of ‘Animal Spirits’
Investment decisions are forward-looking, based on expected rates of return. Precisely because investment decisions depend primarily on perceptions about future economic conditions–what Keynes referred to as the ‘animal spirits’ driving the optimism (and pessimism) or business people–, they do not depend primarily on the level of GDP in the current year. Thus, on a Keynesian cross diagram, the investment function can be drawn as a horizontal line, at a fixed level of expenditure. Figure 3 shows an investment function where the level of investment is, for the sake of concreteness, set at the specific level of 500. Just as a consumption function shows the relationship between consumption levels and real GDP (or national income), the investment functionshows the relationship between investment levels and real GDP.
The appearance of the investment function as a horizontal line does not mean that the level of investment never moves. It means only that in the context of this two-dimensional diagram, the level of investment on the vertical aggregate expenditure axis does not vary according to the current level of real GDP on the horizontal axis. However, all the other factors that vary investment—new technological opportunities, expectations about near-term economic growth, interest rates, the price of key inputs, and tax incentives for investment—can cause the horizontal investment function to shift up or down.
Government Spending and Taxes as a Function of National Income
In the Keynesian cross diagram, government spending appears as a horizontal line, as in Figure 4, where government spending is set at a level of 1,300. As in the case of investment spending, this horizontal line does not mean that government spending is unchanging. It means only that government spending changes when Congress decides on a change in the budget, rather than shifting in a predictable way with the current size of the real GDP shown on the horizontal axis.
The situation of taxes is different because taxes often rise or fall with the volume of economic activity. For example, income taxes are based on the level of income earned and sales taxes are based on the amount of sales made, and both income and sales tend to be higher when the economy is growing and lower when the economy is in a recession. For the purposes of constructing the basic Keynesian cross diagram, it is helpful to view taxes as a proportionate share of GDP. In the United States, for example, taking federal, state, and local taxes together, government typically collects about 30–35 % of income as taxes.
Table 2 revises the earlier table on the consumption function so that it takes taxes into account. The first column shows national income. The second column calculates taxes, which in this example are set at a rate of 30%, or 0.3. The third column shows after-tax income; that is, total income minus taxes. The fourth column then calculates consumption in the same manner as before: multiply after-tax income by 0.8, representing the marginal propensity to consume, and then add $600, for the amount that would be consumed even if income was zero. When taxes are included, the marginal propensity to consume is reduced by the amount of the tax rate, so each additional dollar of income results in a smaller increase in consumption than before taxes. For this reason, the consumption function, with taxes included, is flatter than the consumption function without taxes, as Figure 5 shows.
Table 2. The Consumption Function Before and After Taxes
| Income | Taxes | After-Tax Income | Consumption | Savings |
| $0 | $0 | $0 | $600 | –$600 |
| $1,000 | $300 | $700 | $1,160 | –$460 |
| $2,000 | $600 | $1,400 | $1,720 | –$320 |
| $3,000 | $900 | $2,100 | $2,280 | –$180 |
| $4,000 | $1,200 | $2,800 | $2,840 | –$40 |
| $5,000 | $1,500 | $3,500 | $3,400 | $100 |
| $6,000 | $1,800 | $4,200 | $3,960 | $240 |
| $7,000 | $2,100 | $4,900 | $4,520 | $380 |
| $8,000 | $2,400 | $5,600 | $5,080 | $520 |
| $9,000 | $2,700 | $6,300 | $5,640 | $660 |
Exports and Imports as a Function of National Income
The export function, which shows how exports change with the level of a country’s own real GDP, is drawn as a horizontal line, as in the example in Figure 6 (a) where exports are drawn at a level of $840. Again, as in the case of investment spending and government spending, drawing the export function as horizontal does not imply that exports never change. It just means that they do not change because of what is on the horizontal axis—that is, a country’s own level of domestic production—and instead are shaped by the level of aggregate demand in other countries. More demand for exports from other countries would cause the export function to shift up; less demand for exports from other countries would cause it to shift down.
Imports are drawn in the Keynesian cross diagram as a downward-sloping line, with the downward slope determined by the marginal propensity to import (MPI), out of national income. In Figure 6 (b), the marginal propensity to import is 0.1. Thus, if real GDP is $5,000, imports are $500; if national income is $6,000, imports are $600, and so on. The import function is drawn as downward sloping and negative, because it represents a subtraction from the aggregate expenditures in the domestic economy. A change in the marginal propensity to import, perhaps as a result of changes in preferences, would alter the slope of the import function.
Building the Combined Aggregate Expenditure Function
All the components of aggregate demand—consumption, investment, government spending, and the trade balance—are now in place to build the Keynesian cross diagram. Figure 7 builds up an aggregate expenditure function, based on the numerical illustrations of C, I, G, X, and M that have been used throughout this text. The first three columns in Table 3 are lifted from the earlier Table 2, which showed how to bring taxes into the consumption function. The first column is real GDP or national income, which is what appears on the horizontal axis of the income-expenditure diagram. The second column calculates after-tax income, based on the assumption, in this case, that 30% of real GDP is collected in taxes. The third column is based on an MPC of 0.8, so that as after-tax income rises by $700 from one row to the next, consumption rises by $560 (700 × 0.8) from one row to the next. Investment, government spending, and exports do not change with the level of current national income. In the previous discussion, investment was $500, government spending was $1,300, and exports were $840, for a total of $2,640. This total is shown in the fourth column. Imports are 0.1 of real GDP in this example, and the level of imports is calculated in the fifth column. The final column,aggregate expenditures, sums up C + I + G + X – M. This aggregate expenditure line is illustrated in Figure 7.
Table 3. National Income-Aggregate Expenditure Equilibrium
| National Income | After-Tax Income | Consumption | Government Spending + Investment + Exports | Imports | Aggregate Expenditure |
| $3,000 | $2,100 | $2,280 | $2,640 | $300 | $4,620 |
| $4,000 | $2,800 | $2,840 | $2,640 | $400 | $5,080 |
| $5,000 | $3,500 | $3,400 | $2,640 | $500 | $5,540 |
| $6,000 | $4,200 | $3,960 | $2,640 | $600 | $6,000 |
| $7,000 | $4,900 | $4,520 | $2,640 | $700 | $6,460 |
| $8,000 | $5,600 | $5,080 | $2,640 | $800 | $6,920 |
| $9,000 | $6,300 | $5,640 | $2,640 | $900 | $7,380 |
The aggregate expenditure function is formed by stacking on top of each other the consumption function (after taxes), the investment function, the government spending function, the export function, and the import function. The point at which the aggregate expenditure function intersects the vertical axis will be determined by the levels of investment, government, and export expenditures—which do not vary with national income. The upward slope of the aggregate expenditure function will be determined by the marginal propensity to save, the tax rate, and the marginal propensity to import. A higher marginal propensity to save, a higher tax rate, and a higher marginal propensity to import will all make the slope of the aggregate expenditure function flatter—because out of any extra income, more is going to savings or taxes or imports and less to spending on domestic goods and services.
The equilibrium occurs where national income is equal to aggregate expenditure, which is shown on the graph as the point where the aggregate expenditure schedule crosses the 45-degree line. In this example, the equilibrium occurs at 6,000. This equilibrium can also be read off the table under the figure; it is the level of national income where aggregate expenditure is equal to national income.
Glossary
autonomous consumption
: the amount of aggregate consumption expenditure that would occur even if income dropped to zero
marginal propensity to consume
: the share of the additional dollar of income a person decides to devote to consumption expenditures
marginal propensity to save
: the share of the additional dollar of income a person decides to save |
1309 | https://medium.com/math-simplified/horners-polynomial-method-step-by-step-with-python-df512d34f128 | Sitemap
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## Math Simplified
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Horner’s Polynomial Method Step-by-Step with Python
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Nov 11, 2022
Yesterday I learned from a colleague about an arithmetic tool I didn’t know about. It’s called Horner’s method, and I find it quite cool. Actually, it is designed to efficiently evaluate polynomials at a given point by hand. But it can also be used for polynomial division, polynomial root finding, or for partial fraction decomposition. In this post, I will show how Horner’s method works and give a step-by-step implementation in terms of Python code.
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The Method
Suppose you have the polynomial
and you want to evaluate it at 𝑥=2 by hand. Horner’s method is based on the observation that you can always rewrite the polynomial like this:
If you have trouble seeing this, read it from right to left!
Then you set up a table and write the coefficients of the polynomial in the first row. In the second row below the first coefficient, you write 0. Then you write the point where to evaluate at the left…
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## Published in Math Simplified
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1310 | https://vixra.org/pdf/1306.0192vD.pdf | Mathematical Proof of Four-Color Theorem By Liu Ran Abstract The method and basic theory are far from traditional graph theory. Maybe they are the key factor of success. K4 regions (every region is adjacent to other 3 regions) are the max adjacent relationship, four-color theorem is true because more than 4 regions, there must be a non-adjacent region existing. Non-adjacent regions can be color by the same color and decrease color consumption. Another important three-color theorem is that the border of regions can be colored by 3 colors. Every region has at least 2 optional colors. Every border region can be permuted without impact on other border regions. 1. Introduce How many different colors are sufficient to color the regions on a map in such a way that no two adjacent regions have the same color? After examining a wide variety of different planar graphs, one discovers the apparent fact that every graph, regardless of size or complexity, can be colored with just four distinct colors. The famous four color theorem, sometimes known as the four-color map theorem or Guthrie's problem. There had been numerous attempts to prove the supposition in mathematical history, but these so-called proofs turned out to be flawed. There had been accepted proofs that a map could be colored in using more colors than four, such as six or five, but proving that only four colors were required was not done successfully until 1976 by mathematicians Appel and Haken, although some mathematicians do not accept it since parts of the proof consisted of an analysis of discrete cases by a computer. But, at the present time, the proof remains viable. It is possible that an even simpler, more elegant, proof will someday be discovered, but many mathematicians think that a shorter, more elegant and simple proof is impossible. In the mathematical field of graph theory, a complete graph is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge. A complete digraph is a directed graph in which every pair of distinct vertices is connected by a pair of unique edges (one in each direction). Kn denotes the complete graph on n vertices. K1 through K4 are all planar graphs. However, every planar drawing of a complete graph with five or more vertices must contain a crossing, and the non-planar complete graph K5 plays a key role in the characterizations of planar graphs. 2. Four color theorem (2.1) For any subdivision of the spherical surface into non-overlapping regions, it is always possible to mark each of the regions with one of the numbers 1, 2, 3, 4, in such a way that no two adjacent regions receive the same number. In fact, if the four-color theorem is true on spherical surface, it is also true on plane surface. Because the map is originate from sphere, and plane surface is part of spherical surface. 3. Strategy K4 regions (every region is adjacent to other 3 regions) are the max adjacent relationship, four-color theorem is true because more than 4 regions, there must be a non-adjacent region existing. Non-adjacent regions can be color by the same color and decrease color consumption. Another important theorem is that the border of regions can be colored by 3 colors. Every region has at least 2 optional colors. Every border region can be permuted without impact on other border regions. 4. Basic axiom (4.1) Coloring the regions on a map has nothing to do with the region shape. This is the only one axiom in proof. It’s obviously true. Color only depends on adjacent relationship. Theorem (4.2) All color solutions for boundary adjacent regions can apply to point adjacent regions or non-adjacent regions. We define adjacent regions as those that share a common boundary of non-zero length. Regions, which meet at a single point or limited points, are not considered to be "adjacent". Because point adjacent regions are not considered to be "adjacent", any color solution can apply to point adjacent regions, include the color solution of boundary adjacent regions. The free degree of non-adjacent region is limitless. So any color solution of boundary adjacent regions can apply to point adjacent regions and non-adjacent regions. For example: Scenario a: non-adjacent and point adjacent Scenario b: boundary adjacent All color solutions for Scenario b can apply to Scenario a. Theorem (4.3) Any irregular regions map can transform into a circle regions map. The color solution for circle regions map can also apply to the irregular regions map Because basic axiom (4.1) Þ any irregular regions map can transform into circle-shaped, ring-shaped or fan-shaped. If circle-shaped, ring-shaped or fan-shaped are point adjacent or non-adjacent, transform into boundary adjacent, finally, to transform into a circle map, ring-shaped and fan-shaped surround circle. Because of Theorem (4.2), the color solution of map transformed can apply to the color solution of map transforming before. For example: This an irregular map. To ensure arbitrary map can be transform into circle map, first select a circle center, second draw a line from center to region, the least region number crossed over is the layer number of ring. From 1 to 6, the least region number is 2, from 1 to 8, the least region number is 2, and so both 6 and 8 are in layer 3 in circle map. Other regions are in layer 2. Layer Region Layer 1 1 Layer 2 2,3,4,5,7 Layer 3 6,8 To ensure to preserve the adjacent relationship in transforming, Region Adjacent region 1 2,3,4,5,7 2 1,5,6,7 3 1,4,7 4 1,3,7,8,5 5 1,2,4,6,7,8 6 2,5,7 7 1,2,3,4,5,6,8 8 4,5,7 In circle map, the necessary condition of adjacent relationship is between 2 adjacent layers, or between 2 adjacent regions (left, right) in the same layer. Such as below: Region 1 (layer 1) is adjacent to 2,3(layer 2). Region 3 (layer 2) is adjacent to 2,4(layer 2). If regions are not in the adjacent layer or more than 2 regions in the same layer, they are sure to be not adjacent. Such as, 6(layer 3) and 1(layer 1) are not adjacent; 2 and 4 are not adjacent in layer 2, because there are 3,4,5 in layer 2, they can’t all adjacent to 1. With the 2 necessary condition of adjacent relationship, check the table one item by one item. 1 2,3,4,5,7 2 1,5,6,7 3 1,4,7 4 1,3,7,8,5 5 1,2,4,6,7,8 6 2,5,7 7 1,2,3,4,5,6,8 8 4,5,7 This is the method to check region 7 and others are similar. First, remove regions in adjacent layers. 7 2,3,4,5 Because 2,3,4,5 are in the same layer and total 4 > 2, the region 7 can’t preserver adjacency. To preserver adjacency, it must to cross layers by region 4, 5 or 7. Go back to original map. Region 7 is across over region 6 to adjacent to 5, and is across over region 3 to adjacent to 4. The final map is below. It equal to the standard circle map below. Transform irregular map to circle map, 1 is circle center, 6 and 8 are in layer 2, 7 is across layer 2,3,4. Other regions are in layer 1. The boundary adjacent relation is never changed, but some point adjacent or non-adjacent relations are changed to boundary adjacent relation to match the circle map transforming. The color solution for circle map transformed can apply to irregular map also. Region 6 has changed color, but there is no same color between boundary adjacent regions. It is a color solution qualified. 5. Terminology To describe conveniently, I have defined some terms in circle map. Solution(G(n), color1,color2,…,colork) is a color solution qualified to color all of n regions of G(n) by color in {color1,color2,…,colork}. G(n) is the n of adjacent regions in map. Non-adjacent regions as those no point met. For example: 1 is non-adjacent to 2 in below circle map. Point adjacent regions as those that meet at a single point or limited points. Point adjacent regions are also non-adjacent. For example: 1 is point adjacent to 2 in below circle map. Boundary adjacent regions as those that share a common boundary of non-zero length. For example: 1, 2, 3 are all boundary adjacent to other 2 regions in below circle map. Covered is that the least regions in upper ring have covered one nation in lower ring. Especially, the least N regions in upper ring covering 1 nation in lower ring calls N Covered, all the regions in upper ring covering 1 nation in lower ring calls full Covered. For example: 1, 2, 3, 4, 5, 6 are covering 7 in below circle map, that is 6 covering. Supported is that the least regions in lower ring have covered one nation in upper ring. Especially, the least N regions in lower ring covering 1 nation in upper ring calls N Supported, all the regions in lower ring covering 1 nation in upper ring calls full Supported. For example: 1, 2, 3, 4, 5, 6 are supporting 7 in below circle map, that is 6 supporting. Color is to color region by one or more than one colors. It is recoded as Color(region) = { color}. If a region can be colored by more than 1 colors, it can be recoded as Color(region) = {color1/color2…/colork}. Such as Color(3) = {yellow/green/gray}. Region 3 is colored by {yellow} now, but region 3 has the freedom to color by { gree} or {gray}. Main color is {color1} in Color(region) = {color1/color2…/colork}, which color the region in fact. Backup color is {color2…/colork} in Color(region) = {color1/color2…/colork}, which doesn’t color the region in fact, but which has the freedom to color by {color2…/colork}. Optional colors are all of main color and backup color. Region number is the total region number of a ring. For example: region number of ring 3 is 7 in below circle map. Record as Region(3) = 7. Border regions are all the boundary regions between colored and uncolored. It’s the frontier of regions colored. Inner regions are the entire boundary regions closed by Border regions. It’s the home of regions colored. For example: Regions 1 to 11 are colored, the Border regions are marked as yellow color, which are close border to seal all regions colored. Inner regions are marked as gray color. Adjacent border regions are a region being adjacent to border regions, there are 1 or 2 regions are adjacent to the region, but not full covered. For example: Region 5 and 9 are adjacent to region 10, but not covered by region 10. Region 5 and 9 are adjacent border regions. Empty region is a point, which is not a real region, only a proving tool. For example, 3 regions is equivalent to 3 regions and a Empty region. Kn regions are k regions are all adjacent. Anyone of k region is adjacent to other k-1 regions. For example: k2 regions are in below circle map. k3 regions are in below circle map. Anyone region is adjacent to other two regions. K4 regions are in below circle map. Anyone region is adjacent to other three regions. Graph theory has proven K4 regions are the max adjacent relationship in planar graph. 6. Preliminary theorem (6.1) K4 regions have only 3 scenarios in circle map. Because in a ring, one region can only adjacent to 2 regions at most => there are at most 3 regions in a ring, but K4 regions have 4 regions > 3 => K4 regions are at least in 2 rings. If total of rings >=3, there must be one ring insulating another ring. =>There must be 2 regions being non-adjacent. => Total of rings <=2, Because total of rings >=2 and total of rings < 2 => total of rings =2. Total of rings =2 and K4 regions have 4 regions => K4 regions have only 3 scenarios in circle map. I.e. (6.1.1) region(1) = 1, region (2) = 3. (6.1.2) region(1) = 2, region (2) = 2. (6.1.3) region(1) = 3, region (2) = 1. Is there any region across rings? No. Because there is only 2 adjacent rings, regions can keep adjacent relationship in adjacent rings, don’t’ need to cross rings. (6.2) K3 regions have only 3 scenarios in circle map. Similarly, we can get 3 scenarios. (6.2.1) region (1) = 3. (6.2.2) region(1) = 1, region (2) = 2. (6.2.3) region(1) = 2, region (2) = 1. To prove conveniently, we can unify (6.1) and (6.2). For K3 regions, we regard there is a empty region in inner regions. Then K3 regions become K4 regions (6.2.1) region (1) = 1 empty region , region (2) = 3. (6.2.2) region (1) = 1+ 1 empty region, region (2) = 2. (6.2.3) region (1) = 2+ 1 empty region, region (2) = 1. (6.3) Three-color theorem G(n) is the n of adjacent regions in map. B(k) is the k of adjacent border regions of G(n). P(n): to a g(n) map with b(k) border regions, there is a Solution(G(n), 1,2,3,4) 1. Three colors {1,2,3} are sufficient to border regions, i.e. there is a Solution(b(k), 1,2,3), which is satisfied Solution(G(n), 1,2,3,4) also; 2. Every region has at least 2 optional colors, i.e. if region color(r(i))={1/2}, there are 2 solutions, Solution1(G(n), 1,2,3,4) and Solution2(G(n), 1,2,3,4) . Solution1 satisfies color(r(i))={1} and Solution2 satisfies color (r(i))={2} 3. Every border region can change into another optional color, which is no impact on other border regions. I.e if border region Color (b(i))={1/2}, there are 2 solutions, Solution1(b(k), 1,2,3), Solution2(b(k), 1,2,3). The only difference of them is Color (b(i))={1} and Color (b(i))={2}. Such as a K4 regions, the possible color is below Change one border region color, other border regions are intact. E.g. When add a region, which is adjacent to 1 border region and has 3 optional colors. When the region changes to other optional colors, which is no impact on other border regions and inner regions. When add a region, which is adjacent to 2 border regions and has 2 optional colors. When the region changes to other optional colors, which is no impact on other border regions and inner regions. When add a region, which is adjacent to 3 border regions and has 3 optional colors. Only explain this scenario, others are similar. 1. To be colored by {4}. 2. To be colored by {3}. 3. To be colored by {2}. The region colored by {1} is intact and {2/3/4} can color the region added. Change the region color into another optional color, other border regions are intact. When add a region, which is full covering all border regions and has 4 optional colors, because there is no border regions except for the region added now. It’s easy to verify every region in border regions has at least 2 optional colors when total of regions is not above 5. (6.3.1) P(k): when total of regions equals to k(k>5), (6.3) Three-color theorem is true. Then add the (k+1)th region, total of regions become k+1. Then we can deduce P(k+1) from P(k). If the (k+1)th region is adjacent to 1 region of border regions, it has 3 optional colors. Change the region color to other optional color, which is no impact on other border regions and inner regions. If the (k+1)th region is adjacent to 2 colors of border regions, it has 2 optional colors. Change the region color to other optional color, which is no impact on other border regions and inner regions. If the (k+1)th region is adjacent to more than 2 colors of border regions, it has at least 2 optional colors. The proof is below. (k+1)th region’s color depends on the adjacent border regions. E.g regions in yellow color. {1,2,3,4} – {1,2} = {3/4} Because (k+1)th region’s color is adjacent to 3 colors, {4} is OK. How to color {3}? First to color (k+1)th region as {3}. Because (6.3.1), border regions of G(k) have at least 2 optional colors and border regions of G(k) being full covered by (k+1)th region and colored by {3} can change to other optional color and this color must in {1,2,4}. Other border regions of G(k) are intact. Then (k+1)th region can be colored by {3/4} and colors of border regions of G(k+1) is still in {1,2,3}. Change the region color to other optional color, which is impact on inner regions, and other border regions are intact. So when total of regions equals to k+1, (6.3) Three-color theorem is true. When n <4, 3 colors are sufficient. When n = 4, 3 colors are sufficient for border regions. When n = 5, we can verify 3 colors are sufficient for border regions. P(k): if n = k, 3 colors are sufficient for border regions, we can deduce P(k+1): n = k+1, 3 colors are sufficient for border regions. So to all of n, in G(n) map, (6.3) Three-color theorem is true. Some people wonder how the inner regions’ color permutes? We can find that the optional colors depend on the adjacent border regions. For example: The region colored by {4/3} = {1,2,3,4} – {1,2}. When color change from {4} to {3}, {3} need to change, it is no impact to its adjacent border regions, but only to inner regions. For example: Red arrow denotes no impact; green arrow denotes impact. It’s the same as next region permuted, which is no impact on adjacent border regions, but only inner regions. Base on the algorithm, we can guarantee the core center is a K4 regions, which can guarantee at least 1 region with 4 optional colors, and any impact will stop here. E.g. All impacts are towards inner regions, can stop at the region with more optional colors. The worst scenario is impact stop at the region with 4 optional colors, which can stop any type of impact. The better scenario is impact stop at the region, which is no impact on inner regions. More impacts are similar to 1 impact, because more impacts can be handled one by one. The region with 4 optional colors still exists. E.g. (6.4) K4 regions are the max adjacent relationship in circle map. Traditional graph theory has proven this theorem. Here is a new proof. (6.4.1) Suppose K5 regions are the max adjacent relationship in circle map. Base on the K4 regions scenario of (6.1.1), let me to add a new country in circle map, which is K5 regions. (6.4.1.1) The new country is in ring 1, like below circle map. Because they are K5 regions, country 1 is full covered by country 2,3,4 and country 5 is also full covered by country 2,3,4. 3 countries cover both country 1 and 5. And countries 2,3,4 are in a ring, so the overlap countries are at most 2 countries. Then the total countries in ring 2 , denoted as Country(2) >= 3+3-2 = 4>3. It’s contradictory. (6.4.1.2) The new country is in ring 2, like below circle map. Because countries 2,3,4,5 are in the same ring 2, country 5 can be only Boundary adjacent with 2 countries in the same ring. But there are 3 other counties in the ring 2, so one country must be Non-adjacent with country 5. (6.4.1.3) The new country is in a new ring 3, like below circle map. Country 5 is a new country in new ring. Obviously, country 5 is always Non-adjacent with country 1, because they are in ring 1 and ring 3, ring 2 has insulated them. (6.4.1.4) The new country is in a new ring k, (k>3), like below circle map. Country 5 is a new country in new ring. Obviously, country 5 is always Non-adjacent with country 1, because they are in ring 1 and ring k, ring 2 has insulated them. (6.4.1.5) The new country is across ring 2, like below circle map. We can divide country 5 into multiple countries in each ring like below. Similar to scenario (6.4.1.2), because countries 2,3,4,52 are in the same ring 2, country 52 can be only Boundary adjacent with 2 countries in the same ring. Select one of them (such as country 4) to find another non-adjacent country. Country 4 can be only Boundary adjacent with 2 countries in the same ring. But there are 3 other counties in the ring 2, so one country must be Non-adjacent with country 4. (6.4.1.6) The new country is not across ring 2, like below circle map. Similar to scenario (6.4.1.3), country 5 is always Non-adjacent with country 1, because ring 2 has insulated them. The same method can prove scenarios of (6.1.2) and (6.1.3). All scenarios are contradictory, so supposition (6.4.1) is false and K4 regions are the max adjacent relationship in circle map. 7. Four color theorem There are N regions. Firstly, search max adjacent relationship. If no K2 regions, all regions are non-adjacent, one color is sufficient. Max adjacent relationship is K2 regions, all regions are at most adjacent to one region, and two colors are sufficient. Max adjacent relationship is K3 regions, we can add a empty region in ring 1, it becomes K4 regions. border regions are formed. Max adjacent relationship is K4 regions, because K4 regions have two rings, border regions are formed now. Base on K4 regions, we can select a region which is adjacent to border regions, and don’t stop coloring these regions one by one until all of the N regions are colored. Because of (6.3) Three-color theorem, we can use 3 colors in {1,2,3} to color border regions and 4 colors in {1,2,3,4} to color inner regions. Above proof indicates border regions’ colors are in {1,2,3}, and inner regions’ colors are in {1,2,3,4}. Finally, four-color theorem is proven now! 8. Verification and Demo One example to verify and explain: To descript clearly, all regions are marked the number by the color order in advance. The order number (region number) map is below, The algorithm bases on (7.4), is simpler than the example in section 7. 1. Search and color max adjacent relationship of complete graph. 2. Find non-adjacent regions and colored with more possibilities. Firstly, search K4 regions and colored by {1,2,3,4} Next, we can select an adjacent region c5, which is adjacent to 3 regions in border. 2 adjacent border regions are {2,3}. So it is colored by color of inner region {4/1}. Next, we can select an adjacent region c6, which is adjacent to 2 regions in border. 2 adjacent border regions are {3,4}, so color is {2/1}. Next, we can select an adjacent region c7, which is adjacent to 3 regions in border regions. adjacent border regions are {2,3}, so color is {1/4}. Next, we can select an adjacent region c8, which is adjacent to all border regions. It is full covered all border regions. So it can be colored by {4/1/2/3}. Every region has at least 2 optional colors. All regions use 4 colors. 9. Conclusion Four-color theorem is an interesting phenomenon, but there is a rule hidden the phenomenon. The max adjacent relationship on a surface decides how many colors are sufficient. More than max adjacent regions, there is a non-adjacent region. The non-adjacent region can decrease color consumption. Every region has at least 2 optional colors, which can be permuted with only impacts on inner regions. REFERENCES AHLFORS, L.V. Complex Analysis, McGraw-Hill Book Company, 1979. DHARWADKER, A. & PIRZADA, S. Graph Theory, Orient Longman and Universities Press of India, 2008. LAM, T.Y. A First Course in Noncommutative Rings, Springer-Verlag, 1991. ROTMAN, J.J. An Introduction to the Theory of Groups, Springer-Verlag, 1995. VAN LINT, J.H. & WILSON, R.M. A Course in Combinatorics, Cambridge University Press, 1992. |
1311 | https://books.google.mw/books?id=Az9RAAAAMAAJ&source=gbs_book_other_versions_r&cad=4 | Advanced Inorganic Chemistry: A Comprehensive Text - Frank Albert Cotton, Geoffrey Wilkinson - Google Books
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Advanced Inorganic Chemistry: A Comprehensive Text
Frank Albert Cotton, Geoffrey Wilkinson
Interscience Publishers, 1966 - 1136 pages
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Contents
The Electronic Structures of Atoms 3
The Nature of Ionic Substances 35
The Nature of Chemical Bonding 57
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Advanced Inorganic Chemistry: A Comprehensive Text
Frank Albert Cotton,Geoffrey Wilkinson
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alkali metalalkylAmerammoniaanionsaqueous solutionatomic orbitalsbond energyboronbridgingcarboncarbonylscationsChemchemicalchemistrychloridecolorlesscomplexescompoundsconfigurationcontainingcoordination numbercovalentcrystallinecrystalsdiamagneticdimericdiscusseddissociationdissolvesdistorteddonorelectron pairselectronegativityelectronic structureelementsequationequilibriumexampleexistfluorineformationgivehalideshalogensheatinghydratedhydrideshydrogen atomhydrogen bondshydrolysishydrolyzedhydroxideimportantInorginsolubleinteractionionicionization potentialsisomerskcal/moleknownlatticelattice energiesligandsliquidmagneticmetal atomsmetal ionsmolecularmolecular orbitalmoleculesnitrogenobtainedoccursoctahedralolefinoverlapoxidationoxygenoxygen atomsparamagneticpolymericpropertiesprotonreactreactionreactivereadilyreducingresonancesaltsshown in Figuresimilarsodiumsolidsolublesolventsspeciesspectraspinstablestoichiometrysulfidesulfursulfuric acidsymmetryTabletemperaturetetrahedraltransition metaltrigonalunpaired electronsvalencevalues
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Title Advanced Inorganic Chemistry: A Comprehensive Text
AuthorsFrank Albert Cotton, Geoffrey Wilkinson
Edition 2
Publisher Interscience Publishers, 1966
Original from the University of Michigan
Digitized 26 Nov 2007
Length 1136 pages
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1312 | https://pubmed.ncbi.nlm.nih.gov/16148731/ | Hepatitis B virus-associated polyarteritis nodosa: clinical characteristics, outcome, and impact of treatment in 115 patients - PubMed
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. 2005 Sep;84(5):313-322.
doi: 10.1097/01.md.0000180792.80212.5e.
Hepatitis B virus-associated polyarteritis nodosa: clinical characteristics, outcome, and impact of treatment in 115 patients
Loïc Guillevin1,Alfred Mahr,Patrice Callard,Pascal Godmer,Christian Pagnoux,Emmanuelle Leray,Pascal Cohen;French Vasculitis Study Group
Affiliations Expand
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1 From Department of Internal Medicine (LG, AM, PG, CP, P Cohen), Hôpital Cochin, Assistance Publique-Hôpitaux de Paris, Paris; and UPRES 3409 "Recherche Clinique et Thérapeutique" (LG, AM, PG, CP, P Cohen), Université Paris-Nord, Bobigny; Department of Pathology (P Callard), Hôpital Tenon, Paris; and Department of Public Health (EL), Faculté de Médecine, Rennes, France.
PMID: 16148731
DOI: 10.1097/01.md.0000180792.80212.5e
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Comparative Study
Hepatitis B virus-associated polyarteritis nodosa: clinical characteristics, outcome, and impact of treatment in 115 patients
Loïc Guillevin et al. Medicine (Baltimore).2005 Sep.
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. 2005 Sep;84(5):313-322.
doi: 10.1097/01.md.0000180792.80212.5e.
Authors
Loïc Guillevin1,Alfred Mahr,Patrice Callard,Pascal Godmer,Christian Pagnoux,Emmanuelle Leray,Pascal Cohen;French Vasculitis Study Group
Affiliation
1 From Department of Internal Medicine (LG, AM, PG, CP, P Cohen), Hôpital Cochin, Assistance Publique-Hôpitaux de Paris, Paris; and UPRES 3409 "Recherche Clinique et Thérapeutique" (LG, AM, PG, CP, P Cohen), Université Paris-Nord, Bobigny; Department of Pathology (P Callard), Hôpital Tenon, Paris; and Department of Public Health (EL), Faculté de Médecine, Rennes, France.
PMID: 16148731
DOI: 10.1097/01.md.0000180792.80212.5e
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Hepatitis B virus-associated polyarteritis nodosa (HBV-PAN) is a typical form of classic PAN whose pathogenesis has been attributed to immune-complex deposition with antigen excess. We conducted the current study to 1) analyze the frequency of HBV infection in patients with PAN, in light of the classification systems described since 1990; 2) describe the clinical characteristics of HBV-PAN; 3) compare the evolution according to conventional or antiviral treatment; and 4) evaluate long-term outcome. One hundred fifteen patients were included in therapeutic trials organized by the French Vasculitis Study Group and/or referred to our department for HBV-PAN between 1972 and 2002. To determine the frequency of HBV-PAN during the 30-year period, we analyzed a control group of patients with PAN without HBV infection, followed during the same period and diagnosed on the same bases. Depending on the year of diagnosis, different treatments were prescribed. Before the antiviral strategy was established, some patients were given corticosteroids (CS) with or without cyclophosphamide (CY). Since 1983, treatment for patients with HBV markers has combined 2 weeks of CS followed by an antiviral agent (successively, vidarabine, interferon-alpha, and lamivudine) combined with plasma exchanges (PE).Ninety-three (80.9%) patients entered remission during this period and 9 (9.7%) of them relapsed; 41 (35.7%) patients died. For the 80 patients given the antiviral strategy as intention-to-treat, 4 (5%) relapsed and 24 (30%) died vs 5 (14.3%) relapses (not significant [NS]) and 17 (48.6%) deaths (NS) among the 35 patients treated with CS alone or with CY or PE. HBe-anti-HBe seroconversion rates for the 2 groups, respectively, were: 49.3% vs 14.7% (p < 0.001). Patients who seroconverted obtained complete remission and did not relapse.Thus, HBV-PAN, a typical form of classic PAN, can be characterized as follows: when renal involvement is present, so is renal vasculitis; glomerulonephritis due to vasculitis is never found; antineutrophil cytoplasmic antibodies (ANCA) are not detected; relapses are rare, and never occur once viral replication has stopped and seroconversion has been obtained. Combining an antiviral drug with PE facilitates seroconversion and prevents the development of long-term hepatic complications of HBV infection. The major cause of death is gastrointestinal tract involvement. Importantly, the frequency of HBV-PAN has decreased in relation to improved blood safety and vaccination campaigns.
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References
Bourgarit A, Le Toumelin P, Pagnoux C, Cohen P, Mahr A, Le Guern V, Mouthon L, Guillevin L. Deaths occurring during the first year after treatment onset for polyarteritis nodosa, microscopic polyangiitis, and Churg-Strauss syndrome. A retrospective analysis of causes and factors predictive of mortality based on 595 patients. Medicine (Baltimore). 2005;84:323-330.
Darras-Joly C, Lortholary O, Cohen P, Brauner M, Guillevin L. Regressing microaneurysms in 5 cases of hepatitis B virus-related polyarteritis nodosa. J Rheumatol. 1995;22:876-880.
Gocke DJ, Hsu K, Morgan C, Bombardieri S, Lockshin M, Christian CL. Association between polyarteritis and Australia antigen. Lancet. 1970;2:1149-1153.
Guillevin L, Jarrousse B, Lok C, Lhote F, Jais JP, Le Thi Huong D, Bussel A. Longterm followup after treatment of polyarteritis nodosa and Churg-Strauss angiitis with comparison of steroids, plasma exchange and cyclophosphamide to steroids and plasma exchange. A prospective randomized trial of 71 patients. The Cooperative Study Group for Polyarteritis Nodosa. J Rheumatol. 1991;18:567-574.
Guillevin L, Lhote F, Cohen P, Sauvaget F, Jarrousse B, Lortholary O, Noel LH, Trepo C. Polyarteritis nodosa related to hepatitis B virus. A prospective study with long-term observation of 41 patients. Medicine (Baltimore). 1995;74:238-253.
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doi 10.17188/1201914
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Energy Above Hull0.727 eV/atom
Space GroupP4_232̅
Band Gap0.00 eV
Predicted Formation Energy0.727 eV/atom
Magnetic OrderingFerromagnetic
Total Magnetization0.50 µB/f.u.
Experimentally ObservedYes
##### Description (Auto-generated) A machine-generated description of this crystal structure written by your friendly Robocrystallographer! 🤖
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Crystal Structure
Lattice (Conventional)
Lengths and angles defining the conventional lattice of this material.
a 5.93 Å
b 5.93 Å
c 5.93 Å
α 90.00 º
β 90.00 º
ɣ 90.00 º
Volume 209.03 ų
Lattice is given in its
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Fractional coordinates of the atoms in the conventional unit cell.
| Wyckoff | Element | x | y | z |
--- ---
| 4b | As | 1/4 | 3/4 | 3/4 |
| 12j | H | 0.320388 | 1/2 | 0 |
Symmetry
Crystal System Cubic
Lattice System Cubic
Hall Number P 4n 2 3
International Number 208
Symbol P4_232̅
Point Group 432
Number of Atoms 16
Density 2.48 g·cm⁻³
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The DOI for this specific material is 10.17188/1201914 and a BibTeX record can be retrieved here. This entry is for this specific material and can be used in addition to the main Materials Project citation.
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1314 | https://artofproblemsolving.com/wiki/index.php/Principle_of_Inclusion-Exclusion?srsltid=AfmBOooC7CIiqh-ycFJtbTxdlalflx9ROlK0aYZREAq7mheKtCr-MINj | Art of Problem Solving
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Principle of Inclusion-Exclusion
The Principle of Inclusion-Exclusion (abbreviated PIE) provides an organized method/formula to find the number of elements in the union of a given group of sets, the size of each set, and the size of all possible intersections among the sets.
Contents
[hide]
1 Important Note(!)
2 Application
2.1 Two Set Example
2.2 Three Set Example
2.3 Four Set Example
2.3.1 Problem
2.3.2 Solution
2.4 Five Set Example
2.4.1 Problem
2.4.2 Solution
3 Statement
4 Proof
5 Remarks
6 Examples
7 See also
Important Note(!)
When using PIE, one should understand how to strategically overcount and undercount, in the end making sure every element is counted once and only once. In particular, memorizing a formula for PIE is a bad idea for problem solving.
Application
Here, we will illustrate how PIE is applied with various numbers of sets.
Two Set Example
Assume we are given the sizes of two sets, and , and the size of their intersection, . We wish to find the size of their union, .
To find the union, we can add and . In doing so, we know we have counted everything in at least once. However, some things were counted twice. The elements that were counted twice are precisely those in . Thus, we have that:
.
Three Set Example
Assume we are given the sizes of three sets, and , the size of their pairwise intersections, , and , and the size their overall intersection, . We wish to find the size of their union, .
Just like in the Two Set Example, we start with the sum of the sizes of the individual sets . We have counted the elements which are in exactly one of the original three sets once, but we've obviously counted other things twice, and even other things thrice! To account for the elements that are in two of the three sets, we first subtract out . Now we have correctly accounted for them since we counted them twice originally, and just subtracted them out once. However, the elements that are in all three sets were originally counted three times and then subtracted out three times. We have to add back in . Putting this all together gives:
.
Four Set Example
Problem
Six people of different heights are getting in line to buy donuts. Compute the number of ways they can arrange themselves in line such that no three consecutive people are in increasing order of height, from front to back. (2015 ARML I10)
Solution
Let be the event that the first, second, and third people are in ordered height, be the event that the second, third, and fourth people are in ordered height, be the event that the third, fourth, and fifth people are in ordered height, and be the event that the fourth, fifth and sixth people are in ordered height. By a combination of complementary counting and PIE, we have that our answer will be . Now for the daunting task of evaluating all of this.
For , we just choose people and there is only one way to put them in order, then ways to order the other three guys for . Same goes for , , and .
Now, for , that's just putting four guys in order. By the same logic as above, this is . Again, would be putting five guys in order, so . is just choosing guys out of , then guys out of for . Now, is just the same as , so , is so , and is so .
Moving on to the next set: is the same as which is , is ordering everybody so , is again ordering everybody which is , and is the same as so .
Finally, is ordering everybody so .
Now, lets substitute everything back in. We get a massive expression of .
Five Set Example
Problem
There are five courses at my school. Students take the classes as follows: 243 take algebra. 323 take language arts. 143 take social studies. 241 take biology. 300 take history. 213 take algebra and language arts. 264 take algebra and social studies. 144 take algebra and biology. 121 take algebra and history. 111 take language arts and social studies. 90 take language arts and biology. 80 take language arts and history. 60 take social studies and biology. 70 take social studies and history. 60 take biology and history. 50 take algebra, language arts, and social studies. 50 take algebra, language arts, and biology. 50 take algebra, language arts, and history. 50 take algebra, social studies, and biology. 50 take algebra, social studies, and history. 50 take algebra, biology, and history. 50 take language arts, social studies, and biology. 50 take language arts, social studies, and history. 50 take language arts, biology, and history. 50 take social studies, biology, and history. 20 take algebra, language arts, social studies, and biology. 15 take algebra, language arts, social studies, and history. 15 take algebra, language arts, biology, and history. 10 take algebra, social studies, biology, and history. 10 take language arts, social studies, biology, and history. 5 take all five. None take none.
How many people are in my school?
Solution
Let A be the subset of students who take Algebra, L-languages, S-Social Studies, B-biology, H-history, M-the set of all students. We have:
Thus, there are people in my school.
Statement
If are finite sets, then:
.
Proof
We prove that each element is counted once.
Say that some element is in sets. Without loss of generality, these sets are
We proceed by induction. This is obvious for
If this is true for we prove this is true for For every set of sets not containing with size there is a set of sets containing with size In PIE, the sum of how many times these sets are counted is There is also one additional set of sets so is counted exactly once.
Remarks
Sometimes it is also useful to know that, if you take into account only the first sums on the right, then you will get an overestimate if is odd and an underestimate if is even. So,
,
,
,
and so on.
Examples
2011 AMC 8 Problems/Problem 6
2017 AMC 10B Problems/Problem 13
2005 AMC 12A Problems/Problem 18
2001 AIME II Problems/Problem 9
2002 AIME I Problems/Problem 1
2020 AIME II Problems/Problem 9
2001 AIME II Problems/Problem 2
2017 AIME II Problems/Problem 1
See also
Combinatorics
Overcounting
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ISEEPre-High SchoolTest Prep
ISEE, SSAT, and HSPT Synonym Questions: Avoiding Traps
ByMatt MadsenOctober 3, 2024 March 14th, 2025No Comments
Synonym questions are a staple of the Independent School Entrance Exam (ISEE), the Secondary School Admission Test (SSAT), and the High School Placement Test (HSPT). These questions ask students to identify a word that means the same, or nearly the same, as the “stem” word provided. While the concept may seem straightforward, test-takers often fall into traps set by incorrect answer choices designed to mislead. Recognizing these pitfalls can drastically improve performance.
Let’s dive into some common traits of incorrect answer choices, how they can trick you, and what strategies to use to avoid them.
1. The Prefix or Suffix Trap
One of the most common tricks in synonym questions is offering an answer choice that shares the same prefix or suffix as the stem word. While these words might look familiar, their meanings are often quite different. Prefixes like “un-,” “dis-,” or “re-” can throw off test-takers who might equate similarity in structure with similarity in meaning.
Example:
Stem Word: “Unlikely”
A) Uncertain
B) Unfinished
C) Impossible
D) Implausible
Explanation: The word “unlikely” means improbable or not expected to happen. The prefix “un-” in “unlikely” can mislead students into picking another “un-” word like B) Unfinished, which means incomplete and doesn’t fit the meaning. The correct answer is D) Implausible, which also means improbable.
How to Avoid It:
Focus on the meaning of the word as a whole rather than getting distracted by familiar prefixes or suffixes. Ask yourself, “What does this word actually mean?”
2. The Rhyme or Sound-Alike Trap
Test writers often include answer choices that rhyme with or sound like the stem word. While these options may feel intuitively correct because of their phonetic similarity, they rarely have any connection to the actual meaning of the word.
Example:
Stem Word: “Frugal”
A) Fragile
B) Cruel
C) Thrifty
D) Fruitful
Explanation: The word “frugal” means economical or sparing in use of resources. Answer A) Fragile sounds similar but means delicate or easily broken, which is unrelated to frugality. Answer D) Fruitful could also rhyme but means productive or useful, also off the mark. C) Thrifty is the correct synonym, as it also refers to being economical or careful with money.
How to Avoid It:
Ignore how the word sounds and focus entirely on its meaning. If a word feels right just because it “sounds right,” take a step back and reevaluate.
3. The Complexity Trap
Sometimes, answer choices are designed to seem equally “complex” as the stem word. This can trick students into assuming that a word with a longer or more sophisticated appearance must be correct simply because it looks impressive. However, complexity doesn’t equal correctness.
Example:
Stem Word: “Perplexed”
A) Confused
B) Complicated
C) Perturbed
D) Presumptuous
Explanation: The word “perplexed” means confused or puzzled. Students might be drawn to D) Presumptuous because it looks complex and shares a similar-sounding “pre-” root, but it means arrogant or overstepping bounds. The correct answer is A) Confused, which directly matches the meaning of perplexed.
How to Avoid It:
Don’t let a word’s length or complexity fool you. Focus solely on the meaning, not on how sophisticated or technical the word might appear.
Tips for Success
Break Down the Stem Word: Before looking at the answer choices, make sure you clearly understand the meaning of the stem word. Try to think of a synonym in your head first.
Eliminate Distractors: Use process of elimination. If a choice shares the same prefix or suffix, sounds similar, or just seems complicated, scrutinize it carefully to make sure it fits the meaning.
Context Is Key: Sometimes thinking of how you’ve seen the word used in context can help clarify its meaning. Consider what situations you’ve encountered the word in before.
Trust Your Instincts, But Verify: If an answer feels right because of how it sounds or looks, take an extra moment to cross-check the actual meaning. Your instincts are helpful but should always be confirmed by logic.
Final Thoughts
Synonym questions on the ISEE, SSAT, and HSPT may seem tricky, but by recognizing common pitfalls and focusing on meaning over appearance, you can avoid the traps. Always remember: prefixes, sounds, and word length can be deceiving. By sharpening your attention to the word’s actual meaning, you’ll navigate these questions more confidently and accurately.
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1316 | https://www.youtube.com/watch?v=m-Qi2naWKLE | Calculus 3: Using the second derivative test to find the local min/max or saddle points
bprp calculus basics
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Posted: 22 Oct 2024
This calculus 3 tutorial covers the second derivative test for a multivariable function f(x,y) to determine whether a critical point is a local minimum, local maximum, or a saddle point. We will first review the second derivative test and then work out two detailed examples to help you understand how it works. Subscribe to @bprpcalculusbasics for more calculus lessons.
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0:00 Revie of the second derivative test (calculus 3 version)
2:20 example 1 (from James Stewart calculus textbook)
8:49 example 2 (from Larson calculus textbook)
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Transcript:
Revie of the second derivative test (calculus 3 version) today let's talk about the calculus 3 version of
the second derivative test and this right here is to help us find the local min local max or
saddle points of a multivariable function let's talk about the steps first and then later I'll
show you guys an example here we go step one we are going to find the critical point that's a
point AB so that the partial derivative of f with with respect to X and the partial derivative
with respect to Y they both equal to zero then we are going to compute the D value at a
and that is just this formula of course we have to use second derivative and to remember this
formula you can just think about putting this into the determinant FXX FY y on this diagonal and
then we have fxy and fyx but really they are the same thing right the mixed partials are equal but
most likely our case are going to be nice so they are going to be equal and then when you do this
times this it's just FXX FY y minus this times that because they're the same we can put that
as this thing Square after that we are going to pay attention to the D value if the D value is
greater than zero then we have to do more work go ahead and compute the second derivative FXX at
AB if the second derivative is greater than zero then we are going to get a local minimum at AB so
it's just like the Cal version you have a function that's concave up so is local Min and then if the
D5 is greater than zero and also FXX is less than zero then we are going to get a local maximum so
this part is very similar to the Cal stuff now though if the D value is less than zero then you
are going to have a saddle point finally if you get the default is equal to zero unfortunately
we cannot draw any conclusion at a it could be a local Min it could be a local Max or maybe it
could be a SLE point we don't know but anyways let's go ahead and take a look at the example so
this is our example first let's just go ahead and example 1 (from James Stewart calculus textbook) do the partial derivative with respect to X first
and then we will have 4x+ power in the X World why it's a constant so that will be zero and then
right here I can just look at the derivative X which is just equal to one and then we maintain
the constant multiple so minus 4 y the RO one Z doesn't matter and then we are going to set this
to be zero but in the meantime we also need to get F suby and that will be 4 y^ the 3 power and then
minus 4X and again set this equal to zero as well now if you look at this equation we can bring the
4 y to the other side and then divide both sides by four so we get X Cub is equal to Y similarly
do the same thing and we will end up with Y Cub is equal to X now perhaps let's put yal x 3
power this thing right here into this y so we are going to get X cub and then to the third
power and that it's equal to X so now multiply the powers we get 9 and put the X to the other
side so x to the 9 - x is equal to Z Factor on X and then look at this as X to 4 Square so we
have a difference of two squares so we get x X to 4 - 1 x to 4 + 1 and then keep going this right
here is X and then this right here is going to be X2 minus 1 x² + 1 and then X to 4 + 1 = to Z and
then this right here Factor it X - 1 x + 1 x to the 2 + 1 and then x to the 4th + 1 that's equal
to Zer now have a look for the first one we know X is equal to Z for the second one we know X is
equal to 1 for the third one we know X is equal to -1 we will get imaginary number imaginary solution
so we have three x values that we need now when X is equal to zero this right here tells us y
will be 0 to the thir power so that's just equal to zero when X is equal to one this right here
tells us we have y is equal to 1 and then when X is equal to 1 again just do the C power so y it's
also going to be 1 so 0 0 1 1 1 Nega one these are our critical numbers critical points now we are
going to do the second derivative so look at this let's go ahead and do FXX that will give us 12 x²
that's it and then let's also do FY y so look at take the derivative with respect to Y 12 y^ 2
now we need fxy so we have to look at FX and then differentiate this with respect to Y that
will give us zero and then take the derivative of this with respect to Y we will get -4 and in fact
this will be the same as fyx now let's look at the D right here is just going to be FXX f y y and
then minus let's put down fxy just that and then square and let's work our Formula First this right
here is 12 X2 and this right here is 12 y^ 2 and then minus this right here is -4 and then Square
so we're looking at 144 x² y^ 2 - 16 so that's a formula for D of XY now we are just be plugging
these numbers into the D formula right here so D of 0 0 Let's see if I put zero here and zero here
of course that will be zero so it just end up with -6 and guess what that's less than zero so you
can just draw a conclusion right away what is it remember if D is less than zero we get sadle point
at the W 0 0 so let's just write this down saddle Point okay sdle point s0 now D of 11 one let's do this in our head D
of 11 one is going to be 144 1 Square 1 square so just that
and then minus 16 okay 144 - 16 no oh my goodness three and then 14 is
eight okay and then that is two so it's 128 all right so D of 1 one is equal
to positive 128 so that means we have to compute FXX at 1 one so FXX at one one
put one and one in here well we get 12 which is greater than zero now just think about
the K situation when the second derivative is greater than zero it's concap up and then you
get a local minimal so I will just put down local Min local minimum at 1 one finally D of -11 if you put that in here you will
see it's also 128 but if you do fxy -11 well okay actually we still get
positive 12 so we also have a local minimum at11 all right now let's do another example
this is our function and here we go first do example 2 (from Larson calculus textbook) the partial with respect to X we will have - 3x^2
Plus plus 4 Y and then the rest will be zero and then of course set this to be zero next FY that
will be zero this right here will give us 4X and then take the derivative this we get min - 4 Y
and that's it and then set this to be zero now if we take a look at the second equation we can
put this to the other side and then divide both sides by four so in place y Y is equal to X and
I can just put the X right here substitute for the Y so we need to say -3x 2 + 4 x is equal to
zero and I solve this and I can factor out an X and for the first one we know X is equal to
zero and for the second one we get X is equal to positive 4 over 3 now because X and Y are
the same so we have two critical points the first one is 0 comma 0 and the second one is 4
over 3 comma 4 over 3 now let's go ahead and do our second derivatives FXX that will give us -
6X that's it and then perhaps that's also do FY y now will give us4 then that's do fxy which
is the same as fyx and now will just give us let look at the X equation and differentiate
with respect to Y we'll just get four okay now let's go ahead and compute the D which is again
let's remember the formula FXX FY y minus fxy and then squared so this times that we get 6X
-4 and then minus 4 and then squared so that's 24 x - 16 oh it actually does not depend on Y
at all yeah let me double check yeah it really doesn't depend on why so now let's just go ahead
and plug in this value so first if we have d 0 0 that's zero and then we just
get -6 so of course that's less than zero that's great because we can
conclude that we have a stle point at 0 0 right and then next d add 4 over
3 4 three so I will have to plug in 43 into here so 20 24 4 over 3
that's 1 that's 8 that's going to be 32 32 - 16 we have positive 16
right here oh okay we have to do FXX add 4 2 4 3 so plugging 4 for 4 into here we will get 6 4 over 3 I reduce this is one this
is two so we get 8 so that's negative so what do we say we have a local maximum
at 4 3 4 over three right so that's it |
1317 | https://www.ncbi.nlm.nih.gov/sites/books/NBK545289/ | Pursed-lip Breathing - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
StatPearls [Internet].
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
Search term
Pursed-lip Breathing
John D. Nguyen; Hieu Duong.
Author Information and Affiliations
Authors
John D. Nguyen 1; Hieu Duong.
Affiliations
1 Florida Atlantic University
Last Update: January 25, 2025.
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Definition/Introduction
Central and peripheral chemoreceptors govern the primary drive for respiration. Central chemoreceptors, located in the anterior medulla of the brainstem, predominantly respond to a decrease in pH caused by the accumulation of carbon dioxide in the cerebrospinal fluid. The blood-brain barrier protects the central nervous system from external stimuli. Still, the lipid-soluble nature of carbon dioxide allows it to diffuse across this barrier, rapidly influencing respiratory drive.
Peripheral chemoreceptors, in contrast, are more sensitive to changes in oxygen levels. These chemoreceptors consist of the carotid and aortic bodies at the bifurcation of the common carotid artery and the aortic arch. Hypoxia stimulates the carotid bodies, prompting cranial nerve IX to send signals to the nucleus tractus solitarius, thereby triggering ventilation. However, the central chemoreceptors dominate the respiratory drive under normal circumstances as they operate within the body's central control system.
Pursed-lip breathing (PLB) is designed to improve control over oxygenation and ventilation.This breathing technique involves a deliberate inhalation through the nose, followed by a slow, controlled exhalation through puckered or pursed lips, which prolongs the expiratory phase compared to the normal inspiration-to-expiration ratio (see Image. Pursed-Lip Breathing). This process generates backpressure, creating a small amount of positive end-expiratory pressure (PEEP). In PLB, respiratory accessory muscles, such as those in the neck and shoulders, remain relaxed while air is inhaled slowly and deeply through the nose and exhaled gently through rounded lips. The positive pressure generated in the upper airway is effectively transferred to the lower airway, helping to prevent bronchial obstruction and the accumulation of secretions, thereby enhancing respiratory efficiency and reducing dyspnea.
During exhalation, airway collapse in noncartilaginous areas increases airway resistance, leading to carbon dioxide trapping. Elevated carbon dioxide levels stimulate the central chemoreceptors to increase the respiratory rate to restore pH balance to approximately 7.4.While this compensatory mechanism clears carbon dioxide, it may exacerbate air trapping and respiratory muscle fatigue. By creating an artificial splint, PEEP prevents airway collapse, supports alveolar patency, increases surface area for gas exchange, and recruits additional alveoli for effective ventilation.
The positive pressure generated by PLB counteracts the inward forces during exhalation, reducing hypercapnia by facilitating the excretion of carbon dioxide as a volatile acid. PLB relieves shortness of breath, reduces the work of breathing, and improves overall gas exchange. Beyond its physiological benefits, PLB empowers individuals by providing a sense of control over their breathing, enhancing relaxation, and alleviating respiratory distress.
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Issues of Concern
While PLB offers significant benefits for managing dyspnea and improving gas exchange in obstructive pulmonary diseases, several issues of concern should be noted. PLB relies on voluntary initiation and proper execution for effectiveness, requiring individuals to master the technique, which includes maintaining prolonged exhalation and coordinated breathing patterns. The method can exacerbate air trapping and carbon dioxide retention without accurate implementation, counteracting its intended therapeutic effects. Overuse or extended application of PLB can lead to fatigue of the respiratory muscles and, in normal individuals, potentially result in lower-than-normal carbon dioxide levels, reducing cerebral perfusion pressure and risking syncope.
The therapeutic effects of PLB are typically short-lived, and its use is often limited to 3 to 5 breaths due to the potential for muscle fatigue and metabolic strain. In conditions like interstitial lung disease (ILD), acute exposure to PLB during activities such as the 6-minute walk test has not shown significant improvement in symptoms, walking distance, muscle oxygenation, or oxygen saturation compared to normal breathing. Due to increased metabolic demand, PLB was found to reduce walking distance in patients with normoxemia and negatively affect exercise capacity or dyspnea in ILD. These findings highlight the variability in PLB's effectiveness across different patient populations and underscore the importance of tailoring its use to specific clinical contexts.
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Clinical Significance
PLB is a simple yet highly effective technique with extensive clinical applications, particularly for individuals experiencing dyspnea and air trapping. When performed correctly, PLB provides significant relief by allowing patients to regain control over their breathing, reduce the work of breathing, and alleviate respiratory distress. PLB can be easily learned and applied with proper teaching and guidance from trained professionals. In addition to its primary role in relieving carbon dioxide retention and improving oxygenation, PLB is an excellent relaxation tool, helping patients manage anxiety and enhance their overall well-being.
As a versatile and clinically significant technique, PLB is a valuable strategy for managing respiratory conditions. The benefits of this technique extend from offering immediate relief from respiratory symptoms to improving lung function, exercise capacity, and quality of life. Whether employed as a standalone intervention or integrated with other therapies, PLB is an effective, low-cost approach that supports respiratory health across various conditions, including chronic obstructive pulmonary disease (COPD), asthma, and COVID-19.PLB's adaptability and proven efficacy make it an indispensable tool in managing and rehabilitating respiratory diseases.
Role of PLB in Chronic Obstructive Pulmonary Disease
PLB commonly helps patients with COPD.PLB may not be voluntary for these patients, but it is a compensatory mechanism to help splint open the airways.Individuals with COPD may have chronic obstruction of their airways from mucus plugging, loss of integrity of the airways, or enlargement of the airways. These changes in the airways can prevent the appropriate driving pressure and airflow to maintain an adequate clearance of carbon dioxide due to increased airway resistance.The increase in airway resistance also affects inhalation, preventing enough oxygen from reaching the alveoli to create a sufficient partial pressure of oxygen needed to drive the diffusion of oxygen across the alveoli-capillary interface adequately.
The defective driving pressure for oxygenation is further exacerbated due to the retention of carbon dioxide, causing less carbon dioxide to diffuse from the blood into the alveoli for excretion. Blunting the proper mechanism to excrete carbon dioxide and adequate oxygenation stimulates the central chemoreceptors to increase respiration until exhaustion. Chronic hypercapnia decreases the sensitivity of the central chemoreceptors, allowing peripheral receptors sensing oxygen levels to become the predominant drive for respiration.Increased purse lip breathing in these patients may indicate impending respiratory failure.
Significant use of PLB in individuals with COPD may indicate respiratory distress or impending respiratory failure. Recognizing this sign allows clinicians to escalate care, potentially including noninvasive or invasive ventilatory support. Even on mechanical ventilation, incorporating prolonged expiration and minimizing auto-PEEP is critical to prevent complications such as inefficient ventilation or oxygenation. Auto-PEEP can be misinterpreted as high peak pressures, delaying appropriate intervention.
Benefits Beyond COPD
For older adults with stable COPD, PLB training has improved lung function and quality of life. The technique is a valuable component of disease rehabilitation, promoting better symptom management and enhancing functional capacity. Evidence supports using PLB alongside other breathing exercises, such as diaphragmatic breathing (DB), in conditions like asthma and other chronic respiratory illnesses. Combining PLB and DB forms a simple, low-cost therapeutic strategy that improves lung function and exercise capacity. These techniques are particularly beneficial for daily respiratory care and as part of comprehensive pulmonary rehabilitation programs.
Applications in COVID-19
In COVID-19, PLB and DB can play a vital role in rehabilitation. Early pulmonary rehabilitation, including breathing exercises, airway clearance, and physical activity training, can help those hospitalized to maintain functional status, relieve dyspnea, and improve health-related quality of life after discharge. For patients recovering from COVID-19, incorporating PLB into pulmonary rehabilitation can help address dyspnea and support their overall recovery process.
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Nursing, Allied Health, and Interprofessional Team Interventions
Implementing PLB as a therapeutic intervention requires coordinated and ethical collaboration among advanced clinicians, nurses, pharmacists, and other health professionals. Advanced clinicians assess the patient’s respiratory condition, prescribe PLB within a broader care plan, and provide evidence-based guidance to the team. Teaching PLB involves explaining its benefits, potential adverse events, and underlying physiology. A proper demonstration should follow, with the trainee performing or explaining the technique to confirm understanding, ensure proper execution, and allow for correction of any mistakes that may occur during the learning process.This process minimizes errors and optimizes therapeutic efficacy.
Nurses play a vital role in teaching and reinforcing PLB. They monitor patients for respiratory distress signs such as increased PLB use, dyspnea, or accessory muscle engagement. Nurses’ ability to promptly recognize distress ensures rapid initiation of response teams and prevents further deterioration. For conditions such as COPD, congestive heart failure, or panic attacks, PLB can calm the patient, alleviate dyspnea, and potentially reduce the need for noninvasive mechanical ventilation.Further, pharmacists optimize medication regimens to alleviate respiratory symptoms, complementing PLB’s benefits. Rehabilitation Specialists integrate PLB into physical therapy sessions, while mental health professionals address anxiety that may exacerbate dyspnea and hinder technique adherence.
Ethical responsibilities include respecting patient autonomy and addressing barriers like literacy or cultural differences during education. Interprofessional communication and care coordination ensure continuity, with regular team huddles and clear documentation fostering collaboration. By actively teaching and reinforcing PLB, health professionals empower patients to manage their symptoms effectively, improving outcomes, safety, and team performance.
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Nursing, Allied Health, and Interprofessional Team Monitoring
Effective patient monitoring using PLB requires coordinated efforts by nursing staff, allied health professionals, and the interprofessional care team to enhance patient outcomes and ensure safety.
Nursing Role
Nurses play a pivotal role in teaching PLB, explaining its benefits and potential adverse events, and ensuring proper technique through demonstration and patient practice. They should also educate trainees or patients about when PLB is useful, such as during episodes of dyspnea related to COPD, congestive heart failure, or panic attacks. With frequent patient contact, nurses are well-positioned to identify respiratory distress signs, including dyspnea, use of accessory muscles, and increased respiratory rate. Recognizing these signs allows nurses to alert the appropriate response team and physicians for timely intervention, potentially preventing deterioration and the need for mechanical ventilation. Nurses should also monitor for respiratory muscle fatigue in COPD patients, a common consequence of compensatory PLB, and promptly report concerns to the care team.
Allied Health Professionals
Respiratory therapists refine PLB techniques, ensuring patients perform them correctly and tailor interventions to individual needs. Physical and occupational therapists integrate PLB into rehabilitation programs, assessing its utility during functional tasks or exertion. Mental health professionals provide support to address anxiety or panic that may interfere with effective technique use.
Interprofessional Team Coordination
Collaboration among all team members is vital for effective monitoring and patient-centered care. Advanced clinicians, nurses, respiratory therapists, and other allied health professionals should communicate regularly to evaluate patient progress and adjust care plans as needed. Pharmacists can optimize medications that improve respiratory function, complementing PLB's efficacy. The team must also thoroughly address patient questions and concerns, ensuring comprehension of the risks and benefits of PLB. By fostering comprehensive education, vigilant monitoring, and seamless communication, the interprofessional team can maximize the effectiveness of PLB, improve patient outcomes, and enhance safety.
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Figure
Pursed-Lip Breathing. This breathing technique involves a deliberate inhalation through the nose, followed by a slow, controlled exhalation through puckered or pursed lips, which prolongs the expiratory phase compared to the normal inspiration-to-expiration (more...)
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Disclosure:John Nguyen declares no relevant financial relationships with ineligible companies.
Disclosure:Hieu Duong declares no relevant financial relationships with ineligible companies.
Definition/Introduction
Issues of Concern
Clinical Significance
Nursing, Allied Health, and Interprofessional Team Interventions
Nursing, Allied Health, and Interprofessional Team Monitoring
Review Questions
References
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1318 | https://arxiv.org/pdf/1906.11965 | Published Time: Mon, 23 Jan 2023 01:23:06 GMT
Some inequalities for tetrahedra
Jin-ichi Itoh, Jo¨ el Rouyer and Costin Vˆ ılcu July 1, 2019
Abstract
We prove inequalities involving intrinsic and extrinsic radii and diameters of tetrahedra.
1 Introduction
A convex surface S is the boundary of a convex body (compact convex set with interior points) in the Euclidean space R3, or a doubly covered planar convex body; in the latter case it is called degenerate . Denote by S the set of all convex surfaces. The intrinsic metric ρ of a convex surface S is defined, for any points x, y
in S, as the length ρ(x, y ) of a geodesic segment (i.e. , shortest path on S)joining x to y.Denote by Diam ( S) the intrinsic diameter of S ∈ S , and by diam ( S) its extrinsic diameter, Diam ( S) := max
x,y ∈S
ρ(x, y ), diam ( S) := max
x,y ∈S
|| x − y|| .
Denote by Rad ( S) the intrinsic radius of S ∈ S , and by rad ( S) its extrinsic radius, Rad ( S) := min
x∈S
max
y∈S
ρ(x, y ), rad ( S) := min
x∈S
max
y∈S
|| x − y|| .
The first three quantities introduced above proved useful for the study of convex surfaces, as one can briefly see in the following. But the fourth one, rad ( S), seems somehow neglected, despite its natural definition. 1
arXiv:1906.11965v1 [math.MG] 27 Jun 2019
N. P. Makuha showed that Diam( S) ≤ π
2 diam( S)holds for any convex surface S, with equality if and only if S is a surface of revolution having constant width, see for instance for definition and fundamental properties. On the other hand, clearly rad( S) ≤ diam( S) ≤ 2rad S,
and the surfaces satisfying rad( S) = diam( S) have constant width. One also has Rad( S) ≤ Diam( S) ≤ 2Rad( S),
and the surfaces satisfying Rad( S) = Diam( S) are studied in , while those satisfying Diam( S) = 2Rad( S) are studied in in relation to critical points for distance functions. The space T of all the tetrahedra in R3, up to isometry and homothety, is studied in with respect to the number of local maxima of intrinsic distance functions. The intrisic diameter and radius of a regular tetrahedron are computed by J. Rouyer in , while V. Dods, C. Traub, and J. Yang studied geodesics on the regular tetrahedron. An old conjecture of A. D. Aleksandrov states that a convex surface with unit intrinsic diameter and largest area is a doubly covered disk. V. A. Zalgaller proved that among all tetrahedra with unit intrinsic diameter, only the regular tetrahedron with edges of length √3/2 has the largest surface area, which is equal to 3 √3/4. In this note we prove new inequalities involving radii and diameters of tetrahedra. In the next section we present some necessary preliminaries. The main result in Section 3 (Theorem 1) concerns the ratio Diam diam , while the main result in Section 4 (Theorem 2) treats the ratio Rad diam , both considered for
T ∈ T . The division of our results into two main sections is just to ease the reading, their topics obviously overlap. Concluding, we have the following inequalities for tetrahedra, either gen-erally known or proven here; to ease the presentation, the functions are given without the argument T ∈ T .21 ≤ Diam diam ≤ 2
√3, 1 < Diam Rad ≤ 2, 1 < diam rad ≤ 2, Rad diam ≤ 1, 1 ≤ Rad rad < 2,
√34 < rad Diam < 1. Of the above inequalities, some are not sharp and could be improved; for example, our Open Problem 1 asks to prove (or disprove) that 2
√3 ≤ Diam ( T )Rad ( T ) .
2 Preliminaries
Let P be (the surface of) a convex polyhedron. A geodesic segment on a P is a shortest path between its extremities. The cut locus C(x) of the point x on P is the set of endpoints (different from x) of all nonextendable geodesic segments (on the surface P ) starting at x. Equivalently, it is the closure of the set of all those points y to which there is more than one shortest path on P from x.The following lemma presents several known properties of cut loci on convex polyhedra, see e.g. .
Lemma 1 (i) C(x) has the structure of a finite 1-dimensional simplicial complex which is a tree. Its leaves (endpoints) are vertices of P , and all vertices of P , excepting x (if the case), are included in C(x). All vertices of
P interior to C(x) are considered as junction points. (ii) Each point y in C(x) is joined to x by as many geodesic segments as the number of connected components of C(x) \ y. For junction points in
C(x), this is precisely their degree in the tree. (iii) The edges of C(x) are geodesic segments on P .(iv) Assume the geodesic segments γ and γ′ from x to y ∈ C(x) are bounding a domain D of P , which intersects no other geodesic segment from
x to y. Then there is an arc of C(x) at y which intersects D and it bisects the angle of D at y.
We shall implicitely use Alexandrov’s Gluing Theorem stated below, see , p.100. 3Lemma 2 Consider a topological sphere S obtained by gluing planar poly-gons (i.e., naturally identifying pairs of sides of the same length) such that at most 2π angle is glued at each point. Then S, endowed with the intrinsic metric induced by the distance in R2, is isometric to a polyhedral convex sur-face P ⊂ R3, possibly degenerated. Moreover, P is unique up to rigid motion and reflection in R3.
In some sense opposite to Alexandrov’s Gluing Theorem is the operation of unfolding. The first two general methods known to unfold the surface P
of any convex polyhedron to a simple (non-overlapping) polygon in the plane are the source unfolding and the star unfolding, both with respect to a point
x ∈ P .Concerning the source unfolding , one cuts P along the cut locus of the point x; this has been studied for polyhedral convex surfaces since (where the cut locus is called the “ridge tree”). Concerning the star unfolding , one cuts P along the shortest paths (sup-posed unique) from x to every vertex of P . The idea goes back to Alexandrov ; the fact that it unfolds P to a non-overlapping polygon was established in . An isosceles tetrahedron is a a tetrahedron whose opposite edges are pair-wise equal. We shall make use of these tetrahedra and of their special prop-erties, see e.g. .
Lemma 3 For any isosceles tetrahedron, the total angle at each vertex is precisely π and its faces are acute triangles. Consequently, the star unfolding of an isosceles tetrahedron with respect to any of its vertices provides an acute planar triangle.
Some extremal cases in our inequalities are attained by what we call ε-thick tetrahedra . Such a tetrahedron is, by definition, a tetrahedron T with one edge included in a ball of radius εdiam( T ) centered at the midpoint of its longest edge. An ε-thick tetrahedron is said to be normal if its longest edge and the one opposite to it are, on the one hand, normal to each other, and on the other hand, normal to the line through their midpoints. For x ∈ S put rad x = max y∈S || x − y|| , hence rad x ≥ rad( S). Also, put Rad x = max y∈S ρ(x, y ), hence Rad x ≥ Rad( S). 4Denote by fx the set of all extrinsic farthest points from x ∈ S; i.e.,
fx = {y ∈ S : || x − y|| = rad x}. Also, denote by Fx the set of all intrinsic farthest points from x ∈ S, and call them antipodes of x; i.e., Fx = {y ∈ S :
ρ(x, y ) = Rad x}.
3 Diameters
A very nice and deep result of J. O’Rourke and C. A. Schevon states the following: if the points x, y in the polyhedral convex surface P realize the intrinsic diameter of P then at least one of them is a vertex of P , or they are joined by at least five distinct geodesic segments. For tetrahedra it directly implies the next lemma.
Lemma 4 If T ∈ T and x ∈ T is a point with Rad x = Diam( T ) then x is either a vertex or an antipode of a vertex.
The extrinsic analog of O’Rourke and Schevon’s criterion for diametral points is a simple result, of some interest in itself.
Proposition 1 Let x, y ∈ P .(1) If y ∈ fx then y is a vertex of P . In particular, if ‖x − y‖ = diam ( P )
then both points are vertices of P .(2) If fx = {y} and ‖x − y‖ = rad ( P ) then x is the foot of y onto a face.
Proof: (1) Assume that y is not a vertex of P . Then there exists some line segment [ uv ] on P containing y in its relative interior. Since ∠xyu + ∠xyv =
π, one of these two angles, say ∠xyu , is at least π/ 2, whence ‖x − y‖ <
‖x − u‖, in contradiction with y ∈ fx.(2) Assume now that rad ( P ) = ‖x − y‖ and fx = {y}, i.e. , ‖x − y‖ >
‖x − v‖ for any vertex v 6 = y. By continuity and (1), there is a neighbourhood
N of x such that, for any point z ∈ N , fx = {y}. Assume now that x is not the foot of y onto a face; then, since P is convex, one can find points
z ∈ N such that ∠zxy < π/ 2, and rad z = ‖z − y‖ < ‖z − x‖ = rad( P ), a contradiction.
In the above Proposition, if || x − y|| = rad ( P ) for y ∈ fx and fx contains at least two points then x may not be the foot of y onto a face. For example, consider a normal ε-thick tetrahedron, with y, z the vertices of the longest edge and x its mid-point. 5Corollary 1 For the regular tetrahedron T of unit edge, diam( T ) = 1 and
rad( T ) =
√23 .
Corollary 1 is the extrinsic analog of Theorem 3.1 in , quoted in the next lemma to clarify the second equality case in Theorem 1.
Lemma 5 For the regular tetrahedron T of unit edge, Diam( T ) = 2√3 is realized between any vertex and the centre of its opposite face, while Rad( T ) = 1 is realized between mid-points of opposite edges.
Lemma 1 immediately implies the next one. An Y-tree is a tree with one junction point and three edges.
Lemma 6 The cut locus of a vertex of T ∈ T is a (possibly degenerate)
Y -tree.
We need one more lemma for our first main result.
Lemma 7 Consider the family I of all planar acute triangles ∆ inscribed in a given circle C, and let l∆ denote the longest side of ∆ ∈ I . Then inf ∆∈I l∆
is achieved for equilateral triangles.
Proof: Under the hypotheses, just note that the edge lengths of the triangle are in the same order as the lengths of the intercepted arc of circles, by the Sine Rule and the monotony of the sine function on [0 , π/ 2]. The longest arc is obviously shortest when all three arcs are equal.
Theorem 1 For any tetrahedron holds
1 ≤ Diam ( T )diam ( T ) ≤ 2
√3
and both inequalities are sharp. The first inequality becomes equality, for example, for ε-thick tetrahedra with ε small enough, while the second inequality becomes equality for the regular tetrahedon.
6Proof: The intrinsic distance between two points is never less than the ex-trinsic distance, so the first inequality is obvious. It is also clear that both diameters are equal for an ε-thick tetrahedron, whenever ε > 0 is sufficiently small. By Lemma 4, there exists a vertex v ∈ T and a point p ∈ T such that
ρ (v, p ) = Diam ( T ). If p is also a vertex, then the [ pv ] is greater than or equal to any other edge, for the lengths of edges are also the intrinsic distance between their endpoints. Hence, by Lemma 1, Diam ( T ) = diam ( T ). Assume now that p is a flat point, that is, the only triple point of C (v), which is a Y-tree be virtue of Lemma 6. Cutting T along the three edges meeting at v and unfolding it onto a plane yields the development shown in plain lines in Figure 1. There, p is on the bisector line of any two of the images of v, by a direct consequence of Lemma 1 (iv). Hence p is the circum-centre of the triangle determined the three images of v. We now conside the triangle ∆ drawn in dots, with vertices at the images of v, as the unfolding of an isosceles tetrahedron denoted by T ′, see Lemma 2. Notice that T ′ has acute triangles as faces, by Lemma 3, even though ∆ might not be acute. We claim that each edge of T ′ is shorter than one edge of T . This is clear if ∆ is acute, because then the edges of T ′ have half-length of the sides of ∆, which in turn have a length less than twice an edge of T by the triangle inequality. If ∆ is not acute, then four of the edges of T ′ have half-length of two sides of ∆. The last two edges of T ′ equal the length of the median line of ∆ with respect to its longest edge, and so, since ∆ is not acute, it is strictly shorter than half-length of that side. Whence diam ( T ′) ≤ diam ( T ). On the other hand, the intrinsic distance between p and x is unchanged, whence Diam ( T ′) ≥ ρ (x, v ) = Diam ( T ). By Lemma 4, there exists a vertex v′ ∈ T ′ and a point p′ ∈ T ′ such that ρ (v′, p ′) = Diam ( T ′). Cutting T ′ along the three edges meeting at v′
and unfolding it onto a plane yields an acute triangle (see Lemma 3) similar to one the shown in doted lines in Figure 1. It is now easy to see that the longest edge of T ′ is longer that the edge of a regular tetrahedron T ′′
whose unfolding is inscribed in same the circle, see Lemma 7. Moreover, this deformation doesn’t change the distance Diam ( T ′) = ρ (p′, v ′) ≤ diam ( T ′′ ), whence Diam ( T )diam ( T ) ≤ Diam ( T ′)diam ( T ) ≤ Diam ( T ′′ )diam ( T ′′ ) = 2
√3.
7pvvvFigure 1: Unfoldings of T and T ′.8The upper bound given by Theorem 1 for tetrahedra, 2√3 ≈ 1.15, is clearly better than the upper bound obtained by N. P. Makuha for general convex surfaces, π
2
≈ 1.57.
4 Diameters and radii
It follows from the triangle inequality, in any compact metric space, that the ratio between diameter and radius belongs to [1 , 2]. In the case of the intrinsic metric of a tetrahedron, we have the following result.
Proposition 2 For any convex polyhedron P holds
1 < Diam ( P )Rad ( P ) ≤ 2.The second inequality is sharp, and achieved by normal ε-thick tetrahedra, for small ε.
Proof: For each convex surface S with Rad ( S) = Diam ( S), the mapping F
is a single-valued involution , and no convex polyhedron has this property , hence Rad ( S) < Diam ( S) in this case. Consider now a normal ε-thick tetrahedron T . Let a, b be the endpoints of its longest edge and c, d the two other vertices. Let m be the midpoint of [ab ], and let q be the midpoint of [ cd ]. Since T is symmetric with respect to the plane abq , the Jordan arc of C (m) between a and b should be included in this plane, and so is the union of [ aq ] and [ bq ]. Similarly, the symmetry with respect to the plane mcd infers that the Jordan arc of C (m) between c
and d is [ cd ]. It follows that q is the only point in C (m) of degree more than two. Hence Fm ⊂ { q, a, b, c, d }. It is clear that for ε small enough ρ (m, q )and ρ (m, c ) = ρ (m, d ) are both less that ρ (m, a ) = ρ (m, b ) = ‖a−b‖
2
, whence Rad m = ‖a−b‖
2
= Rad ( T ). This completes the proof.
We have a similar result for the extrinsic metric.
Proposition 3 For any convex polyhedron P holds
1 < diam ( P )rad ( P ) ≤ 2.The second inequality is sharp, and achieved by ε-thick tetrahedra, for small
ε.
9Proof: The first inequality is strict because no polyhedron has constant width. The case of equality is easy to check.
From Theorem 1 and Proposition 3 directly follows
Corollary 2 For any tetrahedron T holds
√34 < rad Diam < 1.
We return now to the statement of Proposition 1.
Open Problem 1 Prove that
2
√3 ≤ Diam ( T )Rad ( T ) ,with equality for the regular tetrahedon.
Notice that, if solved, the above problem and Theorem 1 would imply Rad ( T ) ≤ diam ( T ), with equality for the regular tetrahedron. However this latter inequality can be proven directly.
Theorem 2 For any tetrahedron T ∈ T we have
Rad ( T ) ≤ diam ( T ) ,with equality if and only if T is regular.
Proof: Let T be a tetrahedron of unit extrinsic diameter. Denote by a and
b the endpoints of its (or one of its) longest edge(s), and by c and d the two other vertices. Let o be the midpoint of [ ab ]. Cutting along the three edges meeting at d and unfolding T onto a plane yields the development G0 shown in plain lines in Figure 2, where da, d b and dc are the images of d.If the triangle bcd a is right or obtuse at da then it is included in the disc of diameter [ bc ], and we set G = G0.If the triangle bcd a is right or obtuse at c then cut along [ bc ] and rotate it around b, to obtain another development G in which the image of the face
bcd is included in the union of the discs of diameters [ bc ] and [ bd c]. 10 a b
dddb a
c
c
oFigure 2: The development of T in the proof of Theorem 2. 11 Otherwise, if the triangle bcd a is acute at da then cut it along the dash line and rotate the outer part around b, to obtain another development G
in which the image of the face bcd is included in the union of the discs of diameters [ bc ] and [ bd c]. In the same way, we can arrange the face acd in the union of the discs of diameters [ ac ] and [ ad c]. We claim that each of those four discs are included in the disc D of center
o and radius 1. Let u be the midpoint of [ ac ]. Since [ ac ] is less than or equal to the longest edge of T , c belongs to the disc of center a and radius 1, and thus u belongs to the disc of diameter [ ab ]: || a − u|| = || a − c|| /2 ≤ 12 ,
|| o − u|| ≤ 12 . Hence the disc of diameter [ ac ] is included in D. The proof is similar for the three other discs. We shall use this fact to conclude the equality case. Notice that the quadrilateral acbd c is convex, because [ ab ] is the longest edge in T .Let o′ belong to Fo; for simplicity, also denote by o′ the image of o′ on G
(or one of its images, if there are more). Assume first that o′ belongs to the convex quadrilateral acbd c. Then
|| o−o′|| ≤ max {|| o−a|| , || o−c|| , || o−b|| , || o−dc||} , and Apollonius’s theorem gives max {|| o − c|| , || o − dc||} ≤ √3/2 < 1. Assume now that o′ belongs to one of the four discs of diameters the sides of the quadrilateral acbd c, say the one centered at u. Then
ρ(o, o ′) ≤ || o − u|| + || u − o′|| ≤ | b − c|| /2 + || a − c|| /2 ≤ 1.
It follows that Rad ( T ) ≤ Rad o = ρ (o, o ′) ≤ 1. Assume now that we have equality, hence Rad ( T ) = Rad o = ρ (o, o ′) = 1. The development has to intersect ∂D . Notice that there are at least three geodesic segments joining o to o′ ∈ Fo on T . So it follows, moreover, that three of the four small circles have to be tangent to ∂D . Assume one of them is that of diameter [ ac ]. This implies that c is one point of intersection of the circles of radius one centered at a and b, and so || a − c|| = || b − c|| =
|| a − b|| = 1. Similarly || a − dc|| = || b − c|| = || b − dc|| = 1. So all edges, except possibly [ cd ], have length one. Now one can repeat the whole argument above replacing [ ab ] by another longest edge to prove that T is actually regular.
The first inequality below is valid for arbitrary convex surfaces; compare it to the first one in Theorem 1. 12 Proposition 4 For any tetrahedron T ∈ T holds
1 ≤ Rad ( T )rad ( T ) < 2.
The first inequality is sharp, and achieved by normal ε-thick tetrahedra, for small ε.
Proof: We have, for all x, y ∈ S, || x − y|| ≤ ρ(x, y ), so rad x ≤ Rad x, hence rad ( S) ≤ Rad ( S). The case of equality is easy to check, see the proof of Proposition 2. Theorem 2 and Proposition 3 yield the last inequality.
References
P. K. Agarwal, B. Aronov, J. O’Rourke and C. A. Schevon, Star un-folding of a polytope with applications , SIAM J. Comput. 26 (1997), 1689-1713 A. D. Alexandrov, Convex Polyhedra , Springer-Verlag, Berlin, 2005. Monographs in Mathematics. Translation of the 1950 Russian edition by N. S. Dairbekov, S. S. Kutateladze, and A. B. Sossinsky B. Aronov and J. O’Rourke, Nonoverlap of the star unfolding , Discrete Comput. Geom. 8 (1992), 219-250. G. D. Chakerian and H. Groemer, Convex Bodies of Constant Width ,in Convexity and its Applications , P. Gruber and J. Wills (Eds.), Birkh¨ auser, Bassel 1983, 49-96 V. Dods, C. Traub, J. Yang, Geodesics on the regular tetrahedron and the cube , Discrete Math. 340 (2017), 3183-3196 J. Leech, Some Properties of the Isosceles Tetrahedron , The Mathemat-ical Gazette 34 (1950), 269-271 N. P. Makuha, A connection between the inner and the outer diameters of a general closed convex surface (in Russian), Ukrain. Geometr. Sb. Vyp. 2 (1966), 49-51 13 J. O’Rourke and C. A. Schevon, Computing the geodesic diameter of a 3-polytope , Proc. 5th ACM Symp. Comput. Geom. (1989), 370-379 J. Rouyer, Antipodes sur un t´ etra` edre r´ egulier , J. Geom. 77 (2003), 152-170 J. Rouyer, Steinhaus conditions for convex polyhedra , in: K. Adiprasito et al. (Eds.), Convexity and Discrete Geometry Including Graph The-ory , Springer Proceedings in Mathematics & Statistics 148, Springer International Publishing 2016, 77–84 J. Rouyer and C. Vˆ ılcu, Sets of tetrahedra, defined by maxima of distance functions , An. S ¸tiint ¸. Univ. “Ovidius” Constant ¸a Ser. Mat. 20 (2012), 197-212. M. Sharir and A. Schorr, On shortest paths in polyhedral spaces , SIAM J. Comput. 15 (1986), 193-215. C. Vˆ ılcu and T. Zamfirescu, Symmetry and the farthest point mapping on convex surfaces , Adv. Geom. 6 (2006), 345-353 C. Vˆ ılcu and T. Zamfirescu, Multiple farthest points on Alexandrov sur-faces , Adv. Geom. 7 (2007), 83-100 V. A. Zalgaller, An isoperimetric problem for tetrahedra , J. Math. Sci.
140 (2007), 511-527 Jin-ichi Itoh
School of Education, Sugiyama Jogakuen University 17-3 Hoshigaoka-motomachi, Chikusa-ku, Nagoya, 464-8662 Japan
j-itoh@sugiyama-u.ac.jp
Jo¨ el Rouyer
16, rue Philippe, 68 200 H´ egenheim - France
Joel.Rouyer@ymail.com
Costin Vˆ ılcu
Simion Stoilow Institute of Mathematics of the Roumanian Academy P.O. Box 1-764, 014700 Bucharest, Roumania
Costin.Vilcu@imar.ro
14 |
1319 | https://upc.aws.openrepository.com/bitstream/handle/10757/662713/EF07_Microeconomia_Aplicada_202102.pdf?sequence=1&isAllowed=y | Microeconomía Aplicada - EF07 - 202102
Item Type info:eu-repo/semantics/report
Authors Bacilio Alarcon, Wilfredo; Carrillo Barnuevo, Manuel Augusto; Gutiérrez Martínez, Mauro Orlando; Macetas Aguilar, Daniela; Mallqui Trejo, Milagros Mercedes; Mora Ruiz, César David; Moreno Pachas, Alexis Fernando; Talledo Carrasco, Manuel Emilio
Publisher Universidad Peruana de Ciencias Aplicadas (UPC)
Rights info:eu-repo/semantics/openAccess
Download date 29/09/2025 09:16:27
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III. INTRODUCCIÓN
Descripción: El curso es obligatorio para las carreras de la Facultad de Economía, y perteneciente a la línea de microeconomía. Se enfoca en proporcionar a los estudiantes elementos teóricos y prácticos intermedios y avanzados en torno a cuatro (4) temas principales: Equilibrio General Competitivo, Distorsiones de mercado: Externalidades, Bienes Públicos, e Información Asimétrica, y Competencia Imperfecta. Propósito: Este curso propone desarrollar las habilidades de abstracción e intuición económica, mediante la aplicación de herramientas matemáticas y gráficas intermedia y avanzadas, y la introducción de la Teoría de Juegos. Es un curso enfocado en el desarrollo de la competencia general Pensamiento crítico, y la competencia específica Investigación económica; ambas en el Nivel 2.
IV. LOGRO (S) DEL CURSO
Al finalizar el curso, el estudiante formula conclusiones a partir de la aplicación de las herramientas (económicas y cuantitativas) del enfoque del Equilibrio General de Mercado, en contextos de competencia perfecta y ante la presencia de distorsiones de mercado; y mediante el análisis de situaciones reales ¿simples y complejas.
I. INFORMACIÓN GENERAL
CURSO
: Microeconomía Aplicada
CÓDIGO
: EF07
CICLO
: 202102
CUERPO ACADÉMICO
: Bacilio Alarcon, Wilfredo Carrillo Barnuevo, Manuel Augusto Gutiérrez Martínez, Mauro Orlando Macetas Aguilar, Daniela Mallqui Trejo, Milagros Mercedes Mora Ruiz, César David Moreno Pachas, Alexis Fernando Talledo Carrasco, Manuel Emilio
CRÉDITOS
: 4
SEMANAS
: 16
HORAS
: 2 H (Práctica) Semanal /3 H (Teoría) Semanal
ÁREA O CARRERA
: Economia y Finanzas
II. MISIÓN Y VISIÓN DE LA UPC
Misión: Formar líderes íntegros e innovadores con visión global para que transformen el Perú. Visión: Ser líder en la educación superior por su excelencia académica y su capacidad de innovación.
2
Competencias: Pensamiento crítico Nivel de logro: Nivel 2 Definición: Capacidad para conceptualizar, aplicar, analizar y/o evaluar activa y hábilmente, información recogida de, o generada por, la observación, experiencia, reflexión o razonamiento, orientado hacia el desarrollo de una creencia o acción. Competencias: Investigación económica Nivel de logro: Nivel 2 Definición: Diseña y ejecuta proyectos de investigación, tomando en cuenta la literatura académica relevante, las técnicas cuantitativas y softwares especializados, así como el buen uso de la información públicamente disponible.
UNIDAD Nº: 1 Equilibrio General Competitivo LOGRO
Competencias: Pensamiento crítico, e Investigación económica. Logro de la unidad: Al finalizar la unidad, el estudiante presenta un análisis sustentado en situaciones reales ¿simples y complejas, mediante el uso de las herramientas de razonamiento económico y cuantitativo intermedio y avanzado, basadas en el enfoque del Equilibrio General de Mercado.
TEMARIO
Contenido: 1) Introducción. 2) Equilibrio con Intercambio Puro. 3) Equilibrio con Producción. 4) Economía del Bienestar. Actividades de aprendizaje: - Exposición participativa. - Estudio de casos. - Preguntas guía. - Resolución de ejercicios. Evidencias de aprendizaje: - PC1. Infiere conclusiones a partir de la aplicación de las herramientas de razonamiento (económico y cuantitativo intermedio y avanzado) a situaciones reales, provistas por el enfoque de Equilibrio General de Mercado con Intercambio Puro. - PC2. Infiere conclusiones a partir de la aplicación de las herramientas de razonamiento (económico y cuantitativo intermedio y avanzado) a situaciones reales, provistas por el enfoque de Equilibrio General de Mercado, y el análisis de Economía del Bienestar. Referencias: - Nicholson y Snyder (2011). Microeconomía intermedia y su aplicación. Cap. 10. Equilibrio general y bienestar. - Varian (2011). Microeconomía intermedia: Un enfoque actual. Cap. 32. El intercambio, Cap. 33. La producción, y Cap. 34. El bienestar.
V. UNIDADES DE APRENDIZAJE 3- Arrow y Maskin (2012). Social Choice and Individual Values. Yale University Press. Caps. 2, 3, 5 y 6.
HORA(S) / SEMANA(S)
Semanas 1 a 4
UNIDAD Nº: 2 Distorsiones de mercado: Externalidades y Bienes públicos LOGRO
Competencias: Pensamiento crítico, e Investigación económica. Logro de la unidad: Al finalizar la unidad, el estudiante evalúa las consecuencias de la inclusión de distorsiones de mercado (externalidades y bienes públicos) en el enfoque de Equilibrio General de Mercado; mediante el análisis de situaciones reales ¿simples y complejas, y el uso de herramientas de razonamiento económico y cuantitativo intermedio y avanzado.
TEMARIO
Contenido: 1) Externalidades. 2) Bienes públicos. Actividades de aprendizaje: - Exposición participativa. - Estudio de casos. - Preguntas guía. - Resolución de ejercicios. Evidencias de aprendizaje: - EA. Evalúa las consecuencias derivadas de la inclusión de distorsiones de mercado (externalidades y bienes públicos) en situaciones reales, y a partir de la aplicación de las herramientas de razonamiento (económico y cuantitativo intermedio y avanzado). Referencias: - Varian (2011). Microeconomía intermedia: Un enfoque actual. Cap. 35. Las externalidades, y Cap. 37. Los bienes públicos. - Nicholson y Snyder (2011). Microeconomía intermedia y su aplicación. Cap. 16. Externalidades y bienes públicos.
HORA(S) / SEMANA(S)
Semanas 5 a 8
UNIDAD Nº: 3 Distorsiones de mercado: Información asimétrica LOGRO
Competencias: Pensamiento crítico, e Investigación económica. Logro de la unidad: Al finalizar la unidad, el estudiante presenta un análisis de las consecuencias de la inclusión de distorsiones de mercado (información asimétrica) en el enfoque de Equilibrio General de Mercado; mediante el análisis de situaciones reales ¿simples y complejas, y el uso de las herramientas de razonamiento económico y cuantitativo intermedio y avanzado. 4
TEMARIO
Contenido: 1) Incertidumbre. 2) Economía de la Información. 3) Teoría de Juegos. Actividades de aprendizaje: - Exposición participativa. - Estudio de casos. - Preguntas guía. - Resolución de ejercicios. Evidencias de aprendizaje: - PC3. Evalúa las consecuencias derivadas de la inclusión de distorsiones de mercado (información asimétrica) en situaciones reales, y a partir de la aplicación de las herramientas de razonamiento (económico y cuantitativo intermedio y avanzado). Referencias: - Varian (2011). Microeconomía intermedia: Un enfoque actual. Cap. 12. La incertidumbre, y Cap. 38. Información asimétrica. - Gravelle y Rees (2006). Microeconomía. Cap. 17. Elección bajo incertidumbre, Cap. 18. Producción en condiciones de incertidumbre, Cap. 19. Seguros, reparto y agrupación de riesgos, y Cap. 20. Teoría de la agencia, Teoría de Contratos y la empresa. - Akerlof (1970). "The Market for "Lemons": Quality Uncertainty and the Market Mechanism". The Quarterly Journal of Economics, The MIT Press. Vol. 84, No. 3 (Aug), pp. 488-500. - Hart y Holmstrom (1986). "The Theory of Contracts" Working papers 418, Massachusetts Institute of Technology (MIT), Department of Economics. - Spence (1973) "Job Market Signaling". The Quarterly Journal of Economics, The MIT Press. Vol. 87, No. 3. (Aug., 1973), pp. 355-374. - Rothschild y Stiglitz (1976). "Equilibrium in Competitive Insurance Markets: An Essay on the Economics of Imperfect Information", The Quarterly Journal of Economics, Oxford University Press, vol. 90(4), pages 629-649.
HORA(S) / SEMANA(S)
Semanas 9 a 12
UNIDAD Nº: 4 Distorsiones de mercado: Competencia imperfecta LOGRO
Competencias: Pensamiento crítico, e Investigación económica. Logro de la unidad: Al finalizar la unidad, el estudiante formula conclusiones a partir de la aplicación de las herramientas (económicas y cuantitativas) del enfoque del Equilibrio General de Mercado, en contextos de competencia perfecta y ante la presencia de distorsiones de mercado; y mediante el análisis de situaciones reales ¿simples y complejas.
TEMARIO
Contenido: 1) Monopolio. 2) Competencia imperfecta. 3) Políticas de promoción de la competencia (Introducción). 5Actividades de aprendizaje: - Exposición participativa. - Estudio de casos. - Preguntas guía. - Resolución de ejercicios. Evidencias de aprendizaje: - TA1. Evalúa las consecuencias derivadas de la inclusión de distorsiones de mercado (competencia imperfecta) en situaciones reales, y a partir de la aplicación de las herramientas de razonamiento (económico y cuantitativo intermedio y avanzado). - EB. Evalúa situaciones reales ¿simples y complejas, mediante el uso de las herramientas (económicas y cuantitativas) del enfoque del Equilibrio General de Mercado, en contextos de competencia perfecta y ante la presencia de distorsiones de mercado. Referencias: - Nicholson y Snyder (2011). Microeconomía intermedia y su aplicación. Cap. 12. Competencia imperfecta. - Perloff (2011). Ch. 11. Monopoly, y Ch. 13. Oligopoly and Monopolistic.
HORA(S) / SEMANA(S)
Semanas 13 a 16
VI. METODOLOGÍA
El Modelo Educativo de la UPC asegura una formación integral, que tiene como pilar el desarrollo de competencias, las que se promueven a través de un proceso de enseñanza-aprendizaje donde el estudiante cumple un rol activo en su aprendizaje, construyéndolo a partir de la reflexión crítica, análisis, discusión, evaluación, exposición e interacción con sus pares, y conectándolo con sus experiencias y conocimientos previos. Por ello, cada sesión está diseñada para ofrecer al estudiante diversas maneras de apropiarse y poner en práctica el nuevo conocimiento en contextos reales o simulados, reconociendo la importancia que esto tiene para su éxito profesional. El curso implica cinco (05) horas semanales, de las cuáles tres (03) horas corresponden a la sesión teórica, y dos (02) horas, a la sesión práctica. En las sesiones teóricas, el docente del curso revisará con los estudiantes los temas previamente propuestos, y brindará a los estudiantes una serie de ejercicios de aplicación teórica a situaciones de la realidad, que permitan consolidar su aprendizaje. Mientras que, en las sesiones prácticas, los asistentes de docencia brindarán pautas y resolverán ejercicios de aplicación teórica a situaciones de la realidad. Ello, sobre la base de las Prácticas Dirigidas (listas de ejercicios, que se irán resolviendo paulatinamente cada semana ¿una por cada Unidad) oportunamente provistas a los estudiantes (vía Aula Virtual). En este sentido, y en ambos tipos de sesiones, se requiere de la participación activa de los estudiantes, mediante la revisión previa de la bibliografía sugerida, resolución individual de los ejercicios planteados, y formulación de consultas en ambos tipos de sesión. En lo referido a la metodología de corroboración de logro de aprendizajes, se contarán actividades evaluadas específicas: Prácticas Calificadas (PC), Tareas Académicas (TA), una Evaluación Parcial y una Evaluación Final. Las PC y las Evaluaciones Parcial y Final (duración: 110 min. c/u) son evaluaciones que enfatizan la resolución de ejercicios, que suponen un dominio intermedio de herramientas matemáticas y gráficas, así como el manejo 6
de la teoría revisada. Asimismo, estas evaluaciones consideran de forma significativa en la calificación: la formulación de justificaciones, explicaciones y argumentaciones; y el dominio de las herramientas matemáticas y gráficas. Especialmente, las Evaluaciones Parcial y Final constituyen una revisión sumaria de aspectos teóricos y prácticos estudiados hasta la fecha de su ejecución.
VII. EVALUACIÓN
FÓRMULA
8.75% (TA1) + 8.75% (PC1) + 30% (EA1) + 8.75% (PC2) + 8.75% (PC3) + 35% (EB1)
TIPO DE NOTA PESO %
TA - TAREAS ACADÉMICAS 8.75 PC - PRÁCTICAS PC 8.75 EA - EVALUACIÓN PARCIAL 30 PC - PRÁCTICAS PC 8.75 PC - PRÁCTICAS PC 8.75 EB - EVALUACIÓN FINAL 35
VIII. CRONOGRAMA
TIPO DE PRUEBA DESCRIPCIÓN NOTA NÚM. DE PRUEBA FECHA OBSERVACIÓN RECUPERABLE
TA TAREAS ACADÉMICAS 1 Semana 3 Unidad 1 NO PC PRÁCTICAS PC 1 Semana 6 Unidad 2 SÍ EA EVALUACIÓN PARCIAL 1 Semana 8 Unidades 1 y 2 SÍ PC PRÁCTICAS PC 2 Semana 11 Unidad 3 SÍ PC PRÁCTICAS PC 3 Semana 14 Unidad 4 SÍ EB EVALUACIÓN FINAL 1 Semana 16 Todas las unidades SÍ
IX. BIBLIOGRAFÍA DEL CURSO
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Sum of the first n elements of the sequence of 9's complement
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Asked 3 years, 7 months ago
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Let's consider the following sequence:
9,8,7,6,5,4,3,2,1,0,89,88,87,86,85,84,83,82,81,80,79,78,77,76,75,74,73,72,71...
9,8,7,6,5,4,3,2,1,0,89,88,87,86,85,84,83,82,81,80,79,78,77,76,75,74,73,72,71...
This is the sequence of 9's complement of a number: that is, a(x)=10 d−1−x where d is the number of digits in x. (A061601 in the OEIS).Your task is to add the first n elements.
Input
A number n∈[0,10000].
Output
The sum of the first n elements of the sequence.
Test cases
0 -> 0
1 -> 9
10 -> 45
100 -> 4050
1000 -> 408600
10000 -> 40904100
Standard code-golf rules apply. The shortest code in bytes wins.
code-golf
sequence
integer
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edited Feb 22, 2022 at 13:53
pxeger
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asked Feb 21, 2022 at 18:32
sinvecsinvec
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Do the test cases make sense? (I guess that's why the 0 confusion in the answers.) 0 -> 0 works if you take 0 as a 0-digit number, and the formula is 10^d - 1 - x as stated. But 1 -> 9 only works if the formula is 10^d - x. I think the 0 case should just be removed, and the formula changed to be without the - 1.Sundar R –Sundar R 2022-02-23 10:26:15 +00:00 Commented Feb 23, 2022 at 10:26
3 @SundarR I presume the confusion may result from indexing the sequence from 0 and then taking sum of first elements. So taking sum of first 0 elements results in 0 and taking first element (of index 0) results in 9.pajonk –pajonk 2022-02-23 10:50:54 +00:00 Commented Feb 23, 2022 at 10:50
@pajonk Ahh thank you. That was a reading fail on my part, I assumed it was "sum up to n's complement" and didn't read where it clearly says "sum of first n elements".Sundar R –Sundar R 2022-02-23 11:05:50 +00:00 Commented Feb 23, 2022 at 11:05
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R, 37 bytes
f=function(n)sum(10^nchar(1:n-1)-1:n)
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note I left out the check for 0 as I do not think it is part of the sequence as refered to A061601 in the OEIS reference. Adding the check would result in the exact same answer of Giuseppe, which would surely be preferable including the correct answer including 0.
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edited Feb 23, 2022 at 7:58
answered Feb 22, 2022 at 14:15
Merijn van TilborgMerijn van Tilborg
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Welcome to Code Golf, and nice answer!rydwolf –rydwolf♦ 2022-02-22 14:26:05 +00:00 Commented Feb 22, 2022 at 14:26
1 This is solid. It doesn't appear that this works for n=0 so you do need to adjust for that. Also, anonymous functions are perfectly fine (as long as they're not recursive), so you can drop the f= from your answer as well :-)Giuseppe –Giuseppe 2022-02-22 14:27:26 +00:00 Commented Feb 22, 2022 at 14:27
We cross posted the same answer, I learned your !!n part which is very elegant to include the n=0. I do think though if an OP asks for a oeis defined sequence, the offset value or in other words the value not part of that defined sequence should not be addressed in the answer.Merijn van Tilborg –Merijn van Tilborg 2022-02-22 14:38:51 +00:00 Commented Feb 22, 2022 at 14:38
2 That's up to the discretion of the question asker, I believe. You can ask in a comment on the main post if you want, though. Also, I recommend looking at the tips for golfing in R; they can be helpful sometimes!Giuseppe –Giuseppe 2022-02-22 16:38:12 +00:00 Commented Feb 22, 2022 at 16:38
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Wolfram Language (Mathematica), 42 bytes
FromDigits[9-IntegerDigits@--n]~Sum~{n,#}&
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-2 bytes from att
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edited Feb 22, 2022 at 18:01
answered Feb 21, 2022 at 18:55
ZaMoCZaMoC
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42 bytesatt –att 2022-02-22 17:45:41 +00:00 Commented Feb 22, 2022 at 17:45
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Julia 1.0, ~~43~~~~38~~ 30 bytes
!n=n>0&&10^ndigits(~-n)-n+!~-n
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-5 dingledooper, -8 MarcMush
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edited Feb 23, 2022 at 19:25
answered Feb 22, 2022 at 15:12
ameliesamelies
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1 38 bytes: !n=n>0&&sum(n->~n+10^ndigits(n),0:n-1)dingledooper –dingledooper 2022-02-22 16:40:05 +00:00 Commented Feb 22, 2022 at 16:40
1 30 bytes with recursion: !n=n>0&&10^ndigits(~-n)-n+!~-nMarcMush –MarcMush 2022-02-23 19:03:13 +00:00 Commented Feb 23, 2022 at 19:03
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R, 39 bytes
function(n)sum(10^nchar(1:n-1)-1:n)!!n
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answered Feb 22, 2022 at 14:14
GiuseppeGiuseppe
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05AB1E, 8 bytes
FNg°<N-O
Try it online or verify all test cases.
Explanation:
F # Loop `N` in the range [0, (implicit) input):
Ng # Push the length of `N`
° # Take 10 to the power this length
< # Decrease it by 1
N- # Decrease it by `N`
O # Take the sum of the values on the stack
# (after which the last sum is output implicitly as result -
# or the implicit input if it was 0 and we never entered the loop)
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answered Feb 22, 2022 at 14:48
Kevin CruijssenKevin Cruijssen
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Vyxalr, 8 bytes
No, I'm not Jo King. I found another use for the r flag!!!
It happens once in a blue moon on average
Okay, yes, I said that
ʁƛL↵-‹;∑
Try it Online!Flagless 9 bytes
Explanation
So the r flag is basically a reverser, which reverses the order of the arguments. That makes it hard to use for long programs, but for short ones like this, it's useful sometimes.
ʁƛL↵-‹;∑
ʁ create list with range(0, n)
ƛ open mapping lambda (with loop item n)
L↵ push 10 to the length of n
- n - the latter (because it's reversed)
‹; decrement by 1 and close lambda
∑ sum list
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answered Feb 22, 2022 at 15:44
math scatmath scat
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Try it Online! for 7 bytes flagless, and 6 bytes with s flag lyxal –lyxal♦ 2023-03-28 23:32:20 +00:00 Commented Mar 28, 2023 at 23:32
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Jelly, 7 bytes
’æċ⁵_)S
A monadic Link accepting a non-negative integer that yields a non-negative integer.
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How?
’æċ⁵_)S - Link: non-negative integer, n
) - for each i in [1..n] (if n is 0 this will yield []):
’ - decrement -> i-1
⁵ - 10
æċ - ceil i-1 to the nearest (positive integer) power of 10
_ - subtract i
S - sum
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edited Feb 22, 2022 at 16:31
answered Feb 22, 2022 at 16:23
Jonathan AllanJonathan Allan
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Python 3.8 (pre-release), ~~50~~ 44 bytes
Gave a fun recursive solution a go :)
edit: -6 bytes thanks to solid.py
f=lambda n:n and f(m:=n-1)-n+10len(str(m))
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edited Feb 22, 2022 at 17:46
answered Feb 22, 2022 at 15:44
friddofriddo
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1 Welcome to Code Golf, and nice answer!rydwolf –rydwolf♦ 2022-02-22 15:45:29 +00:00 Commented Feb 22, 2022 at 15:45
Actually, 44 bytessolid.py –solid.py 2022-02-22 17:38:52 +00:00 Commented Feb 22, 2022 at 17:38
@solid.py yes! haha, that one was painfully obvious friddo –friddo 2022-02-22 17:46:55 +00:00 Commented Feb 22, 2022 at 17:46
Surely repeating n-1 saves a byte, as the assignment takes three bytes.Neil –Neil 2022-02-22 19:55:22 +00:00 Commented Feb 22, 2022 at 19:55
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C (gcc), 79 bytes
g(n,t,s){t=--n/10;s=n%10+1;n=t45+s(19-s)/2+10(t?sg(t+1)+(10-s)g(t)-90:0);}
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answered Feb 22, 2022 at 22:30
attatt
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Zsh, 35 bytes
for ((x=$1;x--;s+=1e$#x+~x)):
<<<$s
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edited Feb 23, 2022 at 15:30
answered Feb 22, 2022 at 20:58
pxegerpxeger
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PowerShell, ~~57~~ 49 bytes
-8 bytes thanks to @mazzy!
0.."$args"-gt0|%{$s-=$_---("1e"+"$_".length)}
+$s
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edited Feb 24, 2022 at 12:54
answered Feb 22, 2022 at 17:19
Zaelin GoodmanZaelin Goodman
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1e2 = 100, 1e3 = 1000, 1e4 = 10000...mazzy –mazzy 2022-02-22 19:59:33 +00:00 Commented Feb 22, 2022 at 19:59
Try it online! ?mazzy –mazzy 2022-02-22 20:13:58 +00:00 Commented Feb 22, 2022 at 20:13
1 @mazzy Ahhhhh, clever, thank you! I totally forgot about scientific notation in powershell!! And nice way to improve on the zero case Zaelin Goodman –Zaelin Goodman 2022-02-24 12:56:03 +00:00 Commented Feb 24, 2022 at 12:56
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Behaviour, ~~36~~ 29 bytes
f=&#(a&10^#(""+a)-1-a;>&a+b)
My own recently made esolang, so bear with me.
ungolfed and commented
f=&
#( // deal with 0 case, convert nil into 0
a & // generate array for n nine complements
10^#(""+a)-1-a; // nine complements logic
>&a+b // reduce array by adding its items
)
Test with:
f:0
f:1
f:10
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edited Feb 24, 2022 at 16:03
answered Feb 23, 2022 at 20:18
Lince AssassinoLince Assassino
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Nibbles, ~~8.5~~ 6.5 bytes
+.,$-^~,`p-$~
Attempt This Online!
-2 bytes thanks to Dominic van Essen
+.,$-^~,`p-$~
+ Sum
. for n in
, range from 1 to
$ input
- subtract
^~ 10 to the power of
, length of
`p convert to string
- ~ subtract 1 from
$ n
n
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edited Mar 31, 2023 at 6:05
answered Mar 28, 2023 at 14:44
AdamátorAdamátor
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1 Use "int to str" (backtick-p) instead of "to base" (backtick-@~) save 1 nibble... (no idea how to type the backticks directly into the comment...)Dominic van Essen –Dominic van Essen 2023-03-30 14:14:57 +00:00 Commented Mar 30, 2023 at 14:14
1 @DominicvanEssen That actually saves 4 nibbles because 0 is no longer a special case. Thanks!Adamátor –Adamátor 2023-03-30 16:02:58 +00:00 Commented Mar 30, 2023 at 16:02
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Japt-mx, 8 bytes
+1 byte to work around a bug in Japt.
aÓApUs l
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answered Feb 22, 2022 at 15:32
ShaggyShaggy
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Python 3, 49 bytes
-1 byte thanks to user friddo
lambda n:sum(~i+10len(str(i))for i in range(n))
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edited Feb 22, 2022 at 17:28
answered Feb 22, 2022 at 14:12
solid.pysolid.py
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Dyalog APL, 17 bytes
+/⍳-⍨1-⍨10≢∘⍕¨∘⍳
⎕IO←0.
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answered Feb 22, 2022 at 18:37
rabbitgrowthrabbitgrowth
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JavaScript, 47 bytes
f=n=>(t=10`${n}`.length)n+n~n/2+~-t~t/11+9
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There isn't any non-recursive implementation here. So maybe it is worth to include one. But it is longer than current answers...
If your language trunk divide result to int automatically, +~-t~t/11 could be changed into -tt/11.
t=10 1+⌊log 10 n⌋t⋅n−n⋅(n+1)2−t 2−100 11
with special case where n=0
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edited Feb 23, 2022 at 7:53
answered Feb 23, 2022 at 7:42
tshtsh
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MATL, 14 bytes
:q"10@Vn^q@-vs
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answered Feb 23, 2022 at 11:11
Sundar RSundar R
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1 You could replace vs by ]vs :-P Luis Mendo –Luis Mendo 2022-02-24 16:00:28 +00:00 Commented Feb 24, 2022 at 16:00
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Desmos, 49 bytes
f(k)=∑_{n=2}^k(10^{1+floor(log(n-1))}-n)+9-0^k9
Would be 44 bytes if we didn't have to deal with 0...
Try It On Desmos!
Try It On Desmos! - Prettified
48 bytes, port of @tsh's answer
t=10^{log(n+0^n)}10
f(n)=tn-(nn+n)/2-(tt-100)/11
Try It On Desmos!
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edited Mar 1, 2022 at 1:03
answered Feb 23, 2022 at 6:11
Aiden ChowAiden Chow
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how the heck do you use desmos?DialFrost –DialFrost 2022-02-23 06:49:09 +00:00 Commented Feb 23, 2022 at 6:49
@DialFrost it's hard to explain fully in a comment, but the gist is that Desmos uses mathematical formulae (like functions or expressions) in order to compute the output. these formulae are expressed through L A T E X (albeit with certain optimizations for byte count), which is the code you see.Aiden Chow –Aiden Chow 2022-02-23 06:57:53 +00:00 Commented Feb 23, 2022 at 6:57
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Raku, 32 bytes
0,|[\+] {{S:g/./9/-$_}($++)}...
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This is an expression for the infinite sequence of sums.
{ ... } ... is a lazy, infinite sequence, where the brackets enclose an expression that generates each sequence element.
$++ postincrements an anonymous state variable each time the generating function is called, producing the numbers 0, 1, 2, .... Each time, that number is passed to the anonymous function enclosed in the inner brackets. (That saves us having to declare a variable to hold the counter value.)
S:g/./9/ converts every digit in the counter to 9.
- $_ subtracts the counter from the previous value.
[\+] produces the sequence of partial sums from the original sequence.
0, | pastes a zero onto the front of the sequence.
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answered Nov 16, 2022 at 20:22
SeanSean
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J-uby, 32 bytes
:|:sum+(-[S|:+@|:&10,:~]|+:+)
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Explanation
```
: | :sum + ( -[ S | :+@ | : & 10, :~ ] | +:+ )
: | # 0...input
:sum + ( ) # Sum by
-[ , ] | +:+ # Sum of
S | :+@ | : & 10 # 10 to the power of the number of digits
:~ # and two's complement
```
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edited Nov 26, 2022 at 5:13
answered Nov 16, 2022 at 16:58
JordanJordan
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Thunno, 11 log 256(96)≈ 9.05 bytes
eDL10@1-_ES
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Explanation
```
Implicit input
e # Map over range(input):
DL # Duplicate and push the length
10 # Push ten
@ # Push 10 length
1- # Subtract one
_ # Subtract the loop variable
ES # After the map, sum
# Implicit output
```
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answered Mar 28, 2023 at 16:23
The ThonnuThe Thonnu
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FunStack alpha, 60 bytes
Cons 9 Reverse From0 over flatmap Times 9 Pow 10 #N Sum Take
Try it at Replit!
Explanation
First, we construct the infinite list [9,8,7,6,5,4,3,2,1,0,89,88,87,...]:
```
N
```
Natural numbers starting from 0.
Times 9 Pow 10
Take 10 to the power of each number, then multiply by 9.
Reverse From0 over flatmap
Create a function that takes a number, generates the range from 0 to that number (exclusive), and reverses it. Then map that function to each number in the above list and flatten the resulting list of lists.
(What we're doing here is really just function composition, but over is fewer bytes than compose and does the same thing in this case.)
Cons 9
Stick a 9 on the front of the list.
Now this value gets appended to the program's argument list (which previously contained the input number, N), and we apply the following functions:
Take
Take the first N values from the infinite list.
Sum
Sum them.
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answered Mar 30, 2023 at 21:03
DLoscDLosc
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JavaScript, 35 bytes
f=n=>n&&f(--n)+10`${n}`.length+~n
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answered Feb 22, 2022 at 14:10
m90m90
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J, 24 bytes
+/@((10&^@#@":->:)"0@i.)
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answered Feb 22, 2022 at 14:31
sinvecsinvec
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Python 2, 42 bytes
f=lambda n:n and-~~n+10len(`n-1`)+f(n-1)
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answered Feb 22, 2022 at 16:40
wasifwasif
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1 Remove ~~ to save two bytes. n is integer, so those two symbols are not needed ophact –ophact 2022-02-22 17:57:02 +00:00 Commented Feb 22, 2022 at 17:57
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Ruby, 39 bytes
->n{n.times.sum{|x|10x.to_s.size+~x}}
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answered Feb 22, 2022 at 17:02
G BG B
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Charcoal, 12 bytes
Ｉ↨¹ＥＮ⁻Ｉ×Ｌι9ι
Try it online! Link is to verbose version of code. Explanation:
Ｎ Input as a number
Ｅ Map over implicit range
9 Literal string `9`
× Repeated by
ι Current value
Ｌ Length (of string representation)
Ｉ Cast to integer
⁻ Subtract
ι Current value
↨¹ Take the sum
Ｉ Cast to string
Implicitly print
(Sum() won't work for an input of 0.)
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answered Feb 22, 2022 at 19:59
NeilNeil
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Burlesque, 19 bytes
-.{Jln'9j.j|-}GZ++
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-. # Decrement (Burlesque runs 0..N inclusive, need to exclude)
{
J # Duplicate
ln # Number of digits
'9 # Character 9
j # Reorder stack
. # Repeat 9 n_digits times as string
j # Reorder stack
|- # String subtract (parse and subtract)
}GZ # Generate for range
++ # Sum
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answered Feb 22, 2022 at 21:19
DeathIncarnateDeathIncarnate
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Pari/GP, 30 bytes
n->sum(i=0,n-1,10^#Str(i)-1-i)
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answered Feb 23, 2022 at 1:43
alephalphaalephalpha
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1
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1321 | https://content.one.lumenlearning.com/calculus1/chapter/maxima-and-minima-learn-it-1/ | Maxima and Minima: Learn It 1 – Calculus I
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Calculus I
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Book Contents Navigation
Book Contents
Faculty Resources
Teaching Resources
Evidence-Based Teaching Practices
Course Resources
Appendix A: Table of Integrals
Glossary of Terms
Students: Additional Lumen Resources
Appendix B: Table of Derivatives
Appendix C: Precalculus Formulas
Basic Functions and Graphs
Basic Functions and Graphs: Cheat Sheet
Basic Functions and Graphs: Background You’ll Need 1
Basic Functions and Graphs: Background You’ll Need 2
Review of Functions: Learn It 1
Review of Functions: Learn It 2
Review of Functions: Learn It 3
Review of Functions: Learn it 4
Review of Functions: Learn it 5
Review of Functions: Apply It
Review of Functions: Fresh Take
Basic Classes of Functions: Learn It 1
Basic Classes of Functions: Learn It 2
Basic Classes of Functions: Learn It 3
Basic Classes of Functions: Learn It 4
Basic Classes of Functions: Learn It 5
Basic Classes of Functions: Learn It 6
Basic Classes of Functions: Apply It
Basic Classes of Functions: Fresh Take
Basic Functions and Graphs: Get Stronger
Basic Functions and Graphs: Get Stronger Answer Key
More Basic Functions and Graphs
More Basic Functions and Graphs: Cheat Sheet
More Basic Functions and Graphs: Background You’ll Need 1
More Basic Functions and Graphs: Background You’ll Need 2
More Basic Functions and Graphs: Background You’ll Need 3
Trigonometric Functions: Learn It 1
Trigonometric Functions: Learn It 2
Trigonometric Functions: Learn It 3
Trigonometric Functions: Learn It 4
Trigonometric Functions: Apply It
Trigonometric Functions: Fresh Take
Inverse Functions: Learn It 1
Inverse Functions: Learn It 2
Inverse Functions: Learn It 3
Inverse Functions: Learn It 4
Inverse Functions: Apply It
Inverse Functions: Fresh Take
Exponential and Logarithmic Functions: Learn It 1
Exponential and Logarithmic Functions: Learn It 2
Exponential and Logarithmic Functions: Learn It 3
Exponential and Logarithmic Functions: Learn It 4
Exponential and Logarithmic Functions: Learn It 5
Exponential and Logarithmic Functions: Apply It
Exponential and Logarithmic Functions: Fresh Take
More Basic Functions and Graphs: Get Stronger
More Basic Functions and Graphs: Get Stronger Answer Key
Understanding Limits
Understanding Limits: Cheat Sheet
Understanding Limits: Background You’ll Need 1
Understanding Limits: Background You’ll Need 2
Understanding Limits: Background You’ll Need 3
A Preview of Calculus: Learn It 1
A Preview of Calculus: Learn It 2
A Preview of Calculus: Learn It 3
A Preview of Calculus: Learn It 4
A Preview of Calculus: Apply It
A Preview of Calculus: Fresh Take
Introduction to the Limit of a Function: Learn It 1
Introduction to the Limit of a Function: Learn It 2
Introduction to the Limit of a Function: Learn It 3
Introduction to the Limit of a Function: Learn It 4
Introduction to the Limit of a Function: Apply It
Introduction to the Limit of a Function: Fresh Take
Understanding Limits: Get Stronger
Understanding Limits: Get Stronger Answer Key
Limits and Continuity
Limits and Continuity: Cheat Sheet
Limits and Continuity: Background You'll Need 1
Limits and Continuity: Background You'll Need 2
The Limit Laws: Learn It 1
The Limit Laws: Learn It 2
The Limit Laws: Learn It 3
The Limit Laws: Learn It 4
The Limit Laws: Apply It
The Limit Laws: Fresh Take
Continuity: Learn It 1
Continuity: Learn It 2
Continuity: Learn It 3
Continuity: Learn It 4
Continuity: Apply It
Continuity: Fresh Take
The Precise Definition of a Limit: Learn It 1
The Precise Definition of a Limit: Learn It 2
The Precise Definition of a Limit: Learn It 3
The Precise Definition of a Limit: Apply It
The Precise Definition of a Limit: Fresh Take
Limits and Continuity: Get Stronger
Limits and Continuity: Get Stronger Answer Key
Introduction to Derivatives
Introduction to Derivatives: Cheat Sheet
Introduction to Derivatives: Background You'll Need 1
Introduction to Derivatives: Background You'll Need 2
Defining the Derivative: Learn It 1
Defining the Derivative: Learn It 2
Defining the Derivative: Learn It 3
Defining the Derivative: Apply It
Defining the Derivative: Fresh Take
The Derivative as a Function: Learn It 1
The Derivative as a Function: Learn It 2
The Derivative as a Function: Learn It 3
The Derivative as a Function: Learn It 4
The Derivative as a Function: Apply It
The Derivative as a Function: Fresh Take
Differentiation Rules: Learn It 1
Differentiation Rules: Learn It 2
Differentiation Rules: Learn It 3
Differentiation Rules: Learn It 4
Differentiation Rules: Learn It 5
Differentiation Rules: Apply It
Differentiation Rules: Fresh Take
Derivatives as Rates of Change: Learn It 1
Derivatives as Rates of Change: Learn It 2
Derivatives as Rates of Change: Learn It 3
Derivatives as Rates of Change: Learn It 4
Derivatives as Rates of Change: Apply It
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Define and Identify the highest and lowest points of a function on a graph, both overall and within specific sections
Locate points on a function within a specific range where the slope is zero or undefined (critical points)
Explain how to use critical points to find the highest or lowest values of a function within a limited range
Extrema and Critical Points
Given a particular function, we are often interested in determining the largest and smallest values of the function. This information is important in creating accurate graphs. Finding the maximum and minimum values of a function also has practical significance because we can use this method to solve optimization problems, such as maximizing profit, minimizing the amount of material used in manufacturing an aluminum can, or finding the maximum height a rocket can reach. In this section, we look at how to use derivatives to find the largest and smallest values for a function.
Absolute Extrema
Consider the function f(x)=x 2+1 f(x)=x 2+1 over the interval (−∞,∞)(−∞,∞). As x→±∞x→±∞, f(x)→∞f(x)→∞. Therefore, the function does not have a largest value. However, since x 2+1≥1 x 2+1≥1 for all real numbers x x and x 2+1=1 x 2+1=1 when x=0 x=0, the function has a smallest value, 1 1, when x=0 x=0.
Figure 1. The given function has an absolute minimum of 1 at x=0 x=0. The function does not have an absolute maximum.
We say that 1 1 is theabsolute minimum of f(x)=x 2+1 f(x)=x 2+1 and it occurs at x=0 x=0. We say that f(x)=x 2+1 f(x)=x 2+1 does not have anabsolute maximum.
absolute maximum and extremum
Let f f be a function defined over an interval I I and let c∈I c∈I.
We say f f has an absolute maximum on I I at c c if f(c)≥f(x)f(c)≥f(x) for all x∈I x∈I.
We say f f has an absolute minimum on I I at c c if f(c)≤f(x)f(c)≤f(x) for all x∈I x∈I.
If f f has an absolute maximum on I I at c c or an absolute minimum on I I at c c, we say f f has anabsolute extremum on I I at c c.
Before proceeding, let’s note two important issues regarding this definition.
First, the term absolute here does not refer to absolute value. An absolute extremum may be positive, negative, or zero.
Second, if a function f f has an absolute extremum over an interval I I at c c, the absolute extremum is f(c)f(c). The real number c c is a point in the domain at which the absolute extremum occurs.
Consider the function f(x)=1(x 2+1)f(x)=1(x 2+1) over the interval (−∞,∞)(−∞,∞). Since
f(0)=1≥1 x 2+1=f(x)f(0)=1≥1 x 2+1=f(x)
for all real numbers x x, we say f f has an absolute maximum over (−∞,∞)(−∞,∞) at x=0 x=0. The absolute maximum is f(0)=1 f(0)=1. It occurs at x=0 x=0, as shown in Figure 2b.
So remember: the maximum/minimum = y y; the location of the maximum/minimum = x x.
A function may have both an absolute maximum and an absolute minimum, just one extremum, or neither. Figure 2 shows several functions and some of the different possibilities regarding absolute extrema.
![Image 2: This figure has six parts a, b, c, d, e, and f. In figure a, the line f(x) = x3 is shown, and it is noted that it has no absolute minimum and no absolute maximum. In figure b, the line f(x) = 1/(x2 + 1) is shown, which is near 0 for most of its length and rises to a bump at (0, 1); it has no absolute minimum, but does have an absolute maximum of 1 at x = 0. In figure c, the line f(x) = cos x is shown, which has absolute minimums of −1 at ±π, ±3π, … and absolute maximums of 1 at 0, ±2π, ±4π, …. In figure d, the piecewise function f(x) = 2 – x2 for 0 ≤ x < 2 and x – 3 for 2 ≤ x ≤ 4 is shown, with absolute maximum of 2 at x = 0 and no absolute minimum. In figure e, the function f(x) = (x – 2)2 is shown on [1, 4], which has absolute maximum of 4 at x = 4 and absolute minimum of 0 at x = 2. In figure f, the function f(x) = x/(2 − x) is shown on 0, 2), with absolute minimum of 0 at x = 0 and no absolute maximum.
Figure 2. Graphs (a), (b), and (c) show several possibilities for absolute extrema for functions with a domain of (−∞,∞)(−∞,∞). Graphs (d), (e), and (f) show several possibilities for absolute extrema for functions with a domain that is a bounded interval.
However, the following theorem, called theextreme value theorem, guarantees that a continuous function f f over a closed, bounded interval [a,b][a,b] has both an absolute maximum and an absolute minimum.
extreme value theorem
If f f is a continuous function over the closed, bounded interval [a,b][a,b], then there is a point in [a,b][a,b] at which f f has an absolute maximum over [a,b][a,b] and there is a point in [a,b][a,b] at which f f has an absolute minimum over [a,b][a,b].
The proof of the extreme value theorem is beyond the scope of this text. Typically, it is proved in a course on real analysis.
There are a couple of key points to note about the statement of this theorem. For the extreme value theorem to apply, the function must be continuous over a closed, bounded interval. If the interval I I is open or the function has even one point of discontinuity, the function may not have an absolute maximum or absolute minimum over I I.
Consider the functions shown in Figure 2(d), (e), and (f).
![Image 3: This figure has six parts a, b, c, d, e, and f. In figure a, the line f(x) = x3 is shown, and it is noted that it has no absolute minimum and no absolute maximum. In figure b, the line f(x) = 1/(x2 + 1) is shown, which is near 0 for most of its length and rises to a bump at (0, 1); it has no absolute minimum, but does have an absolute maximum of 1 at x = 0. In figure c, the line f(x) = cos x is shown, which has absolute minimums of −1 at ±π, ±3π, … and absolute maximums of 1 at 0, ±2π, ±4π, …. In figure d, the piecewise function f(x) = 2 – x2 for 0 ≤ x < 2 and x – 3 for 2 ≤ x ≤ 4 is shown, with absolute maximum of 2 at x = 0 and no absolute minimum. In figure e, the function f(x) = (x – 2)2 is shown on [1, 4], which has absolute maximum of 4 at x = 4 and absolute minimum of 0 at x = 2. In figure f, the function f(x) = x/(2 − x) is shown on 0, 2), with absolute minimum of 0 at x = 0 and no absolute maximum.
Figure 2. Graphs (a), (b), and (c) show several possibilities for absolute extrema for functions with a domain of (−∞,∞)(−∞,∞). Graphs (d), (e), and (f) show several possibilities for absolute extrema for functions with a domain that is a bounded interval.
All three of these functions are defined over bounded intervals. However, the function in graph (e) is the only one that has both an absolute maximum and an absolute minimum over its domain.
The extreme value theorem cannot be applied to the functions in graphs (d) and (f) because neither of these functions is continuous over a closed, bounded interval.
Although the function in graph (d) is defined over the closed interval [0,4][0,4], the function is discontinuous at x=2 x=2. The function has an absolute maximum over [0,4][0,4] but does not have an absolute minimum.
The function in graph (f) is continuous over the half-open interval [0,2)[0,2), but is not defined at x=2 x=2, and therefore is not continuous over a closed, bounded interval. The function has an absolute minimum over [0,2)[0,2), but does not have an absolute maximum over [0,2)[0,2).
These two graphs illustrate why a function over a bounded interval may fail to have an absolute maximum and/or absolute minimum.
Before looking at how to find absolute extrema, let’s examine the related concept of local extrema. This idea is useful in determining where absolute extrema occur.
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Are all pairs of consecutive Fibonacci numbers > 1 co-prime with each other?
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Manas Paldhe
Interested in maths problems! · Author has 143 answers and 2.2M answer views
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Originally Answered: Are all Fibonacci numbers > 1 co-prime with each other? ·
New answer: (Anonymous edited the question to ask if all "consecutive" pairs of Fibonacci numbers are co-prime):
Suppose the two consecutive Fibonacci numbers Fn and F(n+1) are not co-prime.
The series would be:
1, 1, 2, 3, . . . . , F(n-1), F(n), F(n+1)
Now F(n-1) + F(n) = F(n+1)
F(n) and F(n+1) are both divisible by say x where x != 1
Note that F(n-1) is not 0, for any n > 1
So, F(n-1) is divisible by x.
Recursively apply the same logic and F(n-2) is divisible by x.
Keep on applying the logic recursively and we get that 1, 2, 3, are divisible by x.
Ah, a contradiction.
So, our original assumption is i
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New answer: (Anonymous edited the question to ask if all "consecutive" pairs of Fibonacci numbers are co-prime):
Suppose the two consecutive Fibonacci numbers Fn and F(n+1) are not co-prime.
The series would be:
1, 1, 2, 3, . . . . , F(n-1), F(n), F(n+1)
Now F(n-1) + F(n) = F(n+1)
F(n) and F(n+1) are both divisible by say x where x != 1
Note that F(n-1) is not 0, for any n > 1
So, F(n-1) is divisible by x.
Recursively apply the same logic and F(n-2) is divisible by x.
Keep on applying the logic recursively and we get that 1, 2, 3, are divisible by x.
Ah, a contradiction.
So, our original assumption is incorrect.
Thus every pair of consecutive Fibonacci numbers > 1, are co-prime.
Edit: Answer to original question ( Are all pairs of Fibonacci numbers > 1 co-prime with each other?)
1, 1, 2, 3, 5, 8....
2 and 8 are not co-prime.
So, the answer is no!
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Are all positive integer solutions to [math]xy\pm 1=y^2-x^2[/math] pairs of consecutive Fibonacci numbers?
Are 2, 3, and 5 the only successive prime Fibonacci numbers?
Are all co-prime numbers twin primes?
Consider Fibonacci numbers, which are also prime numbers greater than [math]3[/math]. Is it true that all of these numbers are congruent to [math]1 \bmod 4[/math]?
How many Fibonacci numbers are prime?
Alon Amit
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Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) and
Michael Jørgensen
, PhD in mathematics · Author has 8.8K answers and 173.8M answer views
·9y
Originally Answered: Are all Fibonacci numbers > 1 co-prime with each other? ·
Of course not, since 2 is a Fibonacci number and infinitely many Fibonacci numbers are even (why? Because if any two consecutive ones are odd, the next one is even).
Perhaps you meant to ask if consecutive Fibonacci numbers, like 8 and 13 or 34 and 55, are always relatively prime.
The answer is Yes. Why? Suppose we could find two Fibonacci numbers next to each other and both of them divisible by 7. Remember that any Fibonacci number is simply the sum of the two preceding ones. So in other words, the Fibonacci number just before our pair is their difference. And the difference of two numbers th
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Of course not, since 2 is a Fibonacci number and infinitely many Fibonacci numbers are even (why? Because if any two consecutive ones are odd, the next one is even).
Perhaps you meant to ask if consecutive Fibonacci numbers, like 8 and 13 or 34 and 55, are always relatively prime.
The answer is Yes. Why? Suppose we could find two Fibonacci numbers next to each other and both of them divisible by 7. Remember that any Fibonacci number is simply the sum of the two preceding ones. So in other words, the Fibonacci number just before our pair is their difference. And the difference of two numbers that are both divisible be 7 is also divisible by 7.
So our imagined pair that's divisible by 7 can't be first such pair. The pair just before it also qualifies. But then also the pair before this one, and the pair before, and the...
Wait a minute.
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Shai Simonson
30+ years teaching math and CS · Author has 712 answers and 2.6M answer views
·9y
By induction on the number of pairs.
The first consecutive pair 2 and 3 are co-prime.
Assume that the numbers A and B in the kth pair of consecutive Fibonacci numbers, are co-prime. The next pair of consecutive Fibonacci numbers are B and A+B. Now if B and A+B had a common factor, then this factor must also divide A, but A and B have no common factors.
So, once you get a single pair of consecutive Fibonacci numbers that are co-prime, all the rest must be co-prime.
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Yes. Any two consecutive Fibonacci numbers greater than 1 are coprime.
Proof sketch (standard number-theoretic argument using the Euclidean algorithm):
Let F_n denote the nth Fibonacci number with F_1 = 1, F_2 = 1 and F_{n+1} = F_n + F_{n-1}.
Suppose d is a common divisor of two consecutive Fibonacci numbers F_n and F_{n+1}. Then d | F_{n+1} and d | F_n, hence d divides any integer linear combination, in particular d | (F_{n+1} − F_n) = F_{n-1}.
Repeating the same argument, d divides F_{n-2}, then F_{n-3}, and so on. By the Euclidean-algorithm descent, d divides F_2 = 1 (since the sequence steps d
Continue Reading
Yes. Any two consecutive Fibonacci numbers greater than 1 are coprime.
Proof sketch (standard number-theoretic argument using the Euclidean algorithm):
Let F_n denote the nth Fibonacci number with F_1 = 1, F_2 = 1 and F_{n+1} = F_n + F_{n-1}.
Suppose d is a common divisor of two consecutive Fibonacci numbers F_n and F_{n+1}. Then d | F_{n+1} and d | F_n, hence d divides any integer linear combination, in particular d | (F_{n+1} − F_n) = F_{n-1}.
Repeating the same argument, d divides F_{n-2}, then F_{n-3}, and so on. By the Euclidean-algorithm descent, d divides F_2 = 1 (since the sequence steps down to F_2 or F_1).
Any divisor of 1 is 1, so d = 1. Therefore gcd(F_n, F_{n+1}) = 1.
Remarks:
The argument works for every n ≥ 1, so every pair of consecutive Fibonacci numbers (except the trivial initial pair both equal to 1) are coprime.
More generally, gcd(F_m, F_n) = F_{gcd(m,n)}; for consecutive indices gcd(n, n+1) = 1, hence gcd(F_n, F_{n+1}) = F_1 = 1.
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Related questions
Are all positive integer solutions to [math]xy\pm 1=y^2-x^2[/math] pairs of consecutive Fibonacci numbers?
Are 2, 3, and 5 the only successive prime Fibonacci numbers?
Are all co-prime numbers twin primes?
Consider Fibonacci numbers, which are also prime numbers greater than [math]3[/math]. Is it true that all of these numbers are congruent to [math]1 \bmod 4[/math]?
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How do you find a difference of Fibonacci numbers (fibonacci number math)?
What is the biggest known prime number that is also a Fibonacci number?
What is the smallest prime number that is the sum of two consecutive Fibonacci numbers? Hint: The Fibonacci sequence is defined as Fn=Fn−1+Fn−2 , where F0=0 and F1=1?
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Are all positive integer solutions to [math]xy\pm 1=y^2-x^2[/math] pairs of consecutive Fibonacci numbers?
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Counting the number of distinct s-t cuts in a oriented graph
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I am trying to find the number of distinct s-t cuts in a oriented unweighed graph. In an article Enumeration in Graphs p. 45 I found good way how to enumerate those cuts (section 7.3). Is there a faster or simpler algorithm I can use if I am interested only in the number of such cuts and I do not actually need to enumerate it?
The definition of a s-t cut I am using is following. We have a directed graph where two vertices are labelled S and T. Cut is a minimal set of edges of a graph such that by removing those edges there will no longer exist a path from vertex S to vertex T.
I tried to ask this at Stack Overflow and I have been pointed here.
ds.algorithms
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edited May 23, 2017 at 12:40
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asked Dec 13, 2011 at 9:18
user7610user7610
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If your graph is planar, then you can do this in O(n 2)O(n 2) time. See dx.doi.org/10.1016/j.tcs.2011.05.017.Zach Langley –Zach Langley 2012-12-17 05:59:08 +00:00 Commented Dec 17, 2012 at 5:59
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Counting the number of s,t-cuts is #P-complete. This is a result by Provan and Ball.
J. Scott Provan, Michael O. Ball: The Complexity of Counting Cuts and of Computing the Probability that a Graph is Connected. SIAM J. Comput. 12(4): 777-788 (1983)
Therefore, unless some complexity-theoretic collapse happens, you cannot get essentially faster algorithm than listing all of them.
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answered Dec 13, 2011 at 9:28
Yoshio OkamotoYoshio Okamoto
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Is the complexity of counting minimum s,t-cuts also hard?someone –someone 2011-12-13 19:49:27 +00:00 Commented Dec 13, 2011 at 19:49
Yes, that is also hard.Yoshio Okamoto –Yoshio Okamoto 2011-12-13 22:23:09 +00:00 Commented Dec 13, 2011 at 22:23
That's interesting, as there are only n 2 α n 2 α (not s-t) cuts that are within α α of the minimum. Is the exact number of minimum cuts easy to compute?Sasho Nikolov –Sasho Nikolov 2011-12-14 02:39:08 +00:00 Commented Dec 14, 2011 at 2:39
2 @Sasho Nikolov: The number of min-cuts in an n n-vertex undirected graph is at most (n 2)(n 2), and they can be listed in polynomial time. So, the number of min-cuts can be computed in polynomial time.Yoshio Okamoto –Yoshio Okamoto 2011-12-14 05:32:09 +00:00 Commented Dec 14, 2011 at 5:32
1 @YoshioOkamoto, I thought about your "essentially" (at first) I guess you mean some polynomial time difference, but IMHO 2 n>>2 s q r t(n)2 n>>2 s q r t(n) or something similar, this can be a good results and solvable in many cases. (if exists)Saeed –Saeed 2011-12-14 10:14:44 +00:00 Commented Dec 14, 2011 at 10:14
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1324 | https://en.wikipedia.org/wiki/Valence_(chemistry) | Jump to content
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Contents
(Top)
1 Description
2 Definition
3 Historical development
3.1 Electrons and valence
4 Common valences
5 Valence versus oxidation state
5.1 Examples
6 "Maximum number of bonds" definition
7 Maximum valences of the elements
8 See also
9 References
Valence (chemistry)
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From Wikipedia, the free encyclopedia
Combining capacity of elements with other atoms
For other uses, see Valence (disambiguation).
Not to be confused with coordination number, oxidation state, or valence electron.
See also: Polyvalency (chemistry)
In chemistry, the valence (US spelling) or valency (British spelling) of an atom is a measure of its combining capacity with other atoms when it forms chemical compounds or molecules. Valence is generally understood to be the number of chemical bonds that each atom of a given chemical element typically forms. Double bonds are considered to be two bonds, triple bonds to be three, quadruple bonds to be four, quintuple bonds to be five and sextuple bonds to be six. In most compounds, the valence of hydrogen is 1, of oxygen is 2, of nitrogen is 3, and of carbon is 4. Valence is not to be confused with the related concepts of the coordination number, the oxidation state, or the number of valence electrons for a given atom.
Description
The valence is the combining capacity of an atom of a given element, determined by the number of hydrogen atoms that it combines with. In methane, carbon has a valence of 4; in ammonia, nitrogen has a valence of 3; in water, oxygen has a valence of 2; and in hydrogen chloride, chlorine has a valence of 1. Chlorine, as it has a valence of one, can be substituted for hydrogen in many compounds. Phosphorus has a valence 3 in phosphine (PH3) and a valence of 5 in phosphorus pentachloride (PCl5), which shows that an element may exhibit more than one valence. The structural formula of a compound represents the connectivity of the atoms, with lines drawn between two atoms to represent bonds. The two tables below show examples of different compounds, their structural formulas, and the valences for each element of the compound.
| | | | | | |
--- --- --- |
| Compound | H2Hydrogen | CH4Methane | C3H8Propane | C3H6Propylene | C2H2Acetylene |
| Diagram | | | | | |
| Valencies | Hydrogen: 1 | Carbon: 4 Hydrogen: 1 | Carbon: 4 Hydrogen: 1 | Carbon: 4 Hydrogen: 1 | Carbon: 4 Hydrogen: 1 |
| | | | | | | | | |
--- --- --- ---
| Compound | NH3Ammonia | NaCNSodium cyanide | PSCl3Thiophosphoryl chloride | H2SHydrogen sulfide | H2SO4Sulfuric acid | H2S2O6Dithionic acid | Cl2O7Dichlorine heptoxide | XeO4Xenon tetroxide |
| Diagram | | | | | | | | |
| Valencies | Nitrogen: 3 Hydrogen: 1 | Sodium: 1 Carbon: 4 Nitrogen: 3 | Phosphorus: 5 Sulfur: 2 Chlorine: 1 | Sulfur: 2 Hydrogen: 1 | Sulfur: 6 Oxygen: 2 Hydrogen: 1 | Sulfur: 6 Oxygen: 2 Hydrogen: 1 | Chlorine: 7 Oxygen: 2 | Xenon: 8 Oxygen: 2 |
Definition
Valence is defined by the IUPAC as:
: The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.
An alternative modern description is:
: The number of hydrogen atoms that can combine with an element in a binary hydride or twice the number of oxygen atoms combining with an element in its oxide or oxides.
This definition differs from the IUPAC definition as an element can be said to have more than one valence.
Historical development
The etymology of the words valence (plural valences) and valency (plural valencies) traces back to 1425, meaning "extract, preparation", from Latin valentia "strength, capacity", from the earlier valor "worth, value", and the chemical meaning referring to the "combining power of an element" is recorded from 1884, from German Valenz.
The concept of valence was developed in the second half of the 19th century and helped successfully explain the molecular structure of inorganic and organic compounds. The quest for the underlying causes of valence led to the modern theories of chemical bonding, including the cubical atom (1902), Lewis structures (1916), valence bond theory (1927), molecular orbitals (1928), valence shell electron pair repulsion theory (1958), and all of the advanced methods of quantum chemistry.
In 1789, William Higgins published views on what he called combinations of "ultimate" particles, which foreshadowed the concept of valency bonds. If, for example, according to Higgins, the force between the ultimate particle of oxygen and the ultimate particle of nitrogen were 6, then the strength of the force would be divided accordingly, and likewise for the other combinations of ultimate particles (see illustration).
The exact inception, however, of the theory of chemical valencies can be traced to an 1852 paper by Edward Frankland, in which he combined the older radical theory with thoughts on chemical affinity to show that certain elements have the tendency to combine with other elements to form compounds containing 3, i.e., in the 3-atom groups (e.g., NO3, NH3, NI3, etc.) or 5, i.e., in the 5-atom groups (e.g., NO5, NH4O, PO5, etc.), equivalents of the attached elements. According to him, this is the manner in which their affinities are best satisfied, and by following these examples and postulates, he declares how obvious it is that
A tendency or law prevails (here), and that, no matter what the characters of the uniting atoms may be, the combining power of the attracting element, if I may be allowed the term, is always satisfied by the same number of these atoms.
This "combining power" was afterwards called quantivalence or valency (and valence by American chemists). In 1857 August Kekulé proposed fixed valences for many elements, such as 4 for carbon, and used them to propose structural formulas for many organic molecules, which are still accepted today.
Lothar Meyer in his 1864 book, Die modernen Theorien der Chemie, contained an early version of the periodic table containing 28 elements, for the first time classified elements into six families by their valence. Works on organizing the elements by atomic weight, until then had been stymied by the widespread use of equivalent weights for the elements, rather than atomic weights.
Most 19th-century chemists defined the valence of an element as the number of its bonds without distinguishing different types of valence or of bond. However, in 1893 Alfred Werner described transition metal coordination complexes such as [Co(NH3)6]Cl3, in which he distinguished principal and subsidiary valences (German: 'Hauptvalenz' and 'Nebenvalenz'), corresponding to the modern concepts of oxidation state and coordination number respectively.
For main-group elements, in 1904 Richard Abegg considered positive and negative valences (maximal and minimal oxidation states), and proposed Abegg's rule to the effect that their difference is often 8.
An alternative definition of valence, developed in the 1920's and having modern proponents, differs in cases where an atom's formal charge is not zero. It defines the valence of a given atom in a covalent molecule as the number of electrons that an atom has used in bonding:
: valence = number of electrons in valence shell of free atom − number of non-bonding electrons on atom in molecule,
or equivalently:
: valence = number of bonds + formal charge.
In this convention, the nitrogen in an ammonium ion [NH4]+ bonds to four hydrogen atoms, but it is considered to be pentavalent because all five of nitrogen's valence electrons participate in the bonding.
Electrons and valence
The Rutherford model of the nuclear atom (1911) showed that the exterior of an atom is occupied by electrons, which suggests that electrons are responsible for the interaction of atoms and the formation of chemical bonds. In 1916, Gilbert N. Lewis explained valence and chemical bonding in terms of a tendency of (main-group) atoms to achieve a stable octet of 8 valence-shell electrons. According to Lewis, covalent bonding leads to octets by the sharing of electrons, and ionic bonding leads to octets by the transfer of electrons from one atom to the other. The term covalence is attributed to Irving Langmuir, who stated in 1919 that "the number of pairs of electrons which any given atom shares with the adjacent atoms is called the covalence of that atom". The prefix co- means "together", so that a co-valent bond means that the atoms share a valence. Subsequent to that, it is now more common to speak of covalent bonds rather than valence, which has fallen out of use in higher-level work from the advances in the theory of chemical bonding, but it is still widely used in elementary studies, where it provides a heuristic introduction to the subject.
In the 1930s, Linus Pauling proposed that there are also polar covalent bonds, which are intermediate between covalent and ionic, and that the degree of ionic character depends on the difference of electronegativity of the two bonded atoms.
Pauling also considered hypervalent molecules, in which main-group elements have apparent valences greater than the maximal of 4 allowed by the octet rule. For example, in the sulfur hexafluoride molecule (SF6), Pauling considered that the sulfur forms 6 true two-electron bonds using sp3d2 hybrid atomic orbitals, which combine one s, three p and two d orbitals. However more recently, quantum-mechanical calculations on this and similar molecules have shown that the role of d orbitals in the bonding is minimal, and that the SF6 molecule should be described as having 6 polar covalent (partly ionic) bonds made from only four orbitals on sulfur (one s and three p) in accordance with the octet rule, together with six orbitals on the fluorines. Similar calculations on transition-metal molecules show that the role of p orbitals is minor, so that one s and five d orbitals on the metal are sufficient to describe the bonding.
Common valences
For elements in the main groups of the periodic table, the valence can vary between 1 and 8.
| Group | Valence 1 | Valence 2 | Valence 3 | Valence 4 | Valence 5 | Valence 6 | Valence 7 | Valence 8 | Typical valences |
--- --- --- --- --- |
| 1 (I) | NaClKCl | 1 |
| 2 (II) | MgCl2CaCl2 | 2 |
| 13 (III) | InBrTlI | BCl3AlCl3Al2O3 | 3 |
| 14 (IV) | COPbCl2 | CO2CH4SiCl4 | 2 and 4 |
| 15 (V) | NO | NH3PH3As2O3 | NO2 | N2O5PCl5 | 3 and 5 |
| 16 (VI) | H2OH2SSCl2 | SO2SF4 | SO3SF6H2SO4 | 2, 4 and 6 |
| 17 (VII) | HClICl | HClO2ClF3 | ClO2 | IF5HClO3 | IF7Cl2O7HClO4 | 1, 3, 5 and 7 |
| 18 (VIII) | KrF2 | | XeF4 | | XeO3 | XeO4 | 0, 2, 4, 6 and 8 |
Many elements have a common valence related to their position in the periodic table, and nowadays this is rationalised by the octet rule. The Greek/Latin numeral prefixes (mono-/uni-, di-/bi-, tri-/ter-, and so on) are used to describe ions in the charge states 1, 2, 3, and so on, respectively. Polyvalence or multivalence refers to species that are not restricted to a specific number of valence bonds. Species with a single charge are univalent (monovalent). For example, the Cs+ cation is a univalent or monovalent cation, whereas the Ca2+ cation is a divalent cation, and the Fe3+ cation is a trivalent cation. Unlike Cs and Ca, Fe can also exist in other charge states, notably 2+ and 4+, and is thus known as a multivalent (polyvalent) ion. Transition metals and metals to the right are typically multivalent but there is no simple pattern predicting their valency.
Valence adjectives using the -valent suffix†
| Valence | More common adjective‡ | Less common synonymous adjective‡§ |
| 0-valent | zerovalent | nonvalent |
| 1-valent | monovalent | univalent |
| 2-valent | divalent | bivalent |
| 3-valent | trivalent | tervalent |
| 4-valent | tetravalent | quadrivalent |
| 5-valent | pentavalent | quinquevalent, quinquivalent |
| 6-valent | hexavalent | sexivalent |
| 7-valent | heptavalent | septivalent |
| 8-valent | octavalent | — |
| 9-valent | nonavalent | — |
| 10-valent | decavalent | — |
| 11-valent | undecavalent | — |
| 12-valent | dodecavalent | — |
| multiple / many / variable | polyvalent | multivalent |
| together | covalent | — |
| not together | noncovalent | — |
† The same adjectives are also used in medicine to refer to vaccine valence, with the slight difference that in the latter sense, quadri- is more common than tetra-.
‡ As demonstrated by hit counts in Google web search and Google Books search corpora (accessed 2017).
§ A few other forms can be found in large English-language corpora (for example, quintavalent, quintivalent, decivalent), but they are not the conventionally established forms in English and thus are not entered in major dictionaries.
Valence versus oxidation state
Because of the ambiguity of the term valence, other notations are currently preferred. Beside the lambda notation, as used in the IUPAC nomenclature of inorganic chemistry, oxidation state is a more clear indication of the electronic state of atoms in a molecule.
The oxidation state of an atom in a molecule gives the number of valence electrons it has gained or lost. In contrast to the valency number, the oxidation state can be positive (for an electropositive atom) or negative (for an electronegative atom).
Elements in a high oxidation state have an oxidation state higher than +4, and also, elements in a high valence state (hypervalent elements) have a valence higher than 4. For example, in perchlorates ClO−4, chlorine has 7 valence bonds (thus, it is heptavalent, in other words, it has valence 7), and it has oxidation state +7; in ruthenium tetroxide RuO4, ruthenium has 8 valence bonds (thus, it is octavalent, in other words, it has valence 8), and it has oxidation state +8.
In some molecules, there is a difference between valence and oxidation state for a given atom. For example, in disulfur decafluoride molecule S2F10, each sulfur atom has 6 valence bonds (5 single bonds with fluorine atoms and 1 single bond with the other sulfur atom). Thus, each sulfur atom is hexavalent or has valence 6, but has oxidation state +5. In the dioxygen molecule O2, each oxygen atom has 2 valence bonds and so is divalent (valence 2), but has oxidation state 0. In acetylene H−C≡C−H, each carbon atom has 4 valence bonds (1 single bond with hydrogen atom and a triple bond with the other carbon atom). Each carbon atom is tetravalent (valence 4), but has oxidation state −1.
Examples
Variation of valence vs oxidation state for bonds between two different elements
| Compound | Formula | Valence | Oxidation state | Diagram |
| Hydrogen chloride | HCl | H = 1 Cl = 1 | H = +1 Cl = −1 | H−Cl |
| Perchloric acid | HClO4 | H = 1 Cl = 7 O = 2 | H = +1 Cl = +7 O = −2 | |
| Methane | CH4 | C = 4 H = 1 | C = −4 H = +1 | |
| Dichloromethane | CH2Cl2 | C = 4 H = 1 Cl = 1 | C = 0 H = +1 Cl = −1 | |
| Ferrous oxide | FeO | Fe = 2 O = 2 | Fe = +2 O = −2 | Fe=O |
| Ferric oxide | Fe2O3 | Fe = 3 O = 2 | Fe = +3 O = −2 | O=Fe−O−Fe=O |
| Sodium hydride | NaH | Na = 1 H = 1 | Na = +1 H = −1 | Na−H |
The perchlorate ion ClO−4 is monovalent, in other words, it has valence 1. Valences may also be different from absolute values of oxidation states due to different polarity of bonds. For example, in dichloromethane, CH2Cl2, carbon has valence 4 but oxidation state 0. Iron oxides appear in a crystal structure, so no typical molecule can be identified. In ferrous oxide, Fe has oxidation state +2; in ferric oxide, oxidation state +3.
Variation of valence vs oxidation state for bonds between two atoms of the same element
| Compound | Formula | Valence | Oxidation state | Diagram |
| Hydrogen | H2 | H = 1 | H = 0 | H−H |
| Chlorine | Cl2 | Cl = 1 | Cl = 0 | Cl−Cl |
| Hydrogen peroxide | H2O2 | H = 1 O = 2 | H = +1 O = −1 | |
| Hydrazine | N2H4 | H = 1 N = 3 | H = +1 N = −2 | |
| Disulfur decafluoride | S2F10 | S = 6 F = 1 | S = +5 F = −1 | |
| Dithionic acid | H2S2O6 | S = 6 O = 2 H = 1 | S = +5 O = −2 H = +1 | |
| Hexachloroethane | C2Cl6 | C = 4 Cl = 1 | C = +3 Cl = −1 | |
| Ethylene | C2H4 | C = 4 H = 1 | C = −2 H = +1 | |
| Acetylene | C2H2 | C = 4 H = 1 | C = −1 H = +1 | H−C≡C−H |
| Mercury(I) chloride | Hg2Cl2 | Hg = 2 Cl = 1 | Hg = +1 Cl = −1 | Cl−Hg−Hg−Cl |
"Maximum number of bonds" definition
Frankland took the view that the valence (he used the term "atomicity") of an element was a single value that corresponded to the maximum value observed. The number of unused valencies on atoms of what are now called the p-block elements is generally even, and Frankland suggested that the unused valencies saturated one another. For example, nitrogen has a maximum valence of 5, in forming ammonia two valencies are left unattached; sulfur has a maximum valence of 6, in forming hydrogen sulphide four valencies are left unattached.
The International Union of Pure and Applied Chemistry (IUPAC) has made several attempts to arrive at an unambiguous definition of valence. The current version, adopted in 1994:
: The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.
Hydrogen and chlorine were originally used as examples of univalent atoms, because of their nature to form only one single bond. Hydrogen has only one valence electron and can form only one bond with an atom that has an incomplete outer shell. Chlorine has seven valence electrons and can form only one bond with an atom that donates a valence electron to complete chlorine's outer shell. However, chlorine can also have oxidation states from +1 to +7 and can form more than one bond by donating valence electrons.
Hydrogen has only one valence electron, but it can form bonds with more than one atom. In the bifluoride ion ([HF2]−), for example, it forms a three-center four-electron bond with two fluoride atoms:
: [F−H F− ↔ F− H−F]
Another example is the three-center two-electron bond in diborane (B2H6).
Maximum valences of the elements
Maximum valences for the elements are based on the data from list of oxidation states of the elements. They are shown by the color code at the bottom of the table.
| Maximum valences of the elements |
| | 1 | 2 | | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| Group → |
| ↓ Period |
| 1 | 1H | | 2He |
| 2 | 3Li | 4Be | | 5B | 6C | 7N | 8O | 9F | 10Ne |
| 3 | 11Na | 12Mg | | 13Al | 14Si | 15P | 16S | 17Cl | 18Ar |
| 4 | 19K | 20Ca | | 21Sc | 22Ti | 23V | 24Cr | 25Mn | 26Fe | 27Co | 28Ni | 29Cu | 30Zn | 31Ga | 32Ge | 33As | 34Se | 35Br | 36Kr |
| 5 | 37Rb | 38Sr | | 39Y | 40Zr | 41Nb | 42Mo | 43Tc | 44Ru | 45Rh | 46Pd | 47Ag | 48Cd | 49In | 50Sn | 51Sb | 52Te | 53I | 54Xe |
| 6 | 55Cs | 56Ba | | 71Lu | 72Hf | 73Ta | 74W | 75Re | 76Os | 77Ir | 78Pt | 79Au | 80Hg | 81Tl | 82Pb | 83Bi | 84Po | 85At | 86Rn |
| 7 | 87Fr | 88Ra | | 103Lr | 104Rf | 105Db | 106Sg | 107Bh | 108Hs | 109Mt | 110Ds | 111Rg | 112Cn | 113Nh | 114Fl | 115Mc | 116Lv | 117Ts | 118Og |
| |
| | 57La | 58Ce | 59Pr | 60Nd | 61Pm | 62Sm | 63Eu | 64Gd | 65Tb | 66Dy | 67Ho | 68Er | 69Tm | 70Yb | |
| | 89Ac | 90Th | 91Pa | 92U | 93Np | 94Pu | 95Am | 96Cm | 97Bk | 98Cf | 99Es | 100Fm | 101Md | 102No |
| Maximum valences are based on the List of oxidation states of the elements |
| 0 1 2 3 4 5 6 7 8 9 Unknown Background color shows maximum valence of the chemical element Primordial From decay Synthetic Border shows natural occurrence of the element |
See also
Abegg's rule
Oxidation state
References
^ a b Partington, James Riddick (1921). A text-book of inorganic chemistry for university students (1st ed.). OL 7221486M.
^ a b IUPAC Gold Book definition: valence
^ Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. doi:10.1016/C2009-0-30414-6. ISBN 978-0-08-037941-8.
^ Harper, Douglas. "valence". Online Etymology Dictionary.
^ a b Partington, J.R. (1989). A Short History of Chemistry. Dover Publications, Inc. ISBN 0-486-65977-1.
^ Frankland, E. (1852). "On a New Series of Organic Bodies Containing Metals". Philosophical Transactions of the Royal Society of London. 142: 417–444. doi:10.1098/rstl.1852.0020. S2CID 186210604.
^ Alan J. Rocke (1984). Chemical Atomism in the Nineteenth Century: From Dalton to Cannizzaro. Ohio State University Press.
^ a b Parkin, Gerard (May 2006). "Valence, Oxidation Number, and Formal Charge: Three Related but Fundamentally Different Concepts". Journal of Chemical Education. 83 (5): 791. Bibcode:2006JChEd..83..791P. doi:10.1021/ed083p791. ISSN 0021-9584. While the concepts and definitions of valence have been refined over the years, that described by Sidgwick remains the most useful and simple definition for covalent molecules: the valence of an atom in a covalent molecule is simply the number of electrons that an atom has used in bonding.
^ Sidgwick, N. V. (1927). The electronic theory of valency. London: Oxford University Press (Clarendon Press). p. 199. On the whole the best definition of absolute valency seems to be that adopted by Grimm and Sommerfeld, that it is numerically equal to the number of electrons of the atom 'engaged' (beansprucht) in attaching the other atoms.
^ Grimm, H.G.; Sommerfeld, A (1926). "Über den. Zusammenhang des Abschlusses der Elektronengruppen im Atom mit den chemischen Valenzzahlen". Zeitschrift für Physik. 36 (1): 36–59. Bibcode:1926ZPhy...36...36G. doi:10.1007/bf01383924. S2CID 120248399.
^ Smith, Derek W. (2005). "Valence, Covalence, Hypervalence, Oxidation State, and Coordination Number". Journal of Chemical Education. 82 (8): 1202. Bibcode:2005JChEd..82.1202S. doi:10.1021/ed082p1202. ISSN 0021-9584.
^ Langmuir, Irving (1919). "The Arrangement of Electrons in Atoms and Molecules". Journal of the American Chemical Society. 41 (6): 868–934. doi:10.1021/ja02227a002.
^ Magnusson, Eric (1990). "Hypercoordinate molecules of second-row elements: d functions or d orbitals?". J. Am. Chem. Soc. 112 (22): 7940–7951. doi:10.1021/ja00178a014.
^ Frenking, Gernot; Shaik, Sason, eds. (May 2014). "Chapter 7: Chemical bonding in Transition Metal Compounds". The Chemical Bond: Chemical Bonding Across the Periodic Table. Wiley – VCH. ISBN 978-3-527-33315-8.
^ Merriam-Webster, Merriam-Webster's Unabridged Dictionary, Merriam-Webster, archived from the original on 2020-05-25, retrieved 2017-05-11.
^ "Lesson 7: Ions and Their Names". Clackamas Community College. Archived from the original on 21 January 2019. Retrieved 5 February 2019.
^ The Free Dictionary: valence
^ IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "Lambda". doi:10.1351/goldbook.L03418
^ IUPAC, Compendium of Chemical Terminology, 5th ed. (the "Gold Book") (2025). Online version: (2006–) "Oxidation state". doi:10.1351/goldbook.O04365
^ Frankland, E. (1870). Lecture notes for chemical students(Google eBook) (2d ed.). J. Van Voorst. p. 21.
^ Frankland, E.; Japp, F.R (1885). Inorganic chemistry (1st ed.). pp. 75–85. OL 6994182M.
^ Muller, P. (1994). "Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)". Pure and Applied Chemistry. 66 (5): 1077–1184. doi:10.1351/pac199466051077. S2CID 195819485.
Look up valence or valency in Wiktionary, the free dictionary.
| v t e Periodic table |
| Periodic table forms | Alternatives Extended periodic table |
| Sets of elements | | | | | | | | | | --- --- --- --- | | By periodic table structure | | | | --- | | Groups | 1 (Hydrogen and alkali metals) 2 (Alkaline earth metals) 3 4 5 6 7 8 9 10 11 12 13 (Triels) 14 (Tetrels) 15 (Pnictogens) 16 (Chalcogens) 17 (Halogens) 18 (Noble gases) | | Periods | 1 2 3 4 5 6 7 8+ + Aufbau + Fricke + Pyykkö | | Blocks | Aufbau principle | | | By metallicity | | | | --- | | Metals | Lanthanides Actinides Transition metals Post-transition metals | | Metalloids | Lists of metalloids by source Dividing line | | Nonmetals | Noble gases | | | Other sets | Platinum-group metals (PGM) Rare-earth elements Refractory metals Precious metals Coinage metals Noble metals Heavy metals Native metals Transuranium elements Superheavy elements Major actinides Minor actinides | |
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| History | Element discoveries + Dmitri Mendeleev + 1871 table + 1869 predictions Naming + etymology + controversies + for places + for people + in East Asian languages |
| See also | IUPAC + nomenclature + systematic element name Trivial name Dmitri Mendeleev |
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Reumatología Clínica (English Edition)
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Reumatología Clínica is the official publication of scientific Spanish Society of Rheumatology (SER) and the Mexican College of Rheumatology (CMR). Reumatología Clínica publishes original research papers, editorials, reviews, case reports and pictures. Published studies are primarily clinical and epidemiological research but also basic.
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Vol. 11. Issue 2.
Pages 99-107(March - April 2015)
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Abstract
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Introduction
Abstract
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Resumen
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Introduction
Pregnancy and systemic sclerosis
Fertility in systemic sclerosis
Effects of pregnancy on systemic sclerosis
Microchimerism
Systemic sclerosis effects on pregnancy
Effects on the fetus
Management plan
Management of labor
Breastfeeding
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Vol. 11. Issue 2.
Pages 99-107(March - April 2015)
Review Article
DOI: 10.1016/j.reumae.2014.06.005
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Comprehensive Approach to Systemic Sclerosis Patients During Pregnancy
Manejo integral de las pacientes con esclerosis sistémica durante el embarazo
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Alexandra Rueda de León Aguirrea, José Antonio Ramírez Calvob, Tatiana Sofía Rodríguez Reynaa,
aDepartamento de Inmunología y Reumatología, Instituto Nacional de Ciencias Médicas y Nutrición Salvador Zubirán, Mexico City, Mexico
bDepartamento de Medicina Fetal, Instituto Nacional de Perinatología Isidro Espinosa de los Reyes, Mexico City, Mexico
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Tables (3)
Table 1. Effects of Pregnancy on SS.
Table 2. Recommendations for the Management of Pregnancy in Patients With SS.
Table 3. General Recommendations on the Use of Drugs During Puerperium and Lactation in Women With SS.
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Abstract
Systemic sclerosis (SSc) is a connective tissue disease that usually affects women, with a male:female ratio of 1:4–10.
It was thought that there was a prohibitive risk of fatal complications in the pregnancies of patients with SSc. It is now known that the majority of these women undergo a normal progression of pregnancy if the right time is chosen and a close obstetric care is delivered. The obstetric risk will depend on the subtype and clinical stage of the disease, and the presence and severity of the internal organ involvement during the pregnancy.
The management of these pregnancies should be provided in a specialized center, with a multidisciplinary team capable of identifying and promptly treating complications.
Treatment should be limited to drugs with no teratogenic potential, except when renal crises or severe cardiovascular complications develop.
Keywords:
Systemic sclerosis
Pregnancy
Obstetric complications
Resumen
La esclerosis sistémica (ES) es una enfermedad del tejido conectivo poco común que afecta principalmente a mujeres (relación mujer:hombre de 4–10:1).
En el pasado se pensaba que existía gran riesgo de complicaciones fatales en los embarazos de pacientes con ES. Actualmente, se sabe que muchas de estas mujeres pueden llevar a buen término un embarazo si se elige el momento adecuado y se lleva monitorización obstétrica estrecha. El riesgo obstétrico dependerá del subtipo y la fase clínica de la enfermedad y de la presencia y la gravedad de la afección de órganos internos durante el embarazo.
El manejo del embarazo de las pacientes con ES debe realizarse en un centro de atención especializada, con un equipo multidisciplinario capaz de detectar y tratar las complicaciones tempranamente.
El tratamiento debe limitarse a fármacos sin potencial teratogénico, excepto en crisis renales y en complicaciones cardiopulmonares que pongan en peligro la vida de la madre.
Palabras clave:
Esclerosis sistémica
Embarazo
Complicaciones obstétricas
Full Text
Introduction
Systemic sclerosis (SSc) or generalized scleroderma is a relatively rare connective tissue disease of unknown etiology affecting skin, joints, blood vessels, heart, lungs, gastrointestinal tract and kidneys. It has an incidence of 2–10 cases per million and prevalence of 150–300 cases per million worldwide. It mainly affects women, with a peak incidence between the fifth and sixth decades of life with a female to male ratio of 4–10:1; this ratio increases during reproductive age (15–50 years), in which the relationship can reach 15:1.1,2
Survival in SS has improved in recent years, mainly thanks to the introduction of angiotensin-converting enzyme inhibitors (ACEI) in the 1980s, which decreased complications and mortality due to renal crisis. Interstitial lung disease and pulmonary hypertension have replaced renal failure as the most common causes of morbidity and mortality from scleroderma.2–9 The improvement in the prognosis of the patients due to the greater knowledge of the disease as well as increased access to health services and treatment of SS complications has been accompanied by an increase in the number of women with this condition who seek and achieve pregnancy.
This article will discuss the effects of SS on pregnancy, the influence of vascular disease, inflammation and fibrosis on the health of the mother and fetus, and the effects of pregnancy on SS.
Pregnancy and Systemic Sclerosis
In the past there were few reports of pregnancies in patients with SS due, in part, to the fact that the peak age of onset is between 45 and 55 years, and the fact that reports of early cases showed disappointing outcomes for both mothers and babies.10–14 25–30 years ago it was relatively common for doctors to recommend their patients with SS not to conceive and even consider abortion because of the supposed high risk of fatal complications for mother and children. This initial information, based on isolated case reports, was replaced by a series of retrospective and prospective case series which showed large proportion of women with SS who led successful pregnancy with little risk of serious complications when the patient and physician discussed the issue and chose an appropriate time for pregnancy, performing15–18 close obstetric monitoring.
Pregnancy in women with SS should be considered from the onset as a high risk pregnancy due to the increased risk of premature delivery and low birth weight for gestational age. In early pregnancy, each patient should be carefully assessed to establish the subtype of disease (diffuse or limited), the (early or late) phase and the extent and severity of damage to internal organs. Those patients with a duration of symptoms of less than 4 years (scleroderma in early stage), diffuse subtype, or anti-topoisomerase i or anti-RNA polymerase iii antibodies have a higher associated obstetric risk and, if possible, should delay pregnancy until the patient is in a late stage and hence has a less active disease.15
The third trimester of pregnancy is considered the one with the highest risk, since the patient may develop complications secondary to hypertension, renal failure, pulmonary hypertension, interstitial lung disease or heart failure.19,20
Fertility in Systemic Sclerosis
It has been difficult to establish whether fertility is affected in patients with SS, as studies have shown conflicting results. An Italian report made several decades ago suggested that there was no direct association of the disease with decreased fertility and conception problems and concluded that these issues should not be attributed to the SS.21 On the other hand, in two English studies, the authors postulate that fertility may be impaired in patients even years before the onset of the disease, with a twice greater risk of repeated spontaneous abortion and three times more fertility problems, defined by them as an unsuccessful pregnancy before 35 years of age.22,23
Steen et al. have examined this controversy in two retrospective studies in which they compared patients with SSc patients with rheumatoid arthritis (RA) and healthy women, and later performed a prospective study.
The first study demonstrated that patients who already had SS and who become pregnant have similar frequencies of spontaneous abortions (15%) than RA patients (16%) and healthy controls (13%), and conclude that the disease per se does not directly affect total fertility by increasing the proportion of abortions.16
In the second study, the pregnancy outcomes of women whose SS onset occurs during the childbearing years were compared with RA and healthy women. As for fertility, it showed that the percentage of women who had never been pregnant is higher in groups of patients with SS and RA than in healthy controls (P<0.05). However, this does not necessarily mean that patients have fertility problems, as there are additional factors that contribute to these differences. Among the study patients, 8%–10% had no sexual activity; women with SS or RA in this had developed the disease at a younger age compared to those sexually active; A questionnaire study showed that most women with SS (5%) and RA (11%) had chosen not to have children compared with the control group (3%), some of them motivated by the advice of doctors or family. Only 2%–5% of patients with SS had sought unsuccessfully to become pregnant; in most of them, the age of onset of SS was after 40, so they were already probably infertile.17
The study found no significant differences between groups in the number of women who reported a period of at least one year during which they failed to conceive (15% of patients with SS, 12% of patients with RA and 13% of controls). There were 27 women with SSc who were evaluated for infertility, and in that 63% of them already had the disease at the time of the evaluation; the rate of successful pregnancies in patients evaluated for infertility, regardless of the treatment they received for it, was similar in the three groups: 37% for patients with SS, 40% for patients with RA and 43% for healthy women. The analysis found no significant differences in the frequency of infertility or the rate of successful pregnancies in women with RA and SS before or after disease onset.17
Besides the above, this topic is especially related to the sexuality of patients, since vascular lung disease, skin and physical limitations, as well as changes in the appearance and the emotional effects of the disease, have the ability of impact relationships of some of the patients, thus affecting the possibility of conception. The main symptoms that patients reported as affecting their sex life were fatigue, muscle/joint pain, vaginal dryness and dyspareunia. They also associated Raynaud's phenomenon, sore hands, digital ulcers, dyspnea and chest pain.24–26
Patients may also have coexisting secondary antiphospholipid antibody syndrome (APS), so it would be appropriate to identify lupus anticardiolipin antibodies, β2 glycoprotein 1 and lupus anticoagulant in all patients with SS and recurrent fetal loss. In a study by Steen et al. antiphospholipid antibodies were found in 50% of patients with SS and ulcers of the lower limbs, indicating that the association of SS and APS cannot be that uncommon.27 There are other studies indicating that the antiphospholipid antibodies may be independently associated with pulmonary arterial hypertension, macrovascular disease and increased total mortality in patients with systemic sclerosis.28,29
Effects of Pregnancy on Systemic Sclerosis
It is difficult to determine the effects of pregnancy on the disease, as some symptoms are very similar and, therefore, it is not easy to distinguish which of the causes are attributable. Such is the case of GERD, joint pain and swelling, to name a few. The consensus of several studies on this is that there are no significant changes in the disease during pregnancy.10,30 The prospective study by Steen et al. showed that the disease was stable in 60% of the patients, 20% noted improvement, particularly of Raynaud's phenomenon, and 20% worsening, particularly with regard to reflux, arrhythmia, arthritis, skin thickening and development of renal crisis in two cases.18
After delivery, 33% of women noticed an increase in the severity of some symptoms, particularly: Raynaud's phenomenon, arthritis, skin thickening and renal crisis. The 10-year survival of SS patients with and without pregnancy is similar. Most women did not have worsening of symptoms after delivery.18
Pulmonary arterial hypertension is a complication that gets worse during and after pregnancy and in which another pregnancy is not recommended because of the high mortality of this group of patients during pregnancy.31
Table 1 summarizes the changes in clinical signs and symptoms of organ systems in patients with SS during pregnancy.
Table 1.
Effects of Pregnancy on SS.
General The disease remains stable
Raynaud's phenomenon Improvement during pregnancy. It may worsen after delivery, especially in case of complications
Skin disorder It may start during pregnancy, and in some cases there may be DSS progression during the postpartum
Joints Joint pain, similar to pregnancy in patients without SSThere may be arthritis
Gastrointestinal Reflux, nausea, vomiting and constipation. Similar to patients without SSSpecialized care is needed in handling cases of poor intestinal absorption
Cardiopulmonary Symptoms associated with cardiopulmonary conditions aggravated by SS. Management should be equal to that of pregnant patients with cardiopulmonary diseases without SSPoor prognosis in patients with PAH
Renal Renal crises occur more frequently in early diffuse SS patients with or without pregnancy. Due to the high mortality, management with ACE inhibitors should be started early on suspicion of this complication, despite the fetal riskNo evidence of increased incidence of renal crisis has been seen in pregnancies
SS: systemic sclerosis; PAH: pulmonary arterial hypertension; ACE inhibitors, angiotensin converting enzyme inhibitors.
Raynaud's phenomenon and vascular disease. Generally these symptoms improve during pregnancy, presumably by vasodilation associated with it.31 It is common for patients to report exacerbation in the postpartum period, who therefore need to restart treatment at this early stage, particularly those who have had previous ulcers or amputations.
Skin condition. In general, case series report that skin manifestations are usually stable during pregnancy, however there are some reports of the onset of SS during pregnancy. As these are isolated cases, it is not possible to determine whether SS develops more often during pregnancy.15 In the series by Steen and Medsger there were some diffuse SSc patients who worsened in their skin manifestations during pregnancy; in this period, patients had discontinued treatment and exacerbation could also be triggered by this factor.18
Joint pain and arthritis. Joint pains are common in all pregnant women, but there is no evidence that it is more frequent in patients with SS, although it is a common complaint; the physician must also assess whether there is arthritis because it can improve with treatment, as will be discussed later in this text.18
Gastrointestinal. Nausea, vomiting and reflux are more frequent and severe in patients with SS than in healthy pregnant women; poor intestinal absorption due to intestinal dysmotility and bacterial overgrowth can be a serious problem that can threaten the life of the patient and the baby, so its management should be continued throughout pregnancy. Another common problem is constipation.18,32,33
Pulmonary Fibrosis. Evidence in the literature has shown no decline in lung function in patients with mild or no fibrosis.18,31 In mild cases of pulmonary fibrosis, there are reports of successful outcomes with proper handling, but in five cases of women with severe pulmonary fibrosis, forced vital capacity (FVC)<65% was reported by Steen et al.; there were three preterm infants, an abortion and one maternal death from multiple organ failure as well as an18 elective abortion. In addition, dyspnea is a common complaint, which increases particularly during the third quarter of pregnancy.
Pulmonary arterial hypertension. PAH is a serious complication that occurs primarily in patients with SSc and limited long-term evolution of the disease and has been associated with the presence of anti-centromere antibodies, anti-Th/To and anti-U3 RNP (the last 2 with nucleolar pattern on immunofluorescence). However, it can also occur in patients of reproductive age, so there is the risk of concomitant PAH during pregnancy. Women with PAH who become pregnant may develop severe hemodynamic complications due to the low reserve of the pulmonary arterioles, which are not able to handle the increased blood volume and cardiac output occurring physiologically during pregnancy.31
Some studies in pregnant women with PAH, which included patients with PAH from various causes, estimated that maternal mortality ranges from 17 to 50%, with increased risk during labor and in the first 2 weeks postpartum, due to acute cardiovascular collapse.34–36
An analysis of pregnancy outcomes of women with primary PAH showed that these patients have an increased risk of hospitalization and hypertensive disorders of pregnancy.37 Multiple case reports and retrospective studies conducted both before and after the introduction of vasodilators such as sildenafil, prostaglandin analogs and nitric oxide, show high mortality in this group of patients38–48; no controlled studies exist comparing mortality before and after the era of these new vasodilators, and patients with SS and PAH should avoid conceiving, and even can be offered to electively terminate the pregnancy.
Renal disease. Renal crisis is a syndrome characterized by acute, refractory hypertension, progressive proteinuria and acute renal failure associated with microangiopathic changes on the biopsy, with the appearance of “onion skin” lesions of the renal arteries by endothelial proliferation. It affects 5%–10% of Caucasian patients, especially those with diffuse cutaneous SS (DCSS) of rapid evolution in the first 5 years from the onset of symptoms attributable to the disease, with the presence of anti-topoisomerase i or anti-RNA polymerase iii antibodies, and previous exposure to corticosteroids at doses equal to or greater than 15 mg/day, all these in the presence or not of pregnancy.1,15,20 In Mexico, the frequency of renal crisis is likely to be lower; an analysis of complications in internal organs and autoantibodies in our cohort (in a national reference center), mainly in Mexican mestizos from a national referral center, found a prevalence of 2% of renal crisis in the entire cohort, including incident and prevalent cases of these and other patients with DCSS and LSS (Limited systemic sclerosis).49
If there is clinical suspicion, the clinician must perform general laboratory tests including serum creatinine, blood count to look for microangiopathic hemolytic anemia and/or thrombocytopenia, liver function tests for differential diagnosis of HELLP syndrome, as well as consider a urinalysis for proteinuria and other abnormalities.1,15,20
Cases have been reported in which the increase in blood pressure (BP) is not as marked (normotensive renal crisis), so it is necessary to make a clinical differential diagnosis with thrombotic thrombocytopenic purpura (TTP), in which microangiopathic hemolytic anemia, thrombocytopenia and kidney failure are also present. For this, it is useful to measure the plasma levels of the activity of the metalloproteinase enzyme ADAMTS13, since its function is decreased (<10%) or absent in most of TTP patients, while normal in patients with SS and renal crisis.50
Another important differential diagnosis is preeclampsia and eclampsia. In this case, the prevalence of proteinuria further support the diagnosis of preeclampsia and elevated plasma renin levels further support the diagnosis of scleroderma renal crisis.1
The management will be discussed later in the section.
Women with diffuse scleroderma should be advised to become pregnant after their disease has been stable for at least 6 months and the period of greatest risk of developing renal crisis has passed, which is usually 3–5 years after the start of symptoms attributable to the disease.31 The time of this recommendation is based on observations made on the natural history of the disease and published data, as there are no observational longitudinal prospective studies that provide information about it.
Microchimerism
Long term persistence of low levels of DNA from cells from a genetically different individual is called microchimerism. Currently, it is known that during pregnancy there is cellular exchange between mother and fetus through the placenta, with estimates of 2–6 fetal cells per milliliter of maternal blood in the second trimester of pregnancy and lower levels after delivery. However, they may be found in the circulation of up to 64% of healthy women, even decades after delivery.1,51
In the case of SS, the role of microchimerism in the pathogenesis of the disease is of interest because of the similarities the disease has with the graft-versus-host disease (GVHD). It has been considered that microchimerism has the ability to induce autoimmunity through the difference between the major histocompatibility complex (HLA) of the donor cells (fetus) and the host (parent). A prospective study compared the presence of male DNA in women with SSc and healthy controls years after the birth of sons, yielding higher levels in the first group.52 It is possible that certain HLA genes as well as the relationship between the host HLA and the microchimeric cells are determinants of the microchimerism final effect in the host, since there are studies showing enhanced compatibility of HLA Class II, but not in the class i of families with SS. It is proposed that increased compatibility leads to an increase in the exchange of cells through the placenta. These fetal cells may have the ability to react against class I antigens or other minor histocompatibility antigens, altering immunoregulatory mechanisms in the mother, resulting in autoimmunity.1,53,54
This theory does not explain the incidence of SS in nulliparous women and men, but it has been speculated that it could be associated with the presence of maternal cells in these patients. In a study comparing the association of birth order and number of pregnancies/deliveries with the development of SS, using as controls the sisters of affected patients, it was found that the risk of developing SS increased with the order of birth which corresponded to the patients (2.–5° born. OR 1.25, 6.–9° born. 2.22 OR 10.°–15° birth: OR 3.53). It is believed that this could be related to an increase in exposure to chimeric maternal cells of previous pregnancies. This may rely on the fact that researchers have found small amounts of Y chromosome DNA even in nulliparous women.51,54
Furthermore, we have demonstrated the presence of microchimeric maternal cells in men with SS. It has been suggested that microchimeric maternal cells are more aggressive or require less stimulus to activate. One study even found more risk of developing SS in nulliparous patients than in those with parity (OR 1.37).53,55
Systemic Sclerosis Effects on Pregnancy
Studies on the effects of SS in pregnancy are scarce. The groups of Dr. Steen15–18 and Dr. Black have published some22 retrospective and prospective case series that have analyzed this issue.
Abortion: Several studies have recently demonstrated that, although there is a slight increase in the percentage of abortions observed when pregnancy occurs after the onset of the disease with regard to pregnant patients prior to onset of symptoms (15% vs 11%), the incidence of abortions in patients with SS is not higher than the general population or to a control group of patients with RA.17
Placental pathology: Due to the high vascular affection existing in SS patients, research has been performed on whether these pathophysiological changes also affect the placental circulation. A study that analyzed three placentas showed that there is decidual vasculopathy with stromal fibrosis and chronic infarction of chorionic villi, even in the babies born without complications between weeks 34 and 38 of pregnancy.56 It has also been observed that there is placental mesenchymal dysplasia, foamy degeneration of endothelial cells with vascular obstruction, foci with decreased vasculature, corioangiosis foci, fibrinoid deposit material, chorioamnionitis and accelerated placental maturation in these patients.57,58 In 4 of the 13 cases reported in the study by Doss et al., placental abnormalities correlated with fetal growth retardation.58 These alterations are similar to those found in pregnancies complicated by pregnancy-induced hypertension and perhaps are responsible for fetal complications, such as intrauterine growth restriction and preterm birth.
Preterm delivery: The study by Steen et al. showed a marked increase in the frequency of preterm infants in patients with SS, with the most frequent complications occurring in patients who already had the disease at the time of pregnancy (15%) versus patients who become pregnant before the onset of the disease (8%). In this study, which had a control group of healthy persons and patients with RA, the RA patients also had a higher frequency of births following disease onset, indicating that the reasons for these outcomes may not be unique to SS. It is possible that this increase in the number of preterm births is secondary to the mother presenting a chronic disease rather than the disease per se.16 There is no current study has been able to find a higher frequency of pregnancy complications in patients with SS justifying premature births.16–18 It is important to note that although women with SS have a higher risk of preterm delivery, this risk is not large enough to discourage pregnancy.
Hypertensive complications: Although an increase in the frequency of hypertensive diseases during pregnancy has not been found, a study of epidemiological data centers including primary care found an increased risk of about four times in most studies controls. The authors associated this increase to a less than optimal monitoring of patients compared to tertiary centers where clinical care protocols37,59 are normally developed.
When BP increases, their origin must be distinguished to determine whether they are secondary to SS (renal crisis), an obstetric event (preeclampsia) or the appearance of overlap with another rheumatic disease like lupus erythematosus with renal involvement, which can be especially complex from the clinical standpoint. To differentiate whether the etiology is of obstetric origin, seizures, elevated transaminases or serum urate may be useful; however, if the diagnosis remains unclear, the usefulness of the measurement of plasma renin levels has been proven. In preeclampsia, serum renin level varies from low to normal while in SS renal crisis, plasma renin levels are elevated as a result of renal cortical ischemia. In situations where both scenarios are indistinguishable, immediate empirical treatment is indicated. If, due to any situation, a definite diagnosis is absolutely necessary to start treatment, a renal biopsy may be indicated.60
Effects on the Fetus
1. Low birth weight: There is an increase in the frequency of low birth weight for gestational age compared to patients without disease (10% vs 2%, approximately). This result is indicative of intrauterine growth restriction during fetal development, which has been associated with the vascular disease developed in patients with SS. Term infants with low birth weight were observed more frequently in patients who become pregnant after the onset of SS (11%) compared to women who became pregnant before the onset of the disease (5%). Another article, although it is also retrospective but with better methodology, shows an OR for IUGR of 3.74, a statistically significant association. At this point it is highly recommended to seek association with secondary APS, as this and the presence of nephropathy with proteinuria are prognostic determinants of the fetal outcome.
Fetal death: Even though it is traditionally believed that there is an increased risk of this complication, studies in recent years have shown that there is no significant increase in cases of stillbirth compared with healthy controls (4% vs 1.4%–2%).17–20
Fetal heart block: The presence of anti-Ro/SSA and anti-La/SSB antibodies in maternal serum has been associated with the development of heart block in the fetus or newborn. The prevalence of these antibodies is of 12%–37% and 4%, respectively, in patients with systemic sclerosis.49,61,62 Complete heart block develops in 1%–2% of pregnancies in patients with these antibodies, but the frequency may be increased up to 10 times with recurrent pregnancies, so intentional close fetal surveillance for data indicative of this complication is recommended.
There is no evidence of increased frequency of birth defects in children of mothers with systemic sclerosis.
Management Plan
Ideally, in pre-pregnancy planning, the clinician should perform a detailed assessment of the state of the patient and discuss with her the possible complications and the importance of high-risk obstetric monitoring. It is important that the maternal disease be stable before pregnancy to reduce the likelihood of both maternal and fetal obstetric complications. In addition, the severity of the patient organic involvement and antibodies present (topoisomerase, anti-centromere, anti-RNA polymerase iii, anti-Ro/SSA, anti-La/SSB, anticardiolipin, -β2 glycoprotein 1 lupus anticoagulant) should be documented, to treat and prevent complications adequately.
Here, we present the evaluation and management of these patients according to the different organ systems that may be affected.
Vasculopathy. In patients with Raynaud's phenomenon and/or digital ulcers, general measures to avoid sudden temperature changes must be taken. During pregnancy, most vasodilators are contraindicated, but they may be restarted just after delivery.18,31 5-phosphodiesterase inhibitors such as sildenafil have been used without significant adverse effects on the mother or fetus for the treatment of PAH during pregnancy, particularly in the second half,38,46 and recently in some open studies to delay intrauterine growth retardation and for treatment of preeclampsia,63,64 so theoretically they could be used in severe cases of vascular disease during pregnancy, although so far no studies corroborate this.
Skin. For progression of cutaneous manifestations, there are few treatment options since most drugs used for this purpose are contraindicated (cyclophosphamide, D-penicillamine, methotrexate, mycophenolate mofetil). There are reports of improvement in the skin and articular manifestations using IV immunoglobulin in SS patients65 and it is considered safe in pregnancy,66 but its usefulness is disputed.
Joint manifestations. For the treatment of SS associated arthritis during pregnancy, painkillers such as paracetamol may be used, as well as low doses of corticosteroids (less than 15 mg/day of prednisone or equivalent), hydroxychloroquine or, failing this, chloroquine, at minimum necessary doses and for the shortest possible time, though they have not been associated with an increased rate of birth defects or complications of scleroderma.20,67,68 Non-steroidal anti-inflammatory drugs should be limited de to their association with oligohydramnios.
Gastrointestinal manifestations. Upper gastrointestinal symptoms such as nausea and reflux, antacids as magaldrate, proton pump inhibitors and histamine receptor blockers at the minimal effective dose can be used.20,69 Pregnancy should be discouraged in cases of serious disorders of intestinal absorption; however, in a series of patients, one patient with poor intestinal absorption had three fetal losses and managed two successful pregnancies at term with normal weight babies after antibiotic therapy and hyperalimentation for malabsorption.18
Pulmonary fibrosis. Pulmonary function tests with diffusion of carbon monoxide are recommended before pregnancy and often as needed, as indicated by the cardiopneumologist, discouraging pregnancy in cases of severe disease (FVC<50%).18 During pregnancyç immunosuppressants, such as cyclophosphamide or mycophenolate mofetil, are contraindicated, but have been used to try to stop the progression of pulmonary fibrosis; the use of low-dose corticosteroids and immunosuppressive treatment in the immediate postpartum period may be considered.
Pulmonary arterial hypertension. It is recommended to perform an echocardiogram before pregnancy and during any clinical deterioration data over the course of the same. We suggest that women with PAH do not get pregnant. In case of pregnancy, the patient should be treated in a tertiary level center, with close supervision by a cardiopneumologist and continue specific treatment for PAH as needed, particularly during the second half of pregnancy, during and after delivery.31 At this stage, phosphodiesterase 5 inhibitors including sildenafil, prostaglandin analogs such as iloprost or epoprostenol, nitric oxide and supportive therapy with calcium channel antagonists and heparin anticoagulation38,43–46 can be used. The use of endothelin receptor antagonists is contraindicated during pregnancy.47–49
Cardiac manifestations. As for heart disease, there may be new or worsening heart failure, which can be documented with echocardiography. It is recommended to avoid pregnancy in severe cases (ejection fraction<30%). In the few case reports of women with SS and heart failure who became pregnant, favorable outcomes but with preterm births were seen, with frequent monitoring and aggressive management during and after birth with drugs commonly used for heart failure70; in the case of electrocardiographically diagnosed conduction blocks, intervention is rarely required.
Renal disorders. As previously discussed throughout the text, the presence of any elevation in the BP with respect to previous measurements has to be considered potentially serious. Because it is a true medical emergency in which life is at stake both for the mother and baby, as well as renal function, treatment should not wait for diagnostic confirmation and consists of high doses of ACE inhibitors orally or IV. The most widely used has been enalapril, which should be continued indefinitely,15 despite the possibility of fetal toxicity (anhidramnios, renal atresia, pulmonary hypoplasia and fetal death), as the risk of maternal complications exceeds the fetal risk.
If a woman with a previous renal crisis becomes pregnant, you can try driving the BP down with other antihypertensives and closely monitoring creatinine and BP. However, if any of the two are raised, restarting the ACE inhibitor is indicated. In the prospective case series by Steen et al. there were three women with prior renal crisis who continued ACE inhibitors during pregnancy and each managed a successful pregnancy, and one had an abortion; Also there were two women with early diffuse scleroderma renal crisis that developed during pregnancy; one of them required dialysis temporarily and had a preterm baby, and the other had an elective abortion but required dialysis despite this.18
There is no evidence that elective abortion, when a renal crisis occurs, reverses the changes; however, in cases of great maternal or fetal health compromise, pregnancy termination may be considered, coupled with the initiation of therapy with ACE inhibitors.
The treatment of a renal crisis may require dialysis temporarily; there are cases of patients who required dialysis for up to 2 years after the renal crisis and eventually recovered enough renal function to stop dialysis.71
If presented with a worst case scenario during pregnancy, in which the physician is faced with a patient with rapid progression of skin involvement and/or rapid deterioration of organ function, the clinical decision can be guided by the weeks of gestation and the wishes of the patient. In the first quarter, termination of pregnancy should be considered and, during the third trimester, the induction of labor is an option in both cases in order to allow intensive treatment to decrease the rate of disease progression; if these options are not possible, at any time during pregnancy, the patient can be treated symptomatically with permitted drugs, trying to bring the patient to the point where she can undergo preterm labor and receive the same intensive care afterward.31
Obstetric surveillance. As previously discussed, there are certain most common complications in women with SS, especially when these are associated with APS or nephropathy, which are important determinants of perinatal outcome. Therefore, in these women, it is widely recommended to first determine the gestational age by performing an ultrasound during the first trimester of pregnancy; in particular, it is important to carry it out ultrasound at 11–13 weeks, 6 days, because it allows us not only to identify chromosomal abnormalities and some birth defects, but also to screen for placental disease by measuring the pulsatility index of uterine arteries and thereby detect which patients are at risk for preeclampsia and intrauterine growth retardation. According to a systematic review by Bujold,72 the use of aspirin at low doses during pregnancy, as long as one starts before week 16, can reduce by 52% the likelihood of preeclampsia when patients are at high risk for the disease, and considered appropriate in this group of patients, with low doses (80–160 mg/day), from the end of the first quarter.
Obstetric Surveillance should include a structural ultrasound in the second quarter and also closer monitoring of fetal growth, with an ultrasound during the second half of pregnancy.
Management of Labor
At the time of delivery, the physician should consider that patients with SS can be a challenge for the anesthesiologist because of skin changes and the consequent difficulty for peripheral access, placement of a spinal block, achieving position during labor and monitoring with pulse oximeters and sphygmomanometers. In general, these patients are considered candidates for epidural block, which provides adequate anesthesia, peripheral vasodilation and increased skin perfusion.73–75 Some authors indicate lower dosage of anesthetic normally used because they may have longer postpartum sensory and motor block.74 General anesthesia is not recommended due to the difficulty for intubation and the risk of aspiration. It is recommended to monitor the room environment to prevent Raynaud's phenomenon associated complications. The measures used include: childbirth in a warm room, preheat IV fluids, use warm compresses, gloves, thermal socks, etc.
A central line or even a Swan-Ganz catheter may be required for central access, if cardiopulmonary complications, hypertension or renal failure are contemplated.
Depending on the fetal condition in late pregnancy, should the patient opt for a normal delivery, the fetus must be continuously monitored electronically.
Care of the surgical wound, cesarean or episiotomy, must be considered if there is skin involvement in the abdomen or vulva, but there is no published data on local complications and skin involvement in the abdomen has not been associated with the need to end the pregnancy prematurely.
Postpartum, the patient must be provided with continuous monitoring to address timely complications that may arise, remembering that heart failure and cardiovascular collapse in patients with PAH are the most lethal complications. Drugs that have been suspended during pregnancy should be restarted and the clinician should not wait to see if they will be required again.
Table 2 shows a summary of the recommendations for the management of pregnancy in patients with SS.
Table 2.
Recommendations for the Management of Pregnancy in Patients With SS.
Before pregnancy or during the first trimester Evaluation in the early phase of the disease, of the extent of organ involvement and antibody testing to establish potential risks Discontinuing drugs with teratogenic potential preferably 3 months before pregnancy begins
Throughout pregnancy High-risk obstetric care Using proton pump inhibitors, histamine receptor blockers and magaldrate in the smallest possible dose, for gastrointestinal symptoms Minimize the use of corticosteroids Increase the frequency of monitoring of fetal size and uterine contractions Frequent monitoring of maternal BP Intensive treatment of hypertension according to the following general guidelines Any further increase in the BP should be considered serious and require urgent evaluation Monitor if increased creatinine, proteinuria or microangiopathic hemolytic anemia, elevated liver enzymes or uric acid Do not assume it is preeclampsia Give treatment needed to control BP with medications allowed during pregnancy, but if creatinine increases, add ACE inhibitors at low doses and increase them as needed, despite the risk of fetal toxicity. BP control can save the life of the mother and fetus
Labor and delivery Prefer epidural anesthesia Adequate temperature of the delivery room, patient and IV solutions Antepartum placement of central venous access Special Care performing an episiotomy or cesarean section
Postpartum Monitoring of vital signs the first days Reinstatement of drugs for the treatment of SS and intensive treatment of any hypertensive event Close monitoring for possible heart failure, particularly in patients with heart failure and/or PAH, the first 2–4 weeks postpartum Close monitoring of renal function
SS: systemic sclerosis; PAH: pulmonary arterial hypertension; ACE inhibitors, angiotensin converting enzyme inhibitors; BP: blood pressure.
Breastfeeding
As already mentioned, once the pregnancy is completed, treatment should be reinstated for SS in all patients; however, breastfeeding should be taken into consideration, considering whether the drugs are safe, so this process should be individualized for each patient based on their desires and the need for drugs during this period.20 The general recommendations are listed in Table 3.
Table 3.
General Recommendations on the Use of Drugs During Puerperium and Lactation in Women With SS.
| Drug | Recommendation |
:--- |
| Analgesics and NSAIDs | Considered safe, particularly acetaminophen. Small amounts are excreted in human milk. Avoid using aspirin for risk of bleeding in the neonate |
| Corticosteroids | Secreted in breast milk but safe. Delay breastfeeding 4 h at doses >20 mg prednisolone |
| Antimalarials | Small amounts (2%) in breast milk, but generally considered safe |
| Methotrexate, mycophenolate mofetil, D-penicillamine and cyclophosphamide | Not safe. Should not be used during lactation; might be necessary for delaying disease progression, consider stopping breastfeeding |
| Biological | There is insufficient evidence |
NSAIDs: Non-steroidal anti-inflammatory drugs; SS: systemic sclerosis.
Conclusions
-
A large proportion of women with SS can lead successful pregnancies with little risk of serious complications if the patient and their doctors discuss the issue and choose a suitable time for pregnancy, and close obstetric monitoring is performed.
-
Pregnancy in women with SS should be considered, from the onset, as a high risk pregnancy due to increased risk of premature delivery and low birth weight for gestational age.
-
In early pregnancy, the clinician should carefully assess each patient to establish the subtype of disease being treated (diffuse or limited), the (early or late) phase and the extent and severity of damage to internal organs.
-
Those patients with duration of symptoms less than 4 years (scleroderma in early stage), diffuse subtype, or anti-topoisomerase i or anti-RNA polymerase iii antibodies have higher associated obstetric risk and, if possible, should delay pregnancy until the patient is in a late stage and therefore has a less active disease.
-
The third trimester of pregnancy is considered the one with the highest risk, since the patient may develop complications secondary to hypertension, renal failure, pulmonary hypertension, interstitial lung disease or heart failure.
-
The consensus of many studies is that there are no other significant changes in the activity of the SS during pregnancy.
-
The evidence indicates that patients with SS have no fertility problems resulting from their disease.
Ethical Responsibilities Protection of people and animals
The authors declare that no experiments were performed on humans or animals for this article.
Confidentiality of data
The authors state that no patient data appear in this article.
Right to privacy and informed consent
The authors state that no patient data appear in this article.
Conflict of Interest
The authors declare no conflict of interest.
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☆
Please cite this article as: Rueda de León Aguirre A, Ramírez Calvo JA, Rodríguez Reyna TS. Manejo integral de las pacientes con esclerosis sistémica durante el embarazo. Reumatol Clin. 2015;11:99–107.
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1326 | https://ginasthma.org/wp-content/uploads/2024/05/GINA-2024-Strategy-Report-24_05_22_WMS.pdf | Updated 2024 ©2024 Global Initiative for Asthma Global Strategy for Asthma Management and Prevention COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 1 Global Strategy for Asthma Management and Prevention (2024 update) The reader acknowledges that this report is intended as an evidence-based asthma management strategy, for the use of health professionals and policy-makers. It is based, to the best of our knowledge, on current best evidence and medical knowledge and practice at the date of publication. When assessing and treating patients, health professionals are strongly advised to use their own professional judgment, and to take into account local and national regulations and guidelines. GINA cannot be held liable or responsible for inappropriate health care associated with the use of this document, including any use which is not in accordance with applicable local or national regulations or guidelines. Suggested citation: Global Initiative for Asthma. Global Strategy for Asthma Management and Prevention, 2024. Updated May 2024. Available from: www.ginasthma.org COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 2 Table of contents Tables and figures ................................................................................................................................................ 5 Preface ................................................................................................................................................................. 8 Members of GINA committees (2023–24) ........................................................................................................... 9 Abbreviations ...................................................................................................................................................... 11 Introduction ........................................................................................................................................................ 14 Methodology ...................................................................................................................................................... 15 WHAT’S NEW IN GINA 2024? ........................................................................................................................... 19 1. Definition, description, and diagnosis of asthma in adults, adolescents and children 6–11 years ............... 23 Definition of asthma ....................................................................................................................................... 23 Description of asthma .................................................................................................................................... 23 Making the initial diagnosis ............................................................................................................................ 24 Differential diagnosis...................................................................................................................................... 27 Confirming the diagnosis of asthma in patients already taking ICS-containing treatment ............................ 32 How to make the diagnosis of asthma in other contexts ............................................................................... 32 2. Assessment of asthma in adults, adolescents and children 6–11 years ....................................................... 35 Overview ........................................................................................................................................................ 36 Assessing asthma symptom control .............................................................................................................. 38 Assessing future risk of exacerbations, lung function decline and adverse effects ...................................... 41 Role of lung function in assessing asthma control ........................................................................................ 42 Assessing asthma severity ............................................................................................................................ 43 How to distinguish between uncontrolled asthma and severe asthma ......................................................... 46 3. Principles of asthma management in adults, adolescents and children 6–11 years ..................................... 48 The patient–healthcare provider partnership ................................................................................................. 49 Long-term goal of asthma management ........................................................................................................ 50 Remission of asthma ..................................................................................................................................... 50 Personalized control-based asthma management ........................................................................................ 52 Non-pharmacological strategies .................................................................................................................... 57 Referral for expert advice .............................................................................................................................. 66 4. Medications and strategies for adults, adolescents and children 6–11 years ............................................... 67 Categories of asthma medications ................................................................................................................ 69 Why should ICS-containing medication be commenced from the time of diagnosis? ................................... 72 Adults and adolescents: asthma treatment tracks ......................................................................................... 74 Initial asthma treatment for adults and adolescents ...................................................................................... 75 Asthma treatment steps in adults and adolescents ....................................................................................... 77 Track 1 (preferred): treatment steps 1–4 for adults and adolescents using ICS-formoterol reliever ............. 78 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 3 Track 2 (alternative): treatment steps 1–4 for adults and adolescents using SABA reliever ......................... 86 Step 5 (Tracks 1 and 2) in adults and adolescents ....................................................................................... 91 About asthma treatment for children 6–11 years ........................................................................................... 94 Initial asthma treatment in children 6–11 years ............................................................................................. 94 Asthma treatment steps for children 6–11 years ........................................................................................... 96 Reviewing response and adjusting treatment – adults, adolescents and children 6–11 years ................... 100 Allergen immunotherapy .............................................................................................................................. 104 Vaccinations ................................................................................................................................................. 106 Other therapies ............................................................................................................................................ 106 5. Guided asthma self-management education and skills training .................................................................. 108 Skills training for effective use of inhaler devices ........................................................................................ 108 Shared decision-making for choice of inhaler device .................................................................................. 109 Adherence with medication and with other advice ....................................................................................... 111 Asthma information ....................................................................................................................................... 113 Training in guided asthma self-management................................................................................................ 114 Regular review by a healthcare provider or trained healthcare worker ........................................................ 116 School-based programs for children ............................................................................................................. 116 6. Managing asthma with multimorbidity and in specific populations ............................................................... 117 Managing multimorbidity ............................................................................................................................... 117 Managing asthma during the COVID-19 pandemic ..................................................................................... 121 Managing asthma in specific populations or settings .................................................................................. 123 7. Diagnosis and initial treatment in adults with features of asthma, COPD or both ....................................... 131 Objectives .................................................................................................................................................... 132 Background to diagnosing asthma and/or COPD in adult patients ............................................................. 132 Assessment and management of chronic respiratory symptoms ................................................................ 133 8. Difficult-to-treat and severe asthma in adults and adolescents ................................................................... 139 Definitions: uncontrolled, difficult-to-treat, and severe asthma .................................................................... 140 Prevalence: how many people have severe asthma? ................................................................................. 140 Importance: the impact of severe asthma ................................................................................................... 141 Overview of decision tree for assessment and management of difficult-to-treat and severe asthma ......... 141 Investigate and manage difficult-to-treat asthma in adults and adolescents ............................................... 146 Investigate the severe asthma phenotype and consider non-biologic therapies ......................................... 149 Consider Type 2-targeted biologic therapies ............................................................................................... 152 Assess, manage and monitor ongoing severe asthma treatment ............................................................... 156 9. Management of worsening asthma and exacerbations in adults, adolescents and children 6–11 years .... 158 Overview ...................................................................................................................................................... 159 Diagnosis of exacerbations .......................................................................................................................... 160 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 4 Self-management of exacerbations with a written asthma action plan ....................................................... 161 Primary care management of asthma exacerbations (adults, adolescents, children 6–11 years) .............. 166 Emergency department management of exacerbations (adults, adolescents, children 6–11 years) .......... 171 Discharge planning and follow-up................................................................................................................ 176 10. Diagnosis of asthma in children 5 years and younger ............................................................................... 178 Asthma and wheezing in young children ..................................................................................................... 178 Clinical diagnosis of asthma ........................................................................................................................ 179 Tests to assist in diagnosis .......................................................................................................................... 182 Risk profiles ................................................................................................................................................. 183 Differential diagnosis.................................................................................................................................... 183 11. Assessment and management of asthma in children 5 years and younger .............................................. 185 Goal of asthma management ...................................................................................................................... 185 Assessment of asthma ................................................................................................................................ 186 Remission of childhood wheezing and asthma ........................................................................................... 186 Medications for symptom control and risk reduction ................................................................................... 189 Asthma treatment steps for children aged 5 years and younger ................................................................. 192 Reviewing response and adjusting treatment .............................................................................................. 194 Choice of inhaler device .............................................................................................................................. 194 Asthma self-management education for carers of young children .............................................................. 195 12. Management of worsening asthma and exacerbations in children 5 years and younger ......................... 196 Diagnosis of exacerbations .......................................................................................................................... 196 Initial home management of asthma exacerbations .................................................................................... 197 Primary care or hospital management of acute asthma exacerbations in children 5 years or younger ...... 198 Discharge and follow-up after an exacerbation ........................................................................................... 203 13. Primary prevention of asthma .................................................................................................................... 204 Factors associated with increased or decreased risk of asthma in children ............................................... 204 Advice about primary prevention of asthma ................................................................................................ 207 Prevention of occupational asthma in adults ............................................................................................... 208 14. Implementing asthma management strategies into health systems.......................................................... 209 Introduction .................................................................................................................................................. 209 Adapting and implementing asthma clinical practice guidelines ................................................................. 209 Glossary of asthma medication classes .......................................................................................................... 212 References ....................................................................................................................................................... 217 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 5 Tables and figures DIAGNOSIS Box 1-1. Diagnostic flowchart for adults, adolescents and children 6–11 years in clinical practice .................. 25 Box 1-2. Criteria for initial diagnosis of asthma in adults, adolescents, and children 6–11 years ..................... 26 Box 1-3. Differential diagnosis of asthma in adults, adolescents and children 6–11 years ............................... 27 Box 1-4. Steps for confirming the diagnosis of asthma in a patient already taking ICS-containing treatment .. 30 Box 1-5. How to step-down ICS-containing treatment to help confirm the diagnosis of asthma ...................... 32 ASSESSMENT Box 2-1. Summary of assessment of asthma in adults, adolescents, and children 6–11 years ........................ 36 Box 2-2. GINA assessment of asthma control at clinical visits in adults, adolescents and children 6–11 years37 Box 2-3. Specific questions for assessment of asthma in children 6–11 years ................................................. 40 Box 2-4. Investigating poor symptom control and/or exacerbations despite treatment..................................... 47 ASTHMA MANAGEMENT Box 3-1. Communication strategies for healthcare providers ............................................................................ 49 Box 3-2. Long-term goal of asthma management ............................................................................................. 50 Box 3-3. The asthma management cycle for personalized asthma care .......................................................... 53 Box 3-4. Population-level versus patient-level decisions about asthma treatment ........................................... 54 Box 3-5. Treating potentially modifiable risk factors to reduce exacerbations and minimize OCS use ............ 55 Box 3-6. Non-pharmacological interventions – summary (see following text for details) .................................. 57 Box 3-7. Effectiveness of avoidance measures for indoor allergens ................................................................. 62 Box 3-8. Indications for considering referral for expert advice, where available ............................................... 66 ASTHMA MEDICATIONS Box 4-1. Terminology for asthma medications ................................................................................................... 70 Box 4-2. Low, medium and high daily metered doses of inhaled corticosteroids (alone or with LABA) ............ 71 Adults and adolescents Box 4-3. Asthma treatment tracks for adults and adolescents .......................................................................... 74 Box 4-4. Initial asthma treatment for adults and adolescents with a diagnosis of asthma ................................ 75 Box 4-5. Flowchart for selecting initial treatment in adults and adolescents with a diagnosis of asthma ......... 76 Box 4-6. Personalized management for adults and adolescents to control symptoms and minimize future risk77 Box 4-7. Track 1 (preferred) treatment Steps 1–4 for adults and adolescents .................................................. 78 Box 4-8. Medications and doses for GINA Track 1: anti-inflammatory reliever (AIR) therapy .......................... 84 Box 4-9. Track 2 (alternative) treatment Steps 1–4 for adults and adolescents ................................................ 86 Children 6–11 years Box 4-10. Initial asthma treatment for children aged 6–11 years with a diagnosis of asthma ........................... 94 Box 4-11. Flowchart for selecting initial treatment in children aged 6–11 years with a diagnosis of asthma .... 95 Box 4-12. Personalized management for children 6–11 years to control symptoms and minimize future risk . 96 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 6 Stepping down Box 4-13. Options for stepping down treatment in adults and adolescents once asthma is well controlled ... 102 ASTHMA SELF-MANAGEMENT EDUCATION AND SKILLS TRAINING Box 5-1. Shared decision-making between health professional and patient about choice of inhalers ........... 109 Box 5-2. Choice and effective use of inhaler devices ....................................................................................... 110 Box 5-3. Poor adherence with prescribed maintenance treatment in asthma .................................................. 112 Box 5-4. Asthma information ............................................................................................................................. 113 PATIENTS WITH FEATURES OF ASTHMA AND COPD Box 7-1. Current definitions of asthma and COPD, and clinical description of asthma-COPD overlap .......... 133 Box 7-2. Syndromic approach to initial treatment in patients with asthma and/or COPD ............................... 134 Box 7-3. Spirometric measures in asthma and COPD .................................................................................... 135 Box 7-4. Specialized investigations sometimes used in patients with features of asthma and COPD ........... 137 DIFFICULT-TO-TREAT AND SEVERE ASTHMA Box 8-1. What proportion of adults have difficult-to-treat or severe asthma? ................................................. 140 Box 8-2. Decision tree – investigate and manage difficult to treat asthma in adult and adolescent patients .. 142 Box 8-3. Decision tree – assess and treat severe asthma phenotypes .......................................................... 143 Box 8-4. Decision tree – consider add-on biologic Type 2-targeted treatments .............................................. 144 Box 8-5. Decision tree – monitor and manage severe asthma treatment ....................................................... 145 ASTHMA EXACERBATIONS Box 9-1. Factors associated with increased risk of asthma-related death ...................................................... 160 Box 9-2. Self-management of worsening asthma in adults and adolescents with a written asthma action plan162 Box 9-3. Optimizing asthma treatment to minimize need for OCS .................................................................. 165 Box 9-4. Management of asthma exacerbations in primary care (adults, adolescents, children 6–11 years) 167 Box 9-5. Discharge management after acute care for asthma ........................................................................ 170 Box 9-6. Management of asthma exacerbations in acute care facility (e.g., emergency department) ........... 171 CHILDREN 5 YEARS AND YOUNGER Diagnosis Box 10-1. Probability of asthma diagnosis in children 5 years and younger ................................................... 179 Box 10-2. Features suggesting a diagnosis of asthma in children 5 years and younger ................................ 180 Box 10-3. Questions that can be used to elicit features suggestive of asthma ............................................... 180 Box 10-4. Common differential diagnoses of asthma in children 5 years and younger .................................. 183 Box 10-5. Key indications for referral of a child 5 years or younger for expert advice .................................... 184 Assessment and management Box 11-1. GINA assessment of asthma control in children 5 years and younger ........................................... 188 Box 11-2. Personalized management of asthma in children 5 years and younger ......................................... 190 Box 11-3. Low daily doses of inhaled corticosteroids for children 5 years and younger ................................. 191 Box 11-4. Choosing an inhaler device for children 5 years and younger ........................................................ 191 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 7 Exacerbations Box 12-1. Management of acute asthma or wheezing in children 5 years and younger ................................ 199 Box 12-2. Initial assessment of acute asthma exacerbations in children 5 years and younger ...................... 200 Box 12-3. Indications for immediate transfer to hospital for children 5 years and younger............................. 200 Box 12-4. Emergency department management of asthma exacerbations in children 5 years and younger . 201 PRIMARY PREVENTION OF ASTHMA Box 13-1. Advice about primary prevention of asthma in children 5 years and younger ................................ 207 IMPLEMENTATION Box 14-1. Approach to implementation of the Global Strategy for Asthma Management and Prevention ...... 210 Box 14-2. Essential elements required to implement a health-related strategy .............................................. 210 Box 14-3. Examples of barriers to the implementation of evidence-based recommendations ........................ 211 Box 14-4. Examples of high-impact implementation interventions in asthma management ............................ 211 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 8 Preface Asthma is a serious global health problem affecting all age groups. Its prevalence is increasing in many countries, especially among children. Although some countries have seen a decline in hospitalizations and deaths from asthma, asthma still imposes an unacceptable burden on healthcare systems, and on society through loss of productivity in the workplace and, especially for pediatric asthma, disruption to the family. In 2023 the Global Initiative for Asthma celebrated 30 years of working to improve the lives of people with asthma by translating medical evidence into better asthma care worldwide. Established in 1993 by the National Heart, Lung, and Blood Institute and the World Health Organization, GINA works with healthcare professionals, researchers, patients and public health officials around the world to reduce asthma prevalence, morbidity and mortality. The Global Strategy for Asthma Management and Prevention (‘GINA Strategy Report’) was first published in 1995, and has been updated annually since 2002 by the GINA Science Committee. It contains guidance for primary care practitioners, specialists and allied health professionals, based on the latest high-quality evidence available. More resources and supporting material are provided online at www.ginasthma.org. GINA supports global efforts to achieve environmental sustainability in health care, while ensuring that our guidance reflects an optimal balance between clinical and environmental priorities, with a particular focus on patient safety. GINA also supports efforts to ensure global availability of, and access to, effective quality-assured medications, to reduce the burden of asthma mortality and morbidity. Since 2001, GINA has organized the annual World Asthma Day, a focus for local and national activities to raise awareness of asthma and educate families and healthcare professionals about effective asthma care. GINA is an independent organization funded solely through sale and licensing of its educational publications. Members of the GINA Board of Directors are drawn globally from leaders with an outstanding demonstrated commitment to asthma research, asthma clinical management, public health and patient advocacy. GINA Science Committee members are highly experienced asthma experts from around the world, who continually review and synthesize scientific evidence to provide guidance on asthma prevention, diagnosis and management. The GINA Dissemination Task Group is responsible for promoting GINA resources throughout the world. Members work with an international network of patient representatives and leaders in asthma care (GINA Advocates), to implement asthma education programs and support evidence-based care. GINA support staff comprise the Executive Director and Project Manager. We acknowledge the superlative work of all who have contributed to the success of the GINA program. In particular, we recognize the outstanding long-term dedication of founding Scientific Director Dr Suzanne Hurd and founding Executive Director Dr Claude Lenfant in fostering GINA’s development until their retirement in 2015, and we were said to hear of Dr Lenfant’s passing last year. A tribute to Dr Lenfant is available on the GINA website ( We acknowledge the invaluable commitment and skills of our current Executive Director Rebecca Decker, and Program Director Kristi Rurey. We continue to recognize the contribution of Prof J Mark FitzGerald to GINA for over 25 years until his passing in 2022. We also thank all members of the Science Committee, who receive no honoraria or reimbursement for their many hours of work in reviewing evidence and attending meetings, and the GINA Dissemination Working Group and GINA Advocates. We hope you find this report to be a useful resource in the management of asthma and that it will help you work with each of your patients to provide the best personalized care, Helen K Reddel, MBBS PhD Chair, GINA Science Committee Arzu Yorgancıoğlu, MD Chair, GINA Board of Directors COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 9 Members of GINA committees (2023–24) GINA Scientific Committee Helen K. Reddel, MBBS PhD, Chair Woolcock Institute of Medical Research, Macquarie University Sydney, Australia Leonard B. Bacharier, MD Vanderbilt University Medical Center Nashville, TN, USA Eric D. Bateman, MD University of Cape Town Lung Institute Cape Town, South Africa Matteo Bonini MD, PhD Department of Public Health and Infectious Diseases, Sapienza University of Rome, Italy National Heart and Lung Institute (NHLI), Imperial College London, UK Arnaud Bourdin, MD, PhD Department of Respiratory Diseases, University of Montpellier Montpellier, France Christopher Brightling, FMedSci, PhD Leicester NIHR Biomedical Research Centre, University of Leicester Leicester, UK Guy Brusselle, MD, PhD Ghent University Hospital Ghent, Belgium Roland Buhl, MD PhD Mainz University Hospital Mainz, Germany Jeffrey M. Drazen, MD Brigham and Woman’s Hospital Boston, MA, USA Francine Ducharme, MD Departments of Pediatrics and of Social and Preventive Medicine, Sainte-Justine University Health Centre, University of Montreal Montreal, Quebec, Canada Liesbeth Duijts, MD MSc Phd University Medical Center Rotterdam, The Netherlands Louise Fleming, MBChB MD Royal Brompton Hospital London, United Kingdom Hiromasa Inoue, MD Kagoshima University Kagoshima, Japan Alan Kaplan, MD (from March 2024) University of Toronto Toronto, Ontario, Canada Family Physician Airways Group of Canada Markham, Ontario, Canada Fanny Wai-san Ko, MD The Chinese University of Hong Kong Hong Kong Refiloe Masekela MBBCh, PhD Department of Paediatrics and Child Health, University of KwaZulu Natal Durban, South Africa Paulo Pitrez, MD, PhD Pulmonary Division, Hospital Santa Casa de Porto Alegre Universidade Federal de Ciências da Saúde de Porto Alegre (UFCSPA) Porto Alegre, Brazil Sundeep Salvi MD, PhD Pulmocare Research and Education (PURE) Foundation Pune, India Aziz Sheikh, BSc, MBBS, MSc, MD The University of Edinburgh Edinburgh, United Kingdom GINA Board of Directors Arzu Yorgancioglu, MD Chair Celal Bayar University Department of Pulmonology Manisa, Turkey Keith Allan, CBiol, MRSB Patient Partner University Hospitals of Leicester Leicester, UK COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 10 GINA Program Rebecca Decker, BS, MSJ Kristi Rurey, AS Editorial assistance Charu Grover, PhD Jenni Harman, BVSc, BA Graphics assistance Kate Chisnall Information design for severe asthma decision tree Tomoko Ichikawa, MS Hugh Musick, MBA Institute for Healthcare Delivery Design University of Illinois, Chicago, USA Disclosures for members of GINA committees can be found at www.ginasthma.org Eric D. Bateman, MD University of Cape Town Lung Institute Cape Town, South Africa Guy Brusselle, MD, PhD Ghent University Hospital Ghent, Belgium Muhwa Jeremiah Chakaya, MD Respiratory Society of Kenya Department of Medicine, Therapeutics, Dermatology and Psychiatry, Kenyatta University Nairobi, Kenya Alvaro A. Cruz, MD Federal University of Bahia Salvador, BA, Brazil Hiromasa Inoue, MD Kagoshima University Kagoshima, Japan Jerry A. Krishnan, MD PhD University of Illinois Hospital & Health Sciences System Chicago, IL, USA Mark L. Levy, MD Locum GP London, UK Helen K. Reddel, MBBS PhD Woolcock Institute of Medical Research, Macquarie University Sydney, Australia GINA Dissemination Group Mark L. Levy, MD (Chair) Locum GP London, UK Keith Allan, CBiol, MRSB Patient Partner University Hospitals of Leicester Leicester, UK Hiromasa Inoue, MD Kagoshima University Kagoshima, Japan Helen K. Reddel, MBBS PhD Woolcock Institute of Medical Research, Macquarie University Sydney, Australia Arzu Yorgancioglu, MD Celal Bayar University Manisa, Turkey COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 11 Abbreviations ABPA Allergic bronchopulmonary aspergillosis ACE Angiotensin-converting enzyme ACQ Asthma Control Questionnaire ACT Asthma Control Test (see also cACT) AERD Aspirin-exacerbated respiratory disease AIR Anti-inflammatory reliever (see Box 4-1, p.70) ANCA Antineutrophil cytoplasmic antibody Anti-IL4Rα Anti-interleukin 4 receptor alpha (monoclonal antibody) Anti-IL5 Anti-interleukin 5 (monoclonal antibody) Anti-IL5Rα Anti-interleukin 5 receptor alpha (monoclonal antibody) Anti-TSLP Anti-thymic stromal lymphopoietin (monoclonal antibody) APGAR Activities, Persistent, triGgers, Asthma medications, Response to therapy ATS/ERS American Thoracic Society and European Respiratory Society BDP Beclometasone dipropionate BD Bronchodilator BMI Body mass index BNP B-type natriuretic peptide bpm Beats per minute BTS British Thoracic Society cACT Childhood Asthma Control Test CBC Complete blood count (also known as full blood count [FBC]) CDC Centers for Disease Control and Prevention [USA] CI Confidence interval COPD Chronic obstructive pulmonary disease COVID-19 Coronavirus disease 2019 CRP C-reactive protein CRSwNP Chronic rhinosinusitis with nasal polyps CRSsNP Chronic rhinosinusitis without nasal polyps CT Computerized tomography CXR Chest X-ray DLCO Diffusing capacity in the lung for carbon monoxide DEXA Dual-energy X-ray absorptiometry DPI Dry-powder inhaler ED Emergency department EGPA Eosinophilic granulomatosis with polyangiitis COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 12 FeNO Fractional concentration of exhaled nitric oxide FEV1 Forced expiratory volume in 1 second (measured by spirometry) FVC Forced vital capacity (measured by spirometry) FEV1/FVC Ratio of forced expiratory volume in 1 second to forced vital capacity GERD Gastro-esophageal reflux disease (GORD in some countries) GOLD Global Initiative for Chronic Obstructive Pulmonary Disease GRADE Grading of Recommendations Assessment, Development and Evaluation (an approach to clinical practice guideline development) HDM House dust mite HEPA High-efficiency particulate air HFA Hydrofluoroalkane propellant HIV/AIDS Human immunodeficiency virus/acquired immunodeficiency syndrome ICS Inhaled corticosteroid Ig Immunoglobulin IL Interleukin IM Intramuscular ICU Intensive care unit IV Intravenous LABA Long-acting beta2 agonist LAMA Long-acting muscarinic antagonist (also called long-acting anticholinergic) LMIC Low- and middle-income countries LTRA Leukotriene receptor antagonist (also called leukotriene modifier) MART Maintenance-and-reliever therapy (with ICS-formoterol); in some countries called SMART (single-inhaler maintenance-and-reliever therapy) n.a Not applicable NSAID Nonsteroidal anti-inflammatory drug NO2 Nitrogen dioxide (air pollutant) O2 Oxygen OCS Oral corticosteroids OSA Obstructive sleep apnea PaCO2 Arterial partial pressure of carbon dioxide PaO2 Arterial partial pressure of oxygen PEF Peak expiratory flow PM10 Particulate matter with a particle diameter of 10 micrometers or less (air pollution) pMDI Pressurized metered-dose inhaler QTc Corrected QT interval on electrocardiogram RCT Randomized controlled trial SABA Short-acting beta2 agonist COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 13 SC Subcutaneous SCIT Subcutaneous allergen immunotherapy sIgE Specific immunoglobulin E SLIT Sublingual immunotherapy S02 Sulfur dioxide (air pollutant) T2 Type 2 airway inflammation (an asthma phenotype) TSLP Thymic stromal lymphopoietin URTI Upper respiratory tract infection VCD Vocal cord dysfunction (included in inducible laryngeal obstruction) WHO World Health Organization WHO-PEN The World Health Organization Package of essential noncommunicable disease interventions for primary care COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 14 Introduction Asthma is a serious global health problem, affecting approximately 300 million people around the world, and causing around 1,000 deaths per day. Most of these deaths occur in low- and middle-income countries, and most of them are preventable. Asthma interferes with people’s work, education and family life, especially when children have asthma. Asthma is becoming more prevalent in many economically developing countries, and the cost of asthma treatment for healthcare systems, communities and individuals is increasing. The Global Initiative for Asthma (GINA) was established to increase awareness about asthma among healthcare providers, public health authorities and communities, to improve management of asthma, and to help prevent asthma. Every year GINA publishes a strategy report, containing information and recommendations on asthma, based on the latest medical evidence. GINA’s aim is for these to be available and used throughout the world. GINA also promotes international collaboration on asthma research. GINA Committee members are listed on page 9. Goals of asthma management The population goal of asthma management is to prevent asthma deaths and minimize the burden of asthma on individuals, families, communities, health systems and the environment. For individuals with asthma of all ages, the goal of asthma management is to achieve the patient’s best possible long-term outcomes: • Long-term asthma symptom control, which may include: - Few/no asthma symptoms - No sleep disturbance due to asthma - Unimpaired physical activity • Long-term asthma risk minimization, which may include: - No exacerbations - Improved or stable personal best lung function - No requirement for maintenance OCS - No medication side-effects. The patient’s goals for their asthma may be different from these medical goals; and patients with few or no asthma symptoms can still have severe or fatal exacerbations, including from external triggers such as viral infections, allergen exposure (if sensitized) or pollution. Challenges in global asthma management For healthcare providers, the challenges of managing asthma differ between regions and health systems. Despite laudable efforts to improve asthma care over the past 30 years, and the availability of effective medications, many patients globally have not benefited from advances in asthma treatment and often lack even the rudiments of care. Many of the world’s population live in areas with inadequate medical facilities and meager financial resources. GINA recognizes that the recommendations found in this report must be adapted to fit local practices and the availability of healthcare resources. To improve asthma care and patient outcomes, evidence-based recommendations must also be disseminated and implemented nationally and locally, and integrated into health systems and clinical practice. Implementation requires an evidence-based strategy involving professional groups and stakeholders and considering local cultural and socioeconomic conditions. GINA is a partner organization in the Global Alliance against Chronic Respiratory Diseases (GARD). Through the work of GINA, and in cooperation with GARD and the International Union Against Tuberculosis and Lung Diseases (IUATLD), substantial progress toward better care for all patients with asthma should be achieved in the next decade. At the most fundamental level, patients in many areas do not have access to any inhaled corticosteroid-containing medications, which are the cornerstone of care for asthma patients of all severity. More broadly, medications remain the major contributor to the overall costs of asthma management, so the access to and pricing of high-quality asthma medications continues to be an issue of urgent need and a growing area of research interest.1-3 The safest and most COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 15 effective approach to asthma treatment in adolescents and adults, which also avoids the consequences of starting treatment with short-acting beta2 agonist (SABA) alone, and requires only a single medication, depends on access to the combination of inhaled corticosteroid and formoterol (ICS–formoterol) across all asthma severity levels.4,5 Budesonide-formoterol is included in the World Health Organization (WHO) essential medicines list, so the fundamental change to anti-inflammatory treatment that was first included in the 2019 GINA Strategy Report6 may provide a feasible solution to reduce the risk of severe exacerbations in low- and middle-income countries.5 The urgent need to ensure access to affordable, quality-assured inhaled asthma medications as part of universal health coverage must now be prioritized by all relevant stakeholders, particularly manufacturers of relevant inhalers. GINA is collaborating with IUATLD and other organizations to work towards a World Health Assembly Resolution to improve equitable access to affordable care, including inhaled medicines, for children, adolescents and adults with asthma.3 There is increasing concern globally about climate change, and its impact on the health and security of populations, particularly in low- and middle-income countries. The propellants in pressurized metered-dose inhalers contribute significantly to the carbon footprint of health care, particularly from use of SABAs. The GINA Track 1 approach not only provides a large reduction in exacerbations, in risk of adverse effects of oral corticosteroids, and in urgent health care compared with SABA-only treatment, but also, if implemented with a dry powder inhaler (as in most of the clinical trials), it provides a very large reduction in carbon footprint.7 GINA fully supports initiatives to encourage use of dry-powder inhalers, where they are available and clinically appropriate, and to replace harmful propellants with low-carbon alternatives. At the same time, it is essential to ensure continuity of supply of essential inhaled medicines to people in low-resource areas, to avoid exacerbating the existing serious global inequities in health care for asthma.8 Methodology GINA SCIENCE COMMITTEE The GINA Science Committee was established in 2002 to review published research on asthma management and prevention, to evaluate the impact of this research on recommendations in GINA documents, and to provide yearly updates to these documents. The members are recognized leaders in asthma research and clinical practice, with the scientific expertise to contribute to the task of the Committee. They are invited to serve for a limited period and in a voluntary capacity. The Committee is broadly representative of adult and pediatric disciplines, and members are drawn from diverse geographic regions. The Science Committee normally meets in person three times yearly, in conjunction with the American Thoracic Society (ATS) and European Respiratory Society (ERS) international conferences and at a stand-alone meeting, to review asthma-related scientific literature. During COVID-19, meetings of the Science Committee were held online each month, and online meetings have continued every 1–2 months since then. Statements of interest for Committee members (p.9) are found on the GINA website www.ginasthma.org. PROCESSES FOR UPDATES AND REVISIONS OF THE GINA STRATEGY REPORT Literature search Details are provided on the GINA website (www.ginasthma.org/about-us/methodology). In summary, two PubMed searches are performed each year, each covering the previous 18 months, using filters established by the Science Committee. The search terms include asthma, all ages, only items with abstracts, clinical trial or meta-analysis or systematic review, and human. The search is not limited to specific PICOT (Population, Intervention, Comparison, Outcomes, Time) questions. The ‘clinical trial’ publication type includes not only conventional randomized controlled trials, but also pragmatic, real-life and observational studies. The search for systematic reviews includes, but is not limited to, those conducted using GRADE methodology.9 An additional search is conducted for guidelines documents published by other international organizations. The respiratory community is also invited to submit any other fully published peer-reviewed publications that they believe have been missed, providing that the full paper is submitted in (or translated into) English; however, because of the comprehensive process for literature review, such ad hoc submissions have rarely resulted in substantial changes to the report. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 16 Systematic reviews Unique among evidence-based recommendations in asthma, and rare among clinical practice guidelines in most other therapeutic areas, the GINA report is based on an ongoing twice-yearly cumulative update of the evidence base for its recommendations. GINA does not normally carry out or commission its own GRADE-based reviews, because of the current cost of such reviews, the large number of PICOT questions that would be necessary for a comprehensive practical report of this scope, and because it would limit the responsiveness of the GINA Strategy Report to emerging evidence and new developments in asthma management. However, the Science Committee reviews relevant published systematic reviews conducted with GRADE methodology as part of its normal process. GINA recommendations are constantly being reviewed and considered for update as new evidence (including GRADE-based systematic reviews on specific topics) is identified and indicates the need. With recognition of allergen immunotherapy as an area of the GINA report that needed substantial updating, a GINA working group conducted a systematic review of articles on subcutaneous immunotherapy or sublingual immunotherapy since publication of two recent systematic reviews.10,11 From the period 01/01/2018 to 10/28/2023, the working group screened the titles and abstracts of 350 articles for quality and relevance, and undertook full-text review of 73 publications. On the basis of this systematic review, the section of the GINA report on allergen immunotherapy (p.104) has been extensively updated. Literature screening and review Each article identified by the literature search, after removal of duplicates and those already reviewed, is pre-screened in Covidence for relevance and major quality issues by the Editorial Assistant and by at least two non-conflicted members of the Science Committee. Each publication selected from screening is reviewed for quality and relevance by at least two members of the Science Committee, neither of whom may be an author (or co-author) or declare a conflict of interest in relation to the publication. Articles that have been accepted for publication and are online in advance of print are eligible for full text review if the approved/corrected copy-edited proof is available. All members receive a copy of all abstracts and full text publications, and non-conflicted members have the opportunity to provide comments during the pre-meeting review period. Members evaluate the abstract and the full text publication, and answer written questions in a review template about whether the scientific data impact on GINA recommendations, and if so, what specific changes should be made. In 2020, the Critical Appraisal Skills Programme (CASP) checklist12 was provided in the review template to assist in evaluation of systematic reviews. A list of all publications reviewed by the Committee is posted on the GINA website (www.ginasthma.org). Discussion and decisions during Science Committee meetings Each publication that is assessed by at least one reviewer to potentially impact on the GINA Strategy Report is discussed in a Science Committee meeting (virtual or face-to-face). This process comprises three parts, as follows: 1. Quality and relevance of original research and systematic review publications. First, the Committee considers the relevance of the publication to the GINA Strategy Report, the quality of the study, the reliability of the findings, and the interpretation of the results, based on the responses from reviewers and discussion by members of the Committee. For systematic reviews, GRADE assessments, if available, are considered. However, for any systematic review, GINA members also independently consider the clinical relevance of the question addressed by the review, and the scientific and clinical validity of the included populations and study design. For network meta-analyses, reviewers also consider the appropriateness of the comparisons (e.g., whether differences in background exacerbation risk and ICS dose were taken into account) and the generalizability of the findings. During this discussion, a member who is an author (or was involved in the study) may be requested to provide clarification or respond to questions about the study, but they may not otherwise take part in this discussion about the quality and relevance of the publication. 2. Decision about inclusion of the evidence. During this phase, the Committee decides whether the publication or its findings affect GINA recommendations or statements and should be included in the GINA Strategy Report. These decisions to modify the report or its references are made by consensus by Committee members present and, again, any member with a conflict of interest is excluded from these decisions. If the chair is an author on a publication being reviewed, an alternative chair is appointed to lead the discussion in part 1 and the decision in part 2 for that publication. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 17 3. Discussion about related changes to the GINA Strategy Report. If the committee resolves to include the publication or its findings in the report, an author or conflicted member, if present, is permitted to take part in the subsequent discussions about and decisions on changes to the report, including the positioning of the study findings in the report and the way that they would be integrated with existing (or other new) components of the GINA management strategy. These discussions may take place immediately, or over the course of the year as new evidence emerges or as other changes to the report are agreed and implemented. The approach to managing conflicts of interest, as described above, also applies to members of the GINA Board who, ex-officio, attend GINA Science Committee meetings. As with all previous GINA Strategy Reports, levels of evidence are assigned to management recommendations where appropriate. Current criteria (Table A) are based on those originally developed by the National Heart Lung and Blood Institute. From 2019, GINA has included in ‘Level A’ strong observational evidence that provides a consistent pattern of findings in the population for which the recommendation is made, and has also described the values and preferences that were considered in making major new recommendations. The table was updated in 2021 to avoid ambiguity about the positioning of observational data and systematic reviews. Table A. Description of levels of evidence used in this report Evidence level Sources of evidence Definition A Randomized controlled trials (RCTs), systematic reviews, observational evidence. Rich body of data Evidence is from endpoints of well-designed RCTs, systematic reviews of relevant studies or observational studies that provide a consistent pattern of findings in the population for which the recommendation is made. Category A requires substantial numbers of studies involving substantial numbers of participants. B Randomized controlled trials and systematic reviews. Limited body of data Evidence is from endpoints of intervention studies that include only a limited number of patients, post hoc or subgroup analysis of RCTs or systematic reviews of such RCTs. In general, Category B applies when few randomized trials exist, they are small in size, they were undertaken in a population that differs from the target population of the recommendation, or the results are somewhat inconsistent. C Nonrandomized trials or observational studies Evidence is from non-randomized trials or observational studies. D Panel consensus judgment This category is used only in cases where the provision of some guidance was deemed valuable but the clinical literature addressing the subject was insufficient to justify placement in one of the other categories. The Panel Consensus is based on clinical experience or knowledge that does not meet the above listed criteria. New therapies and indications The GINA Strategy Report is a global strategy document. Since regulatory approvals differ from country to country, and manufacturers do not necessarily make regulatory submissions in all countries, some GINA recommendations are likely to be off-label in some countries. This is a particular issue for pediatrics, where across different diseases, many treatment recommendations for preschool children and for children aged 6–11 years are off-label. For new therapies, GINA’s aim is to provide clinicians with evidence-based guidance about new therapies and their positioning in the overall asthma treatment strategy as soon as possible; otherwise, the gap between regulatory approval and the periodic update of many national guidelines is filled only by advertising or educational material produced by the manufacturer or distributor. For new therapies for which the GINA Science Committee considers there is sufficient good-quality evidence for safety and efficacy or effectiveness in relevant asthma populations, recommendations may be held until after approval for asthma by at least one major regulatory agency (e.g., European Medicines Agency or US Food and Drug Administration), since regulators often receive substantially more safety COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 18 and/or efficacy data on new medications than are available to GINA through peer-reviewed literature. However, decisions by GINA to make or not make a recommendation about any therapy, or about its use in any specific population, are based on the best available peer-reviewed evidence and not on labeling directives from regulators. For existing therapies with evidence for new regimens or in different populations, the Science Committee may, where relevant, make recommendations that are not necessarily covered by regulatory indications in any country at the time, provided the Committee is satisfied with the available evidence around safety and efficacy/effectiveness. Since the GINA Strategy Report is a global strategy, the report does not refer to recommendations as being ‘off-label’. However, readers are advised that, when assessing and treating patients, they should use their own professional judgment and should also consider local and national guidelines and eligibility criteria, as well as locally licensed drug doses. External review Prior to publication each year, the GINA Strategy Report undergoes extensive external review by patient advocates and by asthma care experts from primary and specialist care in multiple countries. There is also continuous external review throughout the year in the form of feedback from end-users and stakeholders through the contact form on the GINA website. Literature reviewed for GINA 2024 update The GINA Strategy Report has been updated in 2024 following the routine twice-yearly review of the literature by the GINA Science Committee. The literature searches for ‘clinical trial’ publication types (see above), systematic reviews and guidelines identified a total of 3423 publications, of which 2961 duplicates/animal studies/non-asthma/pilot studies and protocols were removed. A total of 462 publications underwent screening of title and abstract by at least two reviewers, and 68 were screened out for relevance and/or quality. A total of 64 publications underwent full-text review by at least two members of the Science Committee, and 34 publications were subsequently discussed at meetings of the Science Committee. A list of key changes in GINA 2024 is shown on page 19, and a copy showing tracked changes is archived on the GINA website at www.ginasthma.org/archived-reports. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 19 WHAT’S NEW IN GINA 2024? The GINA Strategy Report has been updated in 2024 following the routine twice-yearly cumulative review of the literature by the GINA Scientific Committee, and extensive discussion about issues relevant to clinical practice and research. A copy showing tracked changes from the GINA 2023 report is archived on the GINA website. KEY CHANGES • Diagnosis of asthma: The diagnostic flowchart for clinical practice (Box 1-1, p.25) has been revised, recognizing that, globally, a large proportion of health professionals do not have access (or timely access) to spirometry in their clinical practice. Although peak expiratory flow (PEF) is less reliable than spirometry, it is better than relying on symptoms alone. The flowchart allows for selection of different initial lung function tests, depending on local resources. The criteria for identifying variable expiratory airflow limitation (Box 1-2, p.26) have also been clarified, and more details provided about bronchodilator withholding. GINA again reviewed, but has not adopted, the recommendation by the American Thoracic Society and European Respiratory Society Technical Standards Committee to change the criterion for bronchodilator responsiveness from an increase from baseline of ≥12% and 200 mL to an increase from baseline of >10% predicted. The Technical Standards Committee based this recommendation on data for survival, and had explicitly avoided making any recommendation about the use of this criterion for diagnostic decisions in clinical practice. This topic will be considered by GINA again when data from additional populations, and for other asthma outcomes, have been published, to inform the implications of the proposed new criterion for diagnosis of asthma in clinical practice (p.29). • Cough variant asthma: more information has been added (p.24 and p.32) about this clinical phenotype of asthma, which is common in some countries. Cough variant asthma may be difficult to distinguish from other causes of chronic cough in clinical practice, as spirometry may be normal and variable airflow limitation may be identified only from bronchial provocation testing. Some patients may later also develop wheezing and bronchodilator responsiveness. The treatment of cough variant asthma is the same as for asthma in general; the cough may return if inhaled corticosteroids (ICSs) are stopped. • Assessment of asthma control: We clarify that assessment of symptom control should not be limited to the most recent 4 weeks, but that there are no validated tools for assessing symptom control over longer periods than this, and that recall-error for symptoms is common. GINA continues to emphasize that assessing symptom control is not enough – the patient’s risk factors for exacerbations (including history of exacerbations), for accelerated decline in lung function and for medication adverse effects must also be assessed (Box 2-2, p.37). While ICS markedly reduce asthma exacerbations and, in patients not taking ICS, serious exacerbations are associated with greater decline in lung function, there is no clear evidence that use of ICS per se prevents long-term development of persistent airflow limitation (p.42). • GINA goal of asthma treatment (Box 3-2, p.50): The goal of asthma treatment is to achieve the best possible long-term asthma outcomes for the individual patient, including both long-term symptom control and long-term minimization of risk of exacerbations, lung function decline and medication adverse effects (including long-term adverse effects of OCS). It is also important to elicit the patient/caregiver’s goals for asthma treatment, as these may differ from medical goals. • Remission of asthma (p.50): There has been extensive recent discussion in the clinical and research community about asthma remission on treatment, in the context of biologic therapy for severe asthma. Several proposed definitions and criteria for their operationalization have been published. A new section of the GINA 2024 report outlines a framework for clinical practice and research about clinical and complete (pathophysiological) remission in children and in adults, both off-treatment and on-treatment. These perspectives should also be considered for discussions with patients and parents/caregivers. The concept of asthma remission on treatment is consistent with the GINA long-term goal of asthma treatment (Box 3-2, p.50), but individual patient goals should be achievable. • Initial asthma treatment in adults and adolescents (Tracks 1 and 2): Key changes have been made to the recommendations about the choice of initial treatment step for adults and adolescents in both Tracks 1 and 2, with updating of Boxes 4-4 (p.75) and 4-5 (p.76) about choice of initial treatment step. The suggested criteria at each step for initial treatment are based on evidence (where available) and on consensus, so the thresholds are not COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 20 precise. The new flowchart for initial asthma treatment (Box 4-5, p.76) includes the GINA cycle of asthma management as a reminder that asthma treatment is not just about medications. • For Track 1, as-needed-only low-dose ICS-formoterol has been the preferred treatment option for both Step 1 and Step 2 since 2021, so together they are called ‘Steps 1–2’. Accordingly, the descriptions of evidence and other considerations are now also presented for Steps 1–2 together. A common question is which patients should instead start treatment at Step 3, i.e., with low-dose ICS-formoterol being taken as maintenance-and-reliever therapy (MART) rather than as-needed-only. There is no specific evidence to guide this choice, but clinical factors that are suggested for consideration of starting with MART (if permitted by local regulators) include symptoms every day, current smoking, low lung function, a recent severe exacerbation or a history of life-threatening exacerbation, impaired perception of bronchoconstriction (e.g. low initial lung function but few symptoms), severe airway hyperresponsiveness, or current exposure to a seasonal allergic trigger (p.78). • For Track 2, the previous description of patients suitable for Step 1 treatment (having asthma symptoms less than twice a month and no risk factors for exacerbations) was introduced in GINA 2014 to limit the use of short-acting beta2 agonist (SABA)-only treatment, as its risks in asthma were already well known.13 This criterion for Step 1 treatment has now been replaced, since GINA has recommended against SABA-only treatment since 2019. Another consideration for choosing between Step 1 and Step 2 treatment is that, although maintenance ICS almost halved the risk of serious exacerbations in patients with symptoms ≤2 days/week in a clinical trial, such patients would be very unlikely to take daily ICS if it was prescribed in clinical practice. Therefore, for patients with such infrequent symptoms, taking ICS whenever SABA is taken (Track 2, Step 1) is preferred over daily ICS plus as-needed SABA (Track 2, Step 2) to ensure that patients receive at least some ICS, rather than taking SABA alone. • GINA 2024 treatment figure for adults and adolescents, Box 4-6, p.77. There are no major changes from 2023 in the main treatment figure. In the arrowed circle (also Box 3-3, p.53), ‘asthma medications’ has been changed to ‘asthma medications including ICS’ as a reminder that all patients with asthma should receive ICS-containing therapy. New short versions of the main treatment figure are shown at the start of the sections of text about Steps 1–4 for Track 1 (Box 4-7, p.78) and Track 2 (Box 4-9, p.86) respectively. • Medications and doses for Track 1 anti-inflammatory reliever (AIR) therapy: Following requests from clinicians, Box 4-8, p.84 has been expanded to show all the relevant ICS-formoterol devices (dry-powder inhalers [DPIs] and pressurized metered-dose inhalers [pMDIs]) and doses for AIR therapy by age-group and treatment step, with the corresponding dosing regimens and maximum number of inhalations in a single day. More devices and doses may become available in the future. • Beclometasone-formoterol for MART (Box 4-7, p.78). There is evidence from randomized controlled trials and meta-analyses in approximately 40,000 patients for the long-term safety and efficacy of as-needed budesonide-formoterol up to a maximum total of 72 mcg formoterol (54 mcg delivered dose) in a single day (total of as-needed and maintenance doses, if used) for adults and adolescents, together with data from earlier randomized controlled trials with as-needed formoterol. Based on this extensive evidence, GINA suggests that the same maximum total dose of formoterol (with ICS) in a single day (72 mcg metered dose) should also apply for adults and adolescents prescribed MART with beclometasone-formoterol 100/6 mcg, i.e. a maximum total of 12 inhalations in a single day. For children 6–11 years prescribed MART with budesonide-formoterol, the maximum recommended total dose of formoterol (with ICS) in a single day is 48 mcg metered dose (36 mcg delivered dose). Most patients need far fewer doses in any day than the maximum doses recommended. • ICS-formoterol as reliever with other ICS-LABAs: GINA previously recommended against use of ICS-formoterol as the reliever for patients using maintenance treatment with a combination of ICS and long-acting beta2 agonist (LABA) with a non-formoterol LABA, because of lack of evidence for safety or efficacy with this approach (p.69). This recommendation is now supported by an analysis suggesting that taking two different LABAs in this way may be associated with increased adverse events (p.82).14 • Leukotriene receptor antagonists: Wherever montelukast is mentioned throughout the report, there is a reminder to advise patients/parents/caregivers about the potential risk of neuropsychiatric adverse events associated with this medication. These include new-onset nightmares and behavioral problems and, in some cases, suicidal ideation. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 21 • High-dose inhaled corticosteroids: Wherever this is suggested as a treatment option throughout the report for adults and adolescents, it is again stated that this is only for short-term use, e.g., 3–6 months, to minimize the potential for adverse effects. • Add-on long-acting muscarinic antagonists (LAMA): Subgroup analyses suggest that the reduction in severe exacerbations requiring OCS associated with triple therapy (ICS+LABA+LAMA) was seen primarily in patients with a history of asthma exacerbations in the previous year (p.91). • Severe asthma with good response to Type 2-targeted therapy: Advice about reduction in asthma therapy in patients who have had a good asthma response to therapy targeting Type 2 inflammation has been updated and clarified, with the highest priority to reduce and cease maintenance oral corticosteroids (OCS), if used. Some previous randomized controlled trials included a rapid ICS dose reduction in patients on biological therapies in order to induce loss of asthma control, but this is not relevant to clinical practice). A randomized controlled trial in adult patients with a good response to benralizumab found that, with randomization to MART, most could have their maintenance ICS-formoterol dose slowly reduced. However, the findings suggest that in patients with severe asthma, maintenance doses of ICS-formoterol should not be stopped15 (p.156). This study also provides support for use of MART in patients taking Step 5 treatment. Additional advice about stepping down treatment once asthma is well controlled is in Box 4-13 (p.102). • Initial asthma treatment in children 6–11 years: Boxes 4-10 (p.94) and 4-11 (p.95) about initial asthma treatment in children 6–11 years have been updated. These recommendations are based on evidence (where available) and on consensus. The flowchart includes the GINA cycle of asthma management, as a reminder that asthma treatment is not just about medications. Symptom levels and lung function prompting a particular starting treatment step are similar to those for adults and adolescents. In the text about treatment steps, additional details about studies, populations and outcomes in the 6–11 years age group have been added, including the ICS doses used in the studies of taking ICS whenever SABA is taken (Step 1, p.97). • Low, medium and high doses of inhaled corticosteroids. Box 4-2 (p.71) lists low, medium and high doses of various ICS, alone or in combination with LABA. GINA has emphasized for many years that this table does not imply potency equivalence, but this continues to be assumed. For clarity, an example has been added: if you switch a patient’s treatment from a ‘medium’ dose of one ICS to a ‘medium’ dose of another ICS, this may represent a decrease (or increase) in potency, so the patient’s asthma may become unstable (or they may be at increased risk of adverse effects). After any change of treatment or inhaler device, patients should be monitored to ensure stability. • Allergen immunotherapy. The section on allergen immunotherapy (p.104) has been updated following a systematic review of publications about subcutaneous immunotherapy (SCIT) and sublingual immunotherapy (SLIT) for asthma by a GINA Science Committee working group. Information is also included about the quality assurance, personnel, training, safety and administrative protocols that must be observed for preparation and safe delivery of SCIT. For patients with severe asthma, allergen immunotherapy may be considered as an add-on treatment, but only after asthma symptoms and exacerbations have been controlled. Other updates in GINA 2024 • Mild asthma: Further advice has been provided on language about mild and severe asthma (p.43). The term ‘mild asthma’ is a retrospective label, so it cannot be used to decide which patients are suitable to receive Step 1 or Step 2 treatment. • Pulmonary rehabilitation for asthma: There is now evidence from a systematic review and meta-analysis for the benefit of structured outpatient pulmonary rehabilitation programs in improving functional exercise capacity and quality of life for patients with asthma (p.60). Pulmonary rehabilitation also continues to be recommended for patients with asthma who have dyspnea due to persistent airflow limitation (Section 7, p.131). • Role of FeNO: Further evidence has emerged of differences in inflammatory biomarkers, including fractional concentration of exhaled nitric oxide (FeNO), in patients with obesity (p.31 and p.72). The largest study to date of FeNO-guided management of asthma, conducted in pregnant women, found no reduction in asthma exacerbations or perinatal outcomes compared with usual care (p.103).16 The main role of FeNO in clinical practice continues to be to help guide treatment decisions in patients with severe asthma (p.143). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 22 • Prevention of respiratory infections: More information is provided about vaccinations against respiratory syncytial virus (RSV), pneumococcus and pertussis (p.106), and interventions to reduce RSV infections in infants (p.206).
• Chronic rhinosinusitis with/without nasal polyps: Information about treatment outcomes in patients with both asthma and chronic rhinosinusitis has been updated based on latest evidence (p.120).
• Acute asthma: Although there is a strong emphasis throughout the GINA report on minimizing OCS use to reduce long-term cumulative adverse effects, OCS are essential in management of acute severe asthma. However, the occurrence of any severe exacerbation should be a prompt to assess the patient thoroughly, optimize their asthma treatment, and consider referral for expert advice, to reduce the risk of another exacerbation occurring (Box 9-3, p.165). Evidence about use of dexamethasone has been updated based on the latest evidence (p.173).
• Prevention of occupational asthma: This new section has been added to section 13 (Primary prevention of asthma, p.208).
Topics still under discussion • Assessment of symptom control: GINA continues to seek evidence relevant to the assessment of symptom control in patients whose reliever is ICS-formoterol.
• Severe asthma in children 6–11 years: a pocket guide and decision tree are in development.
• Efficacy and safety of high dose ICS for exacerbations of asthma or wheezing in preschool children: a systematic review is underway.
• Management of acute asthma in hospital and intensive care unit is under discussion • Digital formats for GINA resources: investigation of digital options is ongoing, with the aim of facilitating access to GINA resources on portable devices and smartphones. Presentation of the GINA Strategy Report as an eBook is not feasible at present because bibliographic referencing programs are not yet compatible with any of the current e-Book platforms, so references would need to be re-entered manually every year.
World Asthma Day 2024 GINA’s theme for World Asthma Day, 7 May 2024 is “Asthma education empowers. Information is key”. Structure and layout We have updated the structure and layout of the report. For asthma medications (Section 4), information is presented first for Track 1 (p.78) then for Track 2 (p.86) in adults and adolescents, followed by medications for children 6–11 years (p.96), with detailed information about difficult-to-treat and severe asthma in Section 8 (p.139). A glossary of medication classes has been added as a Supplement (p.212). For best functionality of the GINA Report, download the pdf. All page numbers and citation numbers are hyperlinked. In your pdf reader, if you add the ‘previous view’ button to your toolbar, it is easy to go back and forth between text, references and linked sections of the report.
Note: clarifications and corrections 22 May 2024 • Box 4-8 (p.85): after launch of the 2024 report, we became aware that some formulations of budesonide-formoterol pMDIs, not available in all countries, were being misread. A note has been added to each of the relevant rows of the table, to emphasize that the stated numbers of inhalations apply ONLY to the listed formulations, which have a lower formoterol dose (3 mcg metered dose, 2.25 mcg delivered dose) than in all other budesonide-formoterol formulations.
• Box 4-8 (p.85): The doses of budesonide-formoterol for GINA Track 1 have been further clarified in the footnote: budesonide-formoterol 400/12 [320/9] mcg should not be used as an anti-inflammatory reliever, and, for adults and adolescents, GINA does not suggest use of budesonide-formoterol 100/6 [80/4.5] mcg as an anti-inflammatory reliever, since most evidence is with 200/6 [160/4.5] mcg.
• Box 8-5 (p.145): “do not stop ICS” corrected to “do not stop maintenance ICS-LABA” • Some minor typographical errors have been corrected.
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 23 1. Definition, description, and diagnosis of asthma in adults, adolescents and children 6–11 years KEY POINTS What is asthma? Asthma is a heterogeneous disease, usually characterized by chronic airway inflammation. It is defined by the history of respiratory symptoms, such as wheeze, shortness of breath, chest tightness and cough, that vary over time and in intensity, together with variable expiratory airflow limitation. One or more symptoms (e.g., cough) may predominate. Airflow limitation may later become persistent. Asthma is usually associated with airway hyperresponsiveness and airway inflammation, but these are not necessary or sufficient to make the diagnosis. Recognizable clusters of demographic and clinical characteristics are called ‘clinical asthma phenotypes’. In most instances, these do not correlate strongly with specific pathological processes or treatment responses. However, biomarkers reflecting pathophysiological processes are useful in the assessment of difficult-to-treat asthma and treatment of severe asthma. How is asthma diagnosed? The diagnosis of asthma is based on the history of characteristic symptom patterns and evidence of variable expiratory airflow limitation. This should be documented from bronchodilator reversibility testing or other tests. More than one test may be needed to confirm asthma or exclude alternative causes of respiratory symptoms. Many health professionals do not have access to spirometry. If so, peak expiratory flow (PEF) should be used, rather than relying on symptoms alone. Test before treating, wherever possible, i.e., document the evidence for the diagnosis of asthma before starting inhaled corticosteroid (ICS)-containing treatment, as it is often more difficult to confirm the diagnosis once asthma control has improved. Additional or alternative strategies may be needed to confirm the diagnosis of asthma in particular populations, including patients already on ICS-containing treatment, the elderly, patients presenting with cough as the only symptom (including cough variant asthma), and patients in low-resource settings. DEFINITION OF ASTHMA Asthma is a heterogeneous disease, usually characterized by chronic airway inflammation. It is defined by the history of respiratory symptoms, such as wheeze, shortness of breath, chest tightness and cough, that vary over time and in intensity, together with variable expiratory airflow limitation. This definition was reached by consensus, based on consideration of the characteristics that are typical of asthma before ICS-containing treatment is commenced, and that distinguish it from other respiratory conditions. However, airflow limitation may become persistent later in the course of the disease. DESCRIPTION OF ASTHMA Asthma is a common, chronic respiratory disease affecting 1–29% of the population in different countries.17,18 Asthma is characterized by variable symptoms of wheeze, shortness of breath, chest tightness and/or cough, and by variable expiratory airflow limitation. Both symptoms and airflow limitation characteristically vary over time and in intensity. These variations are often triggered by factors such as exercise, allergen or irritant exposure, change in weather, or viral respiratory infections. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 24 Symptoms and airflow limitation may resolve spontaneously or in response to medication, and may sometimes be absent for weeks or months at a time. On the other hand, patients can experience episodic flare-ups (exacerbations) of asthma that may be life-threatening and carry a significant burden to patients and the community. The majority of asthma deaths occur in low- and middle-income countries.2 Asthma is usually associated with airway hyperresponsiveness to direct or indirect stimuli, and with chronic airway inflammation. These features usually persist, even when symptoms are absent or lung function is normal, but may normalize with treatment. Asthma phenotypes Asthma is a heterogeneous disease, with different underlying disease processes. Recognizable clusters of demographic, clinical and/or pathophysiological characteristics are often called ‘asthma phenotypes’.19-22 In patients with more severe asthma, some phenotype-guided treatments are available. However, except in patients with severe asthma, no strong relationship has been found between specific pathological features and particular clinical patterns or treatment responses. More research is needed to understand the clinical utility of phenotypic classification in asthma. Many clinical phenotypes of asthma have been identified.19-21 Some of the most common are: • Allergic asthma: this is the most easily recognized asthma phenotype, which often commences in childhood and is associated with a past and/or family history of allergic disease such as eczema, allergic rhinitis, or food or drug allergy. Examination of the induced sputum of these patients before treatment often reveals eosinophilic airway inflammation. Patients with this asthma phenotype usually respond well to ICS treatment. • Non-allergic asthma: some patients have asthma that is not associated with allergy. The cellular profile of the sputum of these patients may be neutrophilic, eosinophilic or contain only a few inflammatory cells (paucigranulocytic). Patients with non-allergic asthma often demonstrate a lesser short-term response to ICS. • Cough variant asthma and cough predominant asthma:23 in some children and adults, cough may be the only symptom of asthma, and evidence of variable airflow limitation may be absent apart from during bronchial provocation testing. Some patients subsequently also develop wheezing and bronchodilator responsiveness. ICS-containing treatment is effective. For more details, see p.32. • Adult-onset (late-onset) asthma: some adults, particularly women, present with asthma for the first time in adult life. These patients tend to be non-allergic, and often require higher doses of ICS or are relatively refractory to corticosteroid treatment. Occupational asthma (i.e., asthma due to exposures at work) should be ruled out in patients presenting with adult-onset asthma. • Asthma with persistent airflow limitation: some patients with long-standing asthma develop airflow limitation that is persistent or incompletely reversible (see p.29). This is thought to be due to airway wall remodeling. See Chapter 5 (p.108) for more details about patients with features of both asthma and chronic obstructive pulmonary disease (COPD). • Asthma with obesity: some obese patients with asthma have prominent respiratory symptoms and a different pattern of airway inflammation, with little eosinophilic inflammation.24 There is little evidence about the natural history of asthma after diagnosis, but one longitudinal study showed that approximately 16% of adults with recently diagnosed asthma may experience clinical remission (no symptoms or asthma medication for at least 1 year) within 5 years.25 See p.50 for more information about remission. MAKING THE INITIAL DIAGNOSIS Making the diagnosis of asthma before treatment is started, as shown in Box 1-1 (p.25) and Box 1-2 (p.26) is based on identifying both a characteristic pattern of respiratory symptoms such as wheezing, shortness of breath (dyspnea), chest tightness or cough, and variable expiratory airflow limitation.26 The pattern of symptoms is important, as respiratory symptoms may be due to acute or chronic conditions other than asthma (see Box 1-3, p.27). If possible, the evidence supporting a diagnosis of asthma (Box 1-2, p.26) should be documented when the patient first presents, as the features that are characteristic of asthma may improve spontaneously or with treatment. As a result, it is often more difficult to confirm a diagnosis of asthma once the patient has been started on ICS-containing treatment, because this reduces variability of both symptoms and lung function (see Box 1-4, p.30). GINA recognizes that, globally, many health professionals lack access (or ready access) to spirometry,27 so advice has also been provided for using PEF in asthma diagnosis. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 25 Box 1-1. Diagnostic flowchart for adults, adolescents and children 6–11 years in clinical practice This flowchart is for patients presenting with chronic or recurrent respiratory symptoms in clinical practice. See Box 9-4 (p.167) and Box 9-6 (p.171) for information on patients presenting with an acute exacerbation. Peak expiratory flow (PEF) is less reliable than spirometry, but it is better than having no objective measurement of lung function. When measuring PEF, use the same meter each time as the value may vary by up to 20% between different meters, and use only the highest of three readings. For other abbreviations see p.11. For more information about diagnosis, see text and Box 1-2, p.26. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 26 Box 1-2. Criteria for initial diagnosis of asthma in adults, adolescents, and children 6–11 years 1. HISTORY OF TYPICAL VARIABLE RESPIRATORY SYMPTOMS Feature Symptoms or features that support the diagnosis of asthma Wheeze, shortness of breath, chest tightness and/or cough (Descriptors may vary between cultures and by age) • Symptoms occur variably over time and vary in intensity • Symptoms are often worse at night or on waking • Symptoms are often triggered by exercise, laughter, allergens, cold air • Symptoms often appear or worsen with viral infections 2. CONFIRMED VARIABLE EXPIRATORY AIRFLOW LIMITATION Feature Considerations, definitions, criteria Excessive variability in expiratory lung function (one or more of the following): The greater the variations, or the more occasions excess variation is seen, the more confident the diagnosis of asthma. If initially negative, tests can be repeated during symptoms or in the early morning. If spirometry is not possible, PEF† may be used, but it is less reliable. Positive bronchodilator (BD) responsiveness (reversibility) test with spirometry (or PEF†) Adults: increase from baseline in FEV1 or FVC of ≥12% and ≥200 mL, with greater confidence if the increase is ≥15% and ≥400 mL; or increase in PEF† ≥20% if spirometry is not available. Children: increase from baseline in FEV1 of ≥12% predicted (or in PEF† of ≥15%). Measure change 10–15 minutes after 200–400 mcg salbutamol (albuterol) or equivalent, compared with pre-BD readings. Positive test more likely if BD withheld before test: SABA ≥4 hours, long-acting bronchodilators 24–48 hours (see below). Excessive variability in twice-daily PEF over 2 weeks Adults: average daily diurnal PEF variability >10% Children: average daily diurnal PEF variability >13% Increase in lung function after 4 weeks of treatment Adults: increase from baseline in FEV1 by ≥12% and ≥200 mL (or PEF† by ≥20%) after 4 weeks of daily ICS-containing treatment Children: increase from baseline in FEV1 of ≥12% predicted (or in PEF† of ≥15%). Positive bronchial challenge test Adults: Fall from baseline in FEV1 of ≥20% with standard doses of methacholine, or ≥15% with standardized hyperventilation, hypertonic saline or mannitol challenge, or >10% and >200 mL with standardized exercise challenge.
Children: fall from baseline in FEV1 of >12% predicted (or fall in PEF† >15%) with standardized exercise challenge. If FEV1 decreases during a challenge test, check that FEV1/FVC ratio has also decreased, since incomplete inhalation, e.g., due to inducible laryngeal obstruction or poor effort, can result in a false reduction in FEV1. Excessive variation in lung function between visits (good specificity but poor sensitivity) Adults: variation in FEV1 of ≥12% and ≥200 mL (or in PEF† of ≥20%) between visits. Children: variation in FEV1 of ≥12% (or ≥15% in PEF†) between visits See list of abbreviations (p.11). See Box 1-3 (p.27) for how to confirm the diagnosis in patients already taking ICS-containing treatment. See p.31 for role of FeNO in asthma diagnosis. For bronchodilator responsiveness testing, use either a SABA or a rapid-acting ICS-LABA; see p.29. Withholding periods: Short-acting beta2 agonists: ≥4 hours; formoterol, salmeterol: 24 hours; indacaterol, vilanterol: 36 hours; tiotropium, umeclidinium, aclidinium, glycopyrronium: 36–48 hours. †For each PEF measurement, use the highest of 3 readings. Use the same PEF meter each time, as PEF may vary by up to 20% between different meters. Daily diurnal PEF variability is calculated from twice daily PEF as (day’s highest minus day’s lowest) COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 27 divided by (mean of day’s highest and lowest), averaged over two weeks. BD responsiveness may be lost temporarily during severe exacerbations or viral infections,28 and airflow limitation may become persistent over time. If reversibility is not present at initial presentation, the next step depends on the availability of other tests and the urgency of the need for treatment. In a situation of clinical urgency, asthma treatment may be commenced and diagnostic testing arranged within the next few weeks (Box 1-4, p.30), but other conditions that can mimic asthma (Box 1-3, p.27) should be considered, and the diagnosis confirmed as soon as possible. Patterns of respiratory symptoms that are characteristic of asthma The following features are typical of asthma and, if present, increase the probability that the patient has asthma.26 Respiratory symptoms of wheeze, shortness of breath, cough and/or chest tightness: • Symptoms are often worse at night or in the early morning. • Symptoms vary over time and in intensity. • Symptoms are triggered by viral infections (colds), exercise, allergen exposure, changes in weather, laughter, or irritants such as car exhaust fumes, smoke or strong smells. The following features decrease the probability that respiratory symptoms are due to asthma: • Chronic production of sputum • Shortness of breath associated with dizziness, light-headedness or peripheral tingling (paresthesia) • Chest pain • Exercise-induced dyspnea with noisy inspiration. DIFFERENTIAL DIAGNOSIS The differential diagnosis in a patient with suspected asthma varies with age (Box 1-3, p.27). Any of these alternative diagnoses may also be found together with asthma. See Section 6 (p.117) for management of multimorbidity. Box 1-3. Differential diagnosis of asthma in adults, adolescents and children 6–11 years Age If the symptoms or signs below are present, consider… Condition 6–11 years Sneezing, itching, blocked nose, throat-clearing Chronic upper airway cough syndrome Sudden onset of symptoms, unilateral wheeze Inhaled foreign body Recurrent infections, productive cough Bronchiectasis Recurrent infections, productive cough, sinusitis Primary ciliary dyskinesia Cardiac murmurs Congenital heart disease Pre-term delivery, symptoms since birth Bronchopulmonary dysplasia Excessive cough and mucus production, gastrointestinal symptoms Cystic fibrosis 12–39 years Sneezing, itching, blocked nose, throat-clearing Chronic upper airway cough syndrome Dyspnea, inspiratory wheezing (stridor) Inducible laryngeal obstruction Dizziness, paresthesia, sighing Hyperventilation, dysfunctional breathing Productive cough, recurrent infections Bronchiectasis Excessive cough and mucus production Cystic fibrosis Cardiac murmurs Congenital heart disease COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 28 Shortness of breath, family history of early emphysema Alpha1-antitrypsin deficiency Sudden onset of symptoms Inhaled foreign body 40+ years Dyspnea, inspiratory wheezing (stridor) Inducible laryngeal obstruction Dizziness, paresthesia, sighing Hyperventilation, dysfunctional breathing Cough, sputum, dyspnea on exertion, smoking or noxious exposure COPD Productive cough, recurrent infections Bronchiectasis Dyspnea with exertion, nocturnal symptoms, ankle edema Cardiac failure Treatment with angiotensin-converting enzyme (ACE) inhibitor Medication-related cough Dyspnea with exertion, non-productive cough, finger clubbing Parenchymal lung disease Sudden onset of dyspnea, chest pain Pulmonary embolism Dyspnea, unresponsive to bronchodilators Central airway obstruction All ages Chronic cough, hemoptysis, dyspnea; and/or fatigue, fever, (night) sweats, anorexia, weight loss Tuberculosis Prolonged paroxysms of coughing, sometimes stridor Pertussis See list of abbreviations (p.11). See Section 7 (p.131). Any of the above conditions may also contribute to respiratory symptoms in patients with confirmed asthma. Why is it important to confirm the diagnosis of asthma? This is important to avoid unnecessary treatment or over-treatment, and to avoid missing other important diagnoses. In adults with an asthma diagnosis in the last 5 years, one-third could not be confirmed as having asthma after repeated testing over 12 months with staged withdrawal of ICS-containing treatment. The diagnosis of asthma was less likely to be confirmed in patients who did not undergo lung function testing at the time of initial diagnosis. Some patients (2%) had serious cardiorespiratory conditions that had been misdiagnosed as asthma.29 It is important to confirm the diagnosis of asthma in people with suggestive respiratory symptoms; a study in Canada found that patients with undiagnosed asthma had worse health-related quality of life and more unscheduled healthcare visits than those without asthma, and similar to those with diagnosed asthma.30 History and family history Commencement of respiratory symptoms in childhood, a history of allergic rhinitis or eczema, or a family history of asthma or allergy, increases the probability that the respiratory symptoms are due to asthma. However, these features are not specific for asthma and are not seen in all asthma phenotypes. Patients with allergic rhinitis or atopic dermatitis should be asked specifically about respiratory symptoms. Physical examination Physical examination in people with asthma is often normal. The most frequent abnormality is expiratory wheezing (rhonchi) on auscultation, but this may be absent or only heard on forced expiration. Wheezing may also be absent during severe asthma exacerbations, due to severely reduced airflow (so called ‘silent chest’), but at such times, other physical signs of respiratory failure are usually present. Wheezing may also be heard with inducible laryngeal obstruction, COPD, respiratory infections, tracheomalacia, or inhaled foreign body (when wheezing may be unilateral). Crackles (crepitations) and inspiratory wheezing are not features of asthma. Examination of the nose may reveal signs of allergic rhinitis or nasal polyps. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 29 Lung function testing to document variable expiratory airflow limitation Asthma is characterized by variable expiratory airflow limitation, i.e., expiratory lung function varies over time, and in magnitude, to a greater extent than in healthy populations. In asthma, lung function may vary over time between completely normal and severely obstructed in the same patient. Poorly controlled asthma is associated with greater variability in lung function than well-controlled asthma.28 Lung function is most reliably assessed by spirometry testing, with assessment of forced expiratory volume in 1 second (FEV1) and the ratio of FEV1 to forced vital capacity (FEV1/FVC). Spirometry testing should be carried out by well-trained operators with well-maintained and regularly calibrated equipment,31 with an inline filter to protect against transmission of infection.32 However, globally, many clinicians do not have ready (or any) access to spirometry. In this context assessment of PEF, although less reliable, is better than no objective measurement of lung function. If PEF is used, the best of 3 measurements should be used each time, and the same meter should be used for follow-up testing, as measurements may differ from meter to meter by up to 20%.33 A reduced FEV1 or PEF may be found with many other lung diseases, or poor technique with inadequate inhalation. This may be due to lack of effort or to inducible laryngeal obstruction. Reduced FEV1/FVC (compared with baseline or compared with the lower limit of normal) indicates expiratory airflow limitation. Many spirometers now include age-specific predicted values for lower limit of normal in their software.34 In clinical practice, once an obstructive defect has been confirmed, variation in airflow limitation is generally assessed from variation in FEV1 or PEF. ‘Variability’ refers to improvement and/or deterioration in symptoms and lung function. Excessive variability may be identified over the course of one day (diurnal variability), from day to day, from visit to visit, or seasonally, or from a responsiveness test. Responsiveness (previously called ‘reversibility’)31 generally refers to rapid improvements in FEV1 (or PEF), measured within minutes after inhalation of a rapid-acting bronchodilator such as 200–400 mcg salbutamol, or more sustained improvement over days or weeks after the introduction of ICS treatment.35 In a patient with typical or suggestive respiratory symptoms, obtaining evidence of excessive variability in expiratory lung function is an essential component of the diagnosis of asthma. Some specific examples are: • An increase in lung function 10–15 minutes after administration of a bronchodilator, or after a trial of ICS-containing treatment; lung function may improve gradually, so it should be assessed after at least 4 weeks • A decrease in lung function after exercise (spontaneous or standardized) or during a bronchial provocation test • Variation in lung function beyond the normal range when it is repeated over time, either on separate visits, or on twice-daily home monitoring over at least 1–2 weeks.
Specific criteria for demonstrating excessive variability in expiratory lung function are listed in Box 1-2 (p.26). A decrease in FEV1 or PEF during a respiratory infection, while commonly seen in asthma, does not necessarily indicate that a person has asthma, as it may also be seen in otherwise healthy individuals or people with COPD. How much variation in expiratory airflow is consistent with asthma? Bronchodilator responsiveness: There is overlap in bronchodilator responsiveness and other measures of variation between health and disease.36 In a patient with respiratory symptoms, the greater the variations in their lung function, or the more times excess variation is seen, the more likely the diagnosis is to be asthma (Box 1-2, p.26). Generally, in adults with respiratory symptoms typical of asthma, an increase or decrease in FEV1 of ≥12% and ≥200 mL from baseline, or (if spirometry is not available) a change in PEF of at least 20%, is accepted as being consistent with asthma. A Technical Standards Committee recommended changing the criterion for a positive bronchodilator responsiveness test from an increase from baseline in FEV1 or FVC of ≥12% and >200 mL (as at present) to an increase from baseline of >10% of the patient’s predicted value.37 This recommendation was based on data for survival, and the Technical Standards Committee avoided making any recommendation about the use of this criterion for diagnostic decisions in clinical practice. This topic will be considered again by GINA when more data are available, including comparison with other diagnostic tests for asthma. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 30 Diurnal PEF variability is calculated from twice daily readings as the daily amplitude percent mean, i.e.: (( [Day’s highest – day’s lowest] / mean of day’s highest and lowest) x 100). Then the average of each day’s value is calculated over 1–2 weeks. The upper 95% confidence limit of diurnal variability (amplitude percent mean) from twice daily readings is 9% in healthy adults,38 and 12.3% in healthy children39 so, in general, diurnal variability >10% for adults and >13% for children is regarded as excessive. If FEV1 is within the predicted normal range when the patient is experiencing symptoms, this reduces the probability that the symptoms are due to asthma. However, patients whose baseline FEV1 is >80% predicted can have a clinically important increase in lung function with bronchodilator or ICS-containing treatment. Predicted normal ranges (especially for PEF) have limitations, so the patient’s own best reading (‘personal best’) is recommended as their ‘normal’ value. Box 1-4. Steps for confirming the diagnosis of asthma in a patient already taking ICS-containing treatment Current status Steps to confirm the diagnosis of asthma Variable respiratory symptoms and variable airflow limitation Diagnosis of asthma is confirmed. Assess the level of asthma control (Box 2-2A and Box 2-2B, p.37) and review ICS-containing treatment (Box 4-6, p.77; Box 4-12, p.96.) Variable respiratory symptoms but no variable airflow limitation Consider repeating spirometry (or PEF) after withholding bronchodilator (4 hrs for SABA, 24–48 hrs for long-acting bronchodilators (see below) or during symptoms. Check between-visit variability of FEV1, and bronchodilator responsiveness. If still normal, consider other diagnoses (Box 1-3, p.27). If FEV1 (or PEF) is >70% predicted: consider stepping down ICS-containing treatment (see Box 1-5, p.32) and reassess in 2–4 weeks, then consider bronchial provocation test or repeating bronchodilator responsiveness test. If FEV1 (or PEF) is <70% predicted: consider starting or stepping up maintenance ICS-containing treatment for 3 months (Box 4-6, p.77), then reassess symptoms and lung function. If no response, resume previous ICS dose and refer patient for diagnosis and investigation. Few respiratory symptoms, normal lung function, and no variable airflow limitation Consider repeating BD responsiveness test again after withholding bronchodilator as above or during symptoms. If normal, consider investigation for alternative diagnoses (Box 1-3, p.27). Consider stepping down ICS-containing treatment (see Box 1-5, p.32): • If symptoms emerge and lung function falls: asthma is confirmed. Step up ICS-containing treatment to previous lowest effective dose. • If no change in symptoms or lung function at lowest controller step: consider ceasing maintenance ICS-containing treatment, or switching to as-needed-only ICS-formoterol, and monitor patient closely for at least 12 months (Box 4-13, p.102). Persistent shortness of breath and persistent airflow limitation Consider stepping up ICS-containing treatment for 3 months (Box 4-6, p.77), then reassess symptoms and lung function. If no response, resume previous ICS dose and refer patient for further investigation and management, or manage as for patients with features of both asthma and COPD (Section 7, p.131). See list of abbreviations (p.11). ‘Variable airflow limitation’ refers to expiratory airflow. Withholding period for long-acting bronchodilators: 24 hours for formoterol, salmeterol; 36 hours for indacaterol, vilanterol; 36-48 hours for tiotropium, umeclidinium, aclidinium, glycopyrronium. If spirometry is not possible, PEF may be used, but it is less reliable. Use the same PEF meter each time, as PEF may vary by up to 20% between different meters. For each PEF measurement, use the highest of 3 readings. When can variable expiratory airflow limitation be documented? If possible, evidence of variable expiratory airflow limitation should be documented before treatment is started. This is because variability usually decreases with ICS treatment as lung function improves. In addition, any increase in lung function after initiating ICS-containing treatment can help to confirm the diagnosis of asthma. Bronchodilator responsiveness may not be present between symptoms, during viral infections or if the patient has used a beta2 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 31 agonist within the previous few hours; and in some patients with asthma, airflow limitation may become persistent or irreversible over time. If neither spirometry nor PEF is available, or variable expiratory airflow limitation is not documented, a decision about whether to investigate further or start ICS-containing treatment immediately depends on clinical urgency and access to other tests.27 Box 1-4 (p.30) describes how to confirm the diagnosis of asthma in a patient already taking ICS-containing treatment. Other tests that may be used in diagnosis of asthma Bronchial provocation tests One option for documenting variable expiratory airflow limitation is to refer the patient for bronchial provocation testing to assess airway hyperresponsiveness. Challenge agents include inhaled methacholine,40 histamine, exercise,41 eucapnic voluntary hyperventilation or inhaled mannitol. These tests are moderately sensitive for a diagnosis of asthma but have limited specificity40,41. For example, airway hyperresponsiveness to inhaled methacholine has been described in patients with allergic rhinitis,42 cystic fibrosis43, bronchopulmonary dysplasia44 and COPD.45 This means that a negative test in a patient not taking ICS can help to exclude asthma, but a positive test does not always mean that a patient has asthma – the pattern of symptoms (Box 1-2, p.26) and other clinical features (Box 1-3, p.27) must also be considered. Allergy tests The presence of atopy increases the probability that a patient with respiratory symptoms has allergic asthma, but this is not specific for asthma nor is it present in all asthma phenotypes. Atopic status can be identified by skin prick testing or by measuring the level of specific immunoglobulin E (sIgE) in serum. Skin prick testing with common environmental allergens is simple and rapid to perform and, when performed by an experienced tester with standardized extracts, is inexpensive and has a high sensitivity. Measurement of sIgE is no more reliable than skin prick tests and is more expensive, but may be preferred for uncooperative patients, those with widespread skin disease, or if the history suggests a risk of anaphylaxis.46 The presence of a positive skin test or positive sIgE, however, does not mean that the allergen is causing symptoms – the relevance of allergen exposure and its relation to symptoms must be confirmed by the patient’s history. Imaging Imaging studies are not routinely used in the diagnosis of asthma, but may be useful to investigate the possibility of comorbid conditions or alternative diagnoses in adults with difficult-to-treat asthma. Imaging may also be used to identify congenital abnormalities in infants with asthma-like symptoms, and alternative diagnoses in children with difficult-to-treat asthma. High-resolution computed tomography (CT) of the lungs can identify conditions such as bronchiectasis, emphysema, lung nodules, airway wall thickening and lung distension, and may assess airway distensibility. The presence of radiographically detected emphysema is considered when differentiating asthma from COPD (Box 7-4, p.137), but there is no accepted threshold, and these conditions can coexist. Moreover, air trapping (which may be present in asthma, and is also a feature of ageing) can be difficult to distinguish from emphysema. Chest imaging is not currently recommended to predict treatment outcomes or lung function decline, or to assess treatment response. CT of the sinuses can identify changes suggestive of chronic rhinosinusitis with or without nasal polyps (p.120), which in patients with severe asthma may help with choice of biologic therapy (see Box 8-4, p.144). Exhaled nitric oxide The fractional concentration of exhaled nitric oxide (FeNO) is modestly associated with levels of sputum and blood eosinophils,47 but this association is lost in obesity.24,48 FeNO has not been established as useful for ruling in or ruling out a diagnosis of asthma (see Definition of asthma, p.23) because, while FeNO is higher in asthma that is characterized by Type 2 airway inflammation with elevated interleukin (IL)-4 and IL-13,49 it is also elevated in non-asthma conditions (e.g., eosinophilic bronchitis, atopy, allergic rhinitis, eczema), and it is not elevated in some asthma phenotypes (e.g., neutrophilic asthma, asthma with obesity).24 FeNO is also lower in smokers and during bronchoconstriction50 and the early phases of allergic response;51 it may be increased or decreased during viral respiratory infections.50 For information on the role of FeNO in asthma treatment, see Section 4 (p.72). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 32 CONFIRMING THE DIAGNOSIS OF ASTHMA IN PATIENTS ALREADY TAKING ICS-CONTAINING TREATMENT If the basis of a patient’s diagnosis of asthma has not previously been documented, confirmation with objective testing should be sought. In primary care, the presence of asthma cannot be confirmed in many patients (25–35%) who have previously received this diagnosis.29,52-55 The process for confirming the diagnosis in patients already on ICS-containing treatment depends on the patient’s symptoms and lung function (Box 1-4, p.30). In some patients, this may include a trial of either a lower or a higher dose of ICS-containing treatment. If the diagnosis of asthma cannot be confirmed, refer the patient for expert investigation and diagnosis. For some patients, it may be necessary to step down the ICS-containing treatment to confirm the diagnosis of asthma. The process is described in Box 1-5 (p.32). Box 1-5. How to step-down ICS-containing treatment to help confirm the diagnosis of asthma 1.
ASSESS • Document the patient’s current status including asthma symptom control and risk factors (Box 2-2, p.37) and lung function. If the patient has risk factors for asthma exacerbations (Box 2-2B), step down treatment only with close supervision.
• Choose a suitable time (e.g., no respiratory infection, not going away on vacation, not pregnant).
• Provide a written asthma action plan (Box 9-2, p.162) so the patient/caregiver knows how to recognize and respond if symptoms worsen. Ensure they will have enough medication to be able to resume their previous dose if their asthma worsens after stepping down.
2.
ADJUST • Show the patient/caregiver how to reduce their ICS dose by 25–50%, or stop other maintenance medication (e.g., LABA) if being used. See step-down options in Box 4-13, p.102. Schedule a review visit for 2–4 weeks.
3.
REVIEW RESPONSE • Repeat assessment of asthma control and lung function tests in 2–4 weeks (Box 1-2, p.26).
• If symptoms increase and variable expiratory airflow limitation is confirmed after stepping down treatment, the diagnosis of asthma is confirmed. The patient should be returned to their lowest previous effective treatment.
• If, after stepping down to a low-dose ICS-containing treatment, symptoms do not worsen and there is still no evidence of variable expiratory airflow limitation to confirm the diagnosis of asthma, consider ceasing ICS-containing treatment and repeating asthma control assessment and lung function tests in 2–3 weeks, but follow the patient for at least 12 months.29 See list of abbreviations (p.11) HOW TO MAKE THE DIAGNOSIS OF ASTHMA IN OTHER CONTEXTS Patients presenting with persistent cough as the only respiratory symptom Common causes of an isolated non-productive cough include cough-variant asthma, chronic upper airway cough syndrome (often called ‘postnasal drip’), cough induced by angiotensin-converting enzyme (ACE) inhibitors, gastroesophageal reflux, chronic sinusitis, post-infectious cough,56 inducible laryngeal obstruction,57,58 and eosinophilic bronchitis. In cough variant asthma, a persistent cough is the only symptom, or, in cough predominant asthma, the most prominent symptom.22,23,59 The cough may be worse at night or with exercise, and in some patients it is productive. Spirometry is usually normal, and the only abnormality in lung function may be airway hyperresponsiveness on COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 33 bronchial provocation testing (Box 1-2, p.26,). Some patients with cough variant asthma may later develop wheeze and significant bronchodilator responsiveness on spirometry.60 Most patients with cough variant asthma have sputum eosinophilia, and they may also have elevated FeNO.22 Cough-variant asthma must also be distinguished from eosinophilic bronchitis in which patients have cough and sputum eosinophilia but normal spirometry and normal airway responsiveness.61 Treatment of cough variant asthma follows usual recommendations for asthma. Occupational asthma and work-exacerbated asthma Asthma acquired in the workplace is frequently missed. Asthma may be induced or (more commonly) aggravated by exposure to allergens or other sensitizing agents at work, or sometimes from a single, massive exposure. Occupational rhinitis may precede asthma by up to a year and early diagnosis is essential, as persistent exposure is associated with worse outcomes.62,63 An estimated 5–20% of new cases of adult-onset asthma can be attributed to occupational exposure.62 Adult-onset asthma requires a systematic inquiry about work history and exposures, including hobbies. Asking patients whether their symptoms improve when they are away from work (weekends or vacation) is an essential screening question.64 It is important to confirm the diagnosis of occupational asthma objectively as it may lead to the patient changing their occupation, which may have legal and socioeconomic implications. Specialist referral is usually necessary, and frequent PEF monitoring at and away from work is often used to help confirm the diagnosis. There is more information about occupational asthma in Section 6 (p.117) and in specific guidelines.62 Athletes The diagnosis of asthma in athletes should be confirmed by lung function tests, usually with bronchial provocation testing.65 Conditions that may either mimic or be associated with asthma, such as rhinitis, laryngeal disorders (e.g., inducible laryngeal obstruction),58 dysfunctional breathing, cardiac conditions and over-training, must be excluded.66 Pregnant women Pregnant women and women planning a pregnancy should be asked whether they have asthma so that appropriate advice about asthma management and medications can be given (p.126).67 If the clinical history is consistent with asthma, and other diagnoses appear unlikely (Box 1-3, p.27) but the diagnosis of asthma is not confirmed on initial bronchodilator responsiveness testing (Box 1-2, p.26), manage as asthma with ICS-containing treatment (p.126) and postpone other diagnostic investigations until after delivery. During pregnancy, bronchial provocation testing is contraindicated, and it is not advisable to step down ICS-containing treatment. The elderly Asthma is frequently undiagnosed in the elderly,68 due to poor perception of airflow limitation; acceptance of dyspnea as being ‘normal’ in old age, lack of fitness, and reduced physical activity. The presence of multimorbidity also complicates the diagnosis. In a large population-based survey of asthma patients older than 65 years, factors associated with a history of asthma hospitalization included co-diagnosis of COPD, coronary artery disease, depression, diabetes mellitus, and difficulty accessing medications or clinical care because of cost.69 Symptoms of wheezing, breathlessness and cough that are worse on exercise or at night can also be caused by cardiovascular disease or left ventricular failure, which are common in this age group. A careful history and physical examination, combined with an electrocardiogram and chest X-ray, will assist in the diagnosis.70 Measurement of plasma brain natriuretic polypeptide and assessment of cardiac function with echocardiography may also be helpful.71 In older people with a history of smoking or biomass fuel exposure, COPD and overlapping asthma and COPD (asthma– COPD overlap) should be considered (Section 7, p.131). Smokers and ex-smokers Asthma and COPD may be difficult to distinguish in clinical practice, particularly in older patients and smokers and ex-smokers, and these conditions may overlap (asthma-COPD overlap). The Global Strategy for Diagnosis, Management and Prevention of COPD (GOLD) 202472 defines COPD on the basis of chronic respiratory symptoms, environmental exposures such as smoking or inhalation of toxic particles or gases, with confirmation by post-bronchodilator FEV1/FVC <0.7. Clinically important bronchodilator responsiveness (>12% and >200 mL) is often found in COPD.73 Low diffusion capacity is more common in COPD than asthma. The history and pattern of symptoms and past records COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 34 can help to distinguish patients with COPD from those with long-standing asthma who have developed persistent airflow limitation. Uncertainty in the diagnosis should prompt early referral for specialized investigation and treatment recommendations, as patients with asthma-COPD overlap have worse outcomes than those with asthma or COPD alone (see Section 7, p.131).74 Obese patients While asthma is more common in obese than non-obese people,75 respiratory symptoms associated with obesity can mimic asthma. In obese patients with dyspnea on exertion, it is important to confirm the diagnosis of asthma with objective measurement of variable expiratory airflow limitation. One study found that non-obese patients were just as likely to be over-diagnosed with asthma as obese patients (around 30% in each group).52 Another study found both over- and under-diagnosis of asthma in obese patients.76 Low- and middle-income countries Diagnosis of asthma in low-resource settings, including low- and middle-income countries (LMICs), presents substantial challenges for clinical practice.27 Access to lung function testing, particularly spirometry, is often very limited. Even when available, lung function testing may be substantially underused (e.g., unaffordable for the patient or health system,77 or too time-consuming in a busy clinic. A single lung function test may not be sufficient to confirm the diagnosis of asthma or indicate an alternative cause, so more than one visit by the patient (with resulting costs of time and travel) may be needed.27 The differential diagnosis of asthma in these countries may often include other endemic respiratory diseases (e.g., tuberculosis, HIV/AIDS-associated lung diseases, and parasitic or fungal lung diseases). As a result of these issues, clinicians often use a syndromic approach to diagnosis and initial management, based on history and clinical findings.78 Practical evidence-based resources have been developed and implemented in several countries.79,80 This approach reduces diagnostic precision but is based on the assumption (valid in most LMICs) that under-diagnosis and under-treatment of asthma is more likely81 than the overdiagnosis and overtreatment often seen in high income countries.29,82 GINA does not recommend that diagnosis of asthma should be solely based on syndromic clinical patterns, and suggests lung function testing with a PEF meter if spirometry is not available.27 The World Health Organization (WHO) Package of essential noncommunicable (PEN) disease interventions for primary care83 lists the PEF meter as an essential tool in the management of chronic respiratory diseases. When spirometry is not available, the presence of variable expiratory airflow limitation (including reversible obstruction) can be confirmed by PEF, as outlined in Box 1-2, p.26. For example, before starting long-term ICS-containing treatment, the following investigations can help to confirm the diagnosis of asthma (or prompt investigation for alternative diagnoses) • ≥20% improvement in PEF 15 minutes after giving 2 puffs of albuterol83 • Improvement in symptoms and PEF after a 4-week therapeutic trial with ICS-containing treatment.27 Either of these findings would increase the likelihood of a diagnosis of asthma versus other diagnoses. A structured algorithmic approach to patients presenting with respiratory symptoms forms part of several strategies developed for improving respiratory disease management in LMICs.5 These strategies are of particular use in countries where, owing to the high prevalence of tuberculosis, large numbers of patients with respiratory symptoms present for assessment at tuberculosis clinics. There is a pressing need for access to affordable diagnostic tools (peak flow meters and spirometry), and training in their use, to be substantially scaled up in LMICs.27 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 35 2. Assessment of asthma in adults, adolescents and children 6–11 years KEY POINTS Asthma control • The level of asthma control is the extent to which the features of asthma can be observed in the patient, or have been reduced or removed by treatment. • Asthma control is assessed in two domains: symptom control and risk of adverse outcomes. Poor symptom control is burdensome to patients and increases the risk of exacerbations, but patients with good symptom control can still have severe exacerbations. Asthma severity • The current definition of asthma severity is based on retrospective assessment, after at least 2–3 months of asthma treatment, from the intensity of treatment required to control symptoms and exacerbations. • This definition is clinically useful for severe asthma, as it identifies patients whose asthma is relatively refractory to high intensity treatment with high-dose inhaled corticosteroids (ICS) and a long-acting beta2 agonist (LABA) and who may benefit from additional treatment such as biologic therapy. It is important to distinguish between severe asthma and asthma that is uncontrolled due to modifiable factors such as incorrect inhaler technique and/or poor adherence. • However, the retrospective definition of mild asthma as ‘easy to treat’ is less useful, as patients with few interval symptoms can have exacerbations triggered by external factors such as viral infections or allergen exposure, and the treatment that was historically regarded as the lowest intensity – short-acting beta2 agonist (SABA) alone – actually increases the risk of exacerbations. • ‘Mild asthma’ is a retrospective label, so it cannot be used to decide which patients are suitable to receive Step 1 or Step 2 treatment. • In clinical practice and in the general community, the term ‘mild asthma’ is often used to mean infrequent or mild symptoms, and it is often assumed that these patients are not at risk and do not need ICS-containing treatment. • For these reasons, GINA suggests that the term ‘mild asthma’ should generally be avoided in clinical practice if possible or, if used, qualified with a reminder that patients with infrequent symptoms can still have severe or fatal exacerbations, and that this risk is substantially reduced with ICS-containing treatment. • GINA is continuing to engage in stakeholder discussions about the definition of mild asthma, to obtain agreement about the implications for clinical practice and clinical research of the changes in knowledge about asthma pathophysiology and treatment since the current definition of asthma severity was published. How to assess a patient’s asthma • Assess symptom control from the frequency of daytime and night-time asthma symptoms, night waking and activity limitation and, for patients using SABA reliever, their frequency of SABA use. Other tools for assessing recent symptom control include Asthma Control Test (ACT) and Asthma Control Questionnaire (ACQ). There are no validated tools for assessing symptom control over a longer period. • Also, separately, assess the patient’s risk factors for exacerbations, even if their symptom control is good. Risk factors for exacerbations that are independent of symptom control include not only a history of ≥1 exacerbation in the previous year, but also SABA-only treatment (without any ICS), over-use of SABA, socioeconomic problems, poor adherence, incorrect inhaler technique, low forced expiratory volume in 1 second (FEV1), exposures such as smoking, and blood eosinophilia. To date, there are no suitable composite tools for assessing exacerbation risk. • Also assess risk factors for persistent airflow limitation and medication side-effects (including from oral corticosteroids), treatment issues such as inhaler technique and adherence, and comorbidities, and ask the COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 36 patient/caregiver about their asthma goals and treatment preferences. • Once the diagnosis of asthma has been made, the main role of lung function testing is in the assessment of future risk. It should be recorded at diagnosis, 3–6 months after starting treatment, and periodically thereafter. • Investigate for impaired perception of bronchoconstriction if there are few symptoms but low lung function, and investigate for alternative diagnoses if there are frequent symptoms despite good lung function. OVERVIEW The long-term goal of asthma treatment is to achieve the best possible long-term outcomes for the patient (see Box 3-2, p.50 for more details about goals of treatment). For every patient, assessment of asthma should include the assessment of asthma control (both symptom control and future risk of adverse outcomes), treatment issues (particularly inhaler technique and adherence), and any comorbidities that could contribute to symptom burden and poor quality of life (Box 2-1, p.36). Lung function, particularly FEV1 as a percentage of predicted value, is an important part of the assessment of future risk. The use of digital technology, telemedicine and telehealthcare in the monitoring of patients with asthma is rapidly increasing, particularly during the COVID-19 pandemic. However, the types of interactions are diverse, and high-quality studies are needed to evaluate their utility and effectiveness. Box 2-1. Summary of assessment of asthma in adults, adolescents, and children 6–11 years 1. Assess asthma control = symptom control AND future risk of adverse outcomes • Assess symptom control over the last 4 weeks (Box 2-2A, p.37) or longer. • Identify any other risk factors for exacerbations, persistent airflow limitation or side-effects (Box 2-2B). • Measure lung function at diagnosis/start of treatment, 3–6 months after starting ICS-containing treatment, then periodically, e.g., at least once every 1–2 years, but more often in at-risk patients and those with severe asthma. 2. Assess treatment issues • Document the patient’s current treatment step (Box 4-6, p.77). • Watch inhaler technique (Box 5-2, p.110), assess adherence (Box 5-3, p.112) and side-effects. • Check that the patient has a written asthma action plan. • Ask about the patient’s attitudes and goals for their asthma and medications. 3. Assess multimorbidity • Rhinitis, rhinosinusitis, gastroesophageal reflux, obesity, obstructive sleep apnea, depression and anxiety can contribute to symptoms and poor quality of life, and sometimes to poor asthma control (see Section 6, p.117). What is meant by ‘asthma control’? The level of asthma control is the extent to which the manifestations of asthma can be observed in the patient, or have been reduced or removed by treatment.38,84 It is determined by the interaction between the patient’s genetic background, underlying disease processes, the treatment that they are taking, environment, and psychosocial factors.84 Asthma control has two domains: symptom control and future risk of adverse outcomes (Box 2-2, p.37). Both should always be assessed. Lung function is an important part of the assessment of future risk; it should be measured at the start of treatment, after 3–6 months of treatment (to identify the patient’s personal best), and periodically thereafter for ongoing risk assessment. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 37 Box 2-2. GINA assessment of asthma control at clinical visits in adults, adolescents and children 6–11 years A. Recent asthma symptom control (but also ask the patient/caregiver about the whole period since last review#) In the past 4 weeks, has the patient had: Well controlled Partly controlled Uncontrolled • Daytime asthma symptoms more than twice/week? Yes No None of these 1–2 of these 3–4 of these • Any night waking due to asthma? Yes No • SABA reliever for symptoms more than twice/week? Yes No • Any activity limitation due to asthma? Yes No B. Risk factors for poor asthma outcomes Assess risk factors at diagnosis and periodically, particularly for patients experiencing exacerbations. Measure FEV1 at start of treatment, after 3–6 months of ICS-containing treatment to record the patient’s personal best lung function, then periodically for ongoing risk assessment. a. Risk factors for exacerbations Uncontrolled asthma symptoms: Having uncontrolled symptoms is an important risk factor for exacerbations.85 Factors that increase the risk of exacerbations even if the patient has few asthma symptoms† SABA over-use: High SABA use (≥3 x 200-dose canisters/year associated with increased risk of exacerbations, increased mortality particularly if ≥1 canister per month)86-89 Inadequate ICS: not prescribed ICS, poor adherence,90 or incorrect inhaler technique91 Other medical conditions: Obesity,92,93 chronic rhinosinusitis,93 GERD,93 confirmed food allergy,94 pregnancy95 Exposures: Smoking,96 e-cigarettes,97 allergen exposure if sensitized,96,98 air pollution99-102 Psychosocial: Major psychological or socioeconomic problems103,104 Lung function: Low FEV1 (especially <60% predicted),96,105 high bronchodilator responsiveness93,106,107 Type 2 inflammatory markers: Higher blood eosinophils,93,108,109 high FeNO (adults with allergic asthma on ICS)110 Exacerbation history: Ever intubated or in intensive care unit for asthma;111 ≥1 severe exacerbation in last year112,113 b. Risk factors for developing persistent airflow limitation History: Preterm birth, low birth weight and greater infant weight gain,114 chronic mucus hypersecretion115,116 Medications: Lack of ICS treatment in patient with history of severe exacerbation117 Exposures: Tobacco smoke,115 noxious chemicals; occupational or domestic exposures62 Investigation findings: Low initial FEV1,116 sputum or blood eosinophilia116 c. Risk factors for medication side-effects Systemic Frequent OCS, long-term, high-dose and/or potent ICS, P450 inhibitors118 Local: High-dose or potent ICS,118,119 poor inhaler technique120 See list of abbreviations (p.11).Based on SABA (as-needed ICS-formoterol reliever not included); excludes reliever taken before exercise (see Assessing asthma symptom control, p.38). #In addition to assessing recent asthma symptom control, also ask the patient about symptom control over the whole period since their last clinical review. There are no validated tools for assessing long-term symptom control, i.e., over periods longer than 4 weeks. †‘Independent’ risk factors are those that are significant after adjustment for the level of symptom control. Cytochrome P450 inhibitors such as ritonavir, ketoconazole, itraconazole may increase systemic exposure to some types of ICS and some LABAs; see drug interaction websites and p.122 for details. For children 6–11 years, also refer to Box 2-3, p.40. See Box 3-5, p.55 for specific risk reduction strategies. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 38 How to describe a patient’s asthma control Asthma control should be described in terms of both symptom control and future risk domains. For example: Ms X has good asthma symptom control, but she is at increased risk of future exacerbations because she has had a severe exacerbation within the last year. Mr Y has poor asthma symptom control. He also has several additional risk factors for future exacerbations including low lung function, current smoking, and poor medication adherence. What does the term ‘asthma control’ mean to patients? Many studies describe discordance between the patient’s and health provider’s assessment of the patient’s level of asthma control. This does not necessarily mean that patients ‘over-estimate’ their level of control or ‘under-estimate’ its severity, but that patients understand and use the word ‘control’ differently from health professionals, e.g., based on how quickly their symptoms resolve when they take reliever medication.84,121 If the term ‘asthma control’ is used with patients, the meaning should always be explained. ASSESSING ASTHMA SYMPTOM CONTROL Asthma symptoms such as wheeze, chest tightness, shortness of breath and cough typically vary in frequency and intensity, and contribute to the burden of asthma for the patient. Poor symptom control is also strongly associated with an increased risk of asthma exacerbations.122-124 Asthma symptom control should be assessed at every opportunity, including during routine prescribing or dispensing. Directed questioning is important, as the frequency or severity of symptoms that patients regard as unacceptable or bothersome may vary from current recommendations about the goals of asthma treatment, and may differ from patient to patient. For example, despite having low lung function, a person with a sedentary lifestyle may not experience bothersome symptoms and so may appear to have good symptom control. To assess recent symptom control (Box 2-2A, p.37) ask about the following in the past four weeks: frequency of asthma symptoms (days per week), any night waking due to asthma or limitation of activity and, for patients using a SABA reliever, frequency of its use for relief of symptoms. In general, do not include reliever taken before exercise, because some people take this routinely without knowing whether they need it. Frequency of reliever use Historically, frequency of SABA reliever use (<2 or ≥2 days/week) has been included in the composite assessment of symptom control. This distinction was arbitrary, based on the assumption that if SABA was used on >2 days in a week, the patient needed to start maintenance ICS-containing therapy or increase the dose. In addition, higher average use of SABA over a year is associated with a higher risk of severe exacerbations,86,87 and in the shorter term, increasing use of as-needed SABA is associated with an increased likelihood of a severe exacerbation in subsequent days or weeks.125 However, for patients prescribed an anti-inflammatory reliever (AIR) such as as-needed low-dose ICS-formoterol (GINA Track 1, Box 4-6, p.77), use of this reliever more than 2 days/week is already providing additional ICS therapy, so further dose escalation may not be needed. In addition, increasing use of as-needed ICS-formoterol is associated with a significantly lower risk of severe exacerbation in subsequent days or weeks compared with if the reliever is SABA,126,127 or compared with if the patient is using SABA alone.128 For these reasons, while the assessment of symptom control in Box 2-2A (p.37) includes a criterion for SABA reliever use on ≤2 versus >2 days/week, it does not include a similar criterion for an anti-inflammatory reliever such as as-needed ICS-formoterol. However, the patient’s average frequency of as-needed ICS-formoterol use over the past 4 weeks should be assessed, and considered when the patient’s maintenance ICS dose (or need for maintenance ICS-formoterol) is reviewed. This issue will be reviewed again when more data become available. Tools for assessing recent asthma symptom control in adults and adolescents Simple screening tools: these can be used in primary care to quickly identify patients who need more detailed assessment. Examples include the consensus-based GINA symptom control tool (Part A, Box 2-2A, p.37). This classification correlates with assessments made using numerical asthma control scores.129,130 It can be used, together with a risk assessment (Box 2-2B), to guide treatment decisions (Box 4-6, p.77). Other examples are the Primary Care COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 39 Asthma Control Screening Tool (PACS),131 and the 30-second Asthma Test, which also includes time off work/school.132 Categorical symptom control tools: e.g., the consensus-based ‘Royal College of Physicians (RCP) Three Questions’ tool,133 which asks about difficulty sleeping, daytime symptoms and activity limitation due to asthma in the previous month. The Asthma APGAR tool includes a patient-completed asthma control assessment covering 5 domains: activity limitations, daytime and nighttime symptom frequency (based on US criteria for frequency of night waking), triggers, adherence, and patient-perceived response to treatment. This assessment is linked to a care algorithm for identifying problems and adjusting treatment up or down. A study in the US showed that introduction of the Asthma APGAR tools for patients aged 5–45 years in primary care was associated with improved rates of asthma control; reduced asthma-related urgent care, and hospital visits; and increased practices’ adherence to asthma management guidelines.134 Numerical ‘asthma control’ tools: these tools provide scores and cut points to distinguish different levels of symptom control, validated against healthcare provider assessment. Many translations are available. These scores may be useful for assessing patient progress; they are commonly used in clinical research, but may be subject to copyright restrictions. Numerical asthma control tools are more sensitive to change in symptom control than categorical tools.129 Examples of numerical asthma control tools for assessing recent symptom control are: • Asthma Control Questionnaire (ACQ):135,136 Scores range from 0–6 (higher is worse), with scores calculated as the average from all questions. The authors stated that ACQ ≤0.75 indicated a high probability that asthma was well controlled; 0.75–1.5 as a ‘grey zone’; and ≥1.5 a high probability that asthma was poorly controlled, based on concepts of asthma control at the time; they later added that the crossover point between ‘well-controlled’ and ‘not well-controlled’ asthma was close to 1.00.137 The 5-item ACQ (ACQ-5), comprises five symptom questions. Two additional versions were published: ACQ-6 includes SABA frequency, and ACQ-7 also includes pre-bronchodilator FEV1% predicted. The minimum clinically important difference for all three versions of ACQ is 0.5.137 GINA prefers ACQ 5 over ACQ-6 or 7 because the reliever question assumes regular rather than as-needed use of SABA, there is no option between zero SABA use in a week and SABA use every day, and ACQ has not been validated with ICS-formoterol or ICS-SABA as the reliever. In addition, if ACQ-7 were to be used in adjustment of treatment, the inclusion of FEV1 in the composite score could lead to repeated step-up in ICS dose for patients with persistent airflow limitation. For these reasons, data for ACQ-5, ACQ-6 and ACQ-7 cannot be combined for meta-analysis. • Asthma Control Test (ACT):130,138,139 Scores range from 5–25 (higher is better). Scores of 20–25 are classified as ‘well-controlled’; 16–19 as ‘not well-controlled’; and 5–15 as very poorly controlled asthma. The ACT has four symptom/ reliever questions plus patient self-assessed control. The minimum clinically important difference is 3 points.139 It has not been validated with ICS-formoterol or ICS-SABA reliever. Patients with good symptom control can still be at risk of future severe exacerbations or asthma-related death, and there are many modifiable risk factors for exacerbations that are independent of symptom control (Box 2-2B, p.37), so GINA does not recommend assessment tools that combine symptom control with exacerbation history. When different tools are used for assessing asthma symptom control, the results correlate broadly with each other, but are not identical. Respiratory symptoms may be non-specific so, when assessing changes in symptom control, it is important to clarify that symptoms are due to asthma. Recent symptom control can be assessed over the previous 1–4 weeks using tools such as in GINA Box 2-2A, or ACQ-5 or ACT. There are no validated tools for assessing asthma symptom control over a longer period (e.g., 12 months); in clinical practice, the patient can be asked about previous months with a simple question, but there is likely to be substantial recall error, particularly for mild symptoms. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 40 Box 2-3. Specific questions for assessment of asthma in children 6–11 years Asthma symptom control Day symptoms Ask: How often does the child have cough, wheeze, dyspnea or heavy breathing (number of times per week or day)? What triggers the symptoms? How are symptoms managed? Night symptoms Cough, awakenings, tiredness during the day? (If the only symptom is nocturnal cough, consider other diagnoses such as rhinitis or gastroesophageal reflux disease). Reliever use How often is reliever medication used? (check date on inhaler or last prescription) Distinguish between pre-exercise use (sports) and use for relief of symptoms. Level of activity What sports/hobbies/interests does the child have, at school and in their spare time? How does the child’s level of activity compare with their peers or siblings? How many days is the child absent from school? Try to get an accurate picture of the child’s day from the child without interruption from the parent/caregiver. Risk factors for adverse outcomes Exacerbations Ask: How do viral infections affect the child’s asthma? Do symptoms interfere with school or sports? How long do the symptoms last? How many episodes have occurred since their last medical review? Any urgent doctor/emergency department visits? Is there a written action plan? Risk factors for exacerbations include a history of exacerbations, poor symptom control, poor adherence and poverty,113 and persistent bronchodilator reversibility even if the child has few symptoms.107 Lung function Check spirogram curves and technique. Main focus is on FEV1 and FEV1/FVC ratio. Plot these values as percent predicted to see trends over time. Side-effects Check the child’s height at least yearly, as poorly controlled asthma can affect growth,140 and growth velocity may be lower in the first 1–2 years of ICS treatment.141 Ask about frequency and dose of ICS and OCS. Treatment factors Inhaler technique Ask the child to show how they use their inhaler. Compare with a device-specific checklist. Adherence Is there any of the child’s prescribed maintenance medication (inhalers and/or tablets) in the home at present? On how many days in a week does the child use it (e.g., 0, 2, 4, 7 days)? Is it easier to remember to use it in the morning or evening? Where is the medication kept – is it in plain view to reduce forgetting? Check date on inhaler. Goals/concerns Does the child or their parent or caregiver have any concerns about their asthma (e.g., fear of medication, side-effects, interference with activity)? What are their goals for treatment? Comorbidities Allergic rhinitis Itching, sneezing, nasal obstruction? Can the child breathe through their nose? What medications are being taken for nasal symptoms? Eczema Sleep disturbance, topical corticosteroids? Food allergy Is the child allergic to any foods? (Confirmed food allergy is a risk factor for asthma-related death.)94 Obesity Check age-adjusted BMI. Ask about diet and physical activity. Other investigations (if needed) 2-week diary If no clear assessment can be made based on the above questions, ask the child or parent/caregiver to keep a daily diary of asthma symptoms, reliever use and peak expiratory flow (best of three) for 2 weeks. Formal exercise challenge Provides information about airway hyperresponsiveness and fitness (Box 1-2, p.26). Only perform challenge testing if it is otherwise difficult to assess asthma control. See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 41 Tools for assessing recent asthma symptom control for children aged 6–11 years In children, as in adults, assessment of asthma symptom control is based on symptoms, limitation of activities and use of rescue medication. Careful review of the impact of asthma on a child’s daily activities, including sports, play and social life, and on school absenteeism, is important. Many children with poorly controlled asthma avoid strenuous exercise so their asthma may appear to be well controlled. This may lead to poor fitness and a higher risk of obesity. Children vary considerably in the degree of airflow limitation observed before they complain of dyspnea or use their reliever therapy, and marked reduction in lung function is often seen before it is recognized by the parent or caregiver. They may report irritability, tiredness, and changes in mood in their child as the main problems when the child’s asthma is not controlled. Parents/caregivers have a longer recall period than children, who may recall only the last few days; therefore, it is important to include information from both the parent/caregiver and the child when the level of symptom control is being assessed. Several numeric tools have been developed for assessing recent asthma symptom control for children. These include: • Childhood Asthma Control Test (c-ACT)142 with separate sections for parent/caregiver and child to complete • Asthma Control Questionnaire (ACQ).143,144 Some asthma control scores for children include history of exacerbations with symptoms, but these may have the same limitations as described above for adults. They include the Test for Respiratory and Asthma Control in Kids (TRACK)145-147 and the Composite Asthma Severity Index (CASI).148 The results of these various tests correlate, to some extent, with each other and with the GINA classification of symptom control. Box 2-3 (p.40) provides more details about assessing asthma control in children. ASSESSING FUTURE RISK OF EXACERBATIONS, LUNG FUNCTION DECLINE AND ADVERSE EFFECTS The second component of assessing asthma control (Box 2-2B, p.37) is to identify whether the patient is at risk of adverse asthma outcomes, particularly exacerbations, persistent airflow limitation, and side-effects of medications (Box 2-2B). Asthma symptoms, although an important outcome for patients, and themselves a strong predictor of future risk of exacerbations, are not sufficient on their own for assessing asthma for several reasons: • Asthma symptoms can be controlled by placebo or sham treatments149,150 or by inappropriate use of short-acting SABA or long-acting beta2 agonist (LABA) alone,151 all of which leave airway inflammation untreated. • Respiratory symptoms may be due to other conditions such as lack of fitness, or comorbidities such as inducible laryngeal obstruction.58 • Anxiety or depression may contribute to higher symptom reporting. • Some patients have impaired perception of bronchoconstriction, with few symptoms despite low lung function.152 • In patients with good symptom control, exacerbations can be triggered by environmental exposures such as viral infections, allergen exposure and poor air quality. Asthma symptom control and exacerbation risk should not be simply combined numerically, as poor control of symptoms and of exacerbations may have different causes and may need different treatment approaches. Risk factors for exacerbations Poor asthma symptom control itself substantially increases the risk of exacerbations.122-124 However, several additional independent risk factors have been identified, i.e., factors that, when present, increase the patient’s risk of exacerbations even if symptoms are few. These risk factors (Box 2-2B, p.37) include a history of ≥1 exacerbation in the previous year, poor adherence, incorrect inhaler technique, chronic sinusitis and smoking, all of which can be assessed in primary care.153 The risk of severe exacerbations and mortality increases incrementally with higher SABA use, independent of treatment step.87 Prescribing of three or more 200-dose SABA inhalers in a year, corresponding to more than daily use, is associated with an increased risk of severe exacerbations86,87 and, in one study, increased mortality.87 Risk factors and comorbidities that are modifiable (or potentially modifiable) are sometimes called ‘treatable traits’.154 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 42 In children, the risk of exacerbations is greatly increased if there is a history of previous exacerbations; it is also increased with poor symptom control, suboptimal drug regimen, comorbid allergic disease and poverty.113 Risk factors for development of persistent airflow limitation The average rate of decline in FEV1 in non-smoking healthy adults is 15–20 mL/year.155 People with asthma may have an accelerated decline in lung function and develop airflow limitation that is not fully reversible. This is often associated with more persistent dyspnea. Independent risk factors that have been identified for persistent airflow limitation include exposure to cigarette smoke or noxious agents, chronic mucus hypersecretion, and asthma exacerbations in patients not taking ICS117 (see Box 2-2B, p.37). Children with persistent asthma may have reduced growth in lung function, and some are at risk of accelerated decline in lung function in early adult life.156 There is no clear evidence that treatment with ICS prevents accelerated decline in post-bronchodilator lung function, i.e., that it prevents development of persistent airflow limitation. Risk factors for medication side-effects Choices with any medication are based on the balance of benefit and risk. Most people using asthma medications do not experience any side-effects. The risk of side-effects increases with higher doses of medications, but these are needed in few patients. Systemic side-effects that may be seen with long-term, high-dose ICS include easy bruising, an increase beyond the usual age-related risk of osteoporosis and fragility fractures, cataracts, glaucoma, and adrenal suppression. Local side-effects of ICS include oral candidiasis (thrush) and dysphonia. Patients are at greater risk of ICS side-effects with higher doses or more potent formulations118,119 and, for local side-effects, with incorrect inhaler technique.120 A glossary of asthma medications has been added as an appendix at the end of this report (p.212). Drug interactions with asthma medications: concomitant treatment with cytochrome P450 inhibitors such as ketoconazole, ritonavir, itraconazole, erythromycin and clarithromycin may increase the risk of ICS adverse effects such as adrenal suppression, and with short-term use, may increase the risk of cardiovascular adverse effects of the LABAs salmeterol and vilanterol (alone or in combination with ICS). Concomitant use of these medications is not recommended (see also p.122).157 ROLE OF LUNG FUNCTION IN ASSESSING ASTHMA CONTROL Does lung function relate to other asthma control measures? Lung function does not correlate strongly with asthma symptoms in adults158 or children.159 In some asthma control tools, lung function is numerically averaged or added with symptoms,135,160 but this is not recommended because if the tool includes several symptom items, these can outweigh clinically important differences in lung function.161 In addition, low FEV1 is a strong independent predictor of risk of exacerbations, even after adjustment for symptom frequency. Lung function should be assessed at diagnosis or start of treatment, after 3–6 months of ICS-containing treatment to assess the patient’s personal best FEV1, and periodically thereafter. For example, in most adult patients, lung function should be recorded at least every 1–2 years, but more frequently in higher risk patients including those with exacerbations and those at risk of decline in lung function (see Box 2-2B, p.37). Lung function should also be recorded more frequently in children based on asthma severity and clinical course (Evidence D). Once the diagnosis of asthma has been confirmed, it is not generally necessary to ask patients to withhold their regular or as-needed medications before visits,38 but preferably the same conditions should apply at each visit. How to interpret lung function test results in asthma A low FEV1 percent predicted: • Identifies patients at risk of asthma exacerbations, independent of symptom levels, especially if FEV1 is <60% predicted96,105,162,163 • Is a risk factor for lung function decline, independent of symptom levels116 • If symptoms are few, suggests limitation of lifestyle, or poor perception of airflow limitation,164 which may be due to untreated airway inflammation.152 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 43 Normal FEV1: A ‘normal’ or near-normal FEV1 in a patient with frequent respiratory symptoms (especially when symptomatic) prompts consideration of alternative causes for the symptoms (e.g., cardiac disease, or cough due to post-nasal drip or gastroesophageal reflux disease; Box 1-3, p.27). Persistent bronchodilator responsiveness: Finding significant bronchodilator responsiveness (increase in FEV1 >12% and >200 mL from baseline)35 in a patient taking ICS-containing treatment, or who has taken a SABA within 4 hours, or a LABA within 12 hours (or 24 hours for a once-daily LABA), suggests uncontrolled asthma, particularly poor adherence and/or incorrect technique. In children, spirometry cannot be reliably obtained until age 5 years or more, and it is less useful than in adults. Many children with uncontrolled asthma have normal lung function between flare-ups (exacerbations). How to interpret changes in lung function in clinical practice With regular ICS treatment, FEV1 starts to improve within days, and reaches a plateau after around 2 months.165 The patient’s highest FEV1 reading (personal best) should be documented, as this provides a more useful comparison for clinical practice than FEV1 percent predicted. If predicted values are used in children, measure their height at each visit. Some patients may have a faster than average decrease in lung function, and develop persistent (incompletely reversible) airflow limitation. While a short-term (e.g., 3 months) trial of higher dose ICS or ICS-LABA may be appropriate to see if FEV1 can be improved, high doses should not be continued longer than this if there is no response. The between-visit variability of FEV1 (up to 12% week-to-week or 15% year-to-year in healthy individuals)35 limits its use in adjusting asthma treatment or identifying accelerated decline in clinical practice. The minimal important difference for improvement and worsening in FEV1 based on patient perception of change has been reported to be about 10%.166,167 The role of short-term and long-term lung function monitoring Once the diagnosis of asthma is made, short-term peak expiratory flow (PEF) monitoring may be used to assess response to treatment, to evaluate triggers (including at work) for worsening symptoms, or to establish a baseline for action plans. After starting ICS, personal best PEF (from twice daily readings) is reached on average within 2 weeks.168 Average PEF continues to increase, and diurnal PEF variability to decrease, for about 3 months.158,168 Excessive variation in PEF suggests suboptimal asthma control, and increases the risk of exacerbations.169 Long-term PEF monitoring is now generally only recommended for patients with severe asthma, or those with impaired perception of airflow limitation (e.g. few symptoms despite low initial lung function).152,170-173 For clinical practice, displaying PEF results on a standardized chart may improve accuracy of interpretation.174 Home spirometric monitoring has been used in some clinical trials; careful training of patients in spirometric technique is essential. Results from clinic-based and home-recorded spirometry are not interchangeable. ASSESSING ASTHMA SEVERITY The current concept of asthma severity is based on ‘difficulty to treat’ The current concept of asthma severity, recommended by an ATS/ERS Task Force38,84 and included in most asthma guidelines, is that asthma severity should be assessed retrospectively from how difficult the patient’s asthma is to treat. This is reflected by the level of treatment required to control the patient’s symptoms and exacerbations, i.e., after at least several months of treatment.38,84,175 This definition is mainly relevant to, and useful for, severe asthma. By this definition: • severe asthma is defined as asthma that remains uncontrolled despite optimized treatment with high-dose ICS-LABA, or that requires high-dose ICS-LABA to prevent it from becoming uncontrolled. Severe asthma must be distinguished from asthma that is difficult to treat due to inadequate or inappropriate treatment, or persistent problems with adherence or comorbidities such as chronic rhinosinusitis or obesity,175 as they need very different treatment compared with if asthma is relatively refractory to high-dose ICS-LABA or even oral corticosteroids COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 44 (OCS).175 See Box 2-4 (p.47) for how to distinguish difficult-to-treat asthma from severe asthma, and Section 8 (p.139) for more detail about assessment, referral and treatment in this population. • moderate asthma is asthma that is well controlled with Step 3 or Step 4 treatment e.g., with low- or medium-dose ICS LABA in either treatment track • mild asthma is asthma that is well controlled with low-intensity treatment, i.e., as needed low-dose ICS-formoterol, or low-dose ICS plus as-needed SABA. The utility of this retrospective definition of asthma severity is limited by the fact that it cannot be assessed unless good asthma control has been achieved and treatment stepped down to find the patient’s minimum effective dose at which their asthma remains well controlled (Box 4-13, p.102), or unless asthma remains uncontrolled despite at least several months of optimized maximal therapy. The terms ‘severe asthma’ and ‘mild asthma’ are often used with different meanings than this In the community and in primary care, the terms ‘severe’ or ‘mild’ asthma are more commonly based on the frequency or severity of symptoms or exacerbations, irrespective of treatment. For example, asthma is commonly called ‘severe’ if patients have frequent or troublesome asthma symptoms, regardless of their treatment, and ‘mild asthma’ is commonly used if patients do not have daily symptoms or if symptoms are quickly relieved. In epidemiological studies and clinical trials, asthma is often classified as ‘mild’, ‘moderate’ or ‘severe’ based only on the prescribed treatment by GINA or BTS Step, regardless of patients’ level of asthma control. This assumes that the prescribed treatment was appropriate for the patient’s needs, but asthma is often under-treated or over-treated. Most clinical trials of biologic therapy enroll patients with asthma that is uncontrolled despite taking medium- or high-dose ICS-LABA, but contributory factors such as incorrect inhaler technique, poor adherence, or comorbidities are rarely assessed and treated before the patient’s eligibility for enrolment is considered.176,177 Some clinical trial participants may therefore have ‘difficult-to-treat’, rather than severe asthma. Some guidelines 178,179 also retain another, older, classification of asthma severity based on symptom and SABA frequency, night waking, lung function and exacerbations before ICS-containing treatment is started.38,84 This classification also distinguishes between ‘intermittent’ and ‘mild persistent’ asthma, but this historical distinction was arbitrary: it was not evidence-based, but was based on an untested assumption that patients with symptoms ≤2 days/week were not at risk and would not benefit from ICS, so should be treated with SABA alone. However, it is now known that patients with so-called ‘intermittent’ asthma can have severe or fatal exacerbations,180,181 and that their risk is substantially reduced by ICS-containing treatment compared with SABA alone.182-184 Although this symptom-based classification is stated to apply to patients not on ICS-containing treatment,178,179 it is often used for patients taking these medications. This can cause confusion, as a patient’s asthma may be classified differently, and they may be prescribed different treatment, depending on which definition the clinician or healthcare system uses. For low-resource countries without access to effective medications such as ICS, the World Health Organization definition of severe asthma185 includes a category of ‘untreated severe asthma’. This category corresponds to uncontrolled asthma in patients not taking any ICS-containing treatment. The patient’s view of asthma severity Patients may perceive their asthma as severe if they have intense or frequent symptoms, but this does not necessarily indicate underlying severe disease, as symptoms and lung function can rapidly become well controlled with commencement of ICS-containing treatment, or improved inhaler technique or adherence.38,84 Likewise, patients often perceive their asthma as mild if they have symptoms that are easily relieved by SABA, or that are infrequent.38,84 Of concern, patients often interpret the term ‘mild asthma’ to mean that they are not at risk of severe exacerbations and do not need to take ICS-containing treatment. This is often described as patients ‘underestimating’ their asthma severity, but instead it reflects their different interpretation of the words ‘severity’ and ‘mild’ compared with the academic usage of these terms.38,84 How useful is the current retrospective definition of asthma severity? The retrospective definition of severe asthma based on ‘difficulty to treat’ has been widely accepted in guidelines and in specialist clinical practice. It has obvious clinical utility as it identifies patients who, because of their burden of disease and incomplete response to optimized conventional ICS-based treatment, may benefit from referral to a COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 45 respiratory physician (if available) for further investigation, phenotyping, and consideration of additional treatment such as biologic therapy (See Section 8, p.139). It is appropriate to classify asthma as ‘difficult-to-treat’ rather than severe if there are modifiable factors such as incorrect inhaler technique, poor adherence or untreated comorbidities, because asthma may become well controlled when such issues are addressed.38,84,175 By contrast, the clinical utility of the retrospective definition of mild asthma is much less clear. There is substantial variation in opinions about the specific criteria that should be used, for example whether FEV1 should be ≥80% predicted in order for asthma to be considered ‘mild’, and whether the occurrence of any exacerbation precludes a patient’s asthma being classified as ‘mild’ for the next 12 months.186 There are too few studies of the underlying pathology to discern whether isolated exacerbations necessarily imply greater inherent severity, especially given the contribution of external triggers such as viral infections or allergen exposure to sporadic exacerbations. Further, by this definition, asthma can be classified as ‘mild’ only after several months of ICS-containing treatment, and only if asthma is well controlled on low-dose ICS or as-needed low-dose ICS-formoterol, so this definition clearly cannot be applied to patients with uncontrolled or partly controlled symptoms who are taking SABA. Finally, retrospective classification of asthma as mild appears of little value in deciding on future treatment. In addition, in the studies of as-needed ICS-formoterol, baseline patient characteristics such as daily reliever use, lower lung function or history of exacerbations (or even baseline blood eosinophils or FeNO) did not identify patients who should instead be treated with daily ICS.187,188 Instead, decisions about ongoing treatment should be based upon the large evidence base about the efficacy and effectiveness of as-needed ICS-formoterol or daily ICS, together with an individualized assessment of the patient’s symptom control, exacerbation risk, predictors of response, and patient preferences (see Box 3-3, p.53). However, the most urgent problem with the term ‘mild asthma’, regardless of how it is defined, is that it encourages complacency, since both patients and clinicians often interpret ‘mild asthma’ to mean that the patient is at low risk and does not need ICS-containing treatment. However, up to 30% of asthma exacerbations and deaths occur in people with infrequent symptoms, for example, less than weekly or only on strenuous exercise.180,181 Interim advice about asthma severity descriptors For clinical practice GINA continues to support the current definition of severe asthma as asthma that remains uncontrolled despite optimized treatment with high-dose ICS-LABA, or that requires high-dose ICS-LABA or biologic therapy to prevent it from becoming uncontrolled. GINA also maintains the clinically important distinction between difficult-to-treat and severe asthma. See Box 2-4 (p.47) and Section 8 (p.139) for more detail about assessment and management of difficult-to-treat and severe asthma. For patients who have had a good asthma response to biologic therapy, it may be helpful for administrative reasons to describe their asthma as, e.g., ‘severe eosinophilic asthma, well controlled on [therapy]’, to indicate that the biologic therapy is needed to maintain their improved status. For discussion about the related concept of asthma remission on treatment, see p.50. We suggest that in clinical practice, the term ‘mild asthma’ should generally be avoided if possible, because of the common but mistaken assumption by patients and clinicians that it equates to low risk, and that ICS treatment is not needed. Instead, assess each patient’s symptom control and risk factors on their current treatment (Box 2-1, p.36), as well as multimorbidity and patient goals and preferences. Explain that patients with infrequent or mild asthma symptoms can still have severe or fatal exacerbations if treated with SABA alone,180,181 and that this risk is reduced by half to two-thirds with low-dose ICS or with as-needed low-dose ICS formoterol.182,183 Ensure that you prescribe ICS-containing therapy to reduce the patient’s risk of severe exacerbations (Box 4-3, p.74), and treat any modifiable risk factors or comorbidities using pharmacologic or non-pharmacologic strategies (see Box 3-5, p.55 and Box 3-6, p.57). ‘Mild asthma’ is a retrospective label, so it cannot be used to decide which treatment patients should receive. Advice has been provided in Section 4 about which patients are suitable for low intensity treatment (Step 1 and 2). For health professional education The term ‘apparently mild asthma’ may be useful to highlight the discordance between symptoms and risk, i.e., that patients with infrequent or mild symptoms, who might therefore appear to have mild asthma, can still have severe or fatal exacerbations. However, ‘apparently mild asthma’ in English can easily be mistranslated into some languages as COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 46 ‘obviously mild asthma’, which is the opposite of the intended meaning. Alternative phrases include ‘asthma that seems to be mild’. Regardless of the term used, explain that ‘asthma control’ tools such as ACQ and ACT assess only one domain of asthma control, and only over a short period of time (see Assessing asthma symptom control, p.38), and that patients with infrequent interval symptoms are over-represented in studies of severe, near-fatal and fatal asthma exacerbations.180,181 Always emphasize the need for and benefit from ICS-containing treatment in patients with asthma, regardless of their symptom frequency or severity, and even if they have no obvious additional risk factors. For epidemiologic studies If clinical details are not available, describe the prescribed (or dispensed) treatment, without imputing severity, e.g., ‘patients prescribed SABA with no ICS’ rather than ‘mild asthma’. Since treatment options change over time, and may differ between guidelines, state the actual treatment class, rather than a treatment Step (e.g., ‘low-dose maintenance-and-reliever therapy with ICS-formoterol’ rather than ‘Step 3 treatment’). For clinical trials Describe the patient population by their level of asthma control and treatment, e.g., ‘patients with uncontrolled asthma despite medium-dose ICS-LABA plus as-needed SABA’ rather than ‘moderate asthma’. Further discussion is clearly needed Given the importance of mild asthma and the discordance between its current academic definition and the various ways that the term is used in clinical practice, GINA is continuing to discuss these issues with a wide range of stakeholders. The aim is to obtain agreement among patients, health professionals, researchers, industry and regulators about the implications for clinical practice and clinical research of current knowledge about asthma pathophysiology and treatment,38,84 and whether/how the term ‘mild asthma’ should be used in the future. Pending the outcomes of this discussion, no change has been made to use of the term ‘mild asthma’ elsewhere in this GINA Strategy Report. HOW TO DISTINGUISH BETWEEN UNCONTROLLED ASTHMA AND SEVERE ASTHMA Although good symptom control and minimal exacerbations can usually be achieved with ICS-containing treatment, some patients will not achieve one or both of these goals even with a long period of high-dose therapy.160,175 In some patients this is due to truly refractory severe asthma, but in many others, it is due to incorrect inhaler technique, poor adherence, over-use of SABA, comorbidities, persistent environmental exposures, or psychosocial factors. It is important to distinguish between severe asthma and uncontrolled asthma, because lack of asthma control is a much more common reason for persistent symptoms and exacerbations, and may be more easily improved. Box 2-4 (p.47) shows the initial steps that can be carried out in primary care to identify common causes of uncontrolled asthma. More details are given in Section 8 (p.139) about investigation and management of difficult-to-treat and severe asthma, including referral to a respiratory physician or severe asthma clinic where possible, and use of add-on treatment including biologic therapy. The most common problems that need to be excluded before making a diagnosis of severe asthma are: • Poor inhaler technique (up to 80% of community patients)91 (Box 5-2, p.110) • Poor medication adherence189,190 (Box 5-3, p.112) • Incorrect diagnosis of asthma, with symptoms due to alternative conditions such as inducible laryngeal obstruction, cardiac failure or lack of fitness (Box 1-3, p.27) • Multimorbidity such as rhinosinusitis, GERD, obesity and obstructive sleep apnea93,191 (Section 6, p.117) • Ongoing exposure to sensitizing or irritant agents in the home or work environment, including tobacco smoke. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 47 Box 2-4. Investigating poor symptom control and/or exacerbations despite treatment See list of abbreviations (p.11). See Section 8 (p.139) for more details about assessment and management of difficult-to-treat and severe asthma. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 48 3. Principles of asthma management in adults, adolescents and children 6–11 years KEY POINTS The patient-health professional partnership • Effective asthma management requires a partnership between the person with asthma (or the parent/caregiver) and their healthcare providers. • Teaching communication skills to healthcare providers may lead to increased patient satisfaction, better health outcomes, and reduced use of healthcare resources. • The patient’s ability to obtain, process and understand basic health information to make appropriate health decisions (‘health literacy’) should be considered. Goals of asthma management The GINA goal of asthma management is to achieve the best possible long-term outcomes for the individual patient. This may include good long-term symptom control (few/no asthma symptoms, no sleep disturbance due to asthma, and unimpaired physical activity), and minimized long-term risk of asthma-related mortality, exacerbations, persistent airflow limitation and side-effects of treatment. The patient’s own goals should also be identified. Remission of asthma • Remission of asthma can be identified in children and in adults, either clinical remission or complete remission, and either off-treatment or on-treatment. Definitions and criteria vary. • The concept of clinical remission on treatment is consistent with the long-term goal of asthma management promoted by GINA, to achieve the best possible long-term asthma outcomes for each patient. • Research among patients who have (or have not) experienced clinical or complete remission of asthma, either off-treatment or on-treatment, provides important opportunities for understanding underlying mechanisms of asthma, to develop new approaches to asthma prevention and management. This will be facilitated by using standardized criteria and assessment tools. • Take care if using the term ‘remission’ in conversations with patients or parents/caregivers, as they may assume it means a cure, or may associate it with cancer or leukemia. Explain what you mean, and that if asthma symptoms have gone quiet for a while, they may recur. Making decisions about asthma treatment • Asthma treatment is adjusted in a continual cycle of assessment, treatment, and review of the patient’s response in both symptom control and future risk (of exacerbations and side-effects), and of patient preferences. • For population-level decisions about asthma medications, e.g., national guidelines, insurers, health maintenance organizations or national formularies, the ‘preferred’ regimens in Steps 1–4 represent the best treatments for most patients, based on evidence from randomized controlled trials, meta-analyses and observational studies about safety, efficacy and effectiveness, with a particular emphasis on symptom burden and exacerbation risk. For Steps 1–5, there are different preferred population-level recommendations for different age-groups (adults/adolescents, children 6–11 years, children 5 years and younger). In Step 5, there are also different preferred population-level recommendations depending on the inflammatory phenotype, Type 2 or non-Type 2. • For individual patients, shared decision-making about treatment should also consider any patient characteristics or phenotype or environmental exposures that predict the patient’s risk of exacerbations or other adverse outcomes, or their likely response to treatment, together with the patient’s goals or concerns and practical issues (inhaler technique, adherence, medication access and cost to the patient). • Optimize asthma management, including inhaled therapy and non-pharmacologic strategies, to reduce the need for oral corticosteroids (OCS) and their multiple associated adverse effects. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 49 THE PATIENT–HEALTHCARE PROVIDER PARTNERSHIP Effective asthma management requires the development of a partnership between the person with asthma (or the parent/caregiver) and healthcare providers.192 This should enable the person with asthma to gain the knowledge, confidence and skills to assume a major role in the management of their asthma. Self-management education reduces asthma morbidity in both adults193 (Evidence A) and children194 (Evidence A). There is emerging evidence that shared decision-making is associated with improved outcomes.195 Patients and caregivers should be encouraged to participate in decisions about treatment, and given the opportunity to express their expectations and concerns. This partnership needs to be individualized for each patient. A person’s willingness and ability to engage in self-management may vary depending on factors such as ethnicity, literacy, understanding of health concepts (health literacy), numeracy, beliefs about asthma and medications, desire for autonomy, and the healthcare system. Good communication Good communication by healthcare providers is essential as the basis for good outcomes (Evidence B).196-198 Teaching healthcare providers to improve their communication skills (Box 3-1) can result in increased patient satisfaction, better health outcomes, and reduced use of healthcare resources196-198 without lengthening consultation times.199 It can also enhance patient adherence.199 Training patients to give information clearly, seek information, and check their understanding of information provided is also associated with improved adherence with treatment recommendations.199 Box 3-1. Communication strategies for healthcare providers Key strategies to facilitate good communication197,198 • A congenial demeanor (friendliness, humor and attentiveness) • Allowing the patient to express their goals, beliefs and concerns • Empathy, reassurance, and prompt handling of any concerns • Giving encouragement and praise • Giving appropriate (personalized) information • Providing feedback and review How to reduce the impact of low health literacy200 • Order information from most to least important. • Speak slowly and use simple words (avoid medical language, if possible). • Simplify numeric concepts (e.g., use numbers instead of percentages). • Frame instructions effectively (use illustrative anecdotes, drawings, pictures, table or graphs). • Confirm understanding by using the ‘teach-back’ method (ask patients to repeat instructions). • Ask a second person (e.g., nurse, family member) to repeat the main messages. • Pay attention to non-verbal communication by the patient. • Make patients feel comfortable about asking questions. Health literacy and asthma There is increasing recognition of the impact of low health literacy on health outcomes, including in asthma.200,201 Health literacy means much more than the ability to read: it is defined as ‘the degree to which individuals have the capacity to obtain, process and understand basic health information and services to make appropriate health decisions’.200 Low health literacy is associated with reduced knowledge and worse asthma control.202 In one study, low numeracy among parents of children with asthma was associated with higher risk of exacerbations.201 Interventions adapted for cultural and ethnicity perspectives have been associated with improved knowledge and significant improvements in inhaler technique.203 Suggested communication strategies for reducing the impact of low health literacy are shown in Box 3-1 (p.49). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 50 LONG-TERM GOAL OF ASTHMA MANAGEMENT The long-term goal of asthma management from a clinical perspective is to achieve the best possible outcomes for the patient, including long-term symptom control and long-term asthma risk minimization (Box 3-3, p.53). This includes preventing exacerbations, accelerated decline in lung function, and medication adverse effects. At a population level, the goals of asthma management also include minimizing asthma deaths, urgent health care utilization, and the socioeconomic impacts of uncontrolled asthma. It is also important to elicit the patient’s (or parent/caregiver’s) goals regarding their asthma, as these may differ from medical goals. Shared goals for asthma management can be achieved in various ways, with consideration of differing healthcare systems, medication availability, and cultural and personal preferences. Box 3-2. Long-term goal of asthma management The goal of asthma management is to achieve the best possible long-term asthma outcomes for the patient: • Long-term asthma symptom control, which may include: - Few/no asthma symptoms - No sleep disturbance due to asthma - Unimpaired physical activity • Long-term asthma risk minimization, which may include: - No exacerbations - Improved or stable personal best lung function - No requirement for maintenance systemic corticosteroids - No medication side-effects. The patient’s goals for their asthma may be different from these medical goals; ask the patient what they want from their asthma treatment. When discussing the best possible asthma outcomes with a patient, consider their goals, their asthma phenotype, clinical features, multimorbidity, risk factors (including severity of airflow limitation), practical issues including the availability and cost of medications, and the potential adverse effects of treatment (Box 3-4, p.54). Assessing symptom control is NOT enough: the patient’s risk factors (Box 2-2B, p.37), including history of exacerbations, should always also be assessed. Symptom control and risk may be discordant: patients with few or no symptoms can still have severe or fatal exacerbations, including from external triggers such as viral infections, allergen exposure (if sensitized) or pollution. REMISSION OF ASTHMA Remission of asthma has been investigated extensively in the past, most commonly remission of childhood asthma off treatment. Definitions and criteria vary, but they commonly refer to either clinical remission (e.g., no asthma symptoms or exacerbations for a specific period) or complete (or pathophysiological) remission (e.g., also including normal lung function, airway responsiveness and/or inflammatory markers). There has been interest in remission off treatment, and remission on treatment, for example with biologic therapy for severe asthma.204-206 The concept of clinical remission on treatment is consistent with the long-term goal of asthma management promoted by GINA, which is to achieve the best possible long-term asthma outcomes for the patient (see Box 3-2, p.50). When discussing the best possible outcomes with a patient, consider their own asthma goals, their asthma phenotype, clinical features, multimorbidity, risk factors (including severity of airflow limitation), practical issues including the availability and cost of medications, and the potential adverse effects of treatment (Box 3-4, p.54). Research in patients who have (or have not) experienced clinical or complete remission of asthma, either off treatment or on treatment, provides important opportunities for understanding the heterogeneous and interconnected underlying mechanisms of asthma, and for developing new approaches to asthma prevention and management. This will be facilitated by using standardized criteria and tools. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 51 Remission of childhood asthma Reported rates of remission off treatment from studies in children with wheezing or asthma vary depending on the populations, definitions, and length of follow-up. For example, in one study, 59% of wheezing preschool children had no wheezing at 6 years,207 whereas in another study, only 15% of children with persistent wheezing at/after 9 years had no wheezing at 26 years.208 Clinical remission is more frequent than pathophysiological remission at all ages.209,210 The most important predictors of asthma remission in school-aged children are fewer, milder or decreasing frequency of symptomatic episodes,211-214 good or improving lung function, and less airway hyperresponsiveness.210 Risk factors for persistence of childhood asthma include atopy, parental asthma/allergy, later onset of symptoms, wheezing without colds, and maternal smoking or tobacco smoke exposure. Remission is not cure: after remission in childhood or adolescence, asthma often recurs later in life. Children whose asthma has remitted have an increased risk of accelerated lung decline in adulthood, independent from, but synergistic with, tobacco smoking; and they may develop persistent airflow limitation, although this is less likely than for those whose asthma has persisted.215 This suggests the importance of monitoring lung function in people with remission of asthma symptoms. To date, there is no evidence that interventions in childhood increase the likelihood of remission of asthma or reduce the risk of recurrence. However, treatment of asthma in childhood with inhaled corticosteroid (ICS) substantially reduces the burden of asthma on the child and family, reduces absence from school and social events, reduces the risk of exacerbations and hospitalizations, and allows the child to participate in normal physical activity. Parents/caregivers often ask if their child will grow out of their asthma, and will not need treatment in the future. Current consensus supports the following advice for discussions like these: • If the child has no reported symptoms, check for evidence of ongoing disease activity (e.g., wheezing; child avoiding physical activity), and check lung function if testing is available. • Use a description like ‘asthma has gone quiet for the present’ to help avoid misunderstandings. If you use the term ‘remission’ with parents/caregivers, explain the medical meaning, because it is often interpreted as meaning a permanent cure. • Advise parents/caregivers that, even if the child’s symptoms resolve completely, their asthma may recur later. • Emphasize the benefits of taking controller treatment for the child’s current health, their risk of asthma attacks, and their ability to participate in school and sporting activities, while avoiding claims about effect of therapy on future asthma outcomes. Research needs: clinical questions about remission off treatment in children focus on the risk factors for asthma persistence and recurrence (including clinical, pathological, and genetic factors), the effect of risk reduction strategies on the likelihood of remission, whether monitoring after remission to allow early identification of asthma recurrence improves outcomes, and whether progression to persistent airflow limitation can be prevented. Clinical questions about remission on treatment (e.g., in children with severe asthma treated with biologic therapy) include investigating whether inhaled anti-inflammatory therapy can be down-titrated. Remission of adult asthma Clinical or complete remission off treatment has been observed in some adults, either spontaneously or after cessation of controller treatment. For example, 15.9% of patients with adult-onset asthma experienced clinical remission (no asthma symptoms and no asthma medications) within 5 years.25 Remission is sometimes seen in people with occupational asthma after cessation of exposure.216 Clinical remission of asthma in adult life is more common with childhood-onset asthma than adult-onset asthma. However, persistence of airway hyperresponsiveness and/or airway inflammation is found in most adults with clinical remission of asthma.209 In recent years, there has been increasing interest in asthma remission on treatment, particularly with biologic therapy for severe asthma. Various definitions have been proposed. For clinical remission, these often include criteria such as no asthma symptoms, no exacerbations, no use of OCS, and stable or improving lung function, over a defined prolonged period. For complete remission, normalization of airway responsiveness and/or inflammatory markers has been proposed. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 52 For patients with severe asthma treated with biological therapy and medium- or high-dose ICS in combination with a long-acting beta2 agonist (LABA), remission rates will vary depending on the baseline characteristics of the populations studied and the criteria for and duration of, remission (including how ‘no symptoms’ is assessed).204-206,217, 218 Baseline predictors of remission on treatment with various biologic therapies for severe asthma include better short-term asthma symptom control scores (ACT or ACQ), better lung function, fewer comorbidities, earlier asthma onset, and no or lower maintenance OCS use at baseline.206,218 In a study of clinical remission off treatment of adult-onset asthma, the only baseline predictors of clinical persistence were moderate-to-severe airway hyperresponsiveness and nasal polyps.25 Although clinical asthma remission on treatment has been most extensively investigated in adults with severe asthma treated with biologics, the concept is relevant to patients with asthma of any severity and any treatment, including ICS-containing therapy, oral pharmacotherapies, allergen immunotherapy and non-pharmacological interventions (e.g., lifestyle interventions). In the lay media, the word ‘remission’ is most often heard in association with cancer or leukemia, so if it is used in discussion with patients, the medical meaning for asthma should be explained. If the patient experiences clinical remission, explain that this does not mean permanent cure, and that they should not stop taking any of their asthma medications except on medical advice. Research needs: for asthma remission on treatment in adults include the association between clinical criteria with biomarkers, imaging, or pathology samples (including for ‘omics’ analysis) that may reflect the underlying disease processes, and investigation of predictors of long-term remission or recurrence. The framework for validating proposed criteria for remission on treatment will depend on their intended purpose, for example as an assessment tool in clinical practice, for prognosis of continued long-term stability, or for identifying new targets for therapy. Clinical and qualitative research with a range of treatments is needed to know whether aiming for remission will improve long-term outcomes for patients with asthma. PERSONALIZED CONTROL-BASED ASTHMA MANAGEMENT Asthma control has two domains: symptom control and risk reduction (see Box 2-2, p.37). In control-based asthma management, pharmacological and non-pharmacological treatment is adjusted in a continual cycle that involves assessment of symptom control and risk factors, treatment and review by appropriately trained personnel (Box 3-3, p.53) to achieve the goals of asthma treatment (Section 3, p.48). Asthma outcomes have been shown to improve after the introduction of control-based guidelines219,220 or practical tools for implementation of control-based management strategies.195,221 The concept of control-based management is also supported by the design of most randomized controlled medication trials, in which patients are identified for a change in asthma treatment based on features of poor symptom control with or without other risk factors such as low lung function or a history of exacerbations. Since 2014, GINA asthma management has focused not only on asthma symptom control, but also on personalized management of the patient’s modifiable risk factors for exacerbations, other adverse outcomes and multimorbidity, while also considering the patient’s preferences and goals. Non-modifiable risk factors, such as a history of past ICU admission, should also be documented. For many patients in primary care, achieving good symptom control is a good guide to a reduced risk of exacerbations.222 When ICSs were introduced into asthma management, large improvements were observed in symptom control and lung function, and exacerbations and asthma-related mortality also decreased. However, patients with few or intermittent symptoms may be still at risk of severe exacerbations182 (Box 2-2B, p.37). In addition, some patients continue to have exacerbations despite well-controlled symptoms, and for patients with ongoing symptoms, side-effects may be an issue if ICS doses continue to be stepped up. Therefore, in control-based management, both domains of asthma control (symptom control and future risk; Box 2-2, p.37) should be considered when choosing asthma treatment and reviewing the response.38,84 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 53 Box 3-3. The asthma management cycle for personalized asthma care Personalized asthma management involves a continual cycle of assessment, adjustment of treatment and review (Box 3-2, p.50): • ASSESS the patient’s symptom control and their risk factors for exacerbations, for decline in lung function and for medication adverse effects (Box 2-2, p.37), with particular attention to inhaler technique and adherence. Assess comorbidities and the patient’s goals and preferences, and confirm the diagnosis of asthma if not yet done. • ADJUST the patient’s management, based on these assessments. This includes treatment of modifiable risk factors (Box 3-5, p.55) and comorbidities (Section 6, p.117), relevant non-pharmacologic strategies (Box 3-6, p.57), education and skills training (Section 5, p.108), and adjustment of medication as required (Section 4, p.67). For adults and adolescents, the preferred controller and reliever treatment across all steps is with combination ICS formoterol, as shown in GINA Track 1 (Box 4-6, p.77). • REVIEW the patient in line with the goals of treatment (Box 3-2, p.50), reassess factors affecting symptoms, risk of adverse outcomes and patient satisfaction, arrange further investigations if needed, and readjust treatment if needed. See list of abbreviations (p.11). Choosing between asthma treatment options At each treatment step in asthma management, different medication options are available that, although not of identical efficacy, may be alternatives for controlling asthma. Different considerations apply to recommendations or choices made for broad populations compared with those for individual patients (Box 3-4, p.54): • Population-level medication choices: Population-level medication choices are often applied by bodies such as national formularies or managed care organizations. Population-level recommendations aim to represent the best option for most patients in the particular population. At each treatment step, ‘preferred’ controller and reliever regimens are recommended that provide the best benefit-to-risk ratio for both symptom control and risk reduction. Choice of the preferred controller and/or preferred reliever is based on evidence from efficacy studies (highly controlled studies in well-characterized populations) and effectiveness studies (from pragmatically controlled studies, or studies in broader populations, or strong observational data),223 with a particular focus on symptoms and exacerbation risk. Safety and relative cost are also considered. In Step 5, there are different population-level recommendations depending on the inflammatory phenotype, Type 2 or non-Type 2. • In the treatment figure for adults and adolescents (Box 4-6, p.77), the options are shown in two ‘tracks’. Track 1, with as-needed low-dose ICS-formoterol as the reliever, is the preferred approach for most patients, based on evidence of overall lower exacerbation risk and similar symptom control, and a simpler regimen for stepping COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 54 treatment up and down as needed, compared with treatments in Track 2 in which the reliever is short-acting beta2 agonist (SABA) or, in some cases, combination ICS-SABA (for more details, see Section 4, p.67). • Patient-level medication choices: Treatment choices for individual patients also take into account any patient characteristics or phenotype, or any environmental exposures, that may predict their risk of exacerbations or other adverse outcomes, or a clinically important difference in their response compared with other patients, together with assessment of multimorbidity, the patient’s goals and preferences, and practical issues such as cost, ability to use the medication and adherence (see Box 3-3, p.53). For factors guiding the choice of inhaler, see Section 5 (p.108).
The extent to which asthma treatment can be individualized according to patient characteristics or phenotypes depends on the health system, the clinical context, the potential magnitude of difference in outcomes, cost and available resources. Box 3-4. Population-level versus patient-level decisions about asthma treatment Choosing between treatment options at a population level (e.g., national formularies, health maintenance organizations, national guidelines) The ‘preferred’ medication at each step is the best treatment for most patients, based on: • Efficacy • Effectiveness • Safety • Availability and cost at the population level.
For Steps 1–5, there are different population-level recommendations by age-group (adults/adolescents, children 6– 11 years, children 5 years and younger). In Step 5, there are also different population-level recommendations depending on the inflammatory phenotype, Type 2 or non-Type 2. Choosing between controller options for individual patients Use shared decision-making with the patient or parent/caregiver to discuss the following: 1.
Preferred treatment (as above) based on evidence for symptom control and risk reduction 2.
Patient characteristics or phenotype: • Does the patient have any features that predict differences in their future risk or treatment response, compared with other patients (e.g., smoker; history of exacerbations, blood eosinophilia or high FeNO; environmental exposures)? (Box 2-2B, p.37) • Are there any modifiable risk factors or multimorbidity that may affect treatment outcomes? (Box 2-2B, p.37) 3.
Patient views: • What are the patient’s goals, beliefs and concerns about asthma and medications?
4.
Practical issues: • For the preferred controller and reliever, which inhaler(s) are available to the patient?
• Inhaler technique – can the patient use the inhaler correctly after training?
• Adherence – how often is the patient likely to take the medication?
• Cost to patient – can the patient afford the medication?
• Which of the available inhalers has the lowest environmental impact? (see p.108).
Mainly based on evidence about symptoms and exacerbations (from randomized controlled trials, pragmatic studies and strong observational data) COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 55 Minimizing adverse effects of medication Reduce the potential for local and/or systemic side-effects of inhaled medications by: • Ensuring correct inhaler technique (Box 5-2, p.110) • Reminding patients to rinse and spit out after using ICS, and, after good asthma control has been maintained for 3 months • Finding each patient’s minimum effective dose of ICS-containing therapy (the lowest dose that will, in conjunction with an action plan, maintain good symptom control and minimize exacerbations, Box 4-13, p.102) • Checking for drug interactions particularly with cytochrome P450 inhibitors (see Risk factors for medication side-effects, p.42).
To reduce the need for OCS, with its multiple cumulative adverse effects,224,225 optimize inhaled therapy, including switching treatment to GINA Track 1 with anti-inflammatory reliever therapy (if available). Anti-inflammatory reliever treatment alone (AIR-only) markedly reduces the risk of severe exacerbations requiring OCS compared with SABA alone, and maintenance-and-reliever therapy (MART) with ICS-formoterol reduces the risk of severe exacerbations requiring OCS compared with the same or higher dose of ICS or ICS-LABA, or compared with usual care.226 Treating modifiable risk factors (Box 3-5, p.55) and comorbidities (Section 6, p.117) may also reduce the risk of exacerbations and use of OCS (Box 9-3, p.165). Managing other modifiable risk factors Some patients continue to experience exacerbations even with maximal doses of current treatment. Having even one exacerbation increases the risk that a patient will have another within the next 12 months.112 There is increasing research interest in identifying at-risk patients (Box 2-2B, p.37), and in investigating new strategies to further reduce exacerbation risk. In clinical practice, exacerbation risk can be reduced both by optimizing asthma medications, and by identifying and treating modifiable risk factors (Box 3-5, p.55). Not all risk factors require or respond to a step up in controller treatment. Box 3-5. Treating potentially modifiable risk factors to reduce exacerbations and minimize OCS use Risk factor Treatment strategy Evidence Any patient with one or more risk factors for exacerbations (including poor symptom control) Ensure patient is prescribed an ICS-containing treatment. A Switch to a regimen with an anti-inflammatory reliever (ICS-formoterol or ICS-SABA) if available, as this reduces the risk of severe exacerbations compared with if the reliever is SABA. A Ensure patient has a written action plan appropriate for their health literacy. A Review patient more frequently than low-risk patients. A Check inhaler technique and adherence frequently; correct as needed. A Identify and manage any modifiable risk factors (Box 2-2, p.37). D ≥1 severe exacerbation in last year Switch to a regimen with an anti-inflammatory reliever (as-needed ICS-formoterol or ICS-SABA) if available, as this reduces the risk of severe exacerbations compared with if the reliever is SABA. A Consider stepping up treatment if no modifiable risk factors. A Identify any avoidable triggers for exacerbations. C Exposure to tobacco smoke or e-cigarettes Encourage smoking cessation by patient/family; provide advice and resources (see Box 3-6, p.57). A Consider higher dose of ICS if asthma poorly controlled. B COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 56 Box 3-5 (continued). Treating potentially modifiable risk factors Risk factor Treatment strategy Evidence Low FEV1, especially if <60% predicted Address problems with adherence and inhaler technique A Consider trial of 3 months’ treatment with high-dose ICS. B Exclude other lung disease, e.g., COPD. D Refer for expert advice if no improvement. D Obesity Provide strategies for weight reduction B Distinguish asthma symptoms from symptoms due to deconditioning, mechanical restriction, and/or sleep apnea. D Major psychological problems Arrange mental health assessment. D Help patient to distinguish between symptoms of anxiety and asthma; provide advice about management of panic attacks. D Major socioeconomic problem Identify most cost-effective ICS-based regimen based on local costs. D Optimize inhaler technique to maximize benefit from available medications. D Confirmed food allergy Appropriate food avoidance; anaphylaxis action plan; injectable epinephrine; refer for expert advice. A Occupational or domestic exposure to irritants Remove from exposure as soon as possible. A Refer for expert advice as soon as possible. D Allergen exposure if sensitized Consider trial of simple avoidance strategies if there is evidence for their effectiveness (see p.61); consider cost. C Consider step up of asthma treatment if exposure is unavoidable. D Consider adding SLIT in symptomatic HDM-sensitive adults or adolescents with partly-controlled asthma despite ICS, provided FEV1 is >70% predicted. A Sputum eosinophilia despite medium/high ICS (few centers) Increase ICS dose, independent of level of symptom control. A See list of abbreviations (p.11). Based on evidence from relatively small studies in selected populations. Also see Box 3-6 (p.57) and Non-pharmacological strategies. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 57 NON-PHARMACOLOGICAL STRATEGIES In addition to pharmacological treatments, other strategies should be considered where relevant, to assist in improving symptom control and/or reducing future risk. The advice and evidence level are summarized in Box 3-6, with more detail on the following pages. Box 3-6. Non-pharmacological interventions – summary (see following text for details) Intervention Advice/recommendation Evidence Cessation of smoking, environmental tobacco exposure (ETS) and vaping • At every visit, strongly encourage people with asthma who smoke or vape to quit. Provide access to counseling and smoking cessation programs (if available).
A • Advise parents/caregivers of children with asthma not to smoke or vape, and not to allow smoking or vaping in rooms or cars that their children use.
A • Strongly encourage people with asthma to avoid environmental smoke exposure.
B • Assess smokers/ex-smokers for COPD or overlapping features of asthma and COPD (asthma+COPD, Section 7, p.131), as additional treatment strategies may be required.
D Physical activity • Encourage people with asthma to engage in regular physical activity for its general health benefits.
A • Provide advice about prevention of exercise-induced bronchoconstriction with low-dose ICS-formoterol used as needed and before exercise, or with regular daily ICS.
A/B • Provide advice about prevention of breakthrough exercise-induced bronchoconstriction with: • warm-up before exercise • SABA (or ICS-SABA) before exercise • low-dose ICS-formoterol before exercise (see Box 4-8, p.84).
A A B • Regular physical activity improves cardiopulmonary fitness, and can have a small benefit for asthma control and lung function, including with swimming in young people with asthma.
B • Physical activity interventions in adults with moderate/severe asthma is associated with improved symptoms and quality of life.
A • There is little evidence to recommend one form of physical activity over another for people with asthma.
D Pulmonary rehabilitation programs • Structured outpatient pulmonary rehabilitation programs can improve functional exercise capacity (6-minute walk) and quality of life.
A Avoidance of occupational or domestic exposures to allergens or irritants • Ask all patients with adult-onset asthma about their work history and other exposures to irritant gases or particles, including at home.
D • In management of occupational asthma, identify and eliminate occupational sensitizers as soon as possible, and remove sensitized patients from any further exposure to these agents.
A • Patients with suspected or confirmed occupational asthma should be referred promptly for expert assessment and advice, if available.
A COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 58 Box 3-6 (continued). Non-pharmacological interventions – summary Intervention • Advice/recommendation Evidence Avoidance of medications that may make asthma worse • Always ask about asthma before prescribing NSAIDs, and advise patients to stop using them if asthma worsens. D • Always ask people with asthma about concomitant medications.
D • Aspirin and NSAIDs (non-steroidal anti-inflammatory drugs) are not generally contraindicated unless there is a history of previous reactions to these agents (see p.128).
A • Decide about prescription of oral or ophthalmic beta-blockers on a case-by-case basis. Initiate treatment under close medical supervision by a specialist.
D • If cardioselective beta-blockers are indicated for acute coronary events, asthma is not an absolute contra-indication, but the relative risks/benefits should be considered.
D Healthy diet • Encourage patients with asthma to consume a diet high in fruit and vegetables for its general health benefits.
A Avoidance of indoor allergens • Allergen avoidance is not recommended as a general strategy in asthma.
A • For sensitized patients, there is limited evidence of clinical benefit for asthma in most circumstances with single-strategy indoor allergen avoidance.
A • Remediation of dampness or mold in homes reduces asthma symptoms and medication use in adults.
A • For patients sensitized to house dust mite and/or pets, there is limited evidence of clinical benefit for asthma with avoidance strategies (only in children).
B • Allergen avoidance strategies are often complicated and expensive, and there are no validated methods for identifying those who are likely to benefit.
D Weight reduction • Include weight reduction in the treatment plan for obese patients with asthma.
B • For obese adults with asthma a weight reduction program plus twice-weekly aerobic and strength exercises is more effective for symptom control than weight reduction alone.
B • The greatest improvement in asthma outcomes with weight reduction is seen with bariatric surgery.
A Breathing exercises • Breathing exercises may be a useful supplement to asthma pharmacotherapy for symptoms and quality of life, but they do not reduce exacerbation risk or have consistent effects on lung function.
A Avoidance of indoor air pollution • Encourage people with asthma to use non-polluting heating and cooking sources, and for sources of pollutants to be vented outdoors where possible.
B Avoidance of outdoor allergens • For sensitized patients, when pollen and mold counts are highest, closing windows and doors, remaining indoors, and using air conditioning may reduce exposure to outdoor allergens.
D COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 59 Box 3-6 (continued). Non-pharmacological interventions - summary Dealing with emotional stress • Encourage patients to identify goals and strategies to deal with emotional stress if it makes their asthma worse.
D • There is insufficient evidence to support one stress-reduction strategy over another, but relaxation strategies and breathing exercises may be helpful.
B • Arrange a mental health assessment for patients with symptoms of anxiety or depression.
D Addressing social risk • In US studies, comprehensive social risk interventions were associated with reduced emergency department visits and hospitalizations for children.
Studies from other countries and settings are needed.
A Avoidance of outdoor air pollutants/weather conditions • During unfavorable environmental conditions (very cold weather or high air pollution) it may be helpful, if feasible, to stay indoors in a climate-controlled environment, and to avoid strenuous outdoor physical activity; and to avoid polluted environments during viral infections, if feasible.
D Avoidance of foods and food chemicals • Food avoidance should not be recommended unless an allergy or food chemical sensitivity has been clearly demonstrated, usually by carefully supervised oral challenges.
D • For patients with confirmed food allergy, refer for specialist advice if available.
D • For patients with confirmed food allergy, food allergen avoidance may reduce asthma exacerbations.
D • If food chemical sensitivity is confirmed, complete avoidance is not usually necessary, and sensitivity often decreases when asthma control improves.
D See list of abbreviations (p.11). Interventions with highest level evidence are shown first. Cessation of smoking and vaping and avoidance of environmental tobacco smoke Cigarette smoking has multiple deleterious effects in people with established asthma, in addition to its other well-known effects such as increased risk of lung cancer, chronic obstructive pulmonary disease (COPD) and cardiovascular disease; and, with exposure in pregnancy, increased risk of asthma and lower respiratory infections in children. In people with asthma (children and adults), exposure to environmental tobacco smoke increases the risk of hospitalization and poor asthma control. Active smoking is associated with increased risk of poor asthma control, hospital admissions and, in some studies, death from asthma; increased rate of decline of lung function and may lead to COPD; and reduced the effectiveness of inhaled and oral corticosteroids.227 After smoking cessation, lung function improves and airway inflammation decreases.228 Reduction of environmental tobacco smoke exposure improves asthma control and reduces hospital admissions in adults and children.229 Use of e-cigarettes (vaping) is associated with an increased risk of asthma symptoms or diagnosis and with an increased risk of asthma exacerbations.97,230 Advice • At every visit, strongly encourage people with asthma who smoke to quit. They should be provided with access to counseling and, if available, to smoking cessation programs (Evidence A).
• Strongly encourage people with asthma who vape to quit.
• Strongly encourage people with asthma to avoid environmental smoke exposure (Evidence B).
• Advise parents/caregivers of children with asthma not to smoke or vape and not to allow smoking or vaping in rooms or cars that their children use (Evidence A).
• Assess patients with a >10 pack-year smoking history for COPD or for asthma+COPD, as additional treatment strategies may be required (see Section 7, p.131).
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 60 Physical activity For people with asthma, as in the general population, regular moderate physical activity has important health benefits including reduced cardiovascular risk and improved quality of life.231 There is some evidence that aerobic exercise training can have a small beneficial effect on asthma symptom control and lung function, although not airway inflammation.232 In physically inactive adults with moderate/severe asthma, physical activity interventions were associated with reduced symptoms and improved quality of life.233 Further studies are needed to identify the optimal regimen. Improved cardiopulmonary fitness may reduce the risk of dyspnea unrelated to airflow limitation being mistakenly attributed to asthma. In one study of non-obese patients with asthma, high intensity interval training together with a diet with high protein and low glycemic index improved asthma symptom control, although no benefit on lung function was seen.234 In young people with asthma, swimming training is well tolerated and leads to increased lung function and cardio-pulmonary fitness;235 369 however, there are some concerns about exposure to chlorine and trichloramine with indoor pools.65 Exercise is an important cause of asthma symptoms for many asthma patients, but EIB can usually be reduced with maintenance ICS.65 Breakthrough exercise-related symptoms can be managed with warm-up before exercise,65 and/or by taking SABA65 or low-dose ICS-formoterol236 before or during exercise. Advice • Encourage people with asthma to engage in regular physical activity because of its general health benefits (Evidence A). However, regular physical activity confers no specific benefit on lung function or asthma symptoms per se, with the exception of swimming in young people with asthma (Evidence B). There is insufficient evidence to recommend one form of physical activity over another (Evidence D).
• Provide patients with advice about prevention and management of exercise-induced bronchoconstriction including with daily treatment with ICS (Evidence A) plus SABA as-needed and pre-exercise (Evidence A), or treatment with low-dose ICS-formoterol as-needed and before exercise (Evidence B), with warm-up before exercise if needed (Evidence A). For doses of ICS-formoterol, see Box 4-8, p.84. For patients prescribed as-needed ICS-SABA, this can also be used before exercise.
Pulmonary rehabilitation A systematic review and meta-analysis found that pulmonary rehabilitation programs of 4–12 weeks’ duration that included aerobic training, nutritional advice, psychological counselling, and education in adults with asthma had little or no effect on asthma symptom control, but they achieved clinically meaningful short-term improvements in functional exercise capacity and quality of life (moderate certainty of evidence). It is not known whether these benefits continue long-term after the completion of the program.237 Advice • For asthma patients who have limited exercise tolerance, or have dyspnoea due to persistent airflow limitation, refer for pulmonary rehabilitation, if available.
Avoidance of occupational or domestic exposures Occupational exposures to allergens or sensitizers account for a substantial proportion of the incidence of adult-onset asthma.238 Once a patient has become sensitized to an occupational allergen, the level of exposure necessary to induce symptoms may be extremely low, and resulting exacerbations become increasingly severe. Attempts to reduce occupational exposure have been successful, especially in industrial settings.62 Cost-effective minimization of latex sensitization can be achieved by using non-powdered low-allergen gloves instead of powdered latex gloves.62 Advice • Ask all patients with adult-onset asthma about their work history and other exposures to inhaled allergens or irritants, including at home (Evidence D).
• In management of occupational asthma, identify and eliminate occupational sensitizers as soon as possible, and remove sensitized patients from any further exposure to these agents (Evidence A).
• Patients with suspected or confirmed occupational asthma should be referred for expert assessment and advice, if available, because of the economic and legal implications of the diagnosis (Evidence A).
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 61 Avoidance of medications that may make asthma worse Aspirin and other NSAIDs can cause severe exacerbations.239 Beta-blocker drugs, including topical ophthalmic preparations, may cause bronchospasm240 and have been implicated in some asthma deaths. However, beta-blockers have a proven benefit in the management of cardiovascular disease. People with asthma who have had an acute coronary event and received beta-blockers within 24 hours of hospital admission have been found to have lower in-hospital mortality rates than those who did not receive beta-blockers.241 Advice • Always ask people with asthma about concomitant medications, including eyedrops (Evidence D).
• Always ask about asthma and previous reactions before prescribing NSAIDs, and advise patients to stop using these medications if asthma worsens.
• Aspirin and NSAIDs are not generally contraindicated in asthma unless there is a history of previous reactions to these agents (Evidence A). (See Aspirin-exacerbated respiratory disease, p.128).
• For people with asthma who may benefit from oral or ophthalmic beta-blocker treatment, a decision to prescribe these medications should be made on a case-by-case basis, and treatment should only be initiated under close medical supervision by a specialist (Evidence D).
• Asthma should not be regarded as an absolute contraindication to use cardioselective beta-blockers when they are indicated for acute coronary events, but the relative risks and benefits should be considered (Evidence D). The prescribing physician and patient should be aware of the risks and benefits of treatment.242 Avoidance of indoor allergens Because many asthma patients react to multiple factors that are ubiquitous in the environment, avoiding these factors completely is usually impractical and very burdensome for the patient. Inhaled corticosteroid-containing medications to maintain good asthma control have an important role because patients are often less affected by environmental factors when their asthma is well controlled. There is conflicting evidence about whether measures to reduce exposure to indoor allergens are effective at reducing asthma symptoms.243,244 The majority of single interventions have failed to achieve a sufficient reduction in allergen load to lead to clinical improvement.243,245,246 It is likely that no single intervention will achieve sufficient benefits to be cost effective (Box 3-7, p.62). One study of insecticidal bait in homes eradicated cockroaches for a year and led to a significant decrease in symptoms, improvement in pulmonary function, and less health care use for children with moderate to severe asthma.247 House dust mites HDM live and thrive in many sites throughout the house, so they are difficult to reduce and impossible to eradicate. A systematic review of multi-component interventions to reduce allergens, including HDM, showed no benefit for asthma in adults and a small benefit for children.248 One study that used a rigorously applied integrated approach to HDM control led to a significant decrease in symptoms, medication use and improvement in pulmonary function for children with HDM sensitization and asthma.249 However, this approach is complicated and expensive and is not generally recommended. A study in HDM-sensitized children recruited after emergency department presentation showed a decrease in emergency department visits, but not oral corticosteroids, with the use of mite-impermeable encasement of the mattress, pillow and duvet.250 Furred pets Complete avoidance of pet allergens is impossible for sensitized patients as these allergens are ubiquitous outside the home251 in schools,252 public transport, and even cat-free buildings, probably transferred on clothes.252 Although removal of such animals from the home of a sensitized patient is encouraged,253 it can be many months before allergen levels decrease,254 and the clinical effectiveness of this and other interventions remains unproven.255 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 62 Box 3-7. Effectiveness of avoidance measures for indoor allergens Allergen and avoidance measure Degree of effectiveness (evidence level) Reduction in allergen levels Clinical benefit House dust mites • Encase bedding in impermeable covers Some (A) Adults - none (A) Children - some (A) • Wash bedding on hot cycle (55–60°C) Some (C) None (D) • Replace carpets with hard flooring Some (B) None (D) • Acaricides and/or tannic acid Little (C) None (D) • Minimize objects that accumulate dust None (D) None (D) • Vacuum cleaners with integral HEPA filter and double-thickness bags Little (C) None (D) • Remove, hot wash, or freeze soft toys None (D) None Pets • Remove cat/dog from the home Little (C) None (D) • Keep pet from the main living areas/bedrooms Little (C) None (D) • HEPA-filter air cleaners Some (B) None (A) • Wash pet Little (C) None (D) • Replace carpets with hard flooring None (D) None (D) • Vacuum cleaners with integral HEPA filter and double-thickness bags None (D) None (D) Cockroaches • Bait plus professional extermination of cockroaches Minimal (D) None (D) • Baits placed in homes Some (B) Some (B) Rodents • Integrated pest management strategies Some (B) Some (B) Fungi • Remediation of dampness or mold in homes A A • Air filters, air conditioning Some (B) None (D) See list of abbreviations (p.11). This table is adapted from Custovic et al.261 Levels of evidence (A–D) defined in Methodology, Table A (p.17) Pest rodents Symptomatic patients suspected of domestic exposure to pest rodents should be evaluated with skin prick tests or specific IgE, as exposure may not be apparent unless there is an obvious infestation.256 High-level evidence for the effectiveness of removing rodents is lacking, as most integrated pest management interventions also remove other allergen sources;256 one non-sham-controlled study showed comparable clinical improvement with pest reduction education and integrated pest management.257 Cockroaches Avoidance measures for cockroaches are only partially effective in removing residual allergens258 and evidence of clinical benefit is lacking. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 63 Fungi Fungal exposure has been associated with asthma exacerbations. The number of fungal spores can best be reduced by removing or cleaning mold-laden objects.259 Air conditioners and dehumidifiers may be used to reduce humidity to less than 50% and to filter large fungal spores. However, air conditioning and sealing of windows have also been associated with increases in fungal and HDM allergens.260 Advice • Allergen avoidance is not recommended as a general strategy for people with asthma (Evidence A).
• For sensitized patients, although it would seem logical to attempt to avoid allergen exposure in the home, there is little evidence for clinical benefit with single avoidance strategies (Evidence A) and only limited evidence for benefit with multi-component avoidance strategies (in children) (Evidence B).
• Although allergen avoidance strategies may be beneficial for some sensitized patients (Evidence B), they are often complicated and expensive, and there are no validated methods for identifying those who are likely to benefit (Evidence D).
Healthy diet In the general population, a diet high in fresh fruit and vegetables has many health benefits, including prevention of many chronic diseases and forms of cancer. Many epidemiological studies report that a high fruit and vegetable diet is associated with a lower risk of asthma and lung function decline. There is some evidence that increasing fruit and vegetable intake leads to an improvement in asthma control and a reduced risk of exacerbations.262 Advice • Encourage patients with asthma to consume a diet high in fruit and vegetables for its general health benefits (Evidence A).
Weight reduction for obese patients Asthma can be more difficult to control in obese patients,263-265 the risk of exacerbations is greater,92,93 and response to ICS may be reduced.266 There is limited evidence about the effect of weight loss on asthma control. Studies have ranged from dietary restriction to multifactorial interventions with exercise training and cognitive behavioral therapy, but populations have generally been small, and interventions and results have been heterogeneous.267 In some studies, weight loss has improved asthma control, lung function and health status, and reduced medication needs in obese patients with asthma.268,269 The most striking results have been observed after bariatric surgery,270-272 but even 5–10% weight loss with diet, with or without exercise, can lead to improved asthma control and quality of life.273 Advice • Include weight reduction in the treatment plan for obese patients with asthma (Evidence B). Increased exercise alone appears to be insufficient (Evidence B).
Breathing exercises A systematic review of studies of breathing and/or relaxation exercises in adults with asthma and/or dysfunctional breathing, including the Buteyko method and the Papworth method, reported improvements in symptoms, quality of life and/or psychological measures, but with no consistent effect on lung function and no reduction in risk of exacerbations.274 Studies of non-pharmacological strategies, such as breathing exercises, can only be considered high quality when control groups are appropriately matched for level of contact with health professionals and for asthma education. A study of two physiologically contrasting breathing exercises, which were matched for contact with health professionals and instructions about rescue inhaler use, showed similar improvements in reliever use and ICS dose after down-titration in both groups.275 This suggests that perceived improvement with breathing exercises may be largely due to factors such as relaxation, voluntary reduction in use of rescue medication, or engagement of the patient in their care. The cost of some commercial programs may be a potential limitation. Breathing exercises used in some of these studies are available at www.breathestudy.co.uk 276 and www.woolcock.org.au/resources/breathing-techniques-asthma.275 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 64 Advice • Breathing exercises may be considered as a supplement to conventional asthma management strategies for symptoms and quality of life, but they do not improve lung function or reduce exacerbation risk (Evidence A).
Avoidance of indoor air pollution In addition to passive and active smoking, other major indoor air pollutants that are known to impact on respiratory health include nitric oxide, nitrogen oxides, carbon monoxide, carbon dioxide, sulfur dioxide, formaldehyde, and biologicals (endotoxin).277,278 Sources include cooking and heating devices using gas and solid biomass fuels, particularly if they are not externally flued (vented). Installation of non-polluting, more effective heating (heat pump, wood pellet burner, flued gas) in the homes of children with asthma does not significantly improve lung function but significantly reduces symptoms of asthma, days off school, healthcare utilization, and pharmacist visits.279 Air filters can reduce fine particle exposure, but there is no consistent effect on asthma outcomes.280,281 Advice • Encourage people with asthma to use non-polluting heating and cooking sources, and for sources of pollutants to be vented outdoors where possible (Evidence B).
Strategies for dealing with emotional stress Emotional stress may lead to asthma exacerbations in children282 and adults. Hyperventilation associated with laughing, crying, anger, or fear can cause airway narrowing.283,284 Panic attacks have a similar effect.285,286 However, it is important to note that asthma is not primarily a psychosomatic disorder. During stressful times, medication adherence may also decrease. Advice • Encourage patients to identify goals and strategies to deal with emotional stress if it makes their asthma worse (Evidence D).
• There is insufficient evidence to support one strategy over another, but relaxation strategies and breathing exercises may be helpful in reducing asthma symptoms (Evidence B).
• Arrange a mental health assessment for patients with symptoms of anxiety or depression (Evidence D).
Interventions addressing social risks A systematic review of social risk intervention studies based in the USA found that interventions that addressed these challenges, including health and health care, neighborhood and built environment, and social and community context, were associated with a marked reduction in pediatric emergency department visits and hospitalizations for asthma.287 Data are needed from studies in other countries and other socioeconomic settings. Avoidance of outdoor allergens For patients sensitized to outdoor allergens such as pollens and molds, these are impossible to avoid completely. Advice • For sensitized patients, closing windows and doors, remaining indoors when pollen and mold counts are highest, and using air conditioning may reduce exposure (Evidence D).
• The impact of providing information in the media about outdoor allergen levels is difficult to assess.
Avoidance of outdoor air pollution Meta-analysis of epidemiological studies showed a significant association between air pollutants such as ozone, nitrogen oxides, acidic aerosols, and particulate matter and symptoms or exacerbations of asthma, including emergency department visits and hospitalizations.100 Use of digital monitoring identified a lag of 0–3 days between higher levels of multiple pollutants and increased asthma medication use.102 Proximity to main roads at home and school is associated with greater asthma morbidity.288 Certain weather and atmospheric conditions like thunderstorms289,290 may trigger asthma exacerbations by a variety of mechanisms, including dust and pollution, by increasing the level of respirable allergens, and causing changes in temperature and/or humidity. Reduction of outdoor air pollutants usually requires national or local policy changes. For example, short-term traffic restrictions imposed in COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 65 Beijing during the 2008 Olympics reduced pollution and was associated with a significant fall in asthma outpatient visits.291 Advice • In general, when asthma is well controlled, there is no need for patients to modify their lifestyle to avoid unfavorable outdoor conditions (air pollutants, weather).
• During unfavorable environmental conditions (very cold weather, low humidity or high air pollution), it may be helpful to avoid strenuous outdoor physical activity and stay indoors in a climate-controlled environment, if possible, and to avoid polluted environments during viral infections (Evidence D).
Avoidance of food and food chemicals Food allergy as an exacerbating factor for asthma is uncommon and occurs primarily in young children. Confirmed food allergy is a risk factor for asthma-related mortality.94 Food chemicals, either naturally occurring or added during processing, may also trigger asthma symptoms especially when asthma is poorly controlled. Sulfites (common food and drug preservatives found in such foods as processed potatoes, shrimp, dried fruits, beer, and wine) have often been implicated in causing severe asthma exacerbations.292 However, the likelihood of a reaction is dependent on the nature of the food, the level and form of residual sulfite, the sensitivity of the patient, and the mechanism of the sulfite-induced reaction.292 There is little evidence to support any general role for other dietary substances including benzoate, the yellow dye, tartrazine, and monosodium glutamate in worsening asthma. Advice • Ask people with asthma about symptoms associated with any specific foods (Evidence D).
• Food avoidance should not be recommended unless an allergy or food chemical sensitivity has been clearly demonstrated (Evidence D), usually by carefully supervised oral challenges.94 • Patients with suspected or confirmed food allergy should be referred for expert advice about management of asthma and anaphylaxis (Evidence D).
• If food allergy is confirmed, food allergen avoidance can reduce asthma exacerbations (Evidence D).
• If food chemical sensitivity is confirmed, complete avoidance is not usually necessary, and sensitivity often decreases when overall asthma control improves (Evidence D).237 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 66 REFERRAL FOR EXPERT ADVICE For most patients asthma can usually be managed in primary care, but some clinical situations warrant referral for expert advice regarding diagnosis and/or management (Box 3-8). This list is based on consensus. Indications for referral may vary, because the level at which asthma care is mainly delivered (primary care or specialist care) varies substantially between countries. Box 3-8. Indications for considering referral for expert advice, where available Difficulty confirming the diagnosis of asthma • Patient has symptoms of chronic infection, or features suggesting a cardiac or other non-pulmonary cause (Box 1-3, p.27) (immediate referral recommended).
• Diagnosis is unclear, even after a trial of therapy with ICS or systemic corticosteroids.
• Patient has features of both asthma and COPD, and there is doubt about priorities for treatment.
Suspected occupational asthma • Refer for confirmatory testing and identification of sensitizing or irritant agent, and specific advice about eliminating exposure and pharmacological treatment. See specific guidelines62 for details.
Persistent or severely uncontrolled asthma or frequent exacerbations • Symptoms remain uncontrolled, or patient has ongoing exacerbations or low lung function despite correct inhaler technique and good adherence with Step 4 treatment (medium-dose ICS-LABA, Box 4-6, p.77).
Before referral, depending on the clinical context, identify and treat modifiable risk factors (Box 2-2, p.37; Box 3-5, p.55) and comorbidities (Section 6, p.117).
• Patient frequently uses asthma-related health care, e.g., multiple ED visits or urgent primary care visits.
• For more information, see Section 8 (p.139) on difficult-to-treat and severe asthma, including a decision tree Any risk factors for asthma-related death (see Box 9-1, p.160) • Near-fatal asthma attack (ICU admission, or mechanical ventilation for asthma) at any time in the past • Suspected or confirmed anaphylaxis or food allergy in a patient with asthma Evidence of, or risk of, significant treatment side-effects • Significant side-effects from treatment • Need for long-term oral corticosteroid use • Frequent courses of oral corticosteroids (e.g., two or more courses a year) Symptoms suggesting complications or sub-types of asthma • e.g., aspirin-exacerbated respiratory disease (p.128); allergic bronchopulmonary aspergillosis (ABPA) (p.129) Additional reasons for referral in children 6–11 years • Doubts about diagnosis of asthma e.g., respiratory symptoms are not responding well to treatment in a child who was born prematurely • Symptoms or exacerbations that remain uncontrolled despite medium-dose ICS (Box 4-2B, p.71) with correct inhaler technique and good adherence • Suspected side-effects of treatment (e.g., growth delay) • Concerns about the child’s welfare or well-being See list of abbreviations (p.11). For indications for referral in children 0–5 years, see p.185. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 67 4. Medications and strategies for adults, adolescents and children 6–11 years KEY POINTS • For safety, GINA does not recommend treatment of asthma in adults, adolescents or children 6–11 years with short-acting beta2 agonist (SABA) alone. Instead, they should receive inhaled corticosteroid (ICS)-containing treatment to reduce their risk of serious exacerbations and to control symptoms.
• ICS-containing treatment can be delivered either with regular daily treatment or, in adults and adolescents who have asthma symptoms less than daily and normal or mildly reduced lung function, with as-needed low-dose ICS-formoterol taken whenever needed for symptom relief. For children not likely to be adherent with maintenance ICS, the ICS can be taken whenever the child uses their SABA reliever.
• Reduction in severe exacerbations is a high priority across treatment steps, to reduce the risk and burden to patients and the burden to the health system, and to reduce the need for oral corticosteroids (OCS), which have cumulative long-term adverse effects.
• Tables of low, medium or high dose ICS do not represent equivalent potency. If a patient is switched from one medication to another, monitor them for stability.
Treatment tracks for adults and adolescents • For clarity, the treatment figure for adults and adolescents shows two ‘tracks’, largely based on the choice of reliever. Treatment may be stepped up or down within a track using the same reliever at each step, or treatment may be switched between tracks, according to the individual patient’s needs.
• Track 1, in which the reliever is low-dose ICS-formoterol, is the preferred approach recommended by GINA.
When a patient at any step has asthma symptoms, they use low-dose ICS-formoterol as needed for symptom relief. In Steps 3–5, they also take ICS-formoterol as regular daily treatment. This approach is preferred because it reduces the risk of severe exacerbations compared with using a SABA reliever, with similar symptom control, and because of the simplicity for patients and clinicians of needing only a single medication across treatment Steps 1– 4.
• Medications and doses for Track 1 are explained in Box 4-8, p.84, including the maximum recommended total formoterol (with ICS) dose in any day for each formulation. Based on extensive evidence with budesonide-formoterol, GINA suggests that the same maximum total daily dose should apply for beclometasone-formoterol.
• Track 2, in which the reliever is an ICS-SABA or SABA, is an alternative if Track 1 is not possible, or if a patient is stable, with good adherence and no exacerbations in the past year on their current therapy. In Step 1, the patient takes a SABA and a low-dose ICS together for symptom relief (in combination if available, or with the ICS taken immediately after the SABA). In Steps 2–5, the reliever is a SABA or combination ICS-SABA. Before considering a SABA reliever, consider whether the patient is likely to be adherent with their ICS-containing treatment, as otherwise they would be at higher risk of exacerbations.
Steps 1 and 2 for adults and adolescents • Track 1: (Steps 1–2 combined) In adults and adolescents who were considered by their clinician to have mild asthma, and were taking SABA alone or had controlled asthma on daily low-dose ICS or LTRA, treatment with as-needed-only low-dose ICS-formoterol reduced the risk of severe exacerbations and emergency department visits or hospitalizations by about two-thirds compared with SABA-only treatment. As-needed-only low-dose ICS-formoterol reduced the risk of emergency department visits and hospitalizations compared with daily ICS, with no clinically important difference in symptom control. In patients previously using SABA alone, as-needed low-dose ICS-formoterol also significantly reduced the risk of severe exacerbations needing OCS, compared with daily ICS.
• Track 2: Treatment with regular daily low-dose ICS plus as-needed SABA (Step 2), if taken, is highly effective in reducing asthma symptoms and reducing the risk of asthma-related exacerbations, hospitalization and death.
However, adherence with ICS in the community is poor, leaving patients taking SABA alone and at increased risk COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 68 of exacerbations. For patients with infrequent symptoms, who are likely to have very poor adherence, as-needed-only ICS-SABA with separate or combination inhalers is the best option for Step 1, although current evidence is limited to small studies that were not powered to detect differences in exacerbation rates. Consider step-up if asthma remains uncontrolled despite good adherence and inhaler technique • Before considering any step up, first confirm that the symptoms are due to asthma and identify and address common problems such as inhaler technique, adherence, allergen exposure and multimorbidity; provide patient education.
• For adults and adolescents, the preferred Step 3 treatment is the Track 1 regimen with low-dose ICS-formoterol as maintenance-and-reliever therapy (MART). This reduces the risk of severe exacerbations, with similar or better symptom control, compared with maintenance treatment using a combination of an ICS and a long-acting beta2 agonist (LABA) as controller, plus as-needed SABA. If needed, the maintenance dose of ICS-formoterol can be increased to medium (i.e., Step 4) by increasing the number of maintenance inhalations. MART is also a preferred treatment option at Steps 3 and 4 for children 6–11 years, with a lower dose ICS-formoterol inhaler.
• ICS-formoterol should not be used as the reliever for patients taking a different ICS-LABA maintenance treatment, because clinical evidence for safety and efficacy is lacking.
• Other Step 3 options for adults and adolescents in Track 2, and in children, include maintenance ICS-LABA plus as-needed SABA or plus as-needed ICS-SABA (if available) or, for children 6–11 years, medium-dose ICS plus as-needed SABA. For children, try other controller options at the same step before stepping up.
Step down to find the minimum effective treatment • Once good asthma control has been achieved and maintained for 2–3 months, consider stepping down gradually to find the patient’s lowest treatment that controls both symptoms and exacerbations.
• Provide the patient with a written asthma action plan, monitor closely, and schedule a follow-up visit.
• Do not completely withdraw ICS unless this is needed temporarily to confirm the diagnosis of asthma.
For all patients with asthma, provide asthma education and training in essential skills • After choosing the right class of medication for the patient, the choice of inhaler device depends on which inhalers are available for the patient for that medication, which of these inhalers the patient can use correctly after training, and their relative environmental impact. Check inhaler technique frequently.
• Provide inhaler skills training: this is essential for medications to be effective, but technique is often incorrect.
• Encourage adherence with ICS-containing medication, even when symptoms are infrequent.
• Provide training in asthma self-management (self-monitoring of symptoms and/or peak expiratory flow (PEF), written asthma action plan and regular medical review) to control symptoms and minimize the risk of exacerbations.
For patients with one or more risk factors for exacerbations • Prescribe ICS-containing medication, preferably from Track 1 options, i.e., with as-needed low-dose ICS-formoterol as reliever; provide a written asthma action plan; and arrange review more frequently than for lower-risk patients.
• Identify and address modifiable risk factors (e.g., smoking, low lung function, over-use of SABA).
• Consider non-pharmacological strategies and interventions to assist with symptom control and risk reduction, (e.g., smoking cessation advice, breathing exercises, some avoidance strategies).
Difficult-to-treat and severe asthma (see Section 8, p.139) • Patients who have poor symptom control and/or exacerbations, despite medium- or high-dose ICS-LABA treatment, should be assessed for contributing factors, and asthma treatment optimized.
• If the problems continue or diagnosis is uncertain, refer to a specialist center for phenotypic assessment and consideration of add-on therapy including biologics.
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 69 Allergen immunotherapy • Allergen-specific immunotherapy may be considered as add-on therapy for patients with asthma who have clinically significant sensitization to aeroallergens.
For all patients, use your own professional judgment, and always check local eligibility and payer criteria. CATEGORIES OF ASTHMA MEDICATIONS The pharmacological options for long-term treatment of asthma fall into the following main categories (Box 4-1, p.70): • Controller medications: in the past, this term mostly referred to medications containing ICS that were used to reduce airway inflammation, control symptoms, and reduce risks such as exacerbations and the associated decline in lung function.117 In GINA Track 1, controller treatment is delivered through an anti-inflammatory reliever (AIR), low-dose ICS-formoterol, taken when symptoms occur and before exercise or allergen exposure; in Steps 3–5, the patient also takes maintenance controller treatment as daily or twice-daily ICS-formoterol. This is called “maintenance-and-reliever therapy” (MART). The dose and regimen of controller medications should be optimized to minimize the risk of medication side-effects, including risks of needing OCS.
• Reliever medications: all patients should be provided with a reliever inhaler for as-needed relief of breakthrough symptoms, including during worsening asthma or exacerbations. They are also recommended for short-term prevention of exercise-induced bronchoconstriction (EIB).
Relievers include the anti-inflammatory relievers ICS-formoterol and ICS-SABA, and SABA. Combination ICS-LABA with non-formoterol LABAs cannot be used as a reliever, due to a slower onset of action (e.g., ICS-salmeterol), or due to lack of safety and/or efficacy with more than once-daily use (e.g., ICS-vilanterol, ICS-indacaterol). ICS-formoterol should not be used as the reliever for patients taking maintenance ICS-LABA with a non-formoterol LABA.14 Over-use of SABA (e.g., dispensing of three or more 200-dose canisters in a year, corresponding to average use more than daily) increases the risk of asthma exacerbations.86,87 Regular SABA also increases the risk of poor symptom control.293 • Add-on therapies including for patients with severe asthma (Section 8, p.139).
When compared with medications used for other chronic diseases, most of the medications used for treatment of asthma have very favorable therapeutic ratios. See Box 4-2 (p.71) for low, medium and high ICS doses. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 70 Box 4-1. Terminology for asthma medications Term Definition Notes Maintenance treatment Asthma treatment that is prescribed for use every day (or on a regularly scheduled basis) Medications intended to be used continuously, even when the person does not have asthma symptoms. Examples include ICS-containing medications (ICS, ICS-LABA, ICS-LABA-LAMA), as well as LTRA† and biologic therapy. The term ‘maintenance’ describes the prescribed frequency of administration, not a particular class of asthma medicine. Controller Medication targeting both domains of asthma control (symptom control and future risk) In the past, ‘controller’ was largely used for ICS-containing medications prescribed for regular daily treatment, so ‘controller’ and ‘maintenance’ became almost synonymous. However, this became confusing after the introduction of combination ICS-containing relievers for as-needed use. To avoid confusion, ‘ICS-containing treatment’ and ‘maintenance treatment’ have been substituted as appropriate where the intended meaning was unclear. Reliever Asthma inhaler taken as needed, for quick relief of asthma symptoms Sometimes called rescue inhalers. As well as being used for symptom relief, reliever inhalers can also be used before exercise, to prevent exercise-induced asthma symptoms. Includes SABAs (e.g., salbutamol [albuterol], terbutaline, ICS-salbutamol), as-needed ICS-formoterol, and as-needed ICS-SABA. SABA-containing relievers should not be used for regular maintenance use, or to be taken when the person does not have asthma symptoms (except before exercise). Anti-inflammatory reliever (AIR) Reliever inhaler that contains both a low-dose ICS and a rapid-acting bronchodilator Includes budesonide-formoterol, beclometasone-formoterol and ICS-salbutamol combinations. Patients can also use AIRs as needed before exercise or allergen exposure to prevent asthma symptoms and bronchoconstriction. Non-formoterol LABAs in combination with ICS cannot be used as relievers. ICS-formoterol should not be used as the reliever with maintenance ICS-non-formoterol LABAs (p.69).14 The anti-inflammatory effect of as-needed ICS-formoterol was demonstrated by reduction in FeNO in several studies.187,188,294 Some anti-inflammatory relievers can be used as-needed at Steps 1–2 as the person’s sole asthma treatment, without a maintenance treatment (‘AIR-only’ treatment). Almost all evidence for this is with ICS-formoterol. Some ICS-formoterol combinations can be used as both maintenance treatment and reliever treatment at Steps 3–5 (see MART, below). For medications and doses, see Box 4-8 (p.84). Maintenance-and-reliever therapy (MART) Treatment regimen in which the patient uses an ICS-formoterol inhaler every day (maintenance dose), and also uses the same medication as needed for relief of asthma symptoms (reliever doses) MART (Maintenance-And-Reliever Therapy) can be used only with combination ICS-formoterol inhalers such as budesonide-formoterol and beclometasone-formoterol. Other ICS-formoterol inhalers can also potentially be used, but combinations of ICS with non-formoterol LABAs, or ICS-SABA, cannot be used for MART. MART is also sometimes called SMART (single-inhaler maintenance-and-reliever therapy); the meaning is the same. For medications and doses, see Box 4-8 (p.84). See list of abbreviations (p.11). †If prescribing LTRA, advise patient/caregiver about risk of neuropsychiatric adverse effects.295 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 71 Box 4-2. Low, medium and high daily metered doses of inhaled corticosteroids (alone or with LABA) This is not a table of equivalence, but suggested total daily doses for ‘low’, ‘medium’ and ‘high’ dose ICS options for adults/adolescents (Box 4-6, p.77) and children 6–11 years (Box 4-12, p.96), based on product information. The table does NOT imply potency equivalence. For example, if you switch treatment from a ‘medium’ dose of one ICS to a ‘medium’ dose of another ICS, this may represent a decrease (or an increase) in potency, and the patient’s asthma may become unstable (or they may be at increased risk of adverse effects). Patients should be monitored to ensure stability after any change of treatment or inhaler device. Doses and potency may also differ by country, depending on local products, inhaler devices, regulatory labelling and clinical guidelines or, for one product, with addition of a LAMA to an ICS-LABA.296 Low-dose ICS provides most of the clinical benefit of ICS for most patients with asthma. However, ICS responsiveness varies between patients, so some patients may need medium-dose ICS if their asthma is uncontrolled, or they have ongoing exacerbations, despite good adherence and correct technique with low-dose ICS (with or without LABA). High-dose ICS (in combination with LABA or separately) is needed by very few patients, and its long-term use is associated with an increased risk of local and systemic side-effects, which must be balanced against the potential benefits. The timing of medication use also affects outcomes, particularly for exacerbations, as seen with an anti-inflammatory reliever in GINA Track 1. For Track 1 medications and doses, see Box 4-8, p.84. Daily doses in this table are shown as metered doses. See product information for delivered doses. Inhaled corticosteroid (alone or in combination with LABA) Total daily ICS dose (mcg) – see notes above Low Medium High Adults and adolescents (12 years and older) Beclometasone dipropionate (pMDI, standard particle, HFA) 200–500 >500–1000 >1000 Beclometasone dipropionate (DPI or pMDI, extrafine particle, HFA) 100–200 >200–400 >400 Budesonide (DPI, or pMDI, standard particle, HFA) 200–400 >400–800 >800 Ciclesonide (pMDI, extrafine particle, HFA) 80–160 >160–320 >320 Fluticasone furoate (DPI) 100 200 Fluticasone propionate (DPI) 100–250 >250–500 >500 Fluticasone propionate (pMDI, standard particle, HFA) 100–250 >250–500 >500 Mometasone furoate (DPI) Depends on DPI device – see product information Mometasone furoate (pMDI, standard particle, HFA) 200–400 >400 Children 6–11 years – see notes above (for children 5 years and younger, see Box 11-3, p.191 Beclometasone dipropionate (pMDI, standard particle, HFA) 100–200 >200–400 >400 Beclometasone dipropionate (pMDI, extrafine particle, HFA) 50–100 >100–200 >200 Budesonide (DPI, or pMDI, standard particle, HFA) 100–200 >200–400 >400 Budesonide (nebules) 250–500 >500–1000 >1000 Ciclesonide (pMDI, extrafine particle, HFA) 80 >80–160 >160 Fluticasone furoate (DPI) 50 n.a.
Fluticasone propionate (DPI) 50–100 >100–200 >200 Fluticasone propionate (pMDI, standard particle, HFA) 50–100 >100–200 >200 Mometasone furoate (pMDI, standard particle, HFA) 100 200 See list of abbreviations (p.11). ICS by pMDI should preferably be used with a spacer. For new preparations, including generic ICS, the manufacturer’s information should be reviewed carefully, as products containing the same molecule may not be clinically equivalent. Combination inhalers that include a long-acting muscarinic antagonist (LAMA) may have different ICS dosing – see product information.
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 72 WHY SHOULD ICS-CONTAINING MEDICATION BE COMMENCED FROM THE TIME OF DIAGNOSIS? For the best outcomes, ICS-containing treatment should be initiated when (or as soon as possible after) the diagnosis of asthma is made. All patients should also be provided with a reliever inhaler for quick symptom relief, preferably an anti-inflammatory reliever (AIR). GINA recommends ICS-containing medication from diagnosis for several reasons: • As-needed low-dose ICS-formoterol reduces the risk of severe exacerbations and emergency department visits or hospitalizations by 65% compared with SABA-only treatment.183 This anti-inflammatory reliever regimen (AIR-only) significantly reduces severe exacerbations regardless of the patient’s baseline symptom frequency, lung function, exacerbation history or inflammatory profile: type 2-high or Type 2-low.188 • Starting treatment with SABA alone trains patients to regard it as their main asthma treatment, and increases the risk of poor adherence when daily ICS is subsequently prescribed. • Early initiation of low-dose ICS in patients with asthma leads to a greater improvement in lung function than if symptoms have been present for more than 2–4 years.297,298 One study showed that after this time, higher ICS doses were required, and lower lung function was achieved.298 • Patients not taking ICS who experience a severe exacerbation have a greater long-term decline in lung function than those who are taking ICS.117 • For patients with occupational asthma, early removal from exposure to the sensitizing agent and early ICS-containing treatment increase the probability of resolution of symptoms, and improvement of lung function and airway hyperresponsiveness.62,63 For adults and adolescents, recommended options for initial asthma treatment, based on evidence (where available) and consensus, are listed in Box 4-4 (p.75) and shown in Box 4-5 (p.76). Treatment for adults and adolescents is shown in two tracks, depending on the reliever inhaler (Box 4-6, p.77). For children 6–11 years, recommendations about initial treatment are shown in Box 4-10 (p.94) and Box 4-11 (p.95). The patient’s response should be reviewed, and treatment stepped down once good control is achieved. Recommendations for a stepwise approach to ongoing treatment are found in Box 4-12 (p.96). Does FeNO help in deciding whether to commence ICS? In studies mainly limited to non-smoking adult patients, fractional concentration of exhaled nitric oxide (FeNO) >50 parts per billion (ppb) was associated with a good short-term (weeks) response to ICS.299,300 However, these studies did not examine the longer-term risk of exacerbations, and the relationship between FeNO and other Type 2 biomarkers is lost in obese patients.24,48 In two 12-month studies in patients with mild asthma or taking SABA alone, severe exacerbations were reduced with as-needed low-dose ICS-formoterol versus as-needed SABA and versus maintenance ICS, independent of baseline inflammatory characteristics including FeNO.187,188 Consequently, in patients with a diagnosis or suspected diagnosis of asthma, high FeNO can support the decision to start ICS, but low FeNO cannot be used to decide against treatment with ICS. Based on past and current evidence, GINA recommends treatment with daily low-dose ICS or as-needed low-dose ICS-formoterol for all adults and adolescents with mild asthma, to reduce the risk of serious exacerbations.6,187,188,301,302 Choice of medication, device and dose In clinical practice, the choice of medication, device and dose for maintenance and for reliever for each individual patient should be based on assessment of symptom control, risk factors, which inhalers are available for the relevant medication class, which of these the patient can use correctly after training, their cost, their environmental impact and the patient’s likely adherence. For more detail about choice of inhaler, see Section 5 (p.108) and Box 5-1 (p.109). It is important to monitor the response to treatment and any side-effects, and to adjust the dose accordingly (Box 4-6, p.77). There is currently insufficient good-quality evidence to support use of extrafine-particle ICS aerosols over others.303 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 73 Once good symptom control has been maintained for 2–3 months, and if the patient has not had any exacerbations, asthma treatment can be carefully down-titrated to the minimum medications and dose that will maintain good symptom control and minimize exacerbation risk, while reducing the potential for side-effects (Box 4-6, p.77). For patients with severe asthma who have had a good asthma response to biologic therapy, a longer period of stability is recommended before the ICS dose is reduced, and reduction and cessation of OCS should be undertaken first. More details are given in Section 8, p.139. Patients who are being considered for a high daily dose of ICS (except for short periods) should be referred for expert assessment and advice, where possible (Section 8, p.139). GINA recommends that all adults and adolescents and all children 6–11 years should receive ICS-containing medication, incorporated in their maintenance and/or anti-inflammatory reliever treatment as part of personalized asthma management. For adults and adolescents, treatment options are shown in Box 4-6 (p.77) and, for children aged 6–11 years, in Box 4-12 (p.96). Clinicians should check local eligibility and payer criteria before prescribing. Adjusting ongoing asthma treatment in adults, adolescents, and children aged 6–11 years Once asthma treatment has begun (Box 4-4, Box 4-5, Box 4-10 and Box 4-11, p.75), ongoing treatment decisions are based on a personalized cycle of assessment, adjustment of treatment, and review of the response. For each patient, in addition to treatment of modifiable risk factors, asthma medication can be adjusted up or down in a stepwise approach (adults and adolescents: Box 4-6, p.77, children 6–11 years, Box 4-12, p.96) to achieve good symptom control and minimize future risk of exacerbations, persistent airflow limitation and medication side-effects. When good asthma control has been maintained for 2–3 months, treatment may be stepped down to find the patient’s minimum effective treatment (Box 4-13, p.102). People’s ethnic and racial backgrounds may be associated with different responses to treatment. These are not necessarily associated with genetic differences.304 The contributors are likely to be multifactorial, including differences in exposures, social disadvantage, diet and health-seeking behavior. If a patient has persisting uncontrolled symptoms and/or exacerbations despite 2–3 months of ICS-containing treatment, assess and correct the following common problems before considering any step up in treatment: • Incorrect inhaler technique • Poor adherence • Persistent exposure at home/work to agents such as allergens, tobacco smoke, indoor or outdoor air pollution, or to medications such as beta-blockers or (in some patients) nonsteroidal anti-inflammatory drugs (NSAIDs) • Comorbidities that may contribute to respiratory symptoms and poor quality of life • Incorrect diagnosis.
The evidence supporting treatment options at each step is summarized below, first for adults and adolescents, then for children 6–11 years. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 74 ADULTS AND ADOLESCENTS: ASTHMA TREATMENT TRACKS The steps below refer to the recommended asthma treatment options shown in Box 4-6 (p.77). Treatment recommendations for adults and adolescents are shown in two treatment Tracks (Box 4-3), for clarity. Suggested low, medium and high doses for a range of ICS formulations are shown in Box 4-2 (p.71). Medication options and doses for GINA Track 1 are listed in Box 4-8 (p.84). Details about treatment steps for children 6–11 years start on p.94. Box 4-3. Asthma treatment tracks for adults and adolescents Asthma treatment for adults and adolescents is in two Tracks For adults and adolescents, the main treatment figure (Box 4-6, p.77), shows the options for ongoing treatment as two treatment ‘tracks’. The key difference is the medication that is used for symptom relief. In Track 1 (preferred), the reliever is as needed low-dose ICS formoterol, and in Track 2, as-needed SABA or as-needed ICS-SABA. The reasons for showing treatment in two tracks are: • to show clinicians how treatment can be stepped up and down using the same reliever at each step • because ICS-formoterol cannot be used as the reliever in patients prescribed a combination ICS with non-formoterol LABA, due to lack of evidence about efficacy and safety (p.69).14 Track 1: The reliever is as-needed low-dose ICS-formoterol This is the preferred approach recommended by GINA for adults and adolescents, because using low-dose ICS formoterol (an anti-inflammatory reliever; AIR) reduces the risk of severe exacerbations compared with regimens that use SABA as reliever, with similar symptom control. In addition, the treatment regimen is simpler, with patients using a single medication for reliever and for maintenance treatment if prescribed, across treatment steps. • With this approach, when a patient at any treatment step has asthma symptoms, they use low-dose ICS-formoterol in a single inhaler for symptom relief. In Steps 1–2, this provides their anti-inflammatory therapy.
• In Steps 3–5, patients also take ICS formoterol as their daily maintenance treatment; together, this is called ‘maintenance-and-reliever therapy’ (MART).
• Medications and doses for GINA Track 1 are shown in Box 4-8 (p.84).
Track 2: The reliever is as-needed SABA or as-needed ICS-SABA This is an alternative approach if Track 1 is not possible, or if a patient’s asthma is stable with good adherence and no exacerbations on their current therapy. However, before prescribing a regimen with SABA reliever, consider whether the patient is likely to be adherent with their maintenance therapy, as otherwise they will be at higher risk of exacerbations. • In Step 1, the patient takes a SABA and a low-dose ICS together for symptom relief when symptoms occur (in a combination inhaler, or with the ICS taken immediately after the SABA).
• In Steps 2–5, a SABA (alone) or combination ICS-SABA is used for symptom relief, and the patient takes maintenance ICS-containing medication regularly every day. If the reliever and maintenance medication are in different devices, make sure that the patient can use each inhaler correctly.
• If changing between steps requires a different inhaler device, train the patient how to use the new inhaler.
Stepping up and down Treatment can be stepped up or down along one track, using the same reliever at each step, or it can be switched between tracks, according to the individual patient’s needs and preferences. Before stepping up, check for common problems such as incorrect inhaler technique, poor adherence, and environmental exposures, and confirm that the symptoms are due to asthma (Box 2-4, p.47). Additional controller options The additional controller options, shown below the two treatment tracks, have either limited indications or less evidence for their safety and/or efficacy, compared with the treatments in Tracks 1 and 2. See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 75 INITIAL ASTHMA TREATMENT FOR ADULTS AND ADOLESCENTS Box 4-4. Initial asthma treatment for adults and adolescents with a diagnosis of asthma These recommendations are based on evidence, where available, and on consensus. Presenting symptoms Preferred INITIAL treatment (Track 1) Alternative INITIAL treatment (Track 2) Infrequent asthma symptoms, e.g., 1–2 days/week or less As-needed low-dose ICS-formoterol (Evidence A) Low-dose ICS taken whenever SABA is taken, in combination or separate inhalers (Evidence B). Such patients are highly unlikely to be adherent with daily ICS. Asthma symptoms less than 3–5 days/week, with normal or mildly reduced lung function Low-dose ICS plus as-needed SABA (Evidence A). Before choosing this option, consider likely adherence with daily ICS. Asthma symptoms most days (e.g., 4–5 days/week or more); or waking due to asthma once a week or more, or low lung function. See p.80 for additional considerations for starting at Step 3. Low-dose ICS-formoterol maintenance-and-reliever therapy (MART) (Evidence A) Low-dose ICS-LABA plus as-needed SABA (Evidence A) or plus as-needed ICS-SABA (Evidence B), OR Medium-dose ICS plus as-needed SABA (Evidence A) or plus as-needed ICS-SABA (Evidence B). Consider likely adherence with daily maintenance treatment. Daily asthma symptoms, waking at night with asthma once a week or more, with low lung function Medium-dose ICS-formoterol maintenance-and-reliever therapy (MART) (Evidence D). Medium- or high-dose ICS-LABA (Evidence D) plus as-needed SABA or plus as-needed ICS-SABA. Consider likely adherence with daily maintenance treatment. High-dose ICS plus as-needed SABA is another option (Evidence A) but adherence is worse than with combination ICS-LABA. Initial asthma presentation is during an acute exacerbation Treat as for exacerbation (Box 9-4, p.167 and Box 9-6, p171), including short course of OCS if severe; commence medium-dose MART (Evidence D). Treat as for exacerbation (Box 9-4, p.167 and Box 9-6, p.171), including short course of OCS if severe; commence medium- or high-dose ICS-LABA plus as-needed SABA (Evidence D). Before starting initial controller treatment • Record evidence for the diagnosis of asthma.
• Record the patient’s level of symptom control and risk factors, including lung function (Box 2-2, p.37).
• Consider factors influencing choice between available treatment options (Box 3-4, p.54), including likely adherence with daily ICS-containing treatment, particularly if the reliever is SABA.
• Choose a suitable inhaler (Box 5-1, p.109) and ensure that the patient can use the inhaler correctly.
• Schedule an appointment for a follow-up visit.
After starting initial controller treatment • Review patient’s response (Box 2-2, p.37) after 2–3 months, or earlier depending on clinical urgency.
• See Box 4-6 (p.77) for recommendations for ongoing treatment and other key management issues.
• Check adherence and inhaler technique frequently.
• Step down treatment once good control has been maintained for 3 months (Box 4-13, p.102).
Also consider cost and likely adherence with maintenance treatment. See Box 4-2 (p.71) for low, medium and high ICS doses, and Box 4-8 (p.84) for Track 1 medications and doses. See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 76 Box 4-5. Flowchart for selecting initial treatment in adults and adolescents with a diagnosis of asthma These recommendations are based on evidence, where available, and on consensus. See list of abbreviations (p.11). See Box 4-2 (p.71) for low, medium and high ICS doses for adults and adolescents. See Box 4-6 (p.77), for Track 1 medications and doses. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 77 ASTHMA TREATMENT STEPS IN ADULTS AND ADOLESCENTS Box 4-6. Personalized management for adults and adolescents to control symptoms and minimize future risk Anti-inflammatory reliever. †If prescribing LTRA, advise patient/caregiver about risk of neuropsychiatric adverse effects. See list of abbreviations (p.11). For recommendations about initial asthma treatment in adults and adolescents, see Box 4-4 (p.75) and Box 4-5 (p.76). See Box 4-2 (p.71) for low, medium and high ICS doses for adults and adolescents. See Box 4-8 (p.84) for Track 1 medications and doses. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 78 TRACK 1 (PREFERRED): TREATMENT STEPS 1–4 FOR ADULTS AND ADOLESCENTS USING ICS-FORMOTEROL RELIEVER Track 1 is the preferred approach recommended by GINA for adults and adolescents with asthma, because using low-dose ICS formoterol (an anti-inflammatory reliever; AIR) reduces the risk of severe exacerbations compared with regimens that use SABA as reliever, with similar symptom control and lung function. In addition, the treatment regimen is simpler, with patients using a single medication for reliever and for maintenance treatment (if prescribed), across treatment steps. With the AIR approach, when a patient at any treatment step has asthma symptoms, they use low-dose ICS-formoterol in a single inhaler for symptom relief. In Steps 1–2, this provides their anti-inflammatory therapy. In Steps 3–5, patients also take ICS formoterol as their daily maintenance treatment; together, this is called ‘maintenance-and-reliever therapy’ (MART). Details about medications and doses for Track 1 are in Box 4-8, p.84. Details below are for Track 1, Steps 1–4. In Step 5, treatment options for Tracks 1 and 2 are similar, so the information is shown for both Tracks together, starting on p.91 and in Section 8, p.139. Box 4-7. Track 1 (preferred) treatment Steps 1–4 for adults and adolescents See Box 4-8 (p.84) for details of medications and doses. AIR: anti-inflammatory reliever; ICS: inhaled corticosteroid; MART: maintenance-and-reliever therapy with ICS-formoterol; SABA: short-acting beta2 agonist Track 1 (preferred) Step 1–2 treatment for adults and adolescents: as-needed low-dose combination ICS-formoterol In Track 1, Steps 1–2, low-dose combination ICS-formoterol is used as needed for symptom relief, and before exercise or before expected allergen exposure. Information about Steps 1 and 2 below is combined, because the recommended treatment (as-needed low-dose ICS-formoterol) is the same. In Track 1, Step 1–2 treatment with as-needed-only low-dose combination ICS-formoterol is recommended for: • Step-down treatment for patients whose asthma is well controlled on low-dose maintenance-and-reliever therapy with ICS-formoterol (Evidence D) or on regular low-dose ICS with as-needed SABA (Evidence A) • Initial asthma treatment for patients previously using SABA alone (or with newly diagnosed asthma), with normal or mildly reduced lung function. Some clinical factors, outlined below, may prompt consideration of starting treatment instead at Step 3, with low-dose ICS-formoterol maintenance-and-reliever therapy.
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 79 Populations studied The populations studied in the large randomized controlled trials of as-needed low-dose ICS-formoterol187,188,301,302 included almost 10,000 adults and adolescents with asthma that was considered to be mild, and was either uncontrolled on SABA alone, or controlled on low-dose ICS or LTRA. In the two largest studies, post-bronchodilator FEV1 was required to be ≥80% predicted at baseline.301,302 Evidence Use of low-dose ICS-formoterol as needed for symptom relief (an anti-inflammatory reliever) for adults and adolescents (Evidence B) is supported by evidence from four randomized controlled trials, and by systematic review and meta-analysis of all four studies for several outcomes.183 The two largest studies were double-blind, and two were pragmatic and open-label, intended to evaluate the treatment as it would be used in clinical practice, without patients required to take a twice-daily maintenance inhaler. The key findings with as-needed low-dose ICS-formoterol, as follows, support the Step 1–2 recommendations: • A large double-blind study found a 64% reduction in severe exacerbations requiring OCS, compared with SABA-only treatment,301 with a similar finding in an open-label study in patients previously taking SABA alone (Evidence A).187 In the Cochrane meta-analysis, as-needed low-dose ICS-formoterol reduced the risk of severe exacerbations requiring OCS by 55%, and reduced the risk of emergency department visits or hospitalizations by 65%, compared with SABA alone (Evidence A).183 • Two large double-blind studies showed as-needed budesonide-formoterol was non-inferior for severe exacerbations, compared with regular ICS.301,302 In two open-label randomized controlled trials, representing the way that patients with mild asthma would use as-needed ICS-formoterol in real life, as-needed budesonide-formoterol was superior to maintenance ICS in reducing the risk of severe exacerbations (Evidence A).187,188 • A Cochrane review provided moderate to high certainty evidence that as-needed ICS-formoterol was clinically effective in adults and adolescents with mild asthma, significantly reducing important clinical outcomes including need for oral corticosteroids, severe exacerbation rates, and emergency department visits or hospital admissions compared with daily ICS plus as-needed SABA (Evidence A).183 • In all four studies, the as-needed low-dose ICS-formoterol strategy was associated with a substantially lower average ICS dose than with maintenance low-dose ICS.187,188,301,302 • Clinical outcomes with as-needed ICS-formoterol were similar in adolescents as in adults.305 • A post hoc analysis of one study301 found that a day with >2 doses of as-needed budesonide-formoterol reduced the short-term (21 day) risk of severe exacerbations compared to as needed terbutaline alone, suggesting that timing of use of ICS-formoterol is important.128 • No new safety signals were seen with as-needed budesonide-formoterol in these studies.187,188,301,302,306 Considerations for recommending as-needed-only low-dose ICS-formoterol as preferred treatment for Steps 1–2 The most important considerations for GINA were: • The need to prevent severe exacerbations in patients with mild or infrequent symptoms; these can occur with unpredictable triggers such as viral infection, allergen exposure, pollution or stress.
• The desire to avoid the need for daily ICS in patients with mild asthma, who in clinical practice are often poorly adherent with prescribed ICS, leaving them exposed to the risks of SABA-only treatment.307 • The greater reduction in severe exacerbations with as-needed ICS-formoterol, compared with daily ICS, among patients previously taking SABA alone, with no significant difference for patients with well-controlled asthma on ICS or LTRA at baseline.187,308 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 80 • The very small differences in FEV1, (approximately 30–50 mL), symptom control (difference in ACQ-5 of approximately 0.15 versus minimal clinically important difference 0.5), and symptom-free days (mean difference 10.6 days per year)301,302 compared with regular ICS were considered to be less important. These differences did not increase over the 12-month studies. The primary outcome variable of one study301 was ‘well-controlled asthma weeks’, but this outcome was not considered reliable because it was based on an older concept of asthma control, and was systematically biased against the as-needed ICS-formoterol treatment group because much less ICS was permitted in a week for patients on ICS-formoterol than those on maintenance ICS before the week was classified as not well controlled. • The similar reduction in FeNO with as-needed budesonide-formoterol as with maintenance ICS, and the lack of significant difference in treatment effect with as-needed budesonide-formoterol by patients’ baseline eosinophils or baseline FeNO.187,188 Considerations for the GINA recommendation against SABA-only treatment of asthma There were several important considerations for extending the recommendation for as-needed-only low-dose ICS-formoterol to adults and adolescents with infrequent asthma symptoms (i.e., eliminating SABA-only treatment):6 • Patients with few interval asthma symptoms can still have severe or fatal exacerbations.180 GINA recommends assessing and addressing risk factors for exacerbations as well as symptom control (Box 2-2, p.37). • The historic distinction between so-called ‘intermittent’ and ‘mild persistent’ asthma is arbitrary, with no evidence of difference in response to ICS.182 A large reduction in risk of severe exacerbations with as-needed ICS-formoterol, compared with as-needed SABA, was seen even in patients with SABA use twice a week or less at baseline.187 • A post hoc analysis of one study found that a single day with increased as-needed budesonide-formoterol reduced the short-term (21-day) risk of severe exacerbations compared to as needed SABA alone, suggesting that timing of use of ICS-formoterol is important.128 • In patients with infrequent symptoms, adherence with prescribed daily ICS is very poor,309 exposing them to risks of SABA-only treatment if they are prescribed daily ICS plus as-needed SABA. • There is a lack of evidence for the safety or efficacy of SABA-only treatment. Historic recommendations for SABA-only treatment were based on the assumption that patients with mild asthma would not benefit from ICS.182 • Taking SABA regularly for as little as one week significantly increases exercise-induced bronchoconstriction, airway hyperresponsiveness and airway inflammation, and decreases bronchodilator response.310,311 • Even modest over-use of SABA (indicated by dispensing of 3 or more 200-dose canisters a year) is associated with increased risk of severe exacerbations86 and, in one study, asthma mortality.87 • GINA places a high priority on avoiding patients becoming reliant on SABA, and on avoiding conflicting messages in asthma education. Previously, patients were initially provided only with SABA for symptom relief, but later, despite this treatment being effective from the patient’s perspective, they were told that to reduce their SABA use, they needed to take a daily maintenance treatment, even when they had no symptoms. Recommending that all patients should be provided with ICS-containing treatment (including, in mild asthma, the option of as-needed ICS-formoterol) from the start of therapy allows consistent messaging about the need for both symptom relief and risk reduction, and may avoid establishing patient reliance on SABA as their main asthma treatment. Considerations for starting treatment with low-dose maintenance-and-reliever therapy (Step 3 MART) instead of as-needed-only ICS-formoterol (Steps 1–2) There is no specific evidence to guide this decision, but by consensus, we suggest starting with Step 3 MART (if permitted by local regulators) if the patient has symptoms most days or is waking at night due to asthma more than once a week (to rapidly reduce symptom burden), or if they are currently smoking, have impaired perception of bronchoconstriction (e.g. low initial lung function but few symptoms), a recent severe exacerbation or a history of a life-threatening asthma exacerbation, have severe airway hyperresponsiveness, or are currently exposed to a seasonal allergic trigger. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 81 Anti-inflammatory reliever treatment with as-needed-only ICS-formoterol (‘AIR-only’) is the preferred treatment for Steps 1 and 2 in adults/adolescents, so these steps have been combined in the treatment figure (Track 1, Box 4-6, p.77) to avoid confusion. Practice points for as-needed-only ICS-formoterol in mild asthma The usual dose of as-needed budesonide-formoterol in mild asthma is a single inhalation of 200/6 mcg (delivered dose 160/4.5 mcg), taken whenever needed for symptom relief. The maximum recommended dose of as-needed budesonide-formoterol in a single day corresponds to a total of 72 mcg formoterol (54 mcg delivered dose). However, in randomized controlled trials (RCTs) in mild asthma, such high usage was rarely seen, with average use around 3–4 doses per week.187,301,302 For more details about medications and doses for as-needed-only ICS-formoterol, see Box 4-8 (p.84). Rinsing the mouth is not generally needed after as-needed doses of low-dose ICS-formoterol, as this was not required in any of the mild asthma studies (or in MART studies), and there was no increase in risk of oral candidiasis.306 Other ICS-formoterol formulations have not been studied for as-needed-only use, but beclometasone-formoterol may also be suitable, as it is well-established for as-needed use within maintenance-and-reliever therapy in GINA Steps 3–5.224 Combinations of ICS with non-formoterol LABA cannot be used as-needed for symptom relief. For pre-exercise use in patients with mild asthma, one 6-week study showed that use of low-dose budesonide-formoterol for symptom relief and before exercise reduced exercise-induced bronchoconstriction to a similar extent as regular daily low-dose ICS with SABA for symptom relief and before exercise.236 This suggests that patients with mild asthma who are prescribed as-needed low-dose ICS-formoterol to prevent exacerbations and control symptoms can use the same medication before exercising, if needed, and do not need to be prescribed a SABA for pre-exercise use (Evidence B). Patient preferences: from qualitative research, the majority of patients in a pragmatic open-label study preferred as-needed ICS-formoterol for ongoing treatment rather than regular daily ICS with a SABA reliever. They reported that shared decision-making would be important in choosing between these treatment options.312 Asthma action plan: Simple action plans for AIR-only and MART are available online.313,314 Track 1 (preferred) Step 3 treatment for adults and adolescents: low-dose ICS-formoterol maintenance-and-reliever therapy (MART) For adults and adolescents, the preferred Step 3 option is low-dose ICS-formoterol as both maintenance and reliever treatment (MART). In this regimen, low-dose ICS-formoterol, either budesonide-formoterol or beclometasone-formoterol, is used as both the daily maintenance treatment and as an anti-inflammatory reliever for symptom relief. The low-dose ICS-formoterol can also be used before exercise, and before expected allergen exposure. Before considering a step up, check for common problems such as incorrect inhaler technique, poor adherence, and environmental exposures, and confirm that the symptoms are due to asthma (Box 2-4, p.47). Populations studied Double-blind studies included adult and adolescent patients with ≥1 exacerbation in the previous year despite maintenance low-dose ICS or ICS-LABA treatment, with poor symptom control. Open-label studies were in patients taking at least low-dose ICS or ICS-LABA, with suboptimal asthma control; they did not require a history of exacerbations.226 Evidence Low-dose ICS-formoterol maintenance-and-reliever therapy reduced severe exacerbations and provided similar levels of asthma control at relatively low doses of ICS, compared with a fixed dose of ICS-LABA as maintenance treatment or a higher dose of ICS, both with as-needed SABA (Evidence A).226,315-319 In a meta-analysis, switching patients with uncontrolled asthma from Step 3 treatment plus SABA reliever to MART was associated with a 29% reduced risk of severe exacerbation, compared with stepping up to Step 4 ICS-LABA maintenance plus SABA reliever, and a 30% reduced risk compared with staying on the same treatment with SABA reliever.320 In open-label studies that did not require a history of COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 82 severe exacerbations, maintenance-and-reliever therapy with ICS-formoterol also significantly reduced severe exacerbations, with a lower average dose of ICS.226,321 The benefit of the MART regimen in reducing the risk of severe exacerbations requiring OCS appears to be due to the increase in doses of both the ICS and the formoterol at a very early stage of worsening asthma. As for patients using as-needed-only ICS-formoterol (p.79), this reduces the risk of progressing to a severe exacerbation in the next 3 weeks.126-128 Other considerations Use of ICS-formoterol as an anti-inflammatory reliever across treatment steps provides a simple regimen with easy transition if treatment needs to be stepped up (e.g., from Step 1–2 to Step 3, or Step 3 to Step 4), without the need for an additional medication or different prescription, or a different inhaler type (see Box 4-8, p.84). Practice points for maintenance-and-reliever therapy (MART) with low-dose ICS-formoterol Medications: ICS-formoterol maintenance-and-reliever therapy for Step 3 treatment can be prescribed with low-dose budesonide-formoterol (≥12 years) or low-dose beclometasone-formoterol (≥18 years). The usual dose for MART with budesonide-formoterol is 200/6 mcg metered dose (160/4.5 mcg delivered dose) and the usual dose for MART with beclometasone-formoterol is 100/6 metered dose (delivered dose 84.6/5 mcg for pMDI and 81.9/5 mcg for DPI). Each of these combinations is prescribed as one inhalation twice-daily plus one inhalation whenever needed for symptom relief. Doses: For MART with budesonide-formoterol, the maximum recommended total dose of formoterol in a single day (total of maintenance-and-reliever doses) gives 72 mcg metered dose (54 mcg delivered dose) of formoterol, with extensive evidence from large studies for its safety and efficacy up to this dose in a single day,224,226 with or without ICS.306,322,323 Based on this evidence, GINA suggests that the same maximum total dose of formoterol in a single day should also apply for MART with beclometasone-formoterol (maximum total 12 inhalations, total metered dose 72 mcg). Most patients need far fewer doses than this. For a summary of medications and doses, see Box 4-8 (p.84). ICS-formoterol should not be used as the reliever for patients taking a different ICS-LABA maintenance treatment, since clinical evidence for safety and efficacy is lacking. Use of ICS-formoterol with other LABAs may be associated with increased adverse effects.14 Rinsing the mouth is not generally needed after as-needed doses of ICS-formoterol, as this was not required in any of the MART studies, and there was no increase in risk of oral candidiasis. Additional practice points can be found in an article describing how to use MART, including a customizable written asthma action plan for use with this regimen.313 Other action plans for MART are available online.313,314 Track 1 (preferred) Step 4 treatment for adults and adolescents: medium-dose ICS-formoterol maintenance-and-reliever therapy (MART) At a population level, most benefit from ICS is obtained at low dose, but individual ICS responsiveness varies, and some patients whose asthma is uncontrolled on low-dose ICS-LABA despite good adherence and correct inhaler technique may benefit from increasing the maintenance dose to medium, usually by taking twice the number of inhalations (see Box 4-8, p.84). High-dose ICS-formoterol is not recommended in Track 1 Step 4. Before stepping up, check for common problems such as incorrect inhaler technique, poor adherence, and environmental exposures, and confirm that the symptoms are due to asthma (Box 2-4, p.47). Patients prescribed MART use low-dose ICS-formoterol as needed for symptom relief, and before exercise or allergen exposure if needed. For adult and adolescent patients, combination ICS-formoterol as both maintenance-and-reliever treatment (MART) is more effective in reducing exacerbations than the same dose of maintenance ICS-LABA or higher doses of ICS318 or ICS-LABA224 (Evidence A). The greatest reduction in risk was seen in patients with a history of severe exacerbations,224 but MART was also significantly more effective than conventional best practice with as-needed SABA in open-label studies in which patients were not selected for greater exacerbation risk.226 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 83 In Step 4, the MART regimen can be prescribed with medium-dose maintenance budesonide-formoterol or beclometasone-formoterol treatment, by increasing the maintenance dose of low-dose ICS-formoterol to 2 inhalations twice-daily. The reliever remains 1 inhalation of low-dose ICS-formoterol as needed. The usual dose for MART with budesonide-formoterol is 200/6 mcg metered dose (160/4.5 mcg delivered dose) and the usual dose for MART with beclometasone-formoterol is 100/6 mcg metered dose (delivered dose 84.6/5 mcg for pMDI and 81.9/5 mcg for DPI). For Step 4, each of these combinations is prescribed as two inhalations twice-daily plus one inhalation whenever needed for symptom relief. As in Step 3, the maximum recommended total dose of budesonide-formoterol in a single day (total of maintenance-and-reliever doses) gives 72 mcg metered dose (54 mcg delivered dose) of formoterol, with extensive evidence from large studies for its safety306,322,323 and efficacy224,226 up to this dose in a single day. Based on this evidence, GINA suggests that the same maximum total dose of formoterol in a single day should also apply for MART with beclometasone-formoterol (maximum total 12 inhalations, total metered dose 72 mcg). Most patients need far fewer doses than this. For practice points, see information for GINA Step 3 and an article for clinicians.313 For a summary of medications and doses, see Box 4-8 (p.84). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 84 Box 4-8. Medications and doses for GINA Track 1: anti-inflammatory reliever (AIR) therapy GINA Track 1 – general principles In GINA Track 1, the reliever inhaler is low-dose ICS-formoterol, with or without maintenance ICS-formoterol. This is the preferred treatment approach for adults and adolescents with asthma, because it reduces severe exacerbations across treatment steps compared with using a SABA reliever; it uses a single medication for both reliever and maintenance treatment (less confusing for patients); and the patient’s treatment can be stepped up and down if needed without changing the medication or inhaler device. This cannot be done with any other ICS-LABA. ICS-formoterol can also be used before exercise and before allergen exposure. Low-dose ICS-formoterol is called an anti-inflammatory reliever (AIR) because it relieves symptoms and reduces inflammation. AIR with ICS-formoterol significantly reduces the risk of severe exacerbations across treatment steps compared with using a SABA reliever, with similar symptom control, lung function and adverse effects. Steps 1–2 (AIR-only): low-dose ICS-formoterol is used as needed for symptom relief without any maintenance treatment. It reduces the risk of severe exacerbations and ED visits/hospitalizations by 65% compared with SABA alone, and reduces ED visits/hospitalizations by 37%, compared with daily ICS plus as-needed SABA.183 Starting treatment with as-needed ICS-formoterol avoids training patients to regard SABA as their main asthma treatment. Steps 3–5 (MART): maintenance-and-reliever therapy with ICS-formoterol reduces the risk of severe exacerbations by 32% compared with the same dose of ICS-LABA,224 by 23% compared with a higher dose of ICS-LABA,224 and by 17% compared with usual care.226 MART is also an option for children 6–11 years in Steps 3–4. Asthma action plan: Simple action plans for AIR-only and MART are available online.313,314 Which medications can be used in GINA Track 1, and how often? Most evidence for MART, and all evidence for AIR-only, is with budesonide-formoterol DPI, usually 200/6 mcg metered dose (160/4.5 mcg delivered dose) for adults/adolescents, and 100/6 mcg (80/4.5 mcg delivered dose) for MART in children 6–11 years. Beclometasone dipropionate (BDP)-formoterol 100/6 mcg (84.6/5.0) is also effective for MART in adults. Other low-dose combination ICS-formoterol products may be suitable but have not been studied. For as-needed use, patients should take either 1 or 2 inhalations (based on the formulation; see below and next page) whenever needed for symptom relief, or before exercise or allergen exposure, instead of a SABA reliever. Patients do not need to wait a certain number of hours before taking more reliever doses (unlike SABA), but in a single day, they should not take more than the maximum total number of inhalations shown below and over (total as-needed plus maintenance doses, if used). Most patients need far less than this. Age Inhalers: mcg/inhalation metered dose [delivered dose] and maximum in any day Dosing frequency by age group and treatment step (see next page for additional inhaler options and doses) 6–11 years Budesonide-formoterol 100/6 DPI [80/4.5] (maximum total 8 inhalations in any day) Step 1–2 AIR-only: no evidence to date Step 3 MART: 1 inhalation once daily plus 1 as needed Step 4 MART: 1 inhalation twice daily plus 1 as needed Step 5 MART: not recommended 12–17 years Budesonide-formoterol 200/6 [160/4.5] mcg DPI or pMDI (maximum total 12 inhalations in any day) Step 1–2 (AIR-only): 1 inhalation as needed Step 3 MART: 1 inhalation twice (or once) daily plus 1 as needed Step 4 MART: 2 inhalations twice daily plus 1 as needed Step 5 MART: 2 inhalations twice daily plus 1 as needed ≥18 years Budesonide-formoterol 200/6 [160/4.5] or BDP-formoterol 100/6 mcg, pMDI or DPI (maximum total 12 inhalations in any day†) Step 1–2 (AIR-only): 1 inhalation as needed† Step 3 MART: 1 inhalation twice (or once) daily plus 1 as needed Step 4 MART: 2 inhalations twice daily plus 1 as needed Step 5 MART: 2 inhalations twice daily plus 1 as needed †For beclometasone (BDP)-formoterol, GINA suggests that the maximum total dose in any day should be 12 inhalations, based on extensive safety data with budesonide-formoterol; it has not been studied as-needed only but may be suitable (see p.82. The delivered dose for BDP-formoterol 100/6 mcg is 84.6/5 mcg for pMDI and 81.9/5 mcg for DPI. See next page for more inhaler doses.
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 85 Box 4-8 (continued). Medications and doses for GINA Track 1 anti-inflammatory reliever (AIR) therapy Medications: mcg/inhalation metered dose [delivered dose] (maximum total inhalations in any day) Dosing frequency for ICS-formoterol formulations suitable for AIR therapy, by age group and treatment step Children 6–11 years Budesonide-formoterol DPI 100/6 [80/4.5] (maximum total 8 inhalations in any day) Step 1–2 AIR-only: no evidence to date Step 3 MART: 1 inhalation once daily plus 1 as needed Step 4 MART: 1 inhalation twice daily plus 1 as needed Step 5 MART: not recommended Budesonide-formoterol pMDI 50/3 [40/2.25] (maximum total 16 inhalations in any day) These doses ONLY for pMDIs with 3 [2.25] mcg formoterol These doses ONLY for pMDIs with 3 [2.25] mcg formoterol Step 1–2 AIR-only: no evidence to date Step 3 MART: 2 inhalations once daily plus 2 as needed Step 4 MART: 2 inhalations twice daily plus 2 as needed Step 5 MART: not recommended Adolescents 12–17 years Budesonide-formoterol DPI or pMDI 200/6 [160/4.5] (maximum total 12 inhalations in any day) Step 1–2 (AIR-only): 1 inhalation as needed Step 3 MART: 1 inhalation twice (or once) daily plus 1 as needed Step 4 MART: 2 inhalations twice daily plus 1 as needed Step 5 MART: 2 inhalations twice daily plus 1 as needed Budesonide-formoterol pMDI 100/3 [80/2.25] (maximum total 24 inhalations in any day) These doses ONLY for pMDIs with 3 [2.25] mcg formoterol These doses ONLY for pMDIs with 3 [2.25] mcg formoterol Step 1–2 (AIR-only): 2 inhalations as needed Step 3 MART: 2 inhalations twice (or once) daily plus 2 as needed Step 4 MART: 4 inhalations twice daily plus 2 as needed Step 5 MART: 4 inhalations twice daily plus 2 as needed Adults 18 years and older Budesonide-formoterol DPI or pMDI 200/6 [160/4.5] (maximum total 12 inhalations in any day) Step 1–2 (AIR-only): 1 inhalation as needed Step 3 MART: 1 inhalation twice (or once) daily plus 1 as needed Step 4 MART: 2 inhalations twice daily plus 1 as needed Step 5 MART: 2 inhalations twice daily plus 1 as needed Budesonide-formoterol pMDI 100/3 [80/2.25] (maximum total 24 inhalations in any day) These doses ONLY for pMDIs with 3 [2.25] mcg formoterol These doses ONLY for pMDIs with 3 [2.25] mcg formoterol Step 1–2 (AIR-only): 2 inhalations as needed Step 3 MART: 2 inhalations twice (or once) daily plus 2 as needed Step 4 MART: 4 inhalations twice daily plus 2 as needed Step 5 MART: 4 inhalations twice daily plus 2 as needed Beclometasone-formoterol pMDI or DPI 100/6 (GINA suggests maximum total 12 inhalations in any day†) Step 1–2 (AIR-only): 1 inhalation as needed Step 3 MART: 1 inhalation twice (or once) daily plus 1 as needed Step 4 MART: 2 inhalations twice daily plus 1 as needed Step 5 MART: 2 inhalations twice daily plus 1 as needed For abbreviations, see p.11. Maximum total inhalations in any day = as-needed doses plus maintenance doses, if used.
†Beclometasone (BDP)-formoterol has not been studied for as-needed-only use (Steps 1–2), but it may be suitable given its efficacy for MART in moderate-severe asthma.316 GINA suggests that the maximum total dose of BDP-formoterol in any day should be 12 inhalations, based on extensive safety data with budesonide-formoterol.322 For more details, see p.82.
Budesonide-formoterol 400/12 [320/4.5] mcg should not be used as an anti-inflammatory reliever. For adults/ adolescents, GINA does not suggest use of budesonide-formoterol 100/6 [80/4.5] as an anti-inflammatory reliever, since most evidence is with budesonide-formoterol 200/6 [160/4.5] mcg.
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 86 TRACK 2 (ALTERNATIVE): TREATMENT STEPS 1–4 FOR ADULTS AND ADOLESCENTS USING SABA RELIEVER This is an alternative approach if Track 1 is not possible, or if a patient’s asthma is stable with good adherence and no exacerbations on their current therapy. However, before prescribing a regimen with SABA reliever, consider whether the patient is likely to be adherent with their maintenance therapy; if not, they will be at higher risk of exacerbations. Box 4-9. Track 2 (alternative) treatment Steps 1–4 for adults and adolescents Anti-inflammatory reliever therapy (AIR); ICS: inhaled corticosteroid; LABA: long-acting beta2 agonist; SABA: short-acting beta2 agonist.
Step 1: can be implemented with separate ICS and SABA inhalers, or with combination ICS-SABA inhaler Track 2 (alternative) Step 1 treatment options for adults and adolescents: low-dose ICS taken whenever SABA is taken, using separate inhalers or combination ICS-SABA Low-dose ICS taken whenever SABA is used (in combination or separate ICS and SABA inhalers) is an option if as-needed ICS-formoterol is not available, and the patient is unlikely to take regular ICS. This regimen avoids SABA-only treatment, and may also be useful in regions where the cost of ICS-formoterol is currently prohibitive. Populations studied All the evidence for taking ICS whenever SABA is taken is from studies in patients whose asthma was controlled or partly controlled on daily low-dose ICS, i.e., it has been evaluated as a step-down treatment option. Evidence The evidence for taking ICS whenever SABA is taken is from two small studies in adults and two small studies in children and adolescents, with separate or combination ICS and SABA inhalers.324-327 These studies showed no difference in exacerbations compared with daily ICS, but in the two studies that included a SABA-only arm, SABA alone was the worst option for treatment failure. All four studies used beclometasone dipropionate (BDP). One study of as-needed combination ICS-SABA used a moderate dose (250 mcg BDP+100 mcg albuterol), and the three studies with separate inhalers used 2 inhalations of BDP 50 mcg [40 mcg delivered dose] for each 2 inhalations of 100 mcg albuterol. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 87 Other considerations In making this recommendation, the most important considerations were reducing the risk of severe exacerbations, and the difficulty of achieving good adherence with regularly prescribed ICS in patients with infrequent symptoms. For definitions of low-dose ICS see Box 4-2 (p.71). Patients with symptoms less than 1–2 days a week are extremely unlikely to take ICS regularly even if prescribed, leaving them exposed to the risks of SABA-only treatment, so taking ICS whenever SABA is taken is likely to be a better option in such patients. Practice points If combination ICS-SABA is not available, the patient needs to carry both ICS and SABA inhalers with them for as-needed use. See Box 4-2 (p.71) for ICS doses. There are no studies with daily maintenance low-dose ICS plus as-needed combination ICS-SABA. Medications not recommended for adults and adolescents with asthma SABA-only treatment is not recommended by GINA for adults, adolescents or children 6–11 years with asthma. Although inhaled SABAs are highly effective for the quick relief of asthma symptoms,328 patients whose asthma is treated with SABA alone (compared with ICS) are at increased risk of asthma-related death (Evidence A)87,329 and of urgent asthma-related healthcare (Evidence A),330 even if they have good symptom control.331 The risk of severe exacerbations requiring urgent health care is substantially reduced in adults and adolescents by either as-needed ICS-formoterol,183 or by regular low-dose ICS with as-needed SABA.307 The risk of asthma exacerbations and mortality increases incrementally with higher SABA use, including in patients treated with SABA alone.87 One long-term study of regular SABA in patients with newly diagnosed asthma showed worse outcomes and lower lung function than in patients who were treated with daily low-dose ICS from the start.332 Starting treatment of asthma with SABA alone encourages patients to regard it as their main (and often only) asthma treatment, leading to poor adherence if ICS-containing therapy is prescribed. Treatment with oral bronchodilators (e.g. salbutamol tablets or syrups; oral theophylline) is not recommended for treatment of asthma in any age group. For additional non-recommended bronchodilators, see p.93. Track 2 (alternative) Step 2 treatment options for adults and adolescents: low-dose maintenance ICS plus as-needed SABA Regular daily low-dose ICS with as-needed SABA was standard of care for mild asthma for the past 30 years. Most guidelines recommended its use only for patients with asthma symptoms more than twice a week, based on an assumption that patients with less frequent symptoms did not need, and would not benefit, from ICS.182 Population studied Most studies of daily low-dose ICS have included patients with symptoms between 3–7 days a week. Evidence Regular daily low-dose ICS plus as-needed SABA is a long-established treatment for mild asthma. There is a large body of evidence from RCTs and observational studies showing that the risks of severe exacerbations, hospitalizations and mortality are substantially reduced with regular low-dose ICS; symptoms and exercise-induced bronchoconstriction are also reduced (Evidence A).307,329,333-335 Severe exacerbations are halved with low-dose ICS even in patients with symptoms 0–1 days a week.182 In a meta-analysis of long-term cohort studies, regular ICS was associated with a very small increase in pre-bronchodilator FEV1% predicted, but there is insufficient evidence that it protects from development of persistent airflow limitation.336 Other considerations Clinicians should be aware that adherence with maintenance ICS in the community is extremely low. They should consider the likelihood that patients with infrequent symptoms who are prescribed daily ICS plus as needed SABA will be poorly COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 88 adherent with the ICS, increasing their risk of severe exacerbations. Over-use of SABA, indicated by dispensing of three or more 200-dose canisters of SABA in a year (i.e., average use more than daily), is associated with an increased risk of severe exacerbations86,87 and, in one study, with increased mortality,87 even in patients also taking ICS-containing treatment. Track 2 (alternative) Step 3 treatment for adults and adolescents: maintenance low-dose ICS-LABA plus as-needed SABA or plus as-needed combination ICS-SABA Before considering a step up, check for common problems such as incorrect inhaler technique, poor adherence, and environmental exposures, and confirm that the symptoms are due to asthma (Box 2-4, p.47). Currently approved combination ICS-LABA inhalers for Step 3 maintenance treatment of asthma include low doses of fluticasone propionate-formoterol, fluticasone furoate-vilanterol, fluticasone propionate-salmeterol, beclometasone-formoterol, budesonide-formoterol, mometasone-formoterol, and mometasone-indacaterol (see Box 4-2, p.71). Effectiveness of fluticasone furoate-vilanterol over usual care was demonstrated for asthma symptom control in a large real-world study, but there was no significant difference in risk of exacerbations.337,338 Maintenance ICS-LABA plus as-needed SABA This is an alternative approach if MART is not possible, or if a patient’s asthma is stable with good adherence and no exacerbations on their current therapy. For patients taking maintenance ICS, changing to maintenance combination ICS-LABA provides additional improvements in symptoms and lung function with a reduced risk of exacerbations compared with the same dose of ICS (Evidence A),339,340 but there is only a small reduction in reliever use.341,342 In these studies, the reliever was as-needed SABA. However, before prescribing a regimen with SABA reliever, consider whether the patient is likely to be adherent with their ICS-containing treatment, as otherwise they will be at higher risk of exacerbations. Maintenance ICS-LABA plus as-needed combination ICS-SABA (≥18 years) Population In the double-blind MANDALA study,343 the population relevant to Step 3 recommendations comprised patients with poor asthma control and a history of severe exacerbations who were taking maintenance low-dose ICS-LABA or medium-dose ICS. In this study, patients were randomized to as-needed ICS-SABA or as-needed SABA, and continued to take their usual maintenance treatment. Evidence In the sub-population taking Step 3 maintenance treatment, as-needed use of 2 inhalations of budesonide-salbutamol (albuterol) 100/100 mcg metered dose (80/90 mcg delivered dose), taken for symptom relief, increased the time to first severe exacerbation by 41% compared with as-needed salbutamol (hazard ratio 0.59; CI 0.42–0.85). The proportion of patients with a clinically important difference in ACQ-5 was slightly higher with the budesonide-salbutamol reliever. A formulation with a lower ICS dose did not significantly reduce severe exacerbations.343 Other considerations There are no head-to-head comparisons between this regimen and ICS-formoterol maintenance-and-reliever therapy (MART), both of which include an anti-inflammatory reliever (AIR). However, as ICS-SABA is not recommended for regular use, and its use as the reliever in Steps 3–5 requires the patient to have different maintenance and reliever inhalers, this regimen is more complex for patients than GINA Track 1 with ICS-formoterol, in which the same medication is used for both maintenance and reliever doses. Transition between treatment steps with as-needed ICS-SABA may also be more complex than in GINA Track 1 as there is only one small study of as-needed-only ICS-SABA (beclometasone-salbutamol) as a Step 2 treatment.324 Practice points A maximum number of 6 as-needed doses (each 2 puffs of 100/100 mcg budesonide-salbutamol [80/90 mcg delivered dose]) can be taken in a day. It is essential to educate patients about the different purpose of their maintenance and COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 89 reliever inhalers, and to train them in correct inhaler technique with both devices if they are different; this also applies to SABA relievers. Track 2 (alternative) Step 4 treatment for adults and adolescents: medium or high-dose ICS-LABA plus as-needed SABA or plus as-needed ICS-SABA Maintenance medium- or high-dose ICS-LABA plus as-needed SABA: This is an alternative approach if MART is not possible, or if a patient’s asthma is stable with good adherence and no exacerbations on their current therapy. As above, individual ICS responsiveness varies, and some patients whose asthma is uncontrolled or who have frequent exacerbations on low-dose ICS-LABA despite good adherence and correct inhaler technique may benefit from maintenance medium-dose ICS-LABA (Evidence B)344 plus as-needed SABA, if MART is not available. However, before prescribing a regimen with SABA reliever, consider whether the patient is likely to be adherent with their ICS-containing treatment, as otherwise they will be at higher risk of exacerbations. Occasionally, high-dose ICS-LABA may be needed. Maintenance ICS-LABA plus as-needed combination ICS-SABA (≥18 years) Population In the double-blind MANDALA study,343 the population relevant to Step 4 recommendations comprised patients with poor asthma control and a history of severe exacerbations who were taking maintenance medium-dose ICS-LABA or high-dose ICS. Evidence In the sub-population of patients who were taking maintenance medium-dose ICS-LABA or high-dose ICS (Step 4 treatment), there was no significant increase in time to first severe exacerbation with as-needed budesonide-salbutamol (albuterol) 2 inhalations of 100/200 mcg metered dose (80/90 mcg delivered dose), compared with as-needed salbutamol (hazard ratio 0.81; CI 0.61–1.07). More studies in this population are needed. Other considerations There are no head-to-head comparisons between this regimen and ICS-formoterol MART, both of which include an anti-inflammatory reliever. However, as ICS-SABA is not recommended for regular use, and its use as the reliever in Steps 3–5 requires the patient to have different maintenance and reliever inhalers, this regimen is more complex for patients than GINA Track 1 with ICS-formoterol in which the same medication is used for both maintenance and reliever doses. Practice points A maximum number of 6 as-needed doses (each 2 puffs of 100/100 mcg budesonide-salbutamol [80/90 mcg delivered dose]) can be taken in a day. It is essential to educate patients about the different purpose of their maintenance and reliever inhalers, and to train them in correct inhaler technique with both devices if they are different; this also applies to SABA relievers. OTHER STEP 1–4 TREATMENTS IN ADULTS AND ADOLESCENTS (TRACKS 1 AND 2) Other Step 1 or 2 treatment options for adults and adolescents These options are shown at the bottom of the main treatment figure (Box 4-6, p.77). They have limited indications, or less evidence for efficacy of safety, than the medications shown in the two treatment tracks. Specific allergen immunotherapy (see p.104): For adult patients sensitized to house dust mite, with suboptimally controlled asthma despite low- to high-dose ICS, consider adding sublingual allergen immunotherapy (SLIT), provided FEV1 is >70% predicted.345,346 Leukotriene receptor antagonists (LTRAs): LTRAs, which include montelukast, zafirlukast and zileuton, are less effective than ICS,347 particularly for exacerbations (Evidence A). Before prescribing montelukast, health professionals should COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 90 consider its benefits and risks, and patients or parents/caregivers should be counselled about the risk of neuropsychiatric events.295 Daily ICS-LABA as initial treatment: Regular daily combination low-dose ICS-LABA as the initial maintenance controller treatment (i.e., in patients previously treated with SABA alone) reduces symptoms and improves lung function, compared with low-dose ICS.348 However, it is more expensive and does not further reduce the risk of exacerbations compared with ICS alone (Evidence A).348 Seasonal ICS-containing treatment: For patients with purely seasonal allergic asthma, e.g., with birch pollen, with no interval asthma symptoms, regular daily ICS or as-needed low-dose ICS-formoterol should be started immediately symptoms commence, and be continued for four weeks after the relevant pollen season ends (Evidence D). Other Step 3 treatment options for adults and adolescents These options are shown at the bottom of the main treatment figure (Box 4-6, p.77). They have limited indications, or less evidence for efficacy of safety, than the medications shown in the two treatment tracks. Specific allergen immunotherapy (see p.104): For adult patients sensitized to house dust mite, with suboptimally controlled asthma despite low- to high-dose ICS, consider adding sublingual allergen immunotherapy (SLIT), provided FEV1 is >70% predicted.345,346 Medium-dose ICS: Another option for adults and adolescents is to increase ICS to medium dose166 (see Box 4-2, p.71) but, at population level, this is less effective than adding a LABA (Evidence A).349,350 Other less efficacious options are low-dose ICS-containing therapy plus either LTRA347 (Evidence A for lower efficacy than ICS) or low-dose, sustained-release theophylline351 (lack of efficacy, and safety concerns). Note the concerns about neuropsychiatric adverse effects with montelukast.295 Other Step 4 treatment options for adults and adolescents These options are shown at the bottom of the main treatment figure (Box 4-6, p.77). They have limited indications, or less evidence for efficacy of safety, than the medications shown in the two treatment tracks. Long-acting muscarinic antagonists: LAMAs may be considered as add-on therapy in a separate inhaler for patients aged ≥6 years (tiotropium), or in a combination (‘triple’) inhaler for patients aged ≥18 years (beclometasone-formoterol-glycopyrronium; fluticasone furoate-vilanterol-umeclidinium; mometasone-indacaterol-glycopyrronium) if asthma is persistently uncontrolled despite medium or high-dose ICS-LABA. Adding a LAMA to medium or high-dose ICS-LABA modestly improved lung function (Evidence A)296,352-356 but with no difference in symptoms. In some studies, adding LAMA to ICS-LABA modestly reduced exacerbations, compared with some medium- or high-dose ICS-LABA comparators.296,353,356 In meta-analyses, there was a 17% reduction in risk of severe exacerbations with addition of LAMA to medium- or high-dose ICS-LABA; sub-group analysis suggested that this benefit was mainly in patients with a history of exacerbations in the previous year.357,358 However, for patients experiencing exacerbations despite low-dose ICS-LABA, the ICS dose should be increased to at least medium, or treatment switched to maintenance-and-reliever therapy with ICS-formoterol, before considering adding a LAMA. In one study, the severe exacerbation rate was lower in patients receiving high-dose fluticasone furoate-vilanterol (ICS-LABA) than with low- to medium-dose fluticasone furoate-vilanterol-umeclidinium (ICS-LABA-LAMA).354 For patients prescribed an ICS-LABA-LAMA with a non-formoterol LABA, the appropriate reliever is SABA or ICS-SABA. In Step 4, there is insufficient evidence to support ICS-LAMA over low- or medium-dose ICS-LABA combination; all studies were with ICS and tiotropium in separate inhalers.352 In one analysis, response to adding LAMA to medium-dose ICS, as assessed by FEV1, ACQ, and exacerbations, was not modified by baseline demographics, body-mass index, FEV1, FEV1 responsiveness, or smoking status (past smoking versus never).359 Allergen immunotherapy (see p.104): Consider adding sublingual allergen immunotherapy (SLIT) for adult patients with sensitization to house dust mite, with suboptimally controlled asthma despite low- to high-dose ICS, but only if FEV1 is >70% predicted.345,346 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 91 Other options: For medium- or high-dose budesonide, efficacy may be improved with dosing four times daily (Evidence B),360,361 but adherence may be an issue. For other ICS, twice-daily dosing is appropriate (Evidence D). Other options for adults or adolescents that can be added to a medium or high-dose ICS, but that are less efficacious than adding LABA, include LTRA (Evidence A),362-366 or low-dose sustained-release theophylline (Evidence B),367 but neither of these has been compared with maintenance-and-reliever therapy with ICS-formoterol. Note the concern about potential neuropsychiatric adverse effects with montelukast.295 STEP 5 (TRACKS 1 AND 2) IN ADULTS AND ADOLESCENTS Preferred treatment at Step 5 in adults and adolescents: refer for expert assessment, phenotyping, and add-on therapy (for more details, see Section 8, p.139) Patients of any age with persistent symptoms or exacerbations despite correct inhaler technique and good adherence with Step 4 treatment, and in whom other controller options have been considered, should be referred to a specialist with expertise in investigation and management of severe asthma, if available (Evidence D).175 In severe asthma, as in mild–moderate asthma,368 participants in randomized controlled trials may not be representative of patients seen in clinical practice. For example, a registry study found that over 80% of patients with severe asthma would have been excluded from major regulatory studies evaluating biologic therapy.369 Recommendations from the GINA Short Guide and decision tree on Diagnosis and Management of difficult-to-treat and severe asthma in adolescent and adult patients are included in Section 8 (p.139). These recommendations emphasize the importance of first optimizing existing therapy and treating modifiable risk factors and comorbidities (see Box 8-2, p.142). If the patient still has uncontrolled symptoms and/or exacerbations, additional treatment options that may be considered may include the following (always check local eligibility and payer criteria). Combination high-dose ICS-LABA Combination high-dose ICS-LABA may be considered in adults and adolescents, but for most patients, the increase in ICS dose generally provides little additional benefit (Evidence A),160,166,350 and there is an increased risk of side-effects, including adrenal suppression.370 A high dose is recommended only on a trial basis for 3–6 months when good asthma control cannot be achieved with medium dose maintenance-and-reliever therapy (MART) with ICS-formoterol or medium-dose ICS plus LABA and/or a third controller (e.g., LTRA or sustained-release theophylline with a SABA reliever (Evidence B).365,367 Note safety concerns with montelukast.295 Maintenance-and-reliever therapy (MART) with ICS-formoterol If a patient treated with medium-dose MART requires addition of biologic therapy, it is not logical to switch them from MART to conventional ICS-LABA plus as-needed SABA, as this may increase the risk of exacerbations. There is little evidence about initiating MART in patients receiving add-on treatment such as LAMA or biologic therapy.15 However, in one study,15 patients with severe eosinophilic asthma that was well controlled on benralizumab and high-dose ICS-LABA were randomized to budesonide-formoterol, either as high dose maintenance plus as-needed SABA, or as medium-dose MART with subsequent 8-weekly options for down-titration. Asthma remained stable after the switch from high-dose ICS-formoterol to medium-dose MART, supporting the safety of MART in this population on Step 5 treatment. Most patients randomized to MART were able to further reduce their ICS-formoterol dose, but there was an increase in FeNO and decrease in FEV1 in those who stepped down to as-needed-only ICS-formoterol,15 suggesting that maintenance doses of ICS-formoterol should not be stopped. Add-on long-acting muscarinic antagonists Add-on long-acting muscarinic antagonists (LAMA) can be prescribed in a separate inhaler (tiotropium), or in a combination (‘triple’) inhaler for patients aged ≥18 years (beclometasone-formoterol-glycopyrronium; fluticasone furoate-vilanterol-umeclidinium; mometasone-indacaterol-glycopyrronium) if asthma is not well controlled with medium or high-dose ICS-LABA. Adding LAMA to ICS-LABA modestly improves lung function (Evidence A),296,352-355,357,359,371 but not COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 92 quality of life, with no clinically important change in symptoms.357,358 Some studies showed a reduction in exacerbation risk; in meta-analyses, overall, there was a 17% reduction in risk of severe exacerbations requiring oral corticosteroids (Evidence A),296,352,353,356,357,371 with subgroup analysis suggesting that this benefit was primarily in patients with a history of exacerbations in the previous year.358 For patients with exacerbations despite ICS-LABA, it is essential that sufficient ICS is given (i.e., at least medium-dose ICS-LABA) before considering adding a LAMA. For patients prescribed an ICS-LABA-LAMA with a non-formoterol LABA, the appropriate reliever is SABA or ICS-SABA; patients prescribed ICS-formoterol-LAMA can continue ICS-formoterol reliever. Azithromycin Add-on azithromycin (three times a week) can be considered after specialist referral for adult patients with persistent symptomatic asthma despite high-dose ICS-LABA. Before considering add-on azithromycin, sputum should be checked for atypical mycobacteria, ECG should be checked for long QTc (and re-checked after a month on treatment), and the risk of increasing antimicrobial resistance should be considered.372 Diarrhea is more common with azithromycin 500 mg 3 times per week.373 Treatment for at least 6 months is suggested, as a clear benefit was not seen by 3 months in the clinical trials.373,374 The evidence for this recommendation includes a meta-analysis of two clinical trials373,374 in adults with persistent asthma symptoms that found reduced asthma exacerbations among those taking medium or high-dose ICS-LABA who had either an eosinophilic or non-eosinophilic profile and in those taking high-dose ICS-LABA (Evidence B).375 The option of add-on azithromycin for adults is recommended only after specialist consultation because of the potential for development of antibiotic resistance at the patient or population level.373 Add-on biologic therapy Options recommended by GINA for patients with uncontrolled severe asthma despite optimized maximal therapy (see more details in Section 8, p.139) include: • Add-on anti-immunoglobulin E (anti-IgE) (subcutaneous (SC) omalizumab) for patients aged ≥ 6 years with severe allergic asthma (Evidence A)376,377 • Add-on anti-interleukin-5/5Rα (SC mepolizumab for ages ≥ 6 years, SC benralizumab for ages ≥12 years, or IV reslizumab for ages ≥18 years) for patients with severe eosinophilic asthma (Evidence A).377-382 • Add-on anti-interleukin-4Rα (SC dupilumab) for patients aged ≥ 6 years with severe eosinophilic/Type 2 asthma, or those requiring treatment with maintenance OCS (Evidence A)377,383-386 • Add-on anti-thymic stromal lymphopoietin (anti-TSLP) (SC tezepelumab) for patients aged ≥12 years with severe asthma (Evidence A).387-389 Sputum-guided treatment For adults with persisting symptoms and/or exacerbations despite high-dose ICS or ICS-LABA, treatment may be adjusted based on eosinophilia (>3%) in induced sputum. In severe asthma, this strategy leads to reduced exacerbations and/or lower doses of ICS (Evidence A),390 but few clinicians currently have access to routine sputum testing. Bronchial thermoplasty Add-on treatment with bronchial thermoplasty may be considered for some adult patients with severe asthma (Evidence B).175,391 Evidence is limited and in selected patients (see Bronchial thermoplasty, p.106). The long-term effects compared with control patients, including for lung function, are not known. Oral corticosteroids As a last resort, add-on low-dose OCS (≤7.5 mg/day prednisone equivalent) may be considered for some adults with severe asthma (Evidence D),175 but maintenance OCS is often associated with substantial cumulative side effects (Evidence A).225,392-394 It should only be considered for adults with poor symptom control and/or frequent exacerbations despite good inhaler technique and adherence with Step 5 treatment, and after exclusion of other contributory factors and trial of other add-on treatments including biologics where available and affordable. Patients should be counseled about potential side-effects.393,394 They should be assessed and monitored for risk of adrenal suppression and corticosteroid-COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 93 induced osteoporosis, and those expected to be treated for ≥3 months should be provided with relevant lifestyle counseling and prescription of therapy for prevention of osteoporosis and fragility fractures (where appropriate).395 NON-RECOMMENDED BRONCHODILATORS Anticholinergic agents in the absence of ICS: In adults, inhaled short-acting muscarinic antagonists (SAMA) like ipratropium are potential alternatives to SABA for routine relief of asthma symptoms; however, these agents have a slower onset of action than inhaled SABA. Like SABAs (p.87) they should not be used without ICS. Use of long-acting muscarinic antagonists (LAMA) in asthma without concomitant ICS is associated with an increased risk of severe exacerbations.396 Oral bronchodilators: Oral SABA and theophylline have a higher risk of side-effects and are not recommended. For clinicians in regions without access to inhaled therapies, advice on minimizing the frequency and dose of these oral medications has been provided elsewhere.27 No long-term safety studies have been performed to assess the risk of severe exacerbations associated with oral SABA or theophylline use in patients not also taking ICS. Formoterol without ICS: The rapid-onset LABA, formoterol, is as effective as SABA as a reliever medication in adults and children,397 and reduces the risk of severe exacerbations by 15–45%, compared with as-needed SABA,323,398,399 but use of regular or frequent LABA without ICS is strongly discouraged because of the risk of exacerbations (Evidence A).151,400 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 94 ABOUT ASTHMA TREATMENT FOR CHILDREN 6–11 YEARS For general principles of asthma treatment, and non-pharmacological strategies, see Section 3, p.48. For flowchart on initial asthma treatment for children 6–11 years, see p.94. For asthma treatment steps in children 6–11 years, see p.96. INITIAL ASTHMA TREATMENT IN CHILDREN 6–11 YEARS Box 4-10. Initial asthma treatment for children aged 6–11 years with a diagnosis of asthma These recommendations are based on evidence, where available, and on consensus. Presenting symptoms Preferred INITIAL treatment Infrequent asthma symptoms, e.g., 1–2 days/week or less Low-dose ICS taken whenever SABA is taken (Evidence B) In combination or in separate inhalers Asthma symptoms 2–5 days/week Low-dose ICS plus as-needed SABA (Evidence A) Other options include taking ICS whenever SABA is taken in combination or separate inhalers (Evidence B), or daily LTRA† (Evidence A for less effectiveness for exacerbations than ICS). Consider likely adherence with maintenance treatment if reliever is SABA. Asthma symptoms most days (e.g., 4–5 days/week); or waking due to asthma once a week or more Low-dose ICS-LABA plus as needed SABA (Evidence A), OR Medium-dose ICS plus as-needed SABA (Evidence A), OR Very-low-dose ICS-formoterol maintenance-and-reliever (Evidence B) Other options include daily low-dose ICS and LTRA†, plus as-needed SABA. Daily asthma symptoms, waking at night once or more a week, and low lung function Medium-dose ICS-LABA plus as-needed SABA, OR low-dose ICS-formoterol maintenance-and-reliever (MART). Initial asthma presentation is during an acute exacerbation. Treat as for exacerbation (Box 9-4, p.167), including a short course of OCS if the exacerbation is severe; commence Step 3 or Step 4 treatment, and arrange follow-up. Before starting initial controller treatment • Record evidence for the diagnosis of asthma, if possible.
• Record the child’s level of symptom control and risk factors, including lung function (Box 2-2, p.37; Box 2-3, p.40).
• Consider factors influencing choice between available treatment options (Box 3-4, p.54).
• Choose a suitable inhaler (Box 5-1, p.109) and ensure that the child can use the inhaler correctly.
• Schedule an appointment for a follow-up visit.
After starting initial controller treatment • Review child’s response (Box 2-2, p.37) after 2–3 months, or earlier depending on clinical urgency.
• See Box 4-12 (p.96) for recommendations for ongoing treatment and other key management issues.
• Step down treatment once good control has been maintained for 3 months (Box 4-13, p.102).
This advice is based on evidence from available studies and from consensus, including considerations of cost. †If prescribing LTRA, advise about the risk of neuropsychiatric adverse effects.295 See Box 4-2 (p.71) for low, medium and high ICS doses in children, and Box 4-8 (p.84) for MART doses in children. See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 95 Box 4-11. Flowchart for selecting initial treatment in children aged 6–11 years with a diagnosis of asthma These recommendations are based on evidence, where available, and on consensus. See list of abbreviations (p.11). See Box 4-2 (p.71) for low, medium and high ICS doses in children. See Box 4-8 (p.84) for medications and doses for MART in children. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 96 ASTHMA TREATMENT STEPS FOR CHILDREN 6–11 YEARS Box 4-12. Personalized management for children 6–11 years to control symptoms and minimize future risk See list of abbreviations (p.11). Anti-inflammatory reliever therapy (AIR); see Box 4-8. †If prescribing leukotriene receptor antagonists, note concerns about potential neuropsychiatric adverse effects.295 For initial asthma treatment in children aged 6–11 years, see Box 4-10 (p.94) and 4-11 (p.95). See Box 4-2 (p.71) for low, medium and high ICS doses in children. See Box 4-8 (p.84) for MART doses for children 6–11 years. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 97 The steps below refer to the recommended asthma treatment options shown in Box 4-12 p.96. Suggested low, medium and high doses for a range of ICS formulations are shown in Box 4-2 (p.71). Preferred Step 1 treatment for children 6–11 years: taking ICS whenever SABA is taken For children 6–11 years with asthma symptoms that are well controlled on low-dose ICS, or who are using SABA alone and have symptoms less than twice a week, the recommended treatment is taking ICS whenever SABA is taken. Populations studied The TREXA study325 was in children 5–18 years, with mild persistent asthma that was well controlled during a 4-week run-in on low-dose ICS with as-needed SABA. The ASIST study327 was in African-American children aged 6–17 years, whose asthma was well controlled on low-dose ICS with as-needed SABA in a run-in period of 2–4 weeks. Results for children 6– 11 years have not been published separately. Evidence Both studies used separate albuterol and beclometasone dipropionate (BDP) 50 mcg [40 mcg delivered dose] inhalers for the intervention, 2 puffs of BDP for each 2 puffs of albuterol (with the inhalers taped together, back-to-back, in the TREXA study. In the TREXA study, the comparators were as-needed SABA and as-needed ICS+SABA, each with or without regular ICS. The highest rate of exacerbations was among the children receiving SABA alone, and there was a significant reduction in treatment failures in the group that took ICS whenever SABA was taken, as well as in the other ICS-containing groups.325 In the ASIST study, symptom-based adjustment of ICS dose was associated with similar outcomes as with physician-adjusted treatment, with lower average ICS dose (Evidence B).327 Exacerbations and symptoms were similar with this regimen as with maintenance ICS plus as-needed SABA. Other considerations Neither of these studies was sufficiently powered to examine severe exacerbations as an outcome. In the TREXA study, there were no differences in asthma symptom control or airway hyperresponsiveness between the treatment groups. The children receiving daily ICS had lower linear growth than those receiving as-needed SABA or as-needed ICS+SABA.325 In the ASIST study, interviews with parents/caregivers indicated that those whose children were randomized to as-needed ICS-SABA felt more in control of their child’s asthma than those whose children were randomized to physician-based adjustment.327 Concerns around SABA-only treatment are also relevant to children, and should be considered when initiating Step 1 treatment (see other controller options for children in Step 2, below). Studies of as needed-only ICS-formoterol in children aged 6–11 years are underway. Not recommended SABA-only treatment is not recommended in children 6–11 years, as for adults and adolescents. Although inhaled SABAs are highly effective for the quick relief of asthma symptoms,328 children whose asthma is treated with SABA alone (compared with ICS) are at increased risk of asthma-related death (Evidence A)87,329 and urgent asthma-related health care (Evidence A),330 even if they have good symptom control.331 In children, dispensing of three or more SABA inhalers in a year is associated with a doubling of risk of emergency department presentation. Oral SABA and theophylline are not recommended because of the higher risk of side-effects and lower efficacy. For clinicians in regions without access to inhaled therapies, advice on minimizing the frequency and dose of these oral medications has been provided elsewhere.27 No long-term safety studies have been performed to assess the risk of severe exacerbations associated with oral SABA or theophylline use in children not also taking ICS. The rapid-onset LABA, formoterol, is as effective as SABA as a reliever medication in children as well as in adults,397 and reduces the risk of severe exacerbations by 15–45%, compared with as-needed SABA,323,398,399 but use of regular or frequent LABA without ICS is strongly discouraged because of the risk of exacerbations (Evidence A).151,400 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 98 Preferred Step 2 treatment for children 6–11 years: regular low-dose ICS plus as-needed SABA The preferred controller option for children at Step 2 is regular low-dose ICS plus as-needed SABA (see Box 4-2, p.71 for ICS dose ranges in children). This reduces the risk of serious exacerbations compared with SABA-only treatment.307 Evidence Evidence in children includes the large long-term START study, in which patients aged 6–66 years with newly diagnosed asthma were provided with placebo or low-dose budesonide (200 mcg/day for children <11 years) for 3 years. Low-dose ICS reduced the risk of serious exacerbations by 40%, improved lung function, increased symptom-free days and decreased days lost from school years).401 Alternative Step 2 treatment option for children 6–11 years: taking low-dose ICS whenever SABA is taken Another alternative option at Step 2 is daily LTRA, which, overall, is less effective than ICS,347 and there are concerns about potential neuropyschiatric adverse events.295 Not recommended Sustained-release theophylline has only weak efficacy in asthma (Evidence B)367,402,403 and side-effects are common, and may be life-threatening at higher doses.404 Chromones (nedocromil sodium and sodium cromoglycate) have been discontinued globally; these had a favorable safety profile but low efficacy (Evidence A),405-407 and their pMDI inhalers required burdensome daily washing to avoid blockage. Preferred Step 3 treatment options for children 6–11 years: regular medium dose ICS or low-dose ICS-LABA plus SABA reliever, or MART with very low-dose ICS-formoterol In children, after checking inhaler technique and adherence, and treating modifiable risk factors, there are three preferred options at a population level: • Increase ICS to medium dose (see Box 4-2, p.71) plus as-needed SABA reliever (Evidence A),408 or • Change to combination low-dose ICS-LABA plus as-needed SABA reliever (Evidence A),409 or • Switch to maintenance-and-reliever therapy (MART) with a very low dose of ICS-formoterol (Evidence B).410 For a summary of medications and doses, see Box 4-8 (p.84). Evidence In a large study of children aged 4–11 years with a history of an exacerbation in the previous year, combination ICS-LABA was non-inferior to the same dose of ICS alone for severe exacerbations, with no difference in symptom control or reliever use.411 In children, a single study of maintenance-and-reliever therapy (MART) with very low-dose budesonide-formoterol (100/6 metered dose, 80/4.5 mcg delivered dose for both maintenance and reliever) showed a large reduction in exacerbations, compared with the same dose of budesonide-formoterol plus SABA reliever, or compared with higher-dose ICS.410 Individual children’s responses vary, so try the other controller options above before considering Step 4 treatment.412 Other Step 3 treatment options for children 6–11 years A 2014 systematic review and meta-analysis did not support the addition of LTRA to low-dose ICS in children.413 Note concerns about the risk of neuropsychiatric adverse effects.295 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 99 Preferred Step 4 treatment options for children 6–11 years: refer for expert advice, or increase treatment to medium-dose ICS-LABA plus as-needed SABA, or MART with low-dose ICS-formoterol For children whose asthma is not adequately controlled by low-dose maintenance ICS-LABA with as-needed SABA, consider referral for expert advice. Alternatively, treatment may be increased to medium-dose ICS-LABA (Evidence B).411 For maintenance-and-reliever therapy (MART) with budesonide-formoterol, the maintenance dose may be increased to 100/6 mcg twice daily (metered dose; 80/4.5 mcg delivered dose) (Evidence D); this is still a low-dose regimen. For a summary of medications and doses, see Box 4-8 (p.84). If asthma is not well controlled on medium-dose ICS (Box 4-2B, p.71), refer the child for expert assessment and advice. Other Step 4 options for children 6–11 years that may be considered after referral include: Increasing ICS-LABA dose: increasing the ICS-LABA dose to a high pediatric ICS dose (Box 4-2B, p.71) can be considered, but adverse effects must be considered. Tiotropium: Tiotropium (a long-acting muscarinic antagonist) by mist inhaler may be used as add-on therapy in children aged 6 years and older. It modestly improves lung function and reduces exacerbations (Evidence A),371 largely independent of baseline IgE or blood eosinophils.414 LTRA: If not trialed before, LTRA could be added (but note the concern about risks of neuropsychiatric adverse effects with montelukast).295 Add-on theophylline is not recommended for use in children due to lack of efficacy and safety data. Preferred treatment at Step 5 in children 6–11 years: refer for expert assessment, phenotyping, and add-on therapy Children with persistent asthma symptoms or exacerbations despite correct inhaler technique and good adherence with Step 4 treatment and in whom other controller options have been considered, should be referred to a specialist with expertise in investigation and management of severe asthma, if available (Evidence D).175 In severe asthma, as in mild–moderate asthma,368 participants in randomized controlled trials may not be representative of patients seen in clinical practice. For example, a registry study found that over 80% of patients with severe asthma would have been excluded from major regulatory studies evaluating biologic therapy.369 Add-on long-acting muscarinic antagonists Tiotropium, a long-acting muscarinic antagonists (LAMA), can be prescribed as an add-on treatment in a separate inhaler for patients aged ≥6 years if asthma is not well controlled with medium or high-dose ICS-LABA.371,414 Add-on biologic therapy Options recommended by GINA for children aged 6–11 years with uncontrolled severe asthma despite optimized maximal therapy (see chapter 3.5 for more details) include: • Add-on anti-immunoglobulin E (anti-IgE) (omalizumab) for patients aged ≥6 years with severe allergic asthma (Evidence A)376 • Add-on anti-interleukin-5/5Rα (subcutaneous mepolizumab for patients aged ≥6 years with severe eosinophilic asthma (Evidence A).381,382 • Add-on anti-interleukin-4Rα (subcutaneous dupilumab) for patients aged ≥6 years with severe eosinophilic/Type 2 asthma.386 Maintenance-and-reliever therapy (MART) with ICS-formoterol There is no direct evidence about initiating MART in children receiving add-on treatment such as LAMA or biologic therapy. Switching a patient from MART to conventional ICS-LABA plus as-needed SABA may increase the risk of exacerbations. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 100 REVIEWING RESPONSE AND ADJUSTING TREATMENT – ADULTS, ADOLESCENTS AND CHILDREN 6–11 YEARS How often should asthma be reviewed? Each patient’s asthma should be reviewed regularly to monitor symptom control, risk factors and occurrence of exacerbations, and to document response to any treatment changes. For most controller medications, improvement in symptoms and lung function begins within days of initiating treatment, but the full benefit may only be reached after 3–4 months,415 or even later in patients with severe and chronically under-treated asthma.416 All healthcare providers should be encouraged to assess asthma control, adherence and inhaler technique at every visit, not just when the patient presents because of their asthma.417 The frequency of visits depends upon the patient’s initial level of control, their response to treatment, and their level of engagement in self-management. Ideally, patients should be seen 1–3 months after starting treatment and every 3–12 months thereafter. After an exacerbation, a review visit within 1 week should be scheduled (Evidence D).418 Stepping up asthma treatment Asthma is a variable condition, and adjustments of controller treatment by the clinician and/or the patient may be needed.419 Day-to-day adjustment using an anti-inflammatory reliever (AIR) For patients whose reliever inhaler is combination budesonide-formoterol or beclometasone-formoterol (with or without maintenance ICS-formoterol), the patient adjusts the number of as needed doses of ICS-formoterol from day to day according to their symptoms. This strategy reduces the risk of developing a severe exacerbation requiring OCS within the next 3–4 weeks.126-128 As-needed combination budesonide-salbutamol is an anti-inflammatory reliever option that has been studied in Steps 3–5.343 Short-term step-up (for 1–2 weeks) A short-term increase in maintenance ICS dose for 1–2 weeks may be necessary (e.g., during viral infections or seasonal allergen exposure). This increase may be initiated by the patient according to their written asthma action plan (Box 9-2, p.162), or by the healthcare provider. Sustained step-up (for at least 2–3 months) Although at a group level most benefit from ICS is obtained at low dose, individual ICS responsiveness varies; some patients whose asthma is uncontrolled on low-dose ICS-LABA despite good adherence and correct technique may benefit from increasing the maintenance dose to medium. A step-up in treatment may be recommended (Box 4-6, p.77) after confirming that the symptoms are due to asthma, inhaler technique and adherence are satisfactory, and modifiable risk factors such as smoking have been addressed (Box 3-5, p.55). Any step-up should be regarded as a therapeutic trial; if there is no response after 2–3 months, treatment should be reduced to the previous level, and alternative treatments or referral considered. Stepping down treatment when asthma is well controlled Once good asthma control has been achieved and maintained for 2–3 months and lung function has reached a plateau, treatment can often be successfully reduced, without loss of asthma control. The aims of stepping down are: • To find the patient’s minimum effective treatment, i.e., to maintain good control of symptoms and exacerbations, and to minimize the costs of treatment and potential for side-effects • To encourage the patient to continue ICS-containing treatment. Patients prescribed maintenance controller treatment in either Track often experiment with intermittent treatment through concern about the risks or costs of daily treatment.420 For patients prescribed GINA Track 1 MART, the ICS-formoterol reliever provides a safety net during planned step-down or if adherence with maintenance doses is poor. However, for patients prescribed maintenance COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 101 controller with a SABA reliever (GINA Track 2, Steps 2–5), poor adherence leaves them exposed to the risks of SABA-only treatment. Step-down options for patients on different treatment steps are shown in Box 4-13 (p.102). Before stepping down The approach to stepping down will differ from patient to patient depending on their current treatment, risk factors and preferences. There are few data on the optimal timing, sequence and magnitude of treatment reductions in asthma. Factors associated with a greater risk of exacerbation after step-down include a history of exacerbations and/or emergency department visit for asthma in the previous 12 months,421,422 and a low baseline FEV1.422 Other predictors of loss of control during dose reduction include airway hyperresponsiveness and sputum eosinophilia,423 but these tests are not readily available in primary care. Any treatment step-down should be considered as a therapeutic trial, evaluating the response in terms of both symptom control and exacerbation frequency. Before stepping down, the patient should be given a written asthma action plan and instructions for how and when to re-start their previous treatment if their symptoms worsen. How to step asthma treatment down Decisions about treatment step-down should be based on individual assessment. In one study of patients with well-controlled asthma on medium-dose ICS-LABA, reducing the ICS dose and removing the LABA had similar effects on a composite treatment failure outcome. However, stopping LABA was associated with lower lung function and more hospitalizations, and decreasing the ICS dose was inferior to maintaining a stable dose of ICS-LABA.424 If treatment is stepped down too far or too quickly, the risk of exacerbations may increase even if symptoms remain reasonably controlled (Evidence B).425 Higher baseline FeNO has not been demonstrated to predict exacerbations following step-down of ICS dose.426,427 A meta-analysis suggested that greater reduction in ICS dose may be able to be achieved in patients with baseline FeNO <50 ppb, but the findings point to the need for further research.427 Complete cessation of ICS is associated with a significantly increased risk of exacerbations (Evidence A).428 Stepping down from daily low-dose ICS plus as-needed SABA to as needed-only ICS-formoterol provides similar or greater protection from severe exacerbations and need for urgent health care, with similar symptom control and lung function and a much lower average daily ICS dose, compared with treatment with daily low-dose ICS plus as-needed SABA.183 Step-down strategies for different controller treatments are summarized in Box 4-13 (p.102); these are based on current evidence, but more research is needed. Few step-down studies have been performed in children. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 102 Box 4-13. Options for stepping down treatment in adults and adolescents once asthma is well controlled General principles of stepping down asthma treatment • Consider stepping down when asthma symptoms have been well controlled and lung function has been stable for at least 3 months (Evidence D). If the patient has risk factors for exacerbations (Box 2-2, p.37), for example a history of exacerbations in the past year,421 or persistent airflow limitation, step down only with close supervision.
• Choose an appropriate time (no respiratory infection, patient not travelling, not pregnant).
• Approach each step as a therapeutic trial: engage the patient in the process, document their asthma status (symptom control, lung function and risk factors, Box 2-2, p.37), provide clear instructions, provide a written asthma action plan (Box 9-2, p.162) and ensure the patient has sufficient medication to resume their previous dose if necessary, monitor symptoms and/or PEF, and schedule a follow-up visit (Evidence D).
• Stepping down ICS doses by 25–50% at 3-month intervals is feasible and safe for most patients (Evidence A).429 Current step Current medication and dose Options for stepping down if asthma is well controlled and lung function stable for ≥3 months Evidence Step 5 High-dose ICS-LABA plus oral corticosteroids (OCS) If Type 2-high severe asthma, add biologic therapy if eligible and reduce OCS (see Box 9-5, p.144 for more details) A Optimize inhaled therapy to reduce OCS dose D Use sputum-guided approach to reducing OCS B For low-dose OCS, use alternate-day dosing D Biologic therapy plus high-dose ICS-LABA Cease other add-on medications especially OCS, then consider reducing ICS-LABA dose15 (see Box 8-5 (p.145) and p.145). B Step 4 Moderate- to high-dose ICS-LABA maintenance treatment Continue combination ICS-LABA and reduce ICS component by 50%, by using available formulations B Caution: Discontinuing LABA may lead to deterioration430 A Switch to maintenance-and-reliever therapy (MART) with ICS-formoterol, with lower maintenance dose320 A Medium-dose ICS-formoterol as maintenance and reliever Reduce maintenance ICS-formoterol to low dose, and continue as-needed low-dose ICS-formoterol reliever D High-dose ICS plus second controller Reduce ICS dose by 50% and continue second controller429 B Step 3 Low-dose ICS-LABA maintenance Reduce ICS-LABA to once daily D Caution: Discontinuing LABA may lead to deterioration430 A Low-dose ICS-formoterol as maintenance and reliever Reduce maintenance ICS-formoterol dose to once daily and continue as needed low-dose ICS-formoterol reliever C Consider stepping down to as-needed-only low-dose ICS-formoterol D Medium- or high-dose ICS Reduce ICS dose by 50%429 A Adding LABA may allow ICS dose to be stepped down431 B Step 2 Low-dose maintenance ICS Once-daily dosing (budesonide, ciclesonide, mometasone, fluticasone furoate)432,433 A Switch to as-needed-only low-dose ICS-formoterol188,301,302,308 A Switch to taking ICS whenever SABA is taken324-327 B Low-dose maintenance ICS Switch to as-needed-only low-dose ICS formoterol188,301,302,308 A Caution: Do not completely stop ICS, because the risk of exacerbations is increased with SABA-only treatment308,428 A See list of abbreviations (p.11). MART: low-dose budesonide-formoterol or beclometasone-formoterol (p.69). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 103 Other strategies for adjusting asthma treatment Some alternative strategies for adjusting asthma maintenance ICS-containing treatment have been evaluated: • Treatment guided by sputum eosinophil count: in adults, this approach, when compared with guidelines-based treatment, leads to a reduced risk of exacerbations and similar levels of symptom control and lung function in patients with frequent exacerbations and moderate-severe asthma.390 However, few clinics have routine access to induced sputum analysis. There is insufficient evidence to assess this approach in children.390 Sputum-guided treatment is recommended for adult patients with moderate or severe asthma who are managed in (or can be referred to) centers experienced in this technique (Evidence A).175,390 • Treatment guided by fractional concentration of exhaled nitric oxide (FeNO): In several studies of FeNO-guided treatment, problems with the design of the intervention and/or control algorithms make comparisons and conclusions difficult.434 Results of FeNO measurement at a single point in time should be interpreted with caution.50,299 The relationship between FeNO and other Type 2 biomarkers is lost or altered in obesity.24,48 In a 2016 meta-analysis, FeNO-guided treatment in children and young adults with asthma was associated with a significant reduction in the number of patients with ≥1 exacerbation (OR 0.67; 95% CI 0.51–0.90) and in exacerbation rate (mean difference –0.27; 95% –0.49 to –0.06 per year) compared with guidelines-based treatment (Evidence A);435 FeNO-guided treatment was associated with similar benefits when compared with non-guidelines-based algorithms.435 However, a subsequent good-quality multicenter clinical trial in children with asthma in secondary and primary care centers found that the addition of FeNO to symptom-guided treatment did not reduce severe exacerbations over 12 months.436 In non-smoking adults with asthma, no significant reduction in risk of exacerbations and in exacerbation rates was observed with FeNO-guided treatment, compared with treatment strategies similar to those in most guidelines; a difference was seen only in studies with other (non-typical) comparator approaches to adjustment of treatment.437 In a large study in pregnant women, there was no reduction in exacerbations with FeNO-guided treatment compared with usual care.16 In adults and in children, no significant differences were seen in symptoms or ICS dose with FeNO-guided treatment compared with other strategies.435,437 • Treatment guided by combination biomarkers: An RCT in patients taking high-dose ICS-LABA compared a treatment adjustment strategy based on a composite of T2 biomarkers only with an algorithm based on ACQ-7 and history of recent exacerbation, but the findings were inconclusive because a substantial proportion of patients did not follow recommendations for treatment change.438 • Selection of add-on treatment for patients with severe asthma: The assessment of severe asthma includes identification of the inflammatory phenotype, based on blood or sputum eosinophils or FeNO, to assess the patient’s eligibility for various add-on treatments including biologic therapy. A higher baseline blood eosinophil count and/or FeNO predicts a good asthma response to some biologic therapies (see Box 8-3, p.143 and Box 8-4, p.144).
Further studies are needed to identify the subpopulations of patients with asthma who are most likely to benefit from biomarker-guided adjustment of maintenance ICS-containing treatment, and the optimal frequency of monitoring, including for corticosteroid de-escalation strategies. Until more definitive evidence is available to support a specific strategy, GINA continues to recommend a comprehensive clinical evaluation that includes patient-reported symptoms as well as modifiable risk factors, environmental exposures comorbidities and patient preferences, when making treatment decisions for individual patients. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 104 ALLERGEN IMMUNOTHERAPY Allergen-specific immunotherapy may be considered as add-on therapy for adults and children with asthma who have clinically significant sensitization to aeroallergens, including in those with allergic rhinitis.10,11,439,440 It involves the identification of clinically relevant allergens and the administration of extracts in precisely calculated doses to induce desensitization and/or tolerance. Allergen immunotherapy is currently the only intervention with both an immune modifying effect and long-term efficacy on the allergic response. There are two approaches: subcutaneous immunotherapy (SCIT) and sublingual immunotherapy (SLIT). Few studies reporting effects of allergen immunotherapy on asthma have compared immunotherapy with pharmacological therapy, or used standardized outcomes such as exacerbations; furthermore, most studies have been performed in patients with mild asthma.441 The allergens most often tested in allergen immunotherapy studies are house dust mite and grass pollens. There is insufficient evidence about the safety and efficacy of allergen immunotherapy in patients with asthma who are sensitized to mold.442 More studies are needed to clarify the role of allergen immunotherapy in the development and progression of asthma, and in clinical asthma management.441 There are two approaches: subcutaneous immunotherapy (SCIT) and sublingual immunotherapy (SLIT). Subcutaneous immunotherapy (SCIT) SCIT involves the administration of extracts in progressively higher doses, usually over a period of 3–5 years. There is considerable variation in the specific SCIT regimens used in clinical practice. Efficacy of SCIT for treatment of asthma Systematic reviews and meta-analysis of SCIT in the treatment of adult and pediatric asthma concluded that addition of SCIT led to a reduction in ICS dose requirement and/or the proportion of patients requiring ICS (moderate strength of evidence) and may improve asthma-specific quality of life and lung function, while reducing reliever use and the need for systemic corticosteroids (low strength of evidence).11,440 Few studies of SCIT to house dust mite have been conducted only in children, or report results for children separately.440 A 2020 systematic review of allergen immunotherapy in children with asthma aged 18 years of age and younger reported that SCIT led to a reduction in ICS requirement (moderate strength of evidence), and improved asthma-related quality of life and lung function (low strength of evidence).10 Safety Safety data, overall, suggest that severe allergic reactions occur in fewer than 0.5–0.7% of patients treated with SCIT.443 Serious adverse effects of SCIT are rare, but may include life-threatening anaphylactic reactions. Asthma, especially severe or uncontrolled asthma, has been identified as a major risk factor for severe and fatal adverse reactions to SCIT.444 Food allergy is also a risk factor for systemic reactions to SCIT. Advice • When considering SCIT for adults or children with asthma, the potential benefits, compared with pharmacological treatment and allergen avoidance, must be weighed against the risk of adverse effects and the inconvenience and cost of the prolonged course of therapy (typically 3–5 years), including the minimum 30 minutes of monitoring required after each injection (Evidence D).
• If allergen immunotherapy is considered for patients with severe asthma, the potential benefits and risks should be carefully identified and discussed as part of a shared decision-making process. To minimize the risk of severe reactions, SCIT should not be initiated until good asthma control (symptom control and risk factors for exacerbations) has been established.
• For each patient, SCIT should be tailored to their specific pattern of allergic sensitization. Given the complexity of making up SCIT extracts, combined with the risk of serious adverse events, SCIT prescription and administration should be limited to practitioners who are specifically trained and experienced in allergy testing and in the COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 105 formulation and administration of SCIT. Injections should be administered only in a healthcare setting with capability for, and personnel skilled in the management of, severe allergic reactions/anaphylaxis. SCIT should be administered only with high-quality extracts, and standardized extracts should be used, where available. • Healthcare professionals who offer SCIT must establish effective safety protocols. The risk of severe adverse events is significantly reduced by systems that ensure appropriate supervision after injections, including training of office staff to track time after injections and monitor patient checkout.444 Sublingual immunotherapy (SLIT) Sublingual immunotherapy involves the administration of extracts either as tablet or drops administered under the tongue, with an induction phase in which the dose is progressively increased. The duration of SLIT depends on the allergens used (house dust mite or grass pollen). Efficacy of SLIT for treatment of asthma Several systematic reviews have examined the effect of SLIT for asthma in adults and children,445,446 but many of the studies were unblinded or used non-standardized outcomes. In general, there is limited evidence demonstrating effects of SLIT on important outcomes such as asthma exacerbations and quality of life,446 and few RCTs have compared SLIT with pharmacological therapy for asthma. A 2020 Cochrane review of 66 trials of SLIT for allergic rhinitis, in which at least 80% of participants also had allergic asthma, concluded that addition of SLIT may reduce the risk of asthma exacerbation requiring OCS or healthcare visits (low strength of evidence), but only one study in adults and one in children reported effects on healthcare visits.446 In a 2023 systematic review focusing on individuals (mainly adults) with allergic rhinitis and asthma, SLIT was associated with a significant reduction in asthma symptoms, compared with placebo, but there was no effect on ICS dose, FeNO, lung function or direct treatment cost.447 House dust mite SLIT: European Academy of Allergy & Clinical Immunology (EAACI) guidelines recommend HDM SLIT as add-on treatment in adults with controlled or partially controlled HDM-driven allergic asthma.440 In a subsequent systematic review, addition of standardized HDM SLIT resulted in reduction in ICS dose in one RCT and improved asthma symptoms in two RCTs but there was no consistent effect on exacerbations in adolescents and adults with well or partly controlled asthma.439 There is no separate evidence for adolescents, but no reason to suppose that effectiveness and/or safety would be different than in adults. Ragweed SLIT: In children with allergic rhinoconjunctivitis and asthma who were sensitized to ragweed, ragweed SLIT reduced SABA use and nocturnal awakenings during peak ragweed season.448 Safety The rate of serious adverse events associated with SLIT, as reported in RCTs, is estimated at ≤1% (moderate certainty of evidence)446 with rare cases of anaphylaxis requiring epinephrine.439 In a real-world study, the incidence of serious adverse events was lower among those receiving SLIT than among those receiving SCIT449. Adverse events due to SLIT for inhalant allergens are mainly limited to oral and gastrointestinal symptoms and usually reported to be transient and mild.446,450-453 Advice • For adult or adolescent patients with asthma who are sensitized to house dust mite, with persisting asthma symptoms despite low- to medium-dose ICS-containing therapy, consider adding HDM SLIT, but only if FEV1 is >70% predicted (Evidence B).
• For children with asthma sensitized to ragweed, consider adding SLIT before and during the ragweed season, provided FEV1 is ≥80% predicted. There is insufficient evidence to make a recommendation about HDM SLIT in children with asthma.
• As for any treatment, the potential benefits of SLIT for individual patients should include shared decision making and be weighed against the risk of adverse events and the cost for the patient and the health system.
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 106 VACCINATIONS Influenza Influenza causes significant morbidity and mortality in the general population, and contributes to some acute asthma exacerbations. In 2020, the first year of the COVID-19 pandemic, many countries reported a reduction in influenza-related illness, likely due to the handwashing, masks and social/physical distancing introduced because of the pandemic.454,455 The risk of influenza infection itself can be reduced by annual vaccination. A 2013 systematic review of placebo-controlled randomized controlled trials of influenza vaccination showed no reduction in asthma exacerbations,456 but no such studies had been performed since 2001. A 2017 systematic review and meta-analysis, which included observational studies with a wide range of study designs, suggested that influenza vaccination reduced the risk of asthma exacerbations, but bias could not be excluded for most of the studies.457 There is no evidence for an increase in asthma exacerbations after influenza vaccination compared with placebo.457 A systematic review of studies in individuals aged 2–49 years with mild– moderate asthma found no significant safety concerns or increased risk for asthma-related outcomes after influenza vaccination with live attenuated virus.458 Respiratory syncytial virus Respiratory syncytial virus (RSV) infection causes lower respiratory tract disease in infants, including bronchiolitis and pneumonia. It also causes lower respiratory tract infections in older children and adults, and may exacerbate asthma. Children and the elderly are more likely to experience severe disease with RSV infection. RSV vaccines prevent RSV-related acute respiratory infection; an adjuvanted RSV-subunit vaccine reduced upper and lower respiratory tract disease in adults 60 years or older, including in those with underlying coexisting conditions such as asthma.459,460 Other vaccines People with asthma, particularly children and the elderly, are at higher risk of pneumococcal disease.461 Pneumococcal vaccine protects against invasive pneumococcal infection, but asthma alone is not a specific indication for pneumococcal vaccination.462 Pertussis infection may trigger or mimic asthma exacerbations, and pertussis vaccination reduces the risk of severe pertussis-related disease, but there is limited evidence on the efficacy and safety of vaccines in preventing asthma exacerbations in adults (and hence for an asthma-specific recommendation). For information about COVID-19 vaccines, see p.122. Advice • Advise patients with moderate to severe asthma to receive an influenza vaccination every year, or at least when vaccination of the general population is advised (Evidence C). Follow local immunization schedules.
• Encourage children, adults and the elderly with asthma to follow their local immunization schedule, including for pneumococcal, pertussis, influenza, RSV and COVID-19 vaccinations. Advice about COVID-19 vaccination is on p.122.
• COVID-19 vaccination and influenza vaccination may be given on the same day.
OTHER THERAPIES Bronchial thermoplasty Bronchial thermoplasty is a potential treatment option at Step 5 in some countries for adult patients whose asthma remains uncontrolled despite optimized therapeutic regimens and referral to an asthma specialty center (Evidence B). Bronchial thermoplasty involves treatment of the airways during three separate bronchoscopies with a localized radiofrequency pulse.150 The treatment is associated with a large placebo effect.150 In patients taking high-dose ICS-LABA, bronchial thermoplasty was associated with an increase in asthma exacerbations during the 3 month treatment period, and a subsequent decrease in exacerbations, but no beneficial effect on lung function or asthma symptoms compared with sham-controlled patients.150 Extended follow-up of some treated patients reported a sustained reduction in COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 107 exacerbations compared with pre-treatment.463 However, there is a need for longer-term follow up of larger cohorts comparing effectiveness and safety, including for lung function, in both active and sham-treated patients. Advice • For adult patients whose asthma remains uncontrolled despite optimization of asthma therapy and referral to a severe asthma specialty center, and who do not have access to biologic therapy or are not eligible for it, bronchial thermoplasty is a potential treatment option at Step 5 in some countries (Evidence B).
• Caution should be used in selecting patients for this procedure. The number of studies is small, people with chronic sinus disease, frequent chest infections or FEV1 <60% predicted were excluded from the pivotal sham-controlled study, and patients did not have their asthma treatment optimized before bronchial thermoplasty was performed.
• Bronchial thermoplasty should be performed in adults with severe asthma only in the context of an independent Institutional Review Board-approved systematic registry or a clinical study, so that further evidence about effectiveness and safety of the procedure can be accumulated.175 Vitamin D Several cross-sectional studies have shown that low serum levels of Vitamin D are linked to impaired lung function, higher exacerbation frequency and reduced corticosteroid response.464 Vitamin D supplementation may reduce the rate of asthma exacerbation requiring treatment with systemic corticosteroids or may improve symptom control in asthma patients with baseline 25(OH)D of less than approximately 25–30 nmol/L.465,466 There is no good-quality evidence that Vitamin D supplementation leads to improvement in asthma control or reduction in exacerbations, particularly in preschool children and people with severe asthma.467 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 108 5. Guided asthma self-management education and skills training KEY POINTS As with other chronic diseases, people with asthma need education and skills training to manage it well. This is most effectively achieved through a partnership between the patient/carer and their healthcare providers. The essential components for this include: • Choosing the most appropriate inhaler for the patient’s asthma treatment: consider available devices, cost, the ability of the patient to use the inhaler after training, environmental impact, and patient satisfaction • Skills training to use inhaler devices effectively • Encouraging adherence with medications, appointments and other advice, within an agreed management strategy • Asthma information • Training in guided self-management, with self-monitoring of symptoms or peak expiratory flow (PEF), a written asthma action plan to show how to recognize and respond to worsening asthma, and regular review by a healthcare provider or trained healthcare worker.
In developing, customizing and evaluating self-management interventions for different cultures, sociocultural factors should be considered.468 SKILLS TRAINING FOR EFFECTIVE USE OF INHALER DEVICES Delivery of respiratory medications by inhalation achieves a high concentration in the airways, more rapid onset of action, and fewer systemic adverse effects than systemic delivery. However, using an inhaler is a skill that must be learnt and maintained in order for the medication to be delivered effectively. Poor inhaler technique leads to poor asthma control, increased risk of exacerbations and increased adverse effects.91 Most patients (up to 70–80%) do not use their inhaler correctly. Unfortunately, many healthcare providers are unable to correctly demonstrate how to use the inhalers they prescribe.469 Most people with incorrect technique are unaware that they have a problem. There is no ‘perfect’ inhaler – patients can have problems using any inhaler device. The several factors that should be considered in the choice of inhaler device for an individual patient are described below and in Box 5-1 (p.109). Strategies for ensuring effective use of inhaler devices are summarized in Box 5-2 (p.110).470 These principles apply to all types of inhaler devices. For patients prescribed pressurized metered-dose inhalers (pMDIs), use of a spacer improves delivery of the medicine to the lungs. For inhaled corticosteroids (ICS) spacers also reduce the potential for local side-effects such as dysphonia and oral candidiasis.471 With ICS, the risk of candidiasis can also be reduced by rinsing and spitting out after use. Checking and correcting inhaler technique using a standardized checklist takes only 2–3 minutes and leads to improved asthma control in adults472,473 and older children470 (Evidence A). A physical demonstration is essential to improve inhaler technique.474 This is easiest if the healthcare provider has placebo inhalers and a spacer. After training, inhaler technique deteriorates with time, so checking and re-training must be repeated regularly. This is particularly important for patients with poor symptom control or a history of exacerbations. Attaching a pictogram475,476 or a list of inhaler technique steps477 to the inhaler substantially increases the retention of correct technique at follow-up. Pharmacists, nurses and trained lay health workers can provide highly effective inhaler skills training.470,478-480 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 109 SHARED DECISION-MAKING FOR CHOICE OF INHALER DEVICE Globally, multiple different devices are available for delivery of inhaled medication, including pMDIs, dry-powder inhalers (DPIs), mist inhalers and nebulizers, although the choice of inhaler device for each medication class in any country is often limited. Reducing the risk of severe exacerbations and asthma deaths is a global priority that is driving initiatives to increase access to ICS-containing inhalers for people with asthma worldwide (see p.123) and, when these inhalers are available, to ensure that patients/carers are trained in how to use them correctly. There is also increasing interest in the potential to reduce the impact of asthma and its care (routine and urgent) on the environment, including from the manufacture and potential recycling of inhaler devices, and from the propellants in pMDIs, which are the inhalers most commonly used worldwide.481-483 For all age-groups, selecting the right inhaler for the individual patient is crucial to asthma care, not only to reduce patients’ symptom burden, but also to reduce the need for emergency health care and hospitalization, which have even greater environmental impacts than use of pMDIs.484,485 Box 5-1. Shared decision-making between health professional and patient about choice of inhalers See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 110 Box 5-2. Choice and effective use of inhaler devices CHOOSE • Choose the most appropriate inhaler device for the patient before prescribing. Consider the preferred medication (Box 4-6, p.77 and Box 4-12, p.96), available devices, patient skills, environmental impact and cost (see Box 5-1, p.109).
• If different options are available, encourage the patient to participate in the choice.
• For pMDIs, use of a spacer improves delivery and (with ICS) reduces the potential for side-effects.
• Ensure that there are no physical barriers, e.g., arthritis, that limit use of the inhaler.
• Avoid use of multiple different inhaler types where possible, to avoid confusion.
CHECK • Check inhaler technique at every opportunity.
• Ask the patient to show you how they use their inhaler (don’t just ask if they know how to use it).
• Identify any errors using a device-specific checklist.
CORRECT • Show the patient how to use the device correctly with a physical demonstration, e.g., using a placebo inhaler.
• Check technique again, paying attention to problematic steps. You may need to repeat this process 2–3 times within the same session for the patient to master the correct technique.472 • Consider an alternative device only if the patient cannot use the inhaler correctly after several repeats of training.
• Re-check inhaler technique frequently. After initial training, errors often recur within 4–6 weeks.486 CONFIRM • Clinicians should be able to demonstrate correct technique for each of the inhalers they prescribe.
• Pharmacists and nurses can provide highly effective inhaler skills training.478,479 See list of abbreviations (p.11). Choosing the medication, inhaler and device Several factors must be considered in shared decision-making about the choice of inhaler device for the individual patient (Box 5-1, p.109), starting with the choice of the medication itself: • Which medication class(es) or individual medication(s) does the patient need to relieve and control symptoms and to prevent asthma exacerbations? The approach in GINA Track 1 (Box 4-3, p.74) is preferred, because the use of ICS-formoterol as an anti-inflammatory reliever reduces the risk of severe exacerbations and urgent healthcare utilization compared with using a short-acting beta2 agonist (SABA) reliever. The Track 1 approach also avoids the risks associated with SABA over-use, and allows simple adjustment across treatment steps with a single medication for both symptom relief and delivery of ICS-containing treatment. Most studies of maintenance-and-reliever therapy (MART) with ICS-formoterol, and all studies of as-needed-only ICS-formoterol have used a DPI.
• Which inhaler devices are available to the patient for these medications? The choice of device for any particular medication class in each country is often limited. Consider local availability, access, and cost to the patient. Where more than one medication is needed, a single (combination) inhaler is preferable to multiple inhalers. Also consider the patient’s age, since DPIs are not suitable for most children aged ≤5 years and some elderly patients; pMDIs with spacers remain essential for such patients.
• Can the patient use the available device(s) correctly after training? This may be determined by factors including physical dexterity, coordination, inspiratory flow, and cognitive status. Different inhaler types require different inhalation COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 111 techniques, so it is preferable to avoid prescribing a pMDI and DPI for the same patient. Incorrect inhaler technique increases risk of severe asthma exacerbations. • What are the environmental implications of the available inhaler(s)? This has become an important part of inhaler selection, with particular consideration of carbon emissions due to the propellants in pMDIs, but also of environmental effects of inhaler manufacture and potential recycling. However, clinicians need to be aware of the potential to place the additional burden of ‘green guilt’ on patients, as this could reduce adherence and increase the risk of exacerbations.
• Is the patient satisfied with the medication and inhaler? The best inhaler for each patient is likely to be the one that they prefer and can use correctly, as this promotes adherence and reduces risk of exacerbations and adverse effects.
In follow-up, review symptom control, asthma exacerbations and adverse events, and check the patient’s ability to use their inhaler(s) correctly, ideally at each visit. ADHERENCE WITH MEDICATION AND WITH OTHER ADVICE Identifying poor adherence Poor adherence is defined as the failure of treatment to be taken as agreed upon by the patient and the healthcare provider. There is increasing awareness of the importance of poor adherence in chronic diseases, and of the potential to develop interventions to improve adherence.487 Approximately 50% of adults and children on long-term therapy for asthma fail to take medications as directed at least part of the time.190 In clinical practice, poor adherence may be identified by an empathic question that acknowledges the likelihood of incomplete adherence and encourages an open discussion. See Box 5-3 (p.112) for examples. Checking the date of the last prescription or the date on the inhaler may assist in identifying poor adherence. In some health systems, pharmacists can assist in identifying poorly adherent patients by monitoring dispensing records. Electronic inhaler monitoring has also been used in clinical practice to identify poor adherence in patients with difficult-to-treat asthma.176,177 In clinical studies assessing factors contributing to poor adherence, methods of measuring adherence include using short adherence behavior questionnaires, analysis of dispensing records, dose or pill counting, electronic inhaler monitoring,488,489 and drug assay (e.g., for prednisolone).490 Factors contributing to poor adherence To understand the reasons behind patients’ medication-taking behavior, it is important to elicit their beliefs and concerns about asthma and asthma medications. Both intentional and unintentional factors contribute to poor adherence (Box 5-3, p.112). Issues of ethnicity,491 health literacy,492,493 and numeracy201 are often overlooked. Patients may be concerned about known side-effects or about perceived harm.420,494 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 112 Box 5-3. Poor adherence with prescribed maintenance treatment in asthma Factors contributing to poor adherence How to identify poor adherence in clinical practice Medication/regimen factors Difficulties using inhaler device (e.g., arthritis) Burdensome regimen (e.g., several times per day) Multiple different inhalers Unintentional poor adherence Misunderstanding about instructions Forgetfulness Absence of a daily routine Cost Intentional poor adherence Perception that treatment is not necessary Denial or anger about asthma or its treatment Inappropriate expectations Concerns about side-effects (real or perceived) Dissatisfaction with healthcare providers Stigmatization Cultural or religious issues Cost For patients prescribed maintenance treatment, ask an empathic question Acknowledge the likelihood of incomplete adherence and encourage an open non-judgmental discussion. Examples are: ‘Many patients don’t use their inhaler as prescribed. In the last 4 weeks, how many days a week have you been taking it – not at all, 1, 2, 3 or more days a week?’495 ‘Do you find it easier to remember your inhaler in the morning or the evening?’ Check medication usage: • Check the date of the last prescription.
• Check the date and dose counter on the inhaler.
• In some health systems, prescribing and dispensing frequency can be monitored electronically by clinicians and/or pharmacists.
• See review articles for more detail.189,496 Examples of successful adherence interventions Shared decision-making for medication/dose choice192,195 Inhaler reminders, either proactively or for missed doses497-499 Prescribing low-dose ICS once-daily versus twice-daily500 Home visits for a comprehensive asthma program by an asthma nurse501 See list of abbreviations (p.11). Interventions that improve adherence in asthma Few adherence interventions have been studied comprehensively in asthma. Some examples of successful interventions have been published: • Shared decision-making for medication/dose choice improved adherence and asthma outcomes.192 • Electronic inhaler reminders, either proactively or for missed doses, improved adherence497-499 and possibly reduced exacerbations and oral corticosteroid use.497-499,502 • In a difficult inner-city environment, home visits for a comprehensive asthma program by an asthma nurse led to improved adherence and reduced prednisone courses over the following several months.501 • Providing adherence information to clinicians did not improve ICS use among patients with asthma unless clinicians chose to view the details of their patients’ medication use.503 • In a health maintenance organization, an automated voice recognition program with messages triggered when refills were due or overdue led to improved ICS adherence relative to usual care, but no difference in urgent care visits.504 • In one study, directly observed administration of maintenance asthma treatment at school, combined with telemedicine oversight, was associated with more symptom-free days and fewer urgent visits than usual care.505 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 113 Digital interventions for adherence A 2022 Cochrane review found that a variety of digital intervention strategies improved adherence to maintenance controller medications, especially in those with poor adherence, reduced exacerbations, and improved asthma control, in studies of up to 2 years’ duration in adults and children.502 Electronic monitoring of maintenance inhaler use, and text messages sent to phones appear to be effective. No harms associated with these technologies were reported. The effects of digital interventions on quality of life, lung function and unscheduled healthcare utilization are unclear. Improving adherence to maintenance ICS-containing medications may not necessarily translate to improved clinical outcomes.506 Further studies are needed of adherence strategies that are feasible for implementation in primary care. ASTHMA INFORMATION While education is relevant to asthma patients of all ages, the information and skills training required by each person may vary, as will their ability or willingness to take responsibility. All individuals will require certain core information and skills, but most education must be personalized and provided over several sessions or stages. For young children, the focus of asthma education will be on the parent/caregiver, but young children can be taught simple asthma management skills. Adolescents may have unique difficulties with adherence, and peer support group education may help in addition to education provided by the healthcare provider.507 These are complex interventions, and there have been few studies. Regional issues and the adolescent’s developmental stage may affect the outcomes of such programs.508 The key features and components of an asthma education program are provided in Box 5-4. Information alone improves knowledge but does not improve asthma outcomes.509 Social and psychological support may also be required to maintain positive behavioral change, and skills are required for effective medication delivery. At the initial consultation, verbal information should be supplemented with written or pictorial510,511 information about asthma and its treatment. Patients and their families should be encouraged to make a note of any questions about their asthma or its treatment, and should be given time to address these. Asthma education and training, for both adults and children, can be delivered effectively by a range of healthcare providers including pharmacists and nurses (Evidence A).478,479,512,513 Trained lay health workers (also known as community health workers) can deliver appropriately defined aspects of respiratory care such as asthma self-management education. Asthma education by trained lay health workers has been found to improve patient outcomes and healthcare utilization compared with usual care,480,514 and to a similar extent as nurse-led education in primary care (Evidence B).515 These findings suggest the need for additional studies to assess applicability in other settings and populations. Box 5-4. Asthma information Goal: To provide the person with asthma, their family and other carers with suitable information and training to manage their asthma in partnership with their healthcare providers Approach Focus on the development of the partnership. Accept that this is a continuing process. Share information. Adapt the approach to the patient’s level of health literacy (Box 3-1, p.49). Fully discuss expectations, fears and concerns. Develop shared goals. Topics to include Asthma diagnosis Rationale for treatment, and differences between relievers and maintenance treatments (if prescribed) Potential side-effects of medications Prevention of symptoms and flare-ups: importance of anti-inflammatory treatment How to recognize worsening asthma and what actions to take; how and when to seek medical attention Management of comorbidities COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 114 TRAINING IN GUIDED ASTHMA SELF-MANAGEMENT Guided self-management may involve varying degrees of independence, ranging broadly from patient-directed self-management to doctor-directed self-management. With patient-directed self-management patients make changes in accordance with a prior written action plan without needing to first contact their healthcare provider. With doctor-directed self-management, patients still have a written action plan, but refer most major treatment decisions to their physician at the time of a planned or unplanned consultation. The essential components of effective guided asthma self-management education are:193 • Self-monitoring of symptoms and/or PEF • A written asthma action plan to show how to recognize and respond to worsening asthma • Regular review of asthma control, treatment and skills by a healthcare provider.
Self-management education that includes these components dramatically reduces asthma morbidity, both in adults (Evidence A)193,480,516 and children (Evidence A).194,516 Benefits include reduction of one-third to two-thirds in asthma-related hospitalizations, emergency department visits and unscheduled doctor or clinic visits, missed work/school days, and nocturnal wakening.193 It has been estimated that the implementation of a self-management program in 20 patients prevents one hospitalization, and successful completion of such a program by 8 patients prevents one emergency department visit.193 517 Less intensive interventions that involve self-management education, but not a written action plan, are less effective,518 and information alone is ineffective.509 A systematic meta-review of 270 RCTs on supported self-management for asthma confirmed that it reduces unscheduled health care use, improves asthma control, is applicable to a wide range of target groups and clinical settings, and does not increase healthcare costs (Evidence A).516 Self-monitoring of symptoms and/or peak expiratory flow (PEF) Patients/careers should be trained to keep track of symptoms (with or without a diary), and notice and take action, if necessary, when symptoms start to worsen. PEF monitoring may sometimes be useful: • In short-term monitoring o After an exacerbation, to monitor recovery o After a change in treatment, to help in assessing whether the patient has responded o When symptoms appear excessive (for objective evidence of degree of lung function impairment) o To assist in identification of occupational or domestic triggers for worsening asthma control • In long-term monitoring o For earlier detection of exacerbations, mainly in patients with poor perception of airflow limitation152 o For patients with a history of sudden severe exacerbations o For patients who have difficult-to-control or severe asthma.
For patients carrying out PEF monitoring, use of a laterally compressed PEF chart (showing 2 months on a landscape format page) allows more accurate identification of worsening asthma than other charts.174 One such chart is available for download from www.woolcock.org.au/resources/asthma-peak-flow-chart. There is increasing interest in internet or phone-based monitoring of asthma. Based on existing studies, the main benefit is likely to be for more severe asthma (Evidence B).519 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 115 Written asthma action plans Personal written asthma action plans show patients how to make short-term changes to their treatment in response to changes in their symptoms and/or PEF. They also describe how and when to access medical care.520,521 ‘Written’ action plans include printed, digital or pictorial plans (i.e., the patient is given a record of the instructions). The benefits of self-management education for asthma morbidity are greater in adults when the action plans include both a step up in ICS and the addition of oral corticosteroids (OCS) and, for PEF-based plans, when they are based on personal best rather than percent predicted PEF (Evidence A).521 The efficacy of self-management education is similar regardless of whether patients self-adjust their medications according to an individual written plan or whether the medication adjustments are made by a doctor (Evidence A).518 Thus, patients who are unable to undertake guided self-management can still achieve benefit from a structured program of regular medical review. Action plans for patients using SABA as their reliever Examples of written asthma action plan templates for asthma treatment with a SABA reliever, including for adult and pediatric patients with low literacy, can be found on several websites (e.g., Asthma UK, www.asthma.org.uk; Asthma Society of Canada, www.asthma.ca; Family Physician Airways Group of Canada, www.fpagc.com; National Asthma Council Australia, www.nationalasthma.org.au) and in research publications.522,523 Action plan for patients using as-needed ICS-formoterol as their reliever A different type of action plan is needed for patients using as-needed ICS-formoterol as their reliever in GINA Track 1, because the initial ‘action’ when asthma worsens is for the patient to increase their as-needed doses of ICS-formoterol, rather than taking a SABA and/or increasing their maintenance treatment. An example of such a customized template can be found in a review article about practical use of maintenance-and reliever-therapy (MART).313 A similar action plan template can be used for patients using as-needed-only ICS-formoterol.314 Healthcare providers should become familiar with action plans that are relevant to their local healthcare system, treatment options, and cultural and literacy context. Details of the specific treatment adjustments that can be recommended for written asthma action plans are described in the next chapter (Box 9-2, p.162). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 116 REGULAR REVIEW BY A HEALTHCARE PROVIDER OR TRAINED HEALTHCARE WORKER The third component of effective asthma self-management education is regular review by a healthcare provider or trained healthcare worker. Follow-up consultations should take place at regular intervals. Regular review should include the following: • Ask the patient if they have any questions or concerns o Discuss issues, and provide additional educational messages as necessary.
o If available, refer the patient to someone trained in asthma education.
• Assess asthma control, risk factors for exacerbations, and comorbidities o Review the patient’s level of symptom control and risk factors (Box 2-2, p.37).
o Ask about flare-ups to identify contributory factors and whether the patient’s response was appropriate (e.g., was an action plan used?).
o Review the patient’s symptom or PEF diary, if they keep one.
o Assess comorbidities.
• Assess treatment issues o Watch the patient use their inhaler, and correct and re-check technique if necessary (Box 5-2, p.110).
o Assess medication adherence and ask about adherence barriers (Box 5-3, p.112).
o Ask about adherence with other interventions (e.g., smoking cessation).
o Review the asthma action plan and update it if level of asthma control or treatment have changed.524 A single-page prompt to clinicians has been shown to improve the provision of preventive care to children with asthma during office visits.525 Follow-up by telehealthcare is unlikely to benefit patients with asthma that is well controlled at a low treatment step, but may be of benefit in those with severe disease at risk of hospital admission.519 SCHOOL-BASED PROGRAMS FOR CHILDREN A systematic review found that school-based studies (most conducted in the US and Canada) that included self-management skills for children aged 5–18 years was associated with a 30% decrease in emergency department visits, and a significant decrease in hospitalizations and in days of reduced activity.526 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 117 6. Managing asthma with multimorbidity and in specific populations KEY POINTS Multimorbidity is common in patients with chronic diseases such as asthma. It is important to identify and manage multimorbidity, as it contributes to impaired quality of life, increased healthcare utilization, and adverse effects of medications. In addition, comorbidities such as rhinosinusitis, obesity and gastro-esophageal reflux disease (GERD) may contribute to respiratory symptoms, and some contribute to poor asthma control. For patients with dyspnea or wheezing on exertion: • Distinguish between exercise-induced bronchoconstriction (EIB) and symptoms that result from obesity or a lack of fitness or are the result of alternative conditions such as inducible laryngeal obstruction.
• Provide advice about preventing and managing EIB.
All adolescents and adults with asthma should receive inhaled corticosteroid (ICS)-containing treatment to reduce their risk of severe exacerbations. It should be taken every day or, as an alternative in mild asthma, by as-needed ICS-formoterol for symptom relief. Refer patients with difficult-to-treat or severe asthma to a specialist or severe asthma service, after addressing common problems such as incorrect diagnosis, incorrect inhaler technique, ongoing environmental exposures, and poor adherence (see Section 8, p.139). Women with asthma who are pregnant or planning pregnancy should be advised not to stop ICS-containing therapy, as exacerbations increase the risk of adverse perinatal outcomes. The advantages of actively treating asthma in pregnancy with ICS-containing therapy markedly outweigh any potential risks of these medications. MANAGING MULTIMORBIDITY Multimorbidity is a common problem in patients with chronic diseases such as asthma. It is associated with worse quality of life, increased healthcare utilization and increased adverse effects of treatment.191 Multimorbidity is particularly common among those with difficult-to-treat or severe asthma.93 Active management of comorbidities such as rhinosinusitis, obesity and GERD is important, as these conditions may also contribute to respiratory symptom burden and lead to medication interactions. Some comorbidities also contribute to poor asthma control.527 The advice below covers some of the most common comorbidities of asthma, but is not an exhaustive list. Obesity Clinical features Being overweight or obese is a risk factor for childhood asthma and wheeze, particularly in girls.528 Asthma is more difficult to control in obese patients.266,529-531 This may be due to a different type of airway inflammation, contributory comorbidities such as obstructive sleep apnea and GERD, mechanical factors, or other as-yet undefined factors. In addition, lack of fitness and reduction in lung volume due to abdominal fat may contribute to dyspnea. Diagnosis Document body-mass index (BMI) for all patients with asthma. Because of other potential contributors to dyspnea and wheeze in obese patients, it is important to confirm the diagnosis of asthma with objective measurement of variable expiratory airflow limitation (Box 1-2, p.26). Asthma is more common in obese than non-obese patients,75 but both over- and under-diagnosis of asthma occur in obesity.52,76 The relationship between biomarkers of Type 2 inflammation is lost in the obese.24,48 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 118 Management As for other patients with asthma, ICS are the mainstay of treatment in obese patients (Evidence B), although their response may be reduced.266 Weight reduction should be included in the treatment plan for obese patients with asthma (Evidence B). Increased exercise alone appears to be insufficient (Evidence B).273 Weight loss can improve asthma control, lung function, health status and reduces medication needs in obese patients,268,269 but the studies have generally been small, quality of some studies is poor, and the interventions and results have been variable.267 The most striking results have been observed after bariatric surgery,270,271,532 but even 5–10% weight loss can lead to improved asthma control and quality of life.273 For patients with comorbid obstructive sleep apnea, one study showed a significant reduction in moderate exacerbations with 6 months of continuous positive airway pressure (CPAP) therapy.533 Gastroesophageal reflux disease (GERD) Clinical features GERD can cause symptoms such as heartburn and epigastric or chest pain, and is also a common cause of dry cough. Symptoms and/or diagnosis of GERD are more common in people with asthma than in the general population,527 but this may be in part due to cough being attributed to asthma; in addition, some asthma medications such as beta2 agonists and theophylline cause relaxation of the lower esophageal sphincter. Asymptomatic gastroesophageal reflux is not a likely cause of poorly controlled asthma.527 Diagnosis In patients with confirmed asthma, GERD should be considered as a possible cause of a dry cough; however, there is no value in screening patients with uncontrolled asthma for GERD (Evidence A). For patients with asthma and symptoms suggestive of reflux, an empirical trial of anti-reflux medication, such as a proton pump inhibitor or motility agent, may be considered, as in the general population. If the symptoms do not resolve, specific investigations such as 24-hour pH monitoring or endoscopy may be considered. Management Clinical trials of proton pump inhibitors in patients with confirmed asthma, most of whom had a diagnosis of GERD, showed small benefits for lung function, but no significant benefit for other asthma outcomes.534,535 In a study of adult patients with symptomatic asthma but without symptoms of GERD, treatment with high-dose proton pump inhibitors did not reduce asthma symptoms or exacerbations.536 In general, benefits of proton pump inhibitors in asthma appear to be limited to patients with both symptomatic reflux and night-time respiratory symptoms.537 Other treatment options include motility agents, lifestyle changes and fundoplication. In summary, symptomatic reflux should be treated, but patients with poorly controlled asthma should not be treated with anti-reflux therapy unless they also have symptomatic reflux (Evidence A).535 Few data are available for children with asthma symptoms and symptoms of GERD.538,539 Anxiety and depression Clinical features Anxiety symptoms and psychiatric disorders, particularly depressive and anxiety disorders, are more prevalent among people with asthma.540,541 Psychiatric comorbidity is also associated with worse asthma symptom control and medication adherence, and worse asthma-related quality of life.542 Anxious and depressive symptoms have been associated with increased asthma-related exacerbations and emergency visits.529 Panic attacks may be mistaken for asthma. Diagnosis Although several tools are available for screening for anxious and depressive symptomatology in primary care, the majority have not been validated in asthma populations. Difficulties in distinguishing anxiety or depression from asthma symptoms may therefore lead to misdiagnosis. It is important to be alert to possible depression and/or anxiety in people with asthma, particularly when there is a previous history of these conditions. Where appropriate, patients should be referred to psychiatrists or evaluated with a disease-specific psychiatric diagnostic tool to identify potential cases of depression and/or anxiety. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 119 Management There have been few good quality pharmacological and non-pharmacological treatment trials for anxiety or depression in patients with asthma, and results are inconsistent. A Cochrane review of 15 randomized controlled trials of psychological interventions for adults with asthma included cognitive behavior therapy, psychoeducation, relaxation, and biofeedback.530 Results for anxiety were conflicting, and none of the studies found significant treatment differences for depression. Drug treatments and cognitive behavior therapy531 have been described as having some potential in patients with asthma; however, current evidence is limited, with a small number of studies and methodological shortcomings. Food allergy and anaphylaxis Clinical features Rarely, food allergy is a trigger for asthma symptoms (<2% of people with asthma). In patients with confirmed food-induced allergic reactions (anaphylaxis), co-existing asthma is a strong risk factor for more severe and even fatal reactions. Food-induced anaphylaxis often presents as life-threatening asthma.94 An analysis of 63 anaphylaxis-related deaths in the United States noted that almost all had a past history of asthma; peanuts and tree nuts were the foods most commonly responsible.543 A UK study of 48 anaphylaxis-related deaths found that most were regularly treated for asthma, and that in most of these, asthma was poorly controlled.544 Diagnosis In patients with confirmed food allergy, it is important to assess for asthma. Children with food allergy have a four-fold increased likelihood of having asthma compared with children without food allergy.545 Refer patients with suspected food allergy or intolerance for specialist allergy assessment. This may include appropriate allergy testing such as skin prick testing and/or blood testing for specific IgE. On occasion, carefully supervised food challenges may be needed. Management Patients who have a confirmed food allergy that puts them at risk for anaphylaxis must have an epinephrine auto-injector available at all times, and be trained how to use it. They, and their family, must be educated in appropriate food avoidance strategies, and in the medical notes, they should be flagged as being at high risk. It is especially important to ensure that their asthma is well controlled, they have a written action plan, understand the difference between asthma and anaphylaxis, and are reviewed on a regular basis. Allergic rhinitis Clinical features Evidence clearly supports a link between diseases of the upper and lower airways.546 Most patients with asthma, either allergic or non-allergic, have concurrent rhinitis, and 10–40% of patients with allergic rhinitis have asthma.547 Depending on sensitization and exposure, allergic rhinitis may be seasonal (e.g., ragweed or grass pollen), or perennial (e.g., HDM allergens, furred pets in the home), or intermittent (e.g., furred pets at other locations).548 Rhinitis is defined as irritation and inflammation of the mucous membranes of the nose. Allergic rhinitis may be accompanied by ocular symptoms (conjunctivitis). Diagnosis Rhinitis can be classified as either allergic or non-allergic depending on whether allergic sensitization is demonstrated. Variation in symptoms by season or with environmental and/or occupational exposure (e.g., furred pets, house dust mite, molds, pollens) suggests allergic rhinitis. Examination of the upper airway should be arranged for patients with severe asthma. Management International evidence-based guidelines546,549 recommend intranasal corticosteroids for treatment of allergic rhinitis. In a case-control study, treatment of rhinitis with intranasal corticosteroids was associated with less need for asthma-related hospitalization and emergency department visits,550 but a meta-analysis found improvement in asthma outcomes only in COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 120 patients not also receiving ICS.551 See p.104 for information about allergen immunotherapy for patients with allergic rhinitis and asthma. Chronic rhinosinusitis with and without nasal polyps (CRSwNP and CRSsNP) Rhinosinusitis is defined as inflammation of the nose and paranasal sinuses characterized by more than two symptoms including nasal blockage/obstruction and/or nasal discharge (anterior/posterior nasal drip).552 Other symptoms may include facial pain/pressure and/or a reduction or loss of smell. Sinusitis rarely occurs in the absence of rhinitis. Rhinosinusitis is defined as acute when symptoms last <12 weeks with complete resolution, and chronic when symptoms occur on most days for at least 12 weeks without complete resolution. Chronic rhinosinusitis is an inflammatory condition of the paranasal sinuses that encompasses two clinically distinct entities: chronic rhinosinusitis without nasal polyps (CRSsNP) and chronic rhinosinusitis with nasal polyps (CRSwNP).553 The heterogeneity of chronic rhinosinusitis may explain the wide variation in prevalence rates in the general population, ranging from 1% to 10% without polyps and 4% with polyps. Chronic rhinosinusitis is associated with more severe asthma, especially in patients with nasal polyps.554 Diagnosis Nasendoscopy and/or computed tomography (CT) of the sinuses can identify changes suggestive of chronic rhinosinusitis with or without nasal polyps. In severe asthma, presence of nasal polyps may help with choice of biologic therapy (see Box 8-4, p.144). Management Chronic rhinosinusitis, with or without nasal polyps, has a significant impact on patients’ quality of life. Guidelines for the management of chronic rhinosinusitis with or without nasal polyps have been published.555,556 A 2022 systematic review of studies reporting treatment outcomes in patients with both asthma and chronic rhinosinusitis found that medical treatments (including intranasal saline irrigations, intranasal corticosteroids delivered by irrigation, drops (only one small study of each) or sprays, oral antibiotics (small studies with erythromycin), and oral corticosteroids) improved sinonasal-specific quality of life in patients with chronic rhinosinusitis (most commonly with nasal polyps) and comorbid asthma. However, people with chronic rhinosinusitis and asthma may have a lesser response to rhinosinusitis treatments, compared with in people who do not have asthma.557 There was limited evidence for improvements in lung function and asthma control, and no data on the effect of intranasal corticosteroids on lung function or asthma control.557 The systematic review found strong RCT evidence that anti-IL4Rα and anti-IL5/5Rα receptor therapies improve rhinosinusitis, including reducing polyp counts, as well as improving asthma outcomes, in patients with asthma and CRSwNP who have experienced inadequate response to non-biologic therapy.557 Biologics were less effective in managing chronic sinusitis without polyps in people with asthma.557 The review found no studies that directly compared biologic therapy with endoscopic sinus surgery in patients with CRSwNP and asthma. There was moderate-to-strong evidence that endoscopic sinus surgery improves sinonasal-specific and asthma-specific quality of life in patients with chronic rhinosinusitis and asthma, and may improve asthma symptom control, but there was insufficient evidence for effects on lung function.557 Current evidence supports stepwise treatment to manage chronic rhinosinusitis in people with asthma, beginning with topical nasal saline irrigations and topical nasal steroids as the main treatment. Oral antibiotics can be used as needed after considering the risks and microbial resistance. Oral corticosteroid treatment is effective, but should be minimized due to adverse effects (Box 9-3, p.165). In patients with CRSwNP, omalizumab,558 mepolizumab559,560 and dupilumab561 improved subjective and objective assessments including nasal symptoms and polyp size, compared with placebo. Endoscopic sinus surgery can be considered in patients with asthma who have inadequate response to medical therapies for chronic rhinosinusitis, but it does not improve asthma outcomes. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 121 MANAGING ASTHMA DURING THE COVID-19 PANDEMIC Are people with asthma at higher risk of COVID-19 or severe COVID-19? People with asthma do not appear to be at increased risk of acquiring COVID-19, and systematic reviews have not shown an increased risk of severe COVID-19 in people with well-controlled mild-to-moderate asthma. Overall, studies to date indicate that people with well-controlled asthma are not at increased risk of COVID-19-related death,562,563 and in one meta-analysis, mortality appeared to be lower than in people without asthma.564 However, the risk of COVID-19 death was increased in people who had recently needed oral corticosteroids (OCS) for their asthma,454,562 and in hospitalized patients with severe asthma.454,565 Therefore, it is important to continue good asthma management (as described in the GINA Strategy Report), with strategies to maintain good symptom control, reduce the risk of severe exacerbations and minimize the need for OCS. In one study of hospitalized patients aged ≥50 years with COVID-19, mortality was lower among those with asthma who were using ICS than in patients without an underlying respiratory condition.565 In 2020 and 2021, many countries recorded a reduction in asthma exacerbations and influenza-related illness. The reasons are not precisely known, but may be due to handwashing, masks and social/physical distancing that reduced the incidence of other respiratory infections, including influenza.455 During pandemic conditions, advise patients with asthma to continue taking their prescribed asthma medications, particularly inhaled corticosteroid (ICS)-containing medications, and OCS if prescribed It is important for patients to continue taking their prescribed asthma medications as usual during the COVID-19 pandemic. This includes ICS-containing medications (alone or in combination with a long-acting beta2 agonist (LABA), and add-on therapy including biologic therapy for severe asthma. Stopping ICS often leads to potentially dangerous worsening of asthma. See Section 4 (p.67) for information about asthma medications and regimens and non-pharmacologic strategies, and Section 5 (p.108) for guided asthma self-management education and skills training. For a small proportion of patients with severe asthma, long-term OCS may sometimes be needed, and it is very dangerous to stop these suddenly. See Section 8 (p.139) for advice about investigation and management of difficult-to-treat and severe asthma, including addition of biologic therapy for minimizing use of OCS. Advise patients to discuss with you before stopping any asthma medication. Make sure that all patients have a written asthma action plan A written action plan (printed, digital or pictorial) tells the patient how to recognize worsening asthma, how to increase their reliever and maintenance medications, and when to seek medical help. A short course of OCS may be needed during severe asthma flare-ups (exacerbations). See Box 9-2 (p.162) for more information about specific action plan options for increasing reliever medications (or reliever and maintenance medications), depending on the patient’s usual therapeutic regimen. At present, there is no clear evidence about how to distinguish between worsening asthma due to respiratory viral infections such as rhinovirus and influenza, and COVID-19. If local risk of COVID-19 is moderate or high, avoid use of nebulizers where possible due to the risk of transmitting infection to other patients/family and to healthcare workers Nebulizers can transmit respiratory viral particles across distances of at least 1 m. Use of nebulizers for delivering bronchodilator therapy is mainly restricted to management of life-threatening asthma in acute care settings. Instead, to deliver short-acting beta2 agonist for acute asthma in adults and children, use a pressurized metered-dose inhaler and spacer, with a mouthpiece or tightly fitting face mask, if required. Check the manufacturer’s instructions about whether a spacer can be autoclaved. If not (as is the case for many types of spacers), or if in doubt, spacers should be restricted to single patient use. If use of a nebulizer is needed in settings where COVID-19 infection is possible, strict infection control procedures should be followed. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 122 Remind patients not to share inhaler devices or spacers with family members, to avoid transmitting infection. Avoid spirometry in patients with confirmed/suspected COVID-19 In healthcare facilities, follow local COVID-19 testing recommendations and infection control procedures if spirometry or peak flow measurement is needed.32 Use of an in-line filter minimizes the risk of transmission during spirometry, but many patients cough after performing spirometry; before performing spirometry, coach the patient to stay on the mouthpiece if they feel the need to cough. The US Centers for Disease Control and Prevention (CDC) recommendations are found here. If spirometry is not available due to local infection control restrictions, and information about lung function is needed, consider asking patients to monitor lung function at home. Follow infection control recommendations if any aerosol-generating procedures are needed Other aerosol-generating procedures include oxygen therapy (including with nasal prongs), sputum induction, manual ventilation, non-invasive ventilation and intubation. CDC recommendations are found here. Follow local health advice about hygiene strategies and use of personal protective equipment, as new information becomes available in your country or region. The CDC website provides up-to-date information about COVID-19 for health professionals here, and for patients here. The website of the World Health Organization (WHO) provides comprehensive advice for health professionals and health systems about prevention and management of COVID-19 here. Management of asthma if the patient acquires COVID-19 People with asthma who acquire COVID-19 are not at higher risk of severe COVID-19. However, be aware that those with poorly controlled asthma (e.g., recent need for OCS) are at higher risk of hospitalization for severe disease if they acquire COVID-19.454,562,565 Advise patients to continue taking their usual asthma medications. Patients with severe asthma should continue biologic therapy or OCS, if prescribed. To reduce the risk of transmitting infection, as above, avoid use of nebulizers where possible (use a pressurized metered-dose inhaler [pMDI] and spacer instead), avoid spirometry, and instruct patients to avoid sharing of inhalers/spacers. Before prescribing antiviral therapies, consult local prescribing guidelines. Check carefully for potential interactions between asthma therapy and COVID-19 therapy. For example, ritonavir-boosted nirmatrelvir (NMV/r) is a potent CYP3A4 inhibitor. While this is unlikely to cause clinically important corticosteroid-related adverse effects, because of the short duration of anti-COVID-19 treatment, be cautious if considering prescribing NMV/r for patients taking ICS-salmeterol or ICS-vilanterol, as the interaction may increase cardiac toxicity of the LABA.157 Product information indicates that for patients taking ICS-salmeterol or ICS-vilanterol, concomitant treatment with CYP3A4 inhibitors is not recommended. Some drug interaction websites advise stopping ICS-salmeterol or ICS-vilanterol during NMV/r treatment and for a few days afterwards, but this may increase the risk of an asthma exacerbation. Instead, consider prescribing alternative antiviral therapy (if available) or switching to ICS alone or ICS-formoterol (if available) for the duration of NVM/r therapy and a further 5 days.157 If switching to a different inhaler, remember to teach correct technique with the new inhaler. Advise people with asthma to be up to date with COVID-19 vaccines Many types of COVID-19 vaccines have been studied and are in use. New evidence about the vaccines, including in people with asthma, will emerge over time. In general, allergic reactions to the vaccines are rare. Patients with a history of severe allergic reaction to a COVID-19 vaccine ingredient (e.g., polyethylene glycol for Pfizer/BioNTech or Moderna, or polysorbate 80 for AstraZeneca or J&J/Janssen) should receive a different COVID-19 vaccine. However, people with anaphylaxis to foods, insect venom, or other medications can safely receive COVID-19 vaccines. More details from the US Advisory Committee on Immunization Practices (ACIP) are here. As always, patients should speak to their healthcare provider if they have concerns. Follow local advice about monitoring patients after COVID-19 vaccination. Usual vaccine precautions apply. For example, ask if the patient has a history of allergy to any components of the vaccine, and if the patient has a fever or another infection, delay vaccination until they are well. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 123 For people with severe asthma, GINA suggests that, if possible, the first dose of biologic therapy and COVID-19 vaccine should not be given on the same day, to allow adverse effects of either to be more easily distinguished. Remind people with asthma to have an annual influenza vaccination (p.106). CDC (advice here) now advises that influenza vaccine and COVID-19 vaccine can be given on the same day. Current advice from the CDC is that where there is substantial transmission of COVID-19, people will be better protected, even if they are fully vaccinated, if they wear a mask in indoor public settings. Further details are here. Additional advice about management of asthma in the context of COVID-19 will be posted on the GINA website (www.ginasthma.org) as it becomes available. MANAGING ASTHMA IN SPECIFIC POPULATIONS OR SETTINGS This section includes brief advice about managing asthma in some of the specific populations or settings in which the usual treatment approach may need to be modified. See also How to make the diagnosis of asthma in other contexts (p.32). Low- and middle-income countries Clinical features In 2019, 96% of asthma deaths and 84% of disability-adjusted life years (DALYs) were in low- and middle-income countries (LMICs).2 Symptoms of asthma are similar world-wide, but patient language may differ, and comorbidities may vary depending on environmental exposures such as smoking and biomass fuel exposure and incidence of chronic respiratory infections from tuberculosis and HIV/AIDS. Management The fundamental principles and aims of asthma treatment are the same in LMICs as in high-income countries, but common barriers to effective long-term asthma care include the lack of availability and affordability of inhaled medicines, and prioritization of acute care over chronic care by healthcare systems.2,5 Recommendations by WHO and the International Union Against Tuberculosis and Lung Disease566 form the basis of treatments offered in many LMICs.5 The WHO Model List of Essential Medicines includes ICS, combination ICS-formoterol, and bronchodilators,567 and the WHO Model List of Essential Medicines for Children includes ICS.568 Spacers are included in the WHO list of essential technology, but are rarely available due to obstacles to their manufacture or purchase, practical issues of cleaning, and inconvenience for ambulatory use. Effective spacers can be made at no cost from plastic drink bottles.569 Medicines selected as ‘essential’ are not necessarily the most effective or convenient, particularly for patients with more severe disease, and a limited choice does not allow for consideration of patient preferences and likelihood of adherence. However, ICS-containing medications, when provided for large populations, have achieved impressive reductions in mortality and morbidity,570 including in LMICs. In Brazil, government policy ensuring nationwide easy access to ICS, at no cost to patients, was associated with a 34% reduction in hospitalizations for asthma.184 Prescribing ICS-formoterol as the symptom reliever, with (GINA Steps 3–5) or without (Steps 1–2) maintenance ICS-formoterol, provides the safest and most effective asthma treatment for adolescents and adults,183,224 and avoids the behavioral consequences of starting treatment with SABA alone. Inclusion of essential asthma medicines in formularies and guidelines does not assure sustained and equitable supply to patients. The supply of medicines in many LMICs tends to be sporadic for a wide variety of reasons, sometimes determined by the ability of governments to pay for supplies, issues relating to procurement, poor administration and record keeping, and problems in the supply chain, particularly to remote dispensaries.3,5 Availability of asthma medicines varies widely between LMICs, with some having only oral bronchodilators (salbutamol and theophylline tablets/solutions), supplemented from time to time with oral corticosteroids.27 Oral bronchodilators have a slow onset of action and more adverse effects than inhaled SABA, and even occasional courses of OCS are associated COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 124 with a significant risk of short-term adverse effects such as pneumonia and sepsis,571 and, in adults, with long-term adverse effects including osteoporosis and fragility fractures, cataract and diabetes.225 The largest (52 countries) survey of the accessibility and affordability of inhaled asthma medicines, conducted in 2011, reported that salbutamol was available in only half of public hospitals; ICS was available in fewer than one in five public pharmacies and not at all in 14 countries.572 Obtaining asthma medicines often represents a catastrophic household expense. A recent systematic review of published data on the availability, cost and affordability of essential medicines for asthma and COPD in LMICs found these to be largely unavailable and unaffordable particularly for ICS and combination ICS-LABA.573 This means that the essential cornerstone of treatment that achieves substantial reductions in morbidity and mortality is out of reach for the great majority of the world’s children, adolescents and adults living with asthma. It is not acceptable in 2023 for clinicians to have to manage asthma with SABAs and oral corticosteroids instead of preventive ICS-containing treatments. The research community must develop and evaluate approaches designed to obviate barriers to care in resource-constrained settings. A World Health Assembly Resolution on equitable access to affordable care, including inhaled medicines, for children, adolescents and adults with asthma, wherever they live in the world, would be a valuable step forward – as was achieved in 2021 for the supply of insulin for diabetes.574 GINA strongly supports this initiative.3 In the meantime, in general, Track 2 treatment, although less effective in reducing asthma exacerbations, may be considered preferable in settings where current availability or affordability constrains the ability to implement Track 1 treatment. The ‘other controller options’ in Box 4-6 (p.77), though potentially less costly, may be considerably less effective (e.g., leukotriene receptor antagonists [LTRAs]) or more harmful (e.g., maintenance OCS), or not well supported by evidence especially in the low-resource setting (e.g., use of a low-dose ICS inhaler whenever a SABA is taken for symptom relief). Of these three other controller options, the third would be closest to the preferred recommendations in Tracks 1 and 2, as it would ensure that an ICS was provided, at least during symptomatic periods.27 Adolescents Clinical features Care of teenagers with asthma should take into account the rapid physical, emotional, cognitive and social changes that occur during adolescence. Asthma control may improve or worsen, although remission of asthma is seen more commonly in males than females.575 Exploratory and risk-taking behaviors such as smoking occur at a higher rate in adolescents with chronic diseases than in healthy adolescents. In a large meta-analysis of adherence with ICS by adolescents and young adults,190 overall adherence was 28%, and slightly higher in those <18 years (36%). However, pharmacy refill data provided lower estimates of adherence than self-report measures. Predictors of adherence included personality, illness perceptions, and treatment beliefs. Management General principles for managing chronic disease in adolescents have been published by WHO.576 Adolescents and their parent/caregivers should be encouraged in the transition towards asthma self-management by the adolescent.577 This may involve the transition from a pediatric to an adult healthcare facility. Transitioning should not be based on chronological age but on developmental stage and readiness, using formal tools to assess readiness at around 11–13 years (ideal timing/age not based on evidence). Clinicians should aim to increase self-management, focusing consultations on areas in which the young person is not confident. Consider using technology to assist with adherence and guide young people to web-based apps and tools to improve knowledge of asthma. Awareness of asthma should be promoted to communities and peers. During consultations, the adolescent should be seen separately from the parent/caregiver so that sensitive issues such as smoking, adherence and mental health can be discussed privately, and confidentiality agreed. Information and self-management strategies should be tailored to the patient’s stage of psychosocial development and desire for autonomy; adolescents are often focused on short-term rather than long-term outcomes. An empathic approach should be used to COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 125 identify beliefs and behaviors that may be barriers to optimal treatment; for example, adolescents may be concerned about the impact of treatment on their physical or sexual capabilities. Medication regimens should be tailored to the adolescent’s needs and lifestyle, and reviews arranged regularly so that the medication regimen can be adjusted for changing needs. Information about local youth-friendly resources and support services should be provided, where available. In adolescents with mild asthma, use of as-needed low-dose ICS formoterol reduced risk of severe exacerbations compared with SABA alone, and without the need for daily treatment. Change in height from baseline in younger adolescents was significantly greater with as-needed ICS-formoterol than with daily low-dose ICS plus as-needed SABA.305 Exercise-induced bronchoconstriction (EIB) Clinical features Physical activity is an important stimulus for asthma symptoms for many patients, with symptoms and bronchoconstriction typically worsening after cessation of exercise. However, shortness of breath or wheezing during exercise may also relate to obesity or a lack of fitness, or to comorbid or alternative conditions such as inducible laryngeal obstruction.58,65 Management Regular treatment with ICS significantly reduces EIB (Evidence A).65 Training and sufficient warm-up reduce the incidence and severity of EIB (Evidence A).65 Taking SABAs, LABAs or chromones prior to exercise prevents EIB (Evidence A), but tolerance to the protective effects of SABAs and LABAs against EIB develops with regular (more than once-daily) use (Evidence A).65 However, in a 6-week study in patients with mild asthma, low-dose budesonide-formoterol, taken as needed for relief of symptoms and before exercise, was non-inferior for reducing EIB to regular daily ICS with as-needed SABA.236 More studies are needed, but this suggests that patients with mild asthma who are prescribed as-needed low-dose ICS-formoterol to prevent exacerbations and control symptoms can use the same medication prior to exercise, if needed, and do not need to be prescribed a SABA for pre-exercise use (Evidence B). Chromone pMDIs have been discontinued globally. Breakthrough EIB often indicates poorly controlled asthma, and stepping up ICS-containing treatment (after checking inhaler technique and adherence) generally results in the reduction of exercise-related symptoms. Athletes Clinical features Athletes, particularly those competing at a high level, have an increased prevalence of various respiratory conditions compared to non-athletes. They experience a higher prevalence of asthma, EIB, allergic or non-allergic rhinitis, chronic cough, inducible laryngeal obstruction, and recurrent respiratory infections. Airway hyperresponsiveness is common in elite athletes, often without reported symptoms. Asthma in elite athletes is commonly characterized by less correlation between symptoms and pulmonary function, higher lung volumes and expiratory flows, less eosinophilic airway inflammation, more difficulty in controlling symptoms, and some improvement in airway dysfunction after cessation of training.66 Management Preventive measures to avoid high exposure to air pollutants, allergens (if sensitized) and chlorine levels in pools, particularly during training periods, should be discussed with the athlete. They should avoid training in extreme cold or pollution (Evidence C), and the effects of any therapeutic trials of asthma medications should be documented. Adequate anti-inflammatory therapy, especially ICS, is advised; minimization of use of beta2 agonists will help to avoid the development of tolerance.65 Information on treatment of exercise-induced asthma in athletes can be found in a Joint Task Force Report prepared by the European Respiratory Society, the European Academy of Allergy and Clinical Immunology, and Global Allergy and Asthma European Network (GA(2)LEN)578 and on the World Anti-Doping Agency website (www.wada-ama.org). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 126 Pregnancy Clinical features Asthma control often changes during pregnancy; in approximately one-third of women asthma symptoms worsen, in one-third they improve, and in the remaining one-third they remain unchanged.579 Exacerbations are common in pregnancy, particularly in the second trimester.95 Exacerbations and poor asthma control during pregnancy may be due to mechanical or hormonal changes, or to cessation or reduction of asthma medications due to concerns by the mother and/or the healthcare provider. Pregnant women appear to be particularly susceptible to the effects of viral respiratory infections,580 including influenza. Exacerbations and poor symptom control are associated with worse outcomes for both the baby (pre-term delivery, low birth weight, increased perinatal mortality) and the mother (pre-eclampsia).95 Risk factors for asthma exacerbations during pregnancy include severe asthma, multiparity, black ethnicity, depression and anxiety, current smoking, age >35 years and obesity. Addressing these risk factors may not only reduce the risk of exacerbations, but also the risk of adverse perinatal outcomes.581 If asthma is well controlled throughout pregnancy there is little or no increased risk of adverse maternal or fetal complications.67 Management Although there is a general concern about any medication use in pregnancy, the advantages of actively treating asthma in pregnancy markedly outweigh any potential risks of usual asthma medications (Evidence A).67 For this reason, using medications to achieve good symptom control and prevent exacerbations is justified even when their safety in pregnancy has not been unequivocally proven. Use of ICS, beta2 agonists, montelukast or theophylline is not associated with an increased incidence of fetal abnormalities.582 Women with asthma who are pregnant or planning pregnancy should be advised not to stop ICS-containing therapy. Importantly, ICS reduce the risk of exacerbations of asthma during pregnancy (Evidence A),67,583,584 and cessation of ICS during pregnancy is a significant risk factor for exacerbations,95 (Evidence A). One study reported that a treatment algorithm in non-smoking pregnant women based on monthly measurement of fractional concentration of exhaled nitric oxide (FeNO) and symptoms using the Asthma Control Questionnaire (ACQ) was associated with significantly fewer exacerbations and better fetal outcomes than an algorithm based only on ACQ.585 However, the ACQ-only algorithm did not reflect current clinical recommendations, as LABA was introduced only after ICS had been increased to medium dose, and ICS could be stopped; 58% of women in the ACQ-only group were being treated without ICS by the end of pregnancy. In a subsequent large randomized controlled trial in pregnant women, there was no reduction in exacerbations with FeNO-guided treatment compared with usual care.16 Use of ICS during pregnancy by women with asthma may also be protective for asthma in their children. A study using administrative data reported that uncontrolled maternal asthma increased the risk of early-onset asthma in the offspring.586 In an intervention study with follow-up for 4–6 years, the prevalence of asthma was over 50% lower in children of women with asthma who took ICS during pregnancy compared with women who did not take ICS, with the largest reduction in prevalence of asthma in children when ICS was being taken in early pregnancy (before weeks 12–20).587 On balance, given the evidence in pregnancy and infancy for adverse outcomes from exacerbations during pregnancy (Evidence A),67 including due to lack of ICS or poor adherence,95 and evidence for safety of usual doses of ICS and LABA (Evidence A),582 a low priority should be placed on stepping down treatment (regardless of the method used to assess control) until after delivery (Evidence D), and ICS should not be stopped in preparation for pregnancy or during pregnancy (Evidence C). Despite lack of evidence for adverse effects of asthma treatment in pregnancy, many women and healthcare providers remain concerned about medication.588 Pregnant patients with asthma should be advised that poorly controlled asthma, and exacerbations, provide a much greater risk to their baby than do current asthma treatments. Educational resources about asthma management during pregnancy may provide additional reassurance.589 During pregnancy, monitoring of asthma every 4–6 weeks is recommended.589 It is feasible for this to be achieved by pharmacist-clinician collaboration, with monthly telephone monitoring of asthma symptom control.590 One observational study found that pregnant women COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 127 whose asthma was well controlled without controller therapy and who had no history of previous exacerbations were at low risk for exacerbations during pregnancy.591 However, such women should still be closely monitored. For women with severe asthma, evidence on use of biologic therapies during pregnancy is scarce.592 A registry study found no evidence of an increased risk of major congenital malformations when mothers received omalizumab during pregnancy. Women should be counselled that the potential risks associated with biologic exposure during pregnancy need to be balanced against the risks for themselves and their children caused by uncontrolled asthma.593 During acute asthma exacerbations, pregnant women may be less likely to be treated appropriately than non-pregnant patients.95 To avoid fetal hypoxia, it is important to manage acute asthma exacerbations during pregnancy aggressively with SABA, oxygen, and early administration of systemic corticosteroids. Respiratory infections should be monitored and managed appropriately during pregnancy.580 During labor and delivery, usual maintenance medications should be taken, with reliever if needed. Acute exacerbations during labor and delivery are uncommon, but bronchoconstriction may be induced by hyperventilation during labor, and should be managed with SABA. Neonatal hypoglycemia may be seen, especially in preterm babies, when high doses of beta-agonists have been given within the last 48 hours prior to delivery. If high doses of SABA have been given during labor and delivery, blood glucose levels should be monitored in the baby (especially if preterm) for the first 24 hours.594 A review of asthma guidelines for the management of asthma during pregnancy highlighted the need for greater clarity in current recommendations and the need for more RCTs among pregnant asthma patients.595 Women – perimenstrual asthma (catamenial asthma) Clinical features In approximately 20% of women, asthma is worse in the premenstrual phase. These women tend to be older, have more severe asthma, a higher BMI, a longer duration of asthma, and a greater likelihood of aspirin-exacerbated respiratory disease (AERD). They more often have dysmenorrhea, premenstrual syndrome, shorter menstrual cycles, and longer menstrual bleeding. The role of hormone levels and systemic inflammation remains unclear.596 Management In addition to the usual strategies for management of asthma, oral contraceptives and/or leukotriene receptor antagonists may be helpful (Evidence D).596 Further research is needed. Occupational asthma Clinical features In the occupational setting, rhinitis often precedes the development of asthma (see p.33 for information on making the diagnosis of occupational asthma). Once a patient has become sensitized to an occupational allergen, the level of exposure necessary to induce symptoms may be extremely low; resulting exacerbations become increasingly severe, and with continued exposure, persistent symptoms and irreversible airflow limitation may result.62 Management Detailed information is available in evidence-based guidelines about management of occupational asthma.62 All patients with adult-onset asthma should be asked about their work history and other exposures (Evidence A). The early identification and elimination of occupational sensitizers and the removal of sensitized patients from any further exposure are important aspects of the management of occupational asthma (Evidence A). Attempts to reduce occupational exposure have been successful, especially in industrial settings.62 Cost-effective minimization of latex sensitization can be achieved by using non-powdered low-allergen gloves instead of powdered latex gloves.62 Patients with suspected or confirmed occupational asthma should be referred for expert assessment and advice, if this is available, because of the economic and legal implications of the diagnosis (Evidence A). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 128 The elderly Clinical features Lung function generally decreases with longer duration of asthma and increasing age, due to stiffness of the chest wall, reduced respiratory muscle function, loss of elastic recoil and airway wall remodeling. Older patients may not report asthma symptoms, and may attribute breathlessness to normal aging or comorbidities such as cardiovascular disease and obesity.597-599 Among the elderly, there is no increased risk of cardiovascular disease among those with asthma, compared with those without asthma, except in current or former smokers.600 Comorbid arthritis may contribute to reduced exercise capacity and lack of fitness, and make inhaler device use difficult. Asthma costs may be higher amongst older patients, because of higher hospitalization rates and medication costs.598 Management Decisions about management of asthma in older people with asthma need to take into account both the usual goals of symptom control and risk minimization and the impact of comorbidities, concurrent treatments and lack of self-management skills.597,598 Data on efficacy of asthma medications in the elderly are limited because these patients are often excluded from major clinical trials. Side-effects of beta2 agonists such as cardiotoxicity, and corticosteroid side-effects such as skin bruising, osteoporosis and fragility fractures,601 and cataracts, are more common in the elderly than in younger adults.597 Clearance of theophylline is also reduced.597 Elderly patients should be asked about all of the other medications they are taking, including eye-drops, and potential drug interactions should be considered. Factors such as arthritis, muscle weakness, impaired vision and inspiratory flow should be considered when choosing inhaler devices for older patients,598,602 and inhaler technique should be checked at each visit. Older patients may have difficulties with complex medication regimens, and prescribing of multiple inhaler devices should be avoided if possible. Large-print versions may be needed for written information such as asthma action plans. Patients with cognitive impairment may require a career to help them use their asthma medications. For diagnosis and initial management of patients with asthma-COPD overlap, see Section 7 (p.131). Aspirin-exacerbated respiratory disease (AERD) Clinical features The clinical picture and course of AERD (previously called aspirin-induced asthma) are well established.239 It starts with nasal congestion and anosmia, and progresses to chronic rhinosinusitis with nasal polyps that re-grow rapidly after surgery. Asthma and hypersensitivity to aspirin and nonsteroidal anti-inflammatory drugs (NSAIDs) develop subsequently. Following ingestion of aspirin or NSAIDs, an acute asthma attack develops within minutes to 1–2 hours. It is usually accompanied by rhinorrhea, nasal obstruction, conjunctival irritation, and scarlet flush of the head and neck, and may sometimes progress to severe bronchospasm, shock, loss of consciousness, and respiratory arrest.603,604 AERD is more likely to be associated with low lung function and severe asthma,605,606 and with increased need for emergency care606. The prevalence of AERD is 7% in general adult asthma populations, and 15% in severe asthma.606,607 Diagnosis A history of exacerbation following ingestion of aspirin or other NSAIDs is highly suggestive of AERD. Aspirin challenge (oral, bronchial or nasal) is the gold standard for diagnosis608,609 as there are no reliable in vitro tests, but oral aspirin challenge tests must only be conducted in a specialized center with cardiopulmonary resuscitation capabilities because of the high risk of severe reactions.608,609 Bronchial (inhalational) and nasal challenges with lysine aspirin are safer than oral challenges and may be safely performed in allergy centers.608,610 Management Patients with AERD should avoid aspirin or NSAID-containing products and other medications that inhibit cyclooxygenase-1 (COX-1), but this does not prevent progression of the disease. Where an NSAID is indicated for other medical conditions, a COX-2 inhibitor (e.g., celecoxib or etoricoxib), or paracetamol (acetaminophen), may be considered,611,612 with appropriate healthcare provider supervision and observation for at least 2 hours after administration (Evidence B).613 ICS are the mainstay of asthma therapy in AERD, but OCS are sometimes required; LTRA may also be useful (Evidence COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 129 B),603,613 but note the concern about potential neuropsychiatric adverse effects with montelukast.295 See section 8 (p.139) for treatment options for patients with severe asthma. An additional option is aspirin desensitization, which may be conducted under specialist care in a clinic or hospital.614 Desensitization to aspirin followed by daily aspirin treatment can significantly improve upper respiratory symptoms and overall quality of life, decrease recurrence of nasal polyps, reduce the need for OCS and sinus surgery, and improve nasal and asthma scores, but few double-blind studies have examined asthma outcomes.608,615,616 Aspirin desensitization is associated with a significantly increased risk of adverse effects such as gastritis and gastrointestinal bleeding.616 Allergic bronchopulmonary aspergillosis (ABPA) Clinical features Allergic bronchopulmonary aspergillosis (ABPA) is a complex pulmonary disease characterized by repeated episodes of wheezing, fleeting pulmonary opacities and development of bronchiectasis, sometimes with malaise, weight loss and hemoptysis. Some patients expectorate brownish sputum plugs. ABPA is most commonly diagnosed in people with asthma or cystic fibrosis, due to a hypersensitivity response to Aspergillus fumigatus, a common indoor and outdoor mold. Diagnosis Diagnosis of ABPA is based on composite criteria including immediate hypersensitivity reaction to A. fumigatus, total serum IgE, specific IgG to A. fumigatus, radiological features and blood eosinophils.617 Sensitization to fungal allergens, without the full picture of ABPA, is often found in asthma, particularly in severe asthma, where it is sometimes called ‘severe asthma with fungal sensitization’. Management Current first-line therapy is with oral corticosteroids (e.g., a 4-month tapering course), with itraconazole reserved for those with exacerbations or requiring long-term OCS.618-620 Clinicians should be aware of the potential for drug interactions between itraconazole (a cytochrome P450 inhibitor) and asthma medications. These interactions may lead to increased risk of ICS adverse effects such as adrenal suppression and Cushing’s syndrome, and may increase the risk of cardiovascular adverse effects of some LABAs (salmeterol and vilanterol).157 Concomitant use is not recommended, so it may be appropriate to switch ICS-LABA treatment to an alternative product such as budesonide-formoterol or mometasone-formoterol for the duration of treatment with itraconazole.157 A randomized double-blind placebo-controlled study in patients with severe asthma and ABPA found significantly fewer exacerbations with omalizumab (anti-IgE) than placebo.621 A systematic review and meta-analysis that included this trial and others, but with a total of only 450 patients in the analysis, provides evidence of moderate quality that patients with ABPA who do not respond to treatment with oral corticosteroids have a favorable response to omalizumab without substantial side effects.622 There have also been case series reports of treatment of ABPA with benralizumab, dupilumab and mepolizumab. Information about use of biologic therapies in severe asthma is covered in Section 8 (p.139). In patients with ABPA and bronchiectasis, regular physiotherapy and daily drainage are recommended. Patients with ABPA should be referred for specialist investigation and care if available. Surgery and asthma Clinical features There is no evidence of increased peri-operative risk for the general asthma population.623 However, there is an increased risk for patients with COPD,623 and this may also apply to asthma patients with reduced FEV1. The incidence of severe peri-operative bronchospasm in people with asthma is low, but it may be life threatening.624 Management For elective surgery, meticulous attention should be paid pre-operatively to achieving good asthma control, as detailed elsewhere in this chapter, especially for patients with more severe asthma, uncontrolled symptoms, exacerbation history, or persistent airflow limitation (Evidence B).624 For patients requiring emergency surgery, the risks of proceeding without COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 130 first achieving good asthma control should be weighed against the need for immediate surgery. Patients taking long-term high-dose ICS or who have received OCS for more than 2 weeks during the previous 6 months should receive hydrocortisone peri-operatively as they are at risk of adrenal crisis in the context of surgery (Evidence B).625 More immediate intra-operative issues relating to asthma management are reviewed in detail elsewhere.624 For all patients, maintaining their prescribed ICS-containing therapy throughout the peri-operative period is important. Air travel and asthma Practical advice for air travel by people with respiratory disease was published by the British Thoracic Society (BTS) in 2022.626 The advice for people with asthma included pre-flight optimization of treatment, carrying all asthma medications and spacer (if used) in the cabin to allow immediate access during the flight (and in case checked luggage is mislaid), and carrying a copy of the patient’s asthma action plan. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 131 7. Diagnosis and initial treatment in adults with features of asthma, COPD or both KEY POINTS Asthma and chronic obstructive pulmonary disease (COPD) are heterogeneous and overlapping conditions • ‘Asthma’ and ‘COPD’ are umbrella labels for heterogeneous conditions characterized by chronic airway and/or lung disease. Asthma and COPD each include several different clinical phenotypes, and are likely to have several different underlying mechanisms, some of which may be common to both asthma and COPD.
• Symptoms of asthma and COPD may be similar, and the diagnostic criteria overlap.
Why are the labels ‘asthma’ and ‘COPD’ still important? • There are extremely important differences in evidence-based treatment recommendations for asthma and COPD. For example, treatment with a long-acting beta2 agonist (LABA) and/or long-acting muscarinic antagonist (LAMA) alone (i.e., without inhaled corticosteroids [ICS]) is recommended as initial treatment in COPD but contraindicated in asthma due to the risk of severe exacerbations and death.
• These risks are also seen in patients who have diagnoses of both asthma and COPD, making it important to identify adult patients who, for safety, should not be treated with long-acting bronchodilators alone. ICS reduce mortality and hospitalizations in patients with asthma, including in those with concomitant COPD.
Many patients have features of both asthma and COPD • Distinguishing asthma from COPD can be difficult, particularly in smokers and older adults, and some patients may have features of both asthma and COPD.
• The terms ‘asthma-COPD overlap’ or ‘asthma+COPD’ are simple descriptors for patients who have features of both asthma and COPD.
• These terms do not refer to a single disease entity. They include patients with several clinical phenotypes that are likely caused by a range of different underlying mechanisms.
• More research is needed to better define these phenotypes and mechanisms, but in the meantime, safety of pharmacologic treatment is a high priority.
Diagnosis • Diagnosis in patients with chronic respiratory symptoms involves a stepwise approach, first, is the patient likely to have chronic airways disease, then syndromic categorization as having typical asthma, typical COPD, features of both, and/or with other conditions such as bronchiectasis.
• Lung function testing is essential for confirming persistent airflow limitation, but variable airflow obstruction can be detected with serial peak flow measurements and/or measurements before and after bronchodilator.
Initial treatment for safety and clinical efficacy • For asthma: ICS are essential, either alone or in combination with a LABA, to reduce the risk of severe exacerbations and death. Do not treat with LABA and/or LAMA alone, without ICS.
• For patients with features of both asthma and COPD, treat as asthma. ICS-containing therapy is important to reduce the risk of severe exacerbations and death. Do not give LABA and/or LAMA alone without ICS.
• For COPD: Treat according to current recommendations from the Global Initiative for Chronic Obstructive Lung Disease (GOLD),72 i.e., initial treatment with LAMA and LABA, plus as-needed SABA; add ICS for patients with COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 132 hospitalizations, ≥2 exacerbations/year requiring oral corticosteroids (OCS), or blood eosinophils ≥300/µl. Avoid high dose ICS because of risk of pneumonia. • All patients: provide structured education especially focusing on inhaler technique and adherence; assess for, and treat, other clinical problems, including advice about smoking cessation, immunizations, physical activity, and management of multimorbidity. • Specialist referral for additional investigations in patients with asthma+COPD is encouraged, as they often have worse outcomes than patients with asthma or COPD alone. OBJECTIVES The objectives of this section of the GINA Strategy Report are: • To assist primary care clinicians to identify typical asthma and typical COPD and to recognize when patients have features of both. This is particularly relevant in patients aged 40 years and older. • To provide advice about safe and effective initial treatment • To provide guidance on indications for referral for specialist assessment. BACKGROUND TO DIAGNOSING ASTHMA AND/OR COPD IN ADULT PATIENTS Why are the labels ‘asthma’ and ‘COPD’ still important? Asthma and COPD are heterogeneous conditions characterized by airway obstruction. Each of these ‘umbrella’ labels includes several different patterns of clinical features (phenotypes) that may overlap. Each may also include different inflammatory patterns and different underlying mechanisms, some of which may be common to both asthma and COPD.627 The most easily recognized phenotypes of asthma and COPD such as allergic asthma in children/young adults and emphysema in older smokers are clearly distinguishable. Regulatory studies of pharmacotherapy in asthma and COPD are largely restricted to patients with very clearly defined asthma or COPD. However, in the community, the features of asthma and COPD may overlap, especially in older adults. There are extremely important differences in treatment recommendations for asthma and COPD. In particular, treatment with long-acting bronchodilators alone (i.e., without ICS) is recommended for initial treatment in COPD72 but is contraindicated in asthma due to the risk of severe exacerbations and death.151,400,628,629 Several studies have also shown that patients with diagnoses of both asthma and COPD are at increased risk of hospitalization or death if they are treated with LABA or LABA-LAMA, compared with ICS-LABA (or ICS-LABA-LAMA).630-632 Challenges in clinical diagnosis of asthma and COPD Although asthma is characterized by variable expiratory airflow limitation, at least initially (Box 1-2, p.26), and COPD is characterized by persistent airflow limitation,72 the definitions of asthma and COPD are not mutually exclusive (Box 7-1, p.133). This means that clinical features are also important in making a diagnosis and treating appropriately. In children and young adults with chronic or recurrent respiratory symptoms, the differential diagnosis is different from that in older adults (see Box 1-3, p.27). Once infectious disease and nonpulmonary conditions (e.g., congenital heart disease, inducible laryngeal obstruction) have been excluded, the most likely chronic airway disease in children and young adults is asthma. However, in adults with a history of long-standing asthma,633,634 persistent airflow limitation may be found.635-639 Distinguishing this from COPD is problematic, especially if they are smokers or have other risk factors for COPD.640-643 On the other hand, patients with COPD may show evidence of reversible airflow obstruction when a rapid-acting bronchodilator is administered, a feature more strongly associated with asthma. In medical records, such patients often are assigned both diagnoses.74,644 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 133 In keeping with common usage of the term ‘overlap’ in other contexts, e.g., for the association between COPD with sleep disorders, and in overlap syndromes of collagen vascular disease, the descriptive term ‘asthma-COPD overlap’ is often used. Another common descriptor is ‘asthma+COPD’. ‘Asthma-COPD overlap’ is a descriptor for patients often seen in clinical practice, who comprise a heterogeneous group. It does not mean a single disease entity. Box 7-1. Current definitions of asthma and COPD, and clinical description of asthma-COPD overlap Asthma (GINA) Asthma is a heterogeneous disease, usually characterized by chronic airway inflammation. It is defined by the history of respiratory symptoms such as wheeze, shortness of breath, chest tightness and cough that vary over time and in intensity, together with variable expiratory airflow limitation. COPD (GOLD) Chronic obstructive pulmonary disease (COPD) is a heterogeneous lung condition characterized by chronic respiratory symptoms (dyspnea, cough, sputum production and/or exacerbations) due to abnormalities of the airways (bronchitis, bronchiolitis) and/or alveoli (emphysema) that cause persistent, often progressive, airflow obstruction.72 Asthma+COPD, also called asthma-COPD overlap (descriptive term) ‘Asthma-COPD overlap’ and ‘asthma +COPD’ are terms used to collectively describe patients who have persistent airflow limitation together with clinical features that are consistent with both asthma and COPD. This is not a definition of a single disease entity, but a descriptive term for clinical use that includes several different clinical phenotypes reflecting different underlying mechanisms. Prevalence and morbidity of asthma-COPD overlap In epidemiological studies, reported prevalence rates for asthma+COPD have ranged between 9% and 55% of those with either diagnosis, with variation by gender and age;638,645-647 the wide range reflects the different criteria that have been used by different investigators. Concurrent doctor-diagnosed asthma and COPD has been reported in between 15 and 32% of patients with one or other diagnosis.644,648,649 There is broad agreement that patients with features of both asthma and COPD have a greater burden of symptoms,650 experience frequent exacerbations,74,636,650 have poor quality of life,74,645,650 a more rapid decline in lung function,650 higher mortality,636,644 and greater use of healthcare resources74,651 compared with patients with asthma or COPD alone. ASSESSMENT AND MANAGEMENT OF CHRONIC RESPIRATORY SYMPTOMS 1: History and clinical assessment Establish: • The nature and pattern of respiratory symptoms (variable and/or persistent) • History of asthma diagnosis; childhood and/or current • Exposure history: smoking and/or other exposures to risk factors for COPD. The features that are most helpful in identifying and distinguishing asthma from COPD, and the features that should prompt a patient to be treated as asthma to reduce the risk of severe exacerbations and death, are shown in Box 7-2 (p.134).
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 134 Box 7-2. Syndromic approach to initial treatment in patients with asthma and/or COPD See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 135 Caution: Consider alternative diagnoses; other airways diseases, such as bronchiectasis and chronic bronchitis, and other forms of lung disease such as interstitial lung disease may present with some of the above features. The above approach to diagnosis does not replace the need for a full assessment in patients presenting with respiratory symptoms, to first exclude non-respiratory diagnoses such as heart failure. Physical examination may provide supportive information. 2: Lung function testing is essential Use lung function testing to confirm: • The presence of persistent expiratory airflow limitation • Variable expiratory airflow limitation. Spirometry is preferably performed at the initial assessment. In cases of clinical urgency it may be delayed to a subsequent visit, but confirmation of diagnosis may be more difficult once patients are started on ICS-containing therapy (see Box 1-4, p.30). Early confirmation (or exclusion) of the presence of persistent expiratory airflow limitation may avoid needless trials of therapy, or delays in initiating other investigations. Spirometry can confirm both persistent airflow limitation and reversibility (Box 7-2, p.134 and Box 7-3, p.135). Measurement of peak expiratory flow (PEF), if performed repeatedly on the same meter over a period of 1–2 weeks, may help to confirm reversible airflow limitation and the diagnosis of asthma by demonstrating excessive variability (Box 1-2, p.26). However, PEF is not as reliable as spirometry, and a normal PEF does not rule out either asthma or COPD. Box 7-3. Spirometric measures in asthma and COPD Spirometric variable Asthma COPD Asthma+COPD Normal FEV1/FVC pre- or post BD Compatible with asthma. If patient is symptomatic at a time when lung function is normal, consider alternative diagnosis. Not compatible with COPD Not compatible Reduced post-BD FEV1/FVC (< lower limit of normal, or <0.7) [GOLD 2024] Indicates airflow limitation but may improve spontaneously or on treatment Required for diagnosis of COPD Required for diagnosis of asthma+COPD Post-BD FEV1 ≥80% predicted Compatible with diagnosis of asthma (good asthma control or interval between symptoms) Compatible with mild persistent airflow limitation if post-BD FEV1/FVC is reduced Compatible with mild persistent airflow limitation if post-BD FEV1/FVC is reduced Post-BD FEV1 <80% predicted Compatible with diagnosis of asthma. Risk factor for asthma exacerbations An indicator of severity of airflow limitation and risk of future events (e.g., mortality and COPD exacerbations) As for COPD and asthma Post-BD increase in FEV1 ≥12% and 200 mL from baseline (reversible airflow limitation). Usual at some time in course of asthma, but may not be present when well controlled or on ICS-containing therapy Common and more likely when FEV1 is low Common and more likely when FEV1 is low Post-BD increase in FEV1 >12% and 400 mL from baseline (marked reversibility) High probability of asthma Unusual in COPD Compatible with asthma+COPD See list of abbreviations (p.11). The 2024 GOLD Report is available at COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 136 3: Select initial treatment according to diagnosis (see Box 7-2 (p.134) For asthma Commence treatment as described in Chapter 4 (Box 4-4, p.75 and Box 4-5, p.76). Pharmacotherapy is based on ICS to reduce the risk of severe exacerbations and death and to improve symptom control, with add-on treatment as required, e.g., add-on LABA and/or LAMA. As-needed low-dose ICS-formoterol may be used as the reliever, on its own in mild asthma or in addition to maintenance ICS-formoterol in patients with moderate-severe asthma prescribed maintenance-and-reliever therapy (see Box 4-6, p.77). Inhaled therapy should be optimized to minimize the need for oral corticosteroids (OCS). For COPD Commence treatment as in the current GOLD strategy report.72 Pharmacotherapy starts with symptomatic treatment with long-acting bronchodilators (LABA and LAMA). Use of ICS is strongly favored as per GOLD 2024 for patients with hospitalizations in the last year, ≥2 exacerbations/year requiring OCS, or blood eosinophils ≥300/µL, or with a history of asthma or concomitant asthma. However, ICS should not be used alone without LABA and/or LAMA. Inhaled therapy should be optimized to reduce the need for OCS. In patients with features of COPD, high-dose ICS should be avoided because of the risk of pneumonia.652,653 For patients with features of asthma and COPD Start treatment as for asthma (Box 4-4, p.75 and Box 4-5, p.76) until further investigations have been performed. ICS are essential in preventing morbidity and even death in patients with uncontrolled asthma symptoms, for whom even seemingly ‘mild’ symptoms (compared to those of moderate or severe COPD) might indicate significant risk of a life-threatening attack.654 For patients with asthma+COPD, ICS should be used initially in a low or medium dose (see Box 4-2, p.71), depending on level of symptoms and risk of adverse effects, including pneumonia. Patients with features or diagnosis of both asthma and COPD will usually also require add-on treatment with LABA and/or LAMA to provide adequate symptom control. Patients with any features of asthma should not be treated with LABA and/or LAMA alone, without ICS. A large case-control study in community patients with newly diagnosed COPD found that those who also had a diagnosis of asthma had a lower risk of COPD hospitalizations and death if treated with combination ICS-LABA than with LABA alone.630 In another large retrospective longitudinal population cohort study of patients aged ≥66 years, those recorded as having asthma with COPD had lower morbidity and hospitalizations if they received ICS treatment; a similar benefit was seen in those with COPD plus concurrent asthma.632 All patients with chronic airflow limitation Provide advice, as described in the GINA and GOLD reports, about: • Treatment of modifiable risk factors including advice about smoking cessation • Treatment of comorbidities • Non-pharmacological strategies including physical activity, and, for COPD or asthma-COPD overlap, pulmonary rehabilitation and vaccinations • Appropriate self-management strategies • Regular follow-up. In a majority of patients, the initial management of asthma and COPD can be satisfactorily carried out at primary care level. However, both the GINA and GOLD strategy reports recommend referral for further diagnostic procedures at relevant points in patient management (see below). This may be particularly important for patients with features of both asthma and COPD, given that this is associated with worse outcomes and greater healthcare utilization. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 137 4: Refer for specialized investigations (if necessary) Referral for expert advice and further diagnostic evaluation is advised in the following contexts: • Patients with persistent symptoms and/or exacerbations despite treatment. • Diagnostic uncertainty, especially if an alternative diagnosis (e.g., bronchiectasis, post-tuberculous scarring, bronchiolitis, pulmonary fibrosis, pulmonary hypertension, cardiovascular diseases and other causes of respiratory symptoms) needs to be investigated. • Patients with suspected asthma or COPD in whom atypical or additional symptoms or signs (e.g., hemoptysis, significant weight loss, night sweats, fever, signs of bronchiectasis or other structural lung disease) suggest an additional pulmonary diagnosis. This should prompt early referral, without waiting for a trial of treatment for asthma or COPD. • When chronic airways disease is suspected but syndromic features of either asthma or COPD are few. • Patients with comorbidities that may interfere with the assessment and management of their airways disease, particularly cardiovascular disease. Referral may also be appropriate for issues arising during ongoing management of asthma, COPD or asthma-COPD overlap, as outlined in the GINA and GOLD strategy reports. Box 7-4 (p.137) summarizes specialized investigations that are sometimes used to distinguish asthma and COPD. Box 7-4. Specialized investigations sometimes used in patients with features of asthma and COPD Asthma COPD Lung function tests DLCO Normal (or slightly elevated) Often reduced Arterial blood gases Normal between exacerbations May be chronically abnormal between exacerbations in more severe forms of COPD Airway hyperresponsiveness (AHR) Not useful on its own in distinguishing asthma from COPD, but higher levels of AHR favor asthma Imaging High resolution CT Scan Usually normal but air trapping and increased bronchial wall thickness may be observed. Low attenuation areas denoting either air trapping or emphysematous change can be quantitated; bronchial wall thickening and features of pulmonary hypertension may be seen. Biomarkers A positive test for atopy (specific IgE and/or skin prick test to aeroallergens) Increases probability of allergic asthma; not essential for diagnosis of asthma Conforms to background prevalence; does not rule out COPD FeNO A high level (>50 ppb) in non-smokers is moderately associated with eosinophilic airway inflammation. Usually normal Low in current smokers Blood eosinophilia Supports diagnosis of eosinophilic airway inflammation May be present in COPD including during exacerbations Sputum inflammatory cells Role in differential diagnosis is not established in large populations. See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 138 Unanswered clinical questions There is an urgent need for more research on this topic to guide better recognition and safe and effective treatment. Patients who do not have ‘classical’ features of asthma or of COPD, or who have features of both, have generally been excluded from randomized controlled trials of most therapeutic interventions for airways disease, and from many mechanistic studies. Future research should include study of clinical and physiological characteristics, biomarkers, outcomes and underlying mechanisms, among broad populations of patients with respiratory symptoms or with chronic airflow limitation. In the meantime, the present chapter provides interim advice about diagnosis and initial treatment, from the perspective of clinicians, particularly those in primary care and nonpulmonary specialties. Further research is needed to inform evidence-based definitions and a more detailed classification of patients who present overlapping features of asthma and COPD, and to encourage the development of specific interventions for clinical use. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 139 8. Difficult-to-treat and severe asthma in adults and adolescents KEY POINTS What are difficult-to-treat asthma and severe asthma? • Difficult-to-treat asthma is asthma that is uncontrolled despite prescribing of medium or high-dose treatment with the combination of inhaled corticosteroid (ICS) and long-acting beta2 agonist (LABA), or that requires high-dose ICS-LABA treatment to maintain good symptom control and reduce exacerbations. It does not mean a ‘difficult patient’. • Severe asthma is asthma that is uncontrolled despite adherence with optimized high-dose ICS-LABA therapy and treatment of contributory factors, or that worsens when high-dose treatment is decreased. Approximately 3–10% of people with asthma have severe asthma. • Severe asthma places a large physical, mental, emotional, social and economic burden on patients. It is often associated with multimorbidity. How should these patients be assessed? • Assess all patients with difficult-to-treat asthma to confirm the diagnosis of asthma, and to identify and manage factors that may be contributing to symptoms, poor quality of life, or exacerbations. • Refer for expert advice at any stage, or if asthma does not improve in response to optimizing treatment. • For patients with persistent symptoms and/or exacerbations despite high-dose ICS-containing therapy (with good adherence and correct inhaler technique), the clinical or inflammatory phenotype should be assessed, as this may guide the selection of add-on treatment. Management of severe asthma • Depending on the inflammatory phenotype and other clinical features, add-on treatments for severe asthma include long-acting muscarinic antagonists (LAMA), leukotriene receptor antagonists (LTRAs), low-dose azithromycin (adults), and biologic agents for severe asthma. • Low-dose maintenance oral corticosteroids (OCS) should be considered only as a last resort if no other options are available, because of their serious cumulative long-term side-effects. • Assess the response to any add-on treatment, stop ineffective treatments, and consider other options. • Utilize specialist multidisciplinary team care for severe asthma, if available. • For patients with severe asthma, continue to optimize patient care in collaboration with the primary care clinician, and considering the patient’s social and emotional needs. • Invite patients with severe asthma to enroll in a registry or clinical trial, if available and relevant, to help fill evidence gaps. • See Boxes 8-2 through 8-5 (starting on p.142) for the GINA severe asthma decision tree. • Although the majority of patients can achieve the goal of long-term well controlled asthma, some patients’ asthma will not be well controlled even with optimal therapy. This section will also be published separately as a GINA short guide for health professionals: Difficult-to-Treat and Severe Asthma in Adolescent and Adult Patients. Diagnosis and Management. V5.0, 2024 (the Severe Asthma Guide), available to download or order from the GINA website (www.ginasthma.org). • Other resources about severe asthma include an online toolkit published by the Australian Centre of Excellence in Severe Asthma (www.toolkit.severeasthma.org.au). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 140 DEFINITIONS: UNCONTROLLED, DIFFICULT-TO-TREAT, AND SEVERE ASTHMA Understanding the definitions of difficult-to-treat and severe asthma starts with the concept of uncontrolled asthma. Uncontrolled asthma includes one or both of the following: • Poor symptom control (frequent symptoms or reliever use, activity limited by asthma, night waking due to asthma) • Frequent exacerbations (≥2/year) requiring OCS, or serious exacerbations (≥1/year) requiring hospitalization. Difficult-to-treat asthma is asthma that is uncontrolled despite prescribing of medium- or high-dose ICS with a second controller (usually a LABA) or with maintenance OCS, or that requires high-dose treatment to maintain good symptom control and reduce the risk of exacerbations.175 It does not mean a ‘difficult patient’. In many cases, asthma may appear to be difficult to treat because of modifiable factors such as incorrect inhaler technique, poor adherence, smoking or comorbidities, or because the diagnosis is incorrect. Severe asthma is a subset of difficult-to-treat asthma (Box 8-1). It means asthma that is uncontrolled despite adherence with maximal optimized high-dose ICS-LABA treatment and management of contributory factors, or that worsens when high-dose treatment is decreased.175 At present, therefore, ‘severe asthma’ is a retrospective label. It is sometimes called ‘severe refractory asthma’,175 because it is defined by being relatively refractory to high-dose inhaled therapy. However, with the advent of biologic therapies, the word ‘refractory’ is no longer appropriate. Asthma is not classified as severe if it markedly improves when contributory factors such as inhaler technique and adherence are addressed.175 PREVALENCE: HOW MANY PEOPLE HAVE SEVERE ASTHMA? A study in the Netherlands estimated that around 3.7% of asthma patients have severe asthma, based on the number of patients prescribed high-dose ICS-LABA, or medium or high-dose ICS-LABA plus long-term OCS, who had poor symptom control (by Asthma Control Questionnaire) and had good adherence and inhaler technique (Box 8-1).655 Box 8-1. What proportion of adults have difficult-to-treat or severe asthma? See list of abbreviations (p.11). Data from the Netherlands, reported by Hekking et al (2015)655 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 141 IMPORTANCE: THE IMPACT OF SEVERE ASTHMA The patient perspective Patients with severe asthma experience a heavy burden of symptoms, exacerbations and medication side-effects. Frequent shortness of breath, wheeze, chest tightness and cough interfere with day-to-day living, sleeping, and physical activity, and patients often have frightening or unpredictable exacerbations (also called attacks or severe flare-ups). Medication side-effects are particularly common and problematic with OCS,393 which in the past were a mainstay of treatment for severe asthma. Adverse effects of long-term or frequent OCS include obesity, diabetes, osteoporosis and fragility fractures,601 cataracts, hypertension and adrenal suppression; psychological side-effects such as depression and anxiety are particularly concerning for patients.656 Even short-term use of OCS is associated with sleep disturbance, and increased risk of infection, fracture and thromboembolism.571 Strategies to minimize need for OCS are therefore a high priority. Severe asthma often interferes with family, social and working life, limits career choices and vacation options, and affects emotional and mental health. Patients with severe asthma often feel alone and misunderstood, as their experience is so different from that of most people with asthma.656 Adolescents with severe asthma The teenage years are a time of great psychological and physiological development which can impact on asthma management. It is vital to ensure that the young person has a good understanding of their condition and treatment and appropriate knowledge to enable supported self-management. The process of transition from pediatric to adult care should help support the young person in gaining greater autonomy and responsibility for their own health and wellbeing. Severe asthma may improve over 3 years in approximately 30% of male and female adolescents; the only predictor of asthma becoming non-severe was higher baseline blood eosinophils.657 Studies with longer follow-up time are needed. Healthcare utilization and costs Severe asthma has very high healthcare costs due to medications, physician visits, hospitalizations, and the costs of OCS side-effects. In a UK study, healthcare costs per patient were higher than for type 2 diabetes, stroke, or COPD.658 In a Canadian study, severe uncontrolled asthma was estimated to account for more than 60% of asthma costs.659 Patients with severe asthma and their families also bear a significant financial burden, not only for medical care and medications, but also through lost earnings and career choices. OVERVIEW OF DECISION TREE FOR ASSESSMENT AND MANAGEMENT OF DIFFICULT-TO-TREAT AND SEVERE ASTHMA The clinical decision tree (from p.142), provides brief information about what should be considered in each phase of diagnosis and management of difficult-to-treat and severe asthma. The decision tree is divided into three broad stages: Stages 1–4 (green) are for use in primary care and/or specialist care. Stages 5–8 (blue) are mainly relevant to respiratory specialists. Stages 9–10 (brown) are about maintaining ongoing collaborative care between the patient, primary care physician, specialist and other health professionals. Development of the Severe Asthma Guide and decision tree included extensive collaboration with experts in human-centered design to enhance the utility of these resources for end-users. This included translating existing high-level flowcharts and text-based information to a more detailed visual format, and applying information architecture and diagramming principles. The decision tree is followed by more detailed information on each stage of assessment and management. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 142 Box 8-2. Decision tree – investigate and manage difficult to treat asthma in adult and adolescent patients See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 143 Box 8-3. Decision tree – assess and treat severe asthma phenotypes See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 144 Box 8-4. Decision tree – consider add-on biologic Type 2-targeted treatments See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 145 Box 8-5. Decision tree – monitor and manage severe asthma treatment See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 146 INVESTIGATE AND MANAGE DIFFICULT-TO-TREAT ASTHMA IN ADULTS AND ADOLESCENTS 1. Confirm the diagnosis (asthma or differential diagnoses) Stages 1–5 can be carried out in primary or specialist care. A patient is classified as having difficult-to-treat asthma if they patient have persistent asthma symptoms and/or exacerbations despite prescribing of medium or high-dose ICS with another controller such as LABA, or maintenance OCS, or require high-dose ICS-LABA treatment to maintain good symptom control and prevent exacerbations. Difficult-to-treat asthma does not mean a ‘difficult patient’. Consider referral to a specialist or severe asthma clinic at any stage, particularly if: • There is difficulty confirming the diagnosis of asthma • Patient has frequent urgent healthcare utilization • Patient needs frequent or maintenance OCS • Occupational asthma is suspected • Food allergy or anaphylaxis, as this increases the risk of death • Symptoms are suggestive of infective or cardiac cause • Symptoms are suggestive of complications such as bronchiectasis • Patient has multimorbidity. Are the symptoms due to asthma? Perform a careful history and physical examination to identify whether symptoms are typical of asthma, or are more likely due to an alternative diagnosis or comorbidity: • Dyspnea: COPD, obesity, cardiac disease, deconditioning • Cough: inducible laryngeal obstruction (also called vocal cord dysfunction, VCD), upper airway cough syndrome (also called post-nasal drip), gastro-esophageal reflux disease (GERD), bronchiectasis, angiotensin-converting enzyme (ACE) inhibitors • Wheeze: obesity, COPD, tracheobronchomalacia, VCD. Investigate according to clinical suspicion and age (see Box 1-3, p.27). How can the diagnosis of asthma be confirmed? Confirmation of the diagnosis is important, because in 12–50% of people assumed to have severe asthma, asthma is not found to be the correct diagnosis.660 Perform spirometry, before and after bronchodilator, to assess baseline lung function and seek objective evidence of variable expiratory airflow limitation. If initial bronchodilator responsiveness testing is negative (≤200 mL or ≤12% increase in FEV1), consider repeating after withholding bronchodilators or when symptomatic, or consider stepping controller treatment up or down before further investigations such as bronchial provocation testing (see Box 1-4, p.30). Check full flow-volume curve to assess for upper airway obstruction. If spirometry is normal or is not available, provide the patient with a peak flow meter and diary for assessing variability; consider bronchial provocation testing if patient is able to withhold bronchodilators (short-acting beta2 agonist [SABA] for at least 6 hours, LABA for up to 2 days depending on duration of action).41 Strategies for confirming the diagnosis of asthma in patients already taking ICS-containing treatment are shown in Box 1-4 (p.30). Airflow limitation may be persistent in patients with long-standing asthma, due to remodeling of the airway walls, or limited lung development in childhood. It is important to document lung function when the diagnosis of asthma is first made. Specialist advice should be obtained if the history is suggestive of asthma but the diagnosis cannot be confirmed by spirometry. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 147 2. Look for factors contributing to symptoms and exacerbations Systematically consider factors that may be contributing to uncontrolled symptoms or exacerbations, or poor quality of life, and that can be treated. The most important modifiable factors include: • Incorrect inhaler technique (seen in up to 80% patients): ask the patient to show you how they use their inhaler; compare with a checklist or video. • Suboptimal adherence (up to 75% asthma patients): ask empathically about frequency of use (e.g., ‘Many patients don’t use their inhaler as prescribed. In the last 4 weeks, how many days a week have you been taking it – not at all, 1 day a week, 2, 3 or more?’ or, ‘Do you find it easier to remember your inhaler in the morning or the evening?’ (see Box 5-3, p.112). Ask about barriers to medication use, including cost, and concerns about necessity or side-effects. Check dates on inhalers and view dispensing data, if available. Electronic inhaler monitoring, if available, can be helpful in screening for poor adherence, in some cases avoiding the need for biologic therapy.661 • Comorbidities: review history and examination for comorbidities that can contribute to respiratory symptoms, exacerbations, or poor quality of life. These include anxiety and depression, obesity, deconditioning, chronic rhinosinusitis, inducible laryngeal obstruction, GERD, COPD, obstructive sleep apnea, bronchiectasis, cardiac disease, and kyphosis due to osteoporosis. Investigate according to clinical suspicion. • Modifiable risk factors and triggers: identify factors that increase the risk of exacerbations, e.g., smoking, environmental tobacco exposure, other environmental exposures at home or work including allergens (if sensitized), indoor and outdoor air pollution, molds and noxious chemicals, and medications such as beta-blockers or non-steroidal anti-inflammatory drugs (NSAIDs). For allergens, check for sensitization using skin prick testing or specific IgE. • Regular or over-use of SABAs: regular SABA use causes beta-receptor down-regulation and reduction in response,662 leading in turn to greater use. Over-use may also be habitual. Dispensing of ≥3 SABA canisters per year (corresponding to average use more than daily) is associated with increased risk of emergency department visit or hospitalization independent of severity,86,87 and dispensing of ≥12 canisters per year (one a month) is associated with substantially increased risk of death.87,89 Risks are higher with nebulized SABA.663 • Anxiety, depression and social and economic problems: these are very common in asthma, particularly in difficult asthma656 and contribute to symptoms, impaired quality of life, and poor adherence. • Medication side-effects: systemic effects, particularly with frequent or continuous OCS, or long-term high-dose ICS may contribute to poor quality of life and increase the likelihood of poor adherence. Local side-effects of dysphonia or candidiasismay occur with high-dose or potent ICS, especially if inhaler technique is poor. Consider drug interactions including risk of adrenal suppression with use of P450 inhibitors such as itraconazole. 3. Review and optimize management Review and optimize treatment for asthma, and for comorbidities and risk factors identified at Stage 2. For more details, see Section 6 (p.117). • Provide asthma self-management education, and confirm that patient has (and knows how to use) a personalized written or electronic asthma action plan. Refer to an asthma educator if available. • Optimize asthma medications: confirm that the inhaler is suitable for the patient; check and correct inhaler technique with a physical demonstration and teach-back method, check inhaler technique again at each visit.664 Address suboptimal adherence, both intentional and unintentional.506 Switch to ICS-formoterol maintenance-and-reliever therapy if available, to reduce the risk of exacerbations.224 • Consider non-pharmacologic add-on therapy, e.g., smoking cessation, physical exercise,233 healthy diet, weight loss, mucus clearance strategies, influenza vaccination, breathing exercises, allergen avoidance, if feasible, for patients who are sensitized and exposed. For details see text following Box 3-6, p.57. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 148 • Treat comorbidities and modifiable risk factors identified in Stage 2 of the decision tree, where there is evidence for benefit; however, there is no evidence to support routine treatment of asymptomatic GERD (see p.118). Avoid medications that make asthma worse (beta-blockers including eye-drops, aspirin and other NSAIDs in patients with aspirin-exacerbated respiratory disease, p.128). Refer for management of mental health problems if relevant. • Consider trial of non-biologic medication added to medium/high dose ICS, e.g., LABA, LAMA, LTRA if not already tried. Note concerns about neuropsychiatric adverse effects with montelukast LTRA.295 • Consider trial of high-dose ICS-LABA if not currently used. 4. Review response after approximately 3–6 months Schedule a review visit to assess the response to the above interventions. Timing of the review visit depends on clinical urgency and what changes to treatment have been made. When assessing the response to treatment, specifically review: • Symptom control (symptom frequency, SABA reliever use, night waking due to asthma, activity limitation) • Exacerbations since previous visit, and how they were managed • Medication side-effects • Inhaler technique and adherence • Lung function • Patient satisfaction and concerns. Is asthma still uncontrolled, despite optimized therapy? YES: if asthma is still uncontrolled, the diagnosis of severe asthma has been confirmed. If not done by now, refer the patient to a specialist or severe asthma clinic if possible. NO: if asthma is now well controlled, consider stepping down treatment. Start by decreasing/ceasing OCS first (if used), checking for adrenal insufficiency, then remove other add-on therapy, then decrease ICS dose, but do not stop ICS. See Box 4-13 (p.102) for how to gradually down-titrate treatment intensity. Does asthma become uncontrolled when treatment is stepped down? YES: if asthma symptoms become uncontrolled or an exacerbation occurs when high-dose treatment is stepped down, the diagnosis of severe asthma has been confirmed. Restore the patient's previous dose to regain good asthma control, and refer to a specialist or severe asthma clinic, if possible, if not done already. NO: if symptoms and exacerbations remain well controlled despite treatment being stepped down, the patient does not have severe asthma. Continue optimizing management. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 149 INVESTIGATE THE SEVERE ASTHMA PHENOTYPE AND CONSIDER NON-BIOLOGIC THERAPIES 5. Investigate further and provide patient support Further assessment and management should be by a specialist, preferably in a multidisciplinary severe asthma clinic if available. The team may include a certified asthma educator and health professionals from fields such as speech pathology, otorhinolaryngology, social work and mental health. What other tests may be considered at the specialist level? Additional investigations may be appropriate for identifying less-common comorbidities and differential diagnoses contributing to symptoms and/or exacerbations. Tests should be based on clinical suspicion, and may include: • Blood tests: complete blood count, CRP, IgG, IgA, IgM, IgE, fungal precipitins including Aspergillus • Allergy testing for clinically relevant allergens: skin prick test or specific IgE, if not already done • Other pulmonary investigations: diffusing capacity of the lungs for carbon monoxide (DLCO), chest X-ray or high-resolution chest computed tomography (CT) • Bone density scan, because of risk of osteoporosis with maintenance or frequent OCS or long-term high-dose ICS395 • Other directed testing based on clinical suspicion, e.g., antineutrophil cytoplasmic antibodies (ANCA), CT sinuses, B-natriuretic peptide (BNP), echocardiogram. If blood eosinophils are ≥300/µL, look for and treat non-asthma causes, including parasites (e.g., Strongyloides serology or stool examination), because parasitic infection may be the cause of the blood eosinophilia, and because OCS or biologic therapy in a patient with untreated parasitic infection could potentially lead to disseminated disease. Strongyloides infection is usually asymptomatic.665 If hypereosinophilia is found, e.g., blood eosinophils ≥1500/µL, consider causes such as eosinophilic granulomatosis with polyangiitis (EGPA). Consider need for social/psychological support Refer patients to support services, where available, to help them deal with the emotional, social and financial burden of asthma and its treatment, including during and after severe exacerbations.656 Consider the need for psychological or psychiatric referral, including for patients with anxiety and/or depression. Involve multidisciplinary team care (if available) Multidisciplinary assessment and treatment of patients with severe asthma increases the identification of comorbidities, and improves outcomes.666 Invite patient to enroll in a registry (if available) or clinical trial (if appropriate) Systematic collection of data will help in understanding the mechanisms and burden of severe asthma. There is a need for pragmatic clinical trials in severe asthma, including studies comparing two or more active treatments. Participants in randomized controlled trials designed for regulatory purposes may not necessarily be representative of patients seen in clinical practice. For example, a registry study found that over 80% of patients with severe asthma would have been excluded from key studies evaluating biologic therapy.369 6. Assess the severe asthma phenotype The next step is to assess the patient’s inflammatory phenotype – is it Type 2 high or low? What is Type 2 inflammation? Type 2 inflammation is found in the majority of people with severe asthma. It is characterized by cytokines such as interleukin (IL)-4, IL-5 and IL-13, which are often produced by the adaptive immune system on recognition of allergens. It may also be activated by viruses, bacteria and irritants that stimulate the innate immune system via production of IL-33, IL-25 and thymic stromal lymphopoietin (TSLP) by epithelial cells. Type 2 inflammation is often characterized by elevated COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 150 eosinophils or increased fractional exhaled nitric oxide (FeNO), and it may be accompanied by atopy and elevated IgE, whereas non-Type 2 inflammation is often characterized by increased neutrophils.667 In many patients with asthma, Type 2 inflammation rapidly improves when ICS are taken regularly and correctly; these patients do not have severe asthma. In severe asthma, Type 2 inflammation may be relatively refractory to high-dose ICS. It may respond to OCS but their serious adverse effects225,393 mean that alternative treatments should be sought. In adult patients with uncontrolled asthma despite medium- or high-dose ICS plus LABA or other controllers, a history of exacerbations in the previous year, higher blood eosinophil counts and higher FeNO levels are associated with a greater risk of severe exacerbations.668 Could the patient have refractory or underlying Type 2 inflammation? The possibility of refractory Type 2 inflammation should be considered if any of the following are found while the patient is taking high-dose ICS or daily OCS: • Blood eosinophils ≥150/μL, and/or • FeNO ≥20 ppb, and/or • Sputum eosinophils ≥2%, and/or • Asthma is clinically allergen-driven. Patients requiring maintenance OCS may also have underlying Type 2 inflammation. However, biomarkers of Type 2 inflammation (blood eosinophils, sputum eosinophils and FeNO) are often suppressed by OCS. If possible, therefore, these tests should be performed before starting OCS (a short course, or maintenance treatment), or at least 1–2 weeks after a course of OCS, or on the lowest possible OCS dose. The above criteria are suggested for initial assessment; those for blood eosinophils and FeNO are based on the lowest levels associated with response to some biologics. They are not the criteria for eligibility for Type 2-targeted biologic therapy, which may differ – see section 8 and local criteria. Consider repeating blood eosinophils and FeNO up to 3 times (e.g., when asthma worsens, before giving OCS, or at least 1–2 weeks after a course of OCS, or on the lowest possible OCS dose), before assuming asthma is non-Type 2. One study of patients with uncontrolled asthma taking medium- to high-dose ICS-LABA found that 65% had a shift in their blood eosinophil category over 48–56 weeks.669 Why is the inflammatory phenotype assessed on high-dose ICS? • Most RCT evidence about Type 2 targeted biologics is in such patients. • Modifiable ICS treatment problems such as poor adherence and incorrect inhaler technique are common causes of uncontrolled Type 2 inflammation. • Currently, the high cost of biologic therapies generally precludes their widespread clinical use in patients whose symptoms or exacerbations and Type 2 biomarkers are found to respond to ICS when it is taken correctly. 7.1. Consider other treatments if there is no evidence of Type 2 inflammation If the patient has no evidence of persistent Type 2 inflammation (section 6): • Review the basics for factors that may be contributing to symptoms or exacerbations: differential diagnosis, inhaler technique, adherence, comorbidities, medication side-effects (Section 2). • Recommend avoidance of relevant exposures (tobacco smoke, pollution, allergens if sensitized and there is evidence of benefit from withdrawal, irritants, infections). Ask about exposures at home and at work. • Consider additional diagnostic investigations (if available and not already done): sputum induction to confirm inflammatory phenotype, high resolution chest CT, bronchoscopy to exclude unusual comorbidities or alternative diagnoses such as tracheobronchomalacia or sub-glottic stenosis, functional laryngoscopy for inducible laryngeal obstruction. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 151 • Consider a trial of add-on treatment if available and not already tried (but check local eligibility and payer criteria for specific therapies as they may vary from those listed): • LAMA352 • Low-dose azithromycin (adults),373,374 but first check sputum for atypical mycobacteria, check ECG for long QTc (and re-check after a month on treatment), and consider potential for antibiotic resistance. • Anti-IL4Rα if taking maintenance OCS (see section 8 for more details) • Anti-TSLP (thymic stromal lymphopoietin) (but insufficient evidence in patients taking maintenance OCS; see section 8 for more details). • As a last resort, consider add-on low-dose OCS, but implement strategies with this such as alternate-day treatment to help reduce the dose further and minimize side-effects. • Consider bronchial thermoplasty, with registry enrollment. However, the evidence for efficacy and long-term safety is limited.150,463 • Stop ineffective add-on therapies. • Continue to optimize treatment, including inhaler technique, adherence, non-pharmacologic strategies and treating comorbidities (see sections 3 and 10). 7.2. Consider non-biologic options if there is evidence of type 2 inflammation For patients with elevated Type 2 biomarkers despite high-dose ICS (see section 5), consider non-biologic options first, given the current high cost of biologic therapy: • Assess adherence objectively by monitoring of prescribing or dispensing records, blood prednisone levels,670 or electronic inhaler monitoring.489 In one study, suppression of high FeNO after 5 days of directly observed therapy was an indicator of past poor adherence.671 • Consider increasing the ICS dose for 3–6 months, and review again. • Consider add-on non-biologic treatment for specific Type 2 clinical phenotypes (see Section 6, p.117). For example, for aspirin-exacerbated respiratory disease (AERD), consider add-on LTRA and possibly aspirin desensitization (p.128). For allergic bronchopulmonary aspergillosis (ABPA), consider add-on OCS ± anti-fungal agent (p.129). For chronic rhinosinusitis with or without nasal polyps, consider intensive intranasal corticosteroids; surgical advice may be needed (p.120). For patients with atopic dermatitis, topical steroidal or non-steroidal therapy may be helpful. Allergen immunotherapy may sometimes be used in severe asthma, but only after asthma has been well controlled, to minimize the risk of severe adverse reactions. Allergen immunotherapy extracts should only be prepared and administered by clinicians skilled in immunotherapy (see p.104). 7.3. Is Type 2-targeted biologic therapy available and affordable? If NOT: • Consider higher dose ICS-LABA, if not used • Consider other add-on therapy, e.g., LAMA, LTRA, low-dose azithromycin if not used • As last resort, consider add-on low-dose OCS, but implement strategies to minimize side-effects • Stop ineffective add-on therapies • Continue to optimize treatment, including inhaler technique, adherence, non-pharmacologic strategies and treating comorbidities (see Stages 3 and 10). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 152 CONSIDER TYPE 2-TARGETED BIOLOGIC THERAPIES 8. Consider add-on biologic type 2-targeted treatments If available and affordable, consider an add-on Type 2 targeted biologic for patients with exacerbations and/or poor symptom control despite taking at least high-dose ICS-LABA, and who have allergic or eosinophilic biomarkers or need maintenance OCS. Where relevant, test for parasitic infection, and treat if present, before commencing treatment (see Stage 5). Consider whether to start first with anti-IgE, anti-IL5/5Rα, anti-IL4Rα or anti-TSLP. When choosing between available therapies, consider the following: • Does the patient satisfy local payer eligibility criteria? • Type 2 comorbidities such as atopic dermatitis, nasal polyps • Predictors of asthma response (see below) • Cost • Dosing frequency • Delivery route (IV or SC; potential for self-administration) • Patient preference. Always check local payer eligibility criteria for biologic therapy, as they may vary substantially. However, GINA recommends the use of biologic therapy only for patients with severe asthma, and only after treatment has been optimized. For any biologic therapy, ensure that the manufacturer’s and/or regulator’s instructions for storage, administration and the duration of monitoring post-administration are followed. Provide the patient with advice about what to do if they experience any adverse effects, including hypersensitivity reactions. Omalizumab injections contain polysorbate, which may induce allergic reactions in some patients. GINA suggests that the first dose of asthma biologic therapy should not be given on the same day as a vaccine such as for COVID-19, so that adverse effects of either can be more easily distinguished. Provide practical advice for patients, e.g., allow the refrigerated syringe or pen to come to room temperature before injecting the biologic, as this reduces pain. There is an urgent need for head-to-head comparisons of different biologics in patients eligible for more than one biologic. Add-on anti-IgE for severe allergic asthma Regulatory approvals may include: omalizumab for ages ≥6 years, given by SC injection every 2–4 weeks, with dose based on weight and serum IgE. May also be indicated for nasal polyps and chronic spontaneous (idiopathic) urticaria. Self-administration may be an option. Check local regulatory and payer criteria, as they may differ from these. Mechanism: binds to Fc part of free IgE, preventing binding of IgE to FcƐR1 receptors, reducing free IgE and down-regulating receptor expression. Eligibility criteria (in addition to criteria for severe asthma) may vary between payers or by age-group, but often include: • Sensitization to inhaled allergen(s) on skin prick testing or specific IgE, and • Total serum IgE and body weight within local dosing range, and • More than a specified number of exacerbations within the last year. Outcomes: Meta-analysis of RCTs in severe allergic asthma: anti-IgE led to 44% decrease in severe exacerbations, and improved quality of life; improvements in symptom control and lung function were statistically significant but less than clinically important differences.376 No double-blind randomized controlled trials of OCS-sparing effect. In a meta-analysis COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 153 of observational studies in patients with severe allergic asthma, there was a 59% reduction in exacerbation rate, a 41% reduction in the proportion of patients receiving maintenance OCS, and a significant improvement in symptom control.672 In patients with nasal polyps, omalizumab improved subjective and objective nasal outcomes.558 Additional details about treatment of chronic rhinosinusitis with nasal polyps (CRSwNP) are found on p.120. A registry study of omalizumab in pregnancy found no increased risk of congenital malformations.593 Potential predictors of good asthma response to omalizumab: • Baseline IgE level does not predict likelihood of response673 • In a post-hoc analysis of one clinical trial, a greater decrease in exacerbations was observed (compared with placebo) with blood eosinophils ≥260/μL674,675 or FeNO ≥19.5 ppb674 (these criteria representing their median value in that study) but in two large observational studies, exacerbations were reduced with both low or high blood eosinophils676-678 or with both low or high FeNO.678 • Childhood-onset asthma • Clinical history suggesting allergen-driven symptoms. Adverse effects: injection site reactions, anaphylaxis in approximately 0.2% patients679 Suggested initial trial: at least 4 months Add-on anti-IL5 or anti-IL5Rα for severe eosinophilic asthma Regulatory approvals may include: For ages ≥12 years: mepolizumab (anti-IL5), 100 mg by SC injection every 4 weeks, or benralizumab (anti-IL5 receptor α), 30 mg by SC injection every 4 weeks for 3 doses then every 8 weeks. For ages ≥18 years: reslizumab (anti-IL5), 3 mg/kg by IV infusion every 4 weeks. For ages 6–11 years, mepolizumab (anti-IL5), 40 mg by SC injection every 4 weeks. Mepolizumab may also be indicated for eosinophilic granulomatosis with polyangiitis (EGPA), hypereosinophilic syndrome and chronic rhinosinusitis with nasal polyps. Self-administration may be an option. Check local regulatory and payer criteria, as they may differ from these. Mechanism: mepolizumab and reslizumab bind circulating IL-5; benralizumab binds to IL-5 receptor alpha subunit leading to apoptosis (cell death) of eosinophils. Eligibility criteria (in addition to criteria for severe asthma): these vary by product and between payers, but usually include: • More than a specified number of severe exacerbations in the last year, and • Blood eosinophils above locally specified level (e.g., ≥150 or ≥300/μL). There is sometimes a different eosinophil cut-point for patients taking OCS. Outcomes: Meta-analysis of RCTs in severe asthma patients with exacerbations in the last year, with varying eosinophil criteria: anti-IL5 and anti-IL5Rα led to 47–54% reduction in severe exacerbations. Improvements in lung function and symptom control were statistically significant,382 but less than clinically important differences. There was a clinically important improvement in quality of life with mepolizumab.382 All anti-IL5/5Rα biologics reduced blood eosinophils; almost completely with benralizumab.680 In post hoc analyses, clinical outcomes with mepolizumab or benralizumab were similar in patients with and without an allergic phenotype.681,682 In patients taking OCS, median OCS dose was able to be reduced by approximately 50% with mepolizumab683 or benralizumab380 compared with placebo. In urban children aged 6 years and older with eosinophilic exacerbation-prone asthma, an RCT showed a reduction in the number of exacerbations with subcutaneous mepolizumab versus placebo.381 No differences were seen in lung function, a composite asthma score (CASI), or physician–patient global assessment.381 In patients with nasal polyps, mepolizumab improved subjective and objective outcomes and reduced the need for surgery,559,560 and in patients with nasal polyps and severe eosinophilic asthma, benralizumab improved subjective outcomes for both conditions and improved quality of life.684 See p.120 for more details about treatment of nasal polyps. Potential predictors of good asthma response to anti-IL5 or anti-IL5Rα: • Higher blood eosinophils (strongly predictive)685 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 154 • Higher number of severe exacerbations in previous year (strongly predictive)685 • Adult-onset asthma686 • Nasal polyps682 • Maintenance OCS at baseline682 • Low lung function (FEV1 <65% predicted) in one study.687 Adverse effects: In adults, injection site reactions, anaphylaxis rare, adverse events generally similar between active and placebo. In children, more skin/subcutaneous tissue and nervous system disorders (e.g., headache, dizziness, syncope) were seen with mepolizumab than placebo.381 Suggested initial trial: at least 4 months Add-on anti-IL4Rα for severe eosinophilic/Type 2 asthma or patients requiring maintenance OCS Regulatory approvals may include: For ages ≥12 years: dupilumab (anti-IL4 receptor α), 200 mg or 300 mg by SC injection every 2 weeks for severe eosinophilic/Type 2 asthma; 300 mg by SC injection every 2 weeks for OCS-dependent severe asthma or if there is concomitant moderate/severe atopic dermatitis. For children 6–11 years with severe eosinophilic/Type 2 asthma by SC injection with dose and frequency depending on weight. May also be indicated for treatment of skin conditions including moderate-to-severe atopic dermatitis, chronic rhinosinusitis with nasal polyps and eosinophilic esophagitis. Self-administration may be an option. Check local regulatory and payer criteria, as they may differ from these. Mechanism: binds to interleukin-4 (IL-4) receptor alpha, blocking both IL-4 and IL-13 signaling Eligibility criteria (in addition to criteria for severe asthma): these may vary between payers or by age-group, but often include: • More than a specified number of severe exacerbations in the last year, and • Type 2 biomarkers above a specified level (e.g., blood eosinophils ≥150/μL and ≤1500/μL, or FeNO ≥25 ppb) OR requirement for maintenance OCS. Outcomes: Meta-analysis of RCTs in patients with uncontrolled severe asthma (ACQ-5 ≥1.5) and at least one exacerbation in the last year: anti-IL4Rα led to 56% reduction in severe exacerbations; improvements in quality of life, symptom control and lung function were statistically significant,385 but less than the clinically important differences. In a post hoc analysis, clinical outcomes were similar in patients with allergic and non-allergic phenotype at baseline.688 In patients with OCS-dependent severe asthma, without minimum requirements for blood eosinophil count or FeNO, the median reduction in OCS dose with anti-IL4Rα versus placebo was 50%.689 In follow-up, changes were maintained through 2 years of follow-up.690 In children 6–11 years with eosinophilic/Type 2 asthma, dupilumab reduced severe exacerbation rate and increased lung function; children taking maintenance OCS were excluded.386 In patients with chronic rhinosinusitis with nasal polyps, dupilumab reduced the size of nasal polyps, improved nasal symptoms and reduced the need for OCS or sinus surgery.561,691 See p.120 for more details about nasal polyps. Potential predictors of good asthma response to dupilumab: • Higher blood eosinophils (strongly predictive)383 • Higher FeNO (strongly predictive).383 Adverse effects: injection-site reactions; transient blood eosinophilia (occurs in 4–13% of patients); rare cases of eosinophilic granulomatosis with polyangiitis (EGPA) may be unmasked following reduction/cessation of OCS treatment on dupilumab. Anti-IL4Rα is not suggested for patients with baseline or historic blood eosinophils >1,500 cells/µL because of limited evidence (such patients were excluded from Phase III trials). Suggested initial trial: at least 4 months COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 155 Add-on anti-TSLP for severe asthma Regulatory approvals may include: For ages ≥12 years: tezepelumab (anti-TSLP), 210 mg by SC injection every 4 weeks. Self-administration may be an option. Check local regulatory and payer criteria, as they may differ from these. Mechanism: tezepelumab binds circulating TSLP, a bronchial epithelial cell-derived alarmin implicated in multiple downstream processes involved in asthma pathophysiology. Eligibility criteria (in addition to criteria for severe asthma): these vary between payers, but usually include: • Severe exacerbations in the last year. Anti-TSLP may also be considered in patients with no elevated Type 2 markers (Stage 7.1). Outcomes: In two RCTs in severe asthma patients with severe exacerbations in the last year, anti-TSLP led to 30–70% reduction in severe exacerbations, and improved quality of life, lung function and symptom control, irrespective of allergic status.387,388 There was a clear correlation between higher baseline blood eosinophils or FeNO and better clinical outcomes.388 In patients taking maintenance OCS, anti-TSLP did not lead to a reduced OCS dose compared with placebo.389 Potential predictors of good asthma response to anti-TSLP: • Higher blood eosinophils (strongly predictive) • Higher FeNO levels (strongly predictive). Adverse effects: injection site reactions, anaphylaxis is rare, adverse events generally similar between active and placebo groups. Suggested initial trial: at least 4 months Review response to an initial trial of add-on Type 2-targeted therapy • At present, there are no well-defined criteria for a good response, but consider exacerbations, symptom control, lung function, side-effects, treatment intensity (including OCS dose), and patient satisfaction. • If the response is unclear, consider extending the trial to 6–12 months. • If there is no response, stop the biologic therapy, and consider switching to a trial of a different Type 2-targeted therapy, if available and the patient is eligible. Also consider the patient’s biomarkers (interval and during exacerbations, if available), and response of any comorbid Type 2 conditions (atopic dermatitis, nasal polyps etc). Review response as above. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 156 ASSESS, MANAGE AND MONITOR ONGOING SEVERE ASTHMA TREATMENT 9. Review response and implications for treatment Review response to add-on biologic therapy after 3–4 months, and every 3–6 months for ongoing care, including: • Asthma: symptom control, both recent e.g., with validated tools such as Asthma Control Test (4 weeks) and Asthma Control Questionnaire (ACQ-5, 1 week), and over the whole period since last review, frequency and severity of exacerbations (including whether OCS were needed), lung function • Any change in relevant Type 2 comorbidities, e.g., nasal polyps, atopic dermatitis • Medications: treatment intensity, including courses of OCS and dose of any maintenance OCS, side-effects, affordability • Patient satisfaction.
If the patient has had a good response to Type 2 targeted therapy: Re-evaluate the need for each asthma medication every 3–6 months, but emphasize to patients and their primary care physician that they should not completely stop ICS-containing therapy. Base the order of reduction or cessation of add-on treatments on potential adverse effects, the observed benefit when the medication was started, patient risk factors, cost, and patient satisfaction. Minimizing the use of OCS is a very high priority. After reducing/ceasing any medication, confirm asthma stability before making any further treatment changes. For oral treatments, gradually decrease or stop OCS first, because of their significant adverse effects. Tapering of OCS in severe asthma may be supported by internet-based monitoring of symptom control and FeNO.692 Monitor patients for risk of adrenal insufficiency by measuring morning serum cortisol, and provide patient and primary care physician with advice about the need for extra corticosteroid doses during injury, illness or surgery for up to 6 months after cessation of long-term OCS. Continue to assess for presence of osteoporosis, and review need for preventative strategies including bisphosphonates.395 If asthma remains well controlled, consider ceasing other therapies, based on the above considerations. For inhaled treatments, consider ceasing add-on inhaled therapy such as LAMA before reducing ICS-LABA dose. Reduction in dose of ICS-containing therapy may be considered after asthma has been well controlled on biologic therapy for at least 3–6 months and stability has been confirmed after any other medication changes. However, do not completely stop ICS-containing therapy. Previous advice based on consensus was to continue at least medium-dose ICS-LABA. In an open-label study in patients with good symptom control on anti-IL5Rα, most of those randomized to MART with ICS-formoterol were able to have their maintenance ICS-formoterol dose gradually reduced (and in some cases stopped, continuing as-needed-only ICS-formoterol) without exacerbations. However, patients who ceased maintenance treatment demonstrated evidence of under-dosing with ICS, with reduction in lung function and increase in FeNO, suggesting that in patients with severe asthma, maintenance ICS-containing therapy should not be stopped completely.15 Any reduction in ICS dose should be considered as a treatment trial and the previous dose reinstated if deterioration occurs (Box 4-13, p.102). Patients should be reminded of the importance of continuing their maintenance ICS-containing treatment.
For biologic treatments, current consensus advice is that, generally, for a patient with a good response, a trial of withdrawal of the biologic should not be considered until after at least 12 months of treatment, and only if asthma remains well controlled on medium-dose ICS-containing therapy, and (for allergic asthma) there is no further exposure to a previous well-documented allergic trigger. There are few studies of cessation of biologic therapy,693,694 in these studies, symptom control worsened and/or exacerbations recurred for many (but not all) patients after cessation of the biologic. For example, in a double-blind randomized controlled trial, significantly more patients who stopped mepolizumab experienced a severe exacerbation within 12 months compared with those who continued treatment. In this study, there was a small increase in ACQ-5 but no significant difference in symptom control between groups.695 In adults, long-term safety over 5 or more years of treatment has been reported for several biologics,696-698 with shorter follow-up to date for others.699 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 157 If the patient has NOT had a good response to any Type 2-targeted therapy: Stop the biologic therapy. Review the basics for factors contributing to symptoms, exacerbations and poor quality of life (see Section 2): diagnosis/differential diagnosis, inhaler technique, adherence, modifiable risk factors and triggers including smoking and other environmental exposures at home or work, comorbidities including obesity, medication side-effects or drug interactions, socio-economic and mental health issues. Consider additional investigations (if not already done): high resolution chest CT, induced sputum to confirm inflammatory phenotype, consider bronchoscopy for alternative or additional diagnoses, consider referral if available, including for diagnosis of alternative conditions. Reassess treatment options (if not already done), such as: • Add-on low-dose azithromycin373,374 (adults only; first check sputum for atypical mycobacteria and check ECG for long QTc (and re-check after a month on treatment); consider potential for antibiotic resistance) • As last resort, consider add-on low-dose maintenance OCS, but implement strategies such as alternate-day therapy; add bisphosphonate to minimize side-effects on bones,395 and alert patient to the need for additional corticosteroid therapy during illness or surgery. • Consider bronchial thermoplasty (+ registry). Stop ineffective add-on therapies, but do not completely stop ICS. 10. Continue collaborative optimization of patient care Ongoing management of a patient with severe asthma involves a collaboration between the patient, the primary care physician, specialist(s), and other health professionals, to optimize clinical outcomes and patient satisfaction. Continue to review the patient every 3–6 months including: • Clinical asthma measures (symptom control, exacerbations, lung function) • Comorbidities • The patient's risk factors for exacerbations • Treatments (check inhaler technique and adherence, review need for add- on treatments, assess side-effects including of OCS, and optimize comorbidity management and non-pharmacologic strategies) • The patient’s social and emotional needs. The optimal frequency and location of review (primary care physician or specialist) will depend on the patient’s asthma control, risk factors and comorbidities, and their confidence in self-management, and may depend on local payer requirements and availability of specialist physicians. Communicate regularly with the family physician and other members of the health care team about: • Outcome of review visits (as above) • Patient concerns • Action plan for worsening asthma or other risks • Changes to medications (asthma and non-asthma), potential side-effects • Indications and contact details for expedited review. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 158 9. Management of worsening asthma and exacerbations in adults, adolescents and children 6–11 years KEY POINTS Terminology • Exacerbations represent an acute or sub-acute worsening in symptoms and lung function from the patient’s usual status, or in some cases, a patient may present for the first time during an exacerbation. • The terms ‘episodes’, ‘attacks’ and ‘acute severe asthma’ are also often used, but they have variable meanings. The term ‘flare-up’ is preferable for use in discussions with most patients. • Patients who are at increased risk of asthma-related death should be identified, and flagged for more frequent review. Written asthma action plans • All patients should be provided with a written (i.e., printed, digital or pictorial) asthma action plan appropriate for their age, their current treatment regimen and their reliever inhaler (short-acting beta2 agonist [SABA] or combination inhaled corticosteroids [ICS]-formoterol), their level of asthma control, and their health literacy, so they know how to recognize and respond to worsening asthma. • On the action plan, state when and how to change reliever and/or maintenance medications, use oral corticosteroids (OCS) if needed, and access medical care if symptoms fail to respond to treatment. • Advise patients who have a history of rapid deterioration to go to an acute care facility or see their doctor immediately their asthma starts to worsen. • Base the action plan on changes in symptoms or (only in adults) peak expiratory flow (PEF). Management of exacerbations in a primary care or acute care facility • Assess exacerbation severity from the degree of dyspnea, respiratory rate, pulse rate, oxygen saturation and lung function, while starting SABA and oxygen therapy. Infection control procedures should be followed. • Arrange immediate transfer to an acute care facility if there are signs of severe exacerbation, or to intensive care if the patient is drowsy, confused, or has a silent chest. During transfer, give inhaled SABA and ipratropium bromide, controlled oxygen and systemic corticosteroids. • Start treatment with repeated administration of SABA (in most patients, by pressurized metered-dose inhaler [pMDI] and spacer), early introduction of oral corticosteroids, and controlled flow oxygen if available. Review response of symptoms, oxygen saturation and lung function after 1 hour. Give ipratropium bromide only for severe exacerbations. Consider intravenous magnesium sulfate for patients with severe exacerbations not responding to initial treatment. • Do not routinely request a chest X-ray, and do not routinely prescribe antibiotics for asthma exacerbations. • Decide about hospitalization based on the patient’s clinical status, lung function, response to treatment, recent and past history of exacerbations, and ability to manage at home. Discharge management • Arrange ongoing treatment before the patient goes home. This should include starting ICS-containing controller treatment or stepping up the dose of existing ICS-containing treatment for 2–4 weeks, and reducing reliever medication to as-needed use. • If the patient was using an anti-inflammatory reliever (e.g., ICS-formoterol) before the exacerbation and this was replaced with SABA during an emergency department or hospital stay, they should resume taking as-needed anti-inflammatory reliever instead of SABA reliever before or on discharge. If the patient was previously using maintenance-and-reliever therapy (MART) with ICS-formoterol, they should resume MART. If the patient was COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 159 previously using as-needed-only ICS-formoterol as needed, they should start MART, i.e., add maintenance ICS-formoterol. There is no need to prescribe or provide SABA for patients prescribed ICS-formoterol reliever. • For adults and adolescents using SABA as reliever before the exacerbation, consider switching to maintenance-and-reliever therapy with ICS-formoterol (MART, Track 1, p.77), to reduce the risk of future exacerbations. Arrange early follow-up after any exacerbation, regardless of where it was managed. At follow-up: • Review the patient’s symptom control and risk factors for further exacerbations. • Prescribe ICS-containing controller therapy to reduce the risk of further exacerbations. If already taking ICS-containing therapy, continue increased doses for 2–4 weeks. • Provide a written asthma action plan and, where relevant, advice about avoiding exacerbation triggers • Check inhaler technique and adherence. For management of asthma exacerbations in children 5 years and younger, see Section 12 (p.196). OVERVIEW Definition of asthma exacerbations Exacerbations of asthma are episodes characterized by a progressive increase in symptoms of shortness of breath, cough, wheezing or chest tightness and progressive decrease in lung function, i.e., they represent a change from the patient’s usual status that is sufficient to require a change in treatment.38 Exacerbations may occur in patients with a pre-existing diagnosis of asthma or, occasionally, as the first presentation of asthma. What triggers asthma exacerbations? Exacerbations usually occur in response to exposure to an external agent (e.g., viral upper respiratory tract infection, pollen98 or pollution) and/or poor adherence with ICS-containing medication; however, a subset of patients present more acutely and without exposure to known risk factors.700,701 Severe exacerbations can occur in patients with mild or well-controlled asthma symptoms.28,307 Box 2-2B (p.37) lists factors that increase a patient’s risk of exacerbations, independent of their level of symptom control. Common exacerbation triggers include: • Viral respiratory infections,702 e.g., rhinovirus, influenza, adenovirus, pertussis, respiratory syncytial virus • Allergen exposure e.g., grass pollen and other pollens,98,703 soybean dust,704 fungal spores • Food allergy94 • Outdoor air pollution101,102,705 • Seasonal changes and/or returning to school in fall (autumn)706 • Poor adherence with ICS707 • Epidemics of severe asthma exacerbations may occur suddenly, putting high pressure on local health system responses. Such epidemics have been reported in association with springtime thunderstorms and either rye grass pollen or fungal spores,708 and with environmental exposure to soybean dust.704 Identifying patients at risk of asthma-related death In addition to factors known to increase the risk of asthma exacerbations (Box 2-2, p.37), some features are specifically associated with an increase in the risk of asthma-related death (Box 9-1, p.160). The presence of one or more of these risk factors should be quickly identifiable in the clinical notes, and these patients should be encouraged to seek urgent medical care early in the course of an exacerbation. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 160 Box 9-1. Factors associated with increased risk of asthma-related death • A history of near-fatal asthma requiring intubation and mechanical ventilation709 • Hospitalization709,710 or emergency care visit for asthma in the past year • Currently using or having recently stopped using oral corticosteroids (a marker of event severity)89,709 • Not currently using inhaled corticosteroids90,709 • Over-use of short-acting beta2 agonists (SABAs), especially use of an average of more than one canister of salbutamol (or equivalent) per month,87,111,711 or using nebulized SABA712 • Poor adherence with ICS-containing medications and/or poor adherence with (or lack of) a written asthma action plan103 • A history of psychiatric disease or psychosocial problems103 • Food allergy in a patient with asthma544,713 • Several comorbidities including pneumonia, diabetes and arrhythmias were independently associated with an increased risk of death after hospitalization for an asthma exacerbation710 See list of abbreviations (p.11) Terminology about exacerbations The academic term ‘exacerbation’ is commonly used in scientific and clinical literature, although hospital-based studies more often refer to ‘acute severe asthma’. However, the term ‘exacerbation’ is not suitable for use in clinical practice, as it is difficult for many patients to pronounce and remember.714,715 The term ‘flare-up’ is simpler, and conveys the sense that asthma is present even when symptoms are absent. The term ‘attack’ is used by many patients and healthcare providers but with widely varying meanings, and it may not be perceived as including gradual worsening.714,715 In pediatric literature, the term ‘episode’ is commonly used, but understanding of this term by parent/caregivers is not known. DIAGNOSIS OF EXACERBATIONS Exacerbations represent a change in symptoms and lung function from the patient’s usual status.38 The decrease in expiratory airflow can be quantified by lung function measurements such as PEF or forced expiratory volume in 1 second (FEV1),716 compared with the patient’s previous lung function or predicted values. In the acute setting, these measurements are more reliable indicators of the severity of the exacerbation than symptoms. The frequency of symptoms may, however, be a more sensitive measure of the onset of an exacerbation than PEF.717 Consider the possibility of pertussis in a patient with an atypical exacerbation presentation, with predominant cough. A minority of patients perceive airflow limitation poorly and can experience a significant decline in lung function without a change in symptoms.152,164,172 This especially affects patients with a history of near-fatal asthma and also appears to be more common in males. Regular PEF monitoring may be considered for such patients. Severe exacerbations are potentially life-threatening, and their treatment requires careful assessment and close monitoring. Patients with severe exacerbations should be advised to see their healthcare provider promptly or, depending on the organization of local health services, to proceed to the nearest facility that provides emergency access for patients with acute asthma. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 161 SELF-MANAGEMENT OF EXACERBATIONS WITH A WRITTEN ASTHMA ACTION PLAN All patients with asthma, and parents/caregivers of children with asthma, should be provided with guided self-management education as described in Section 5 (p.108). The definition of guided self-management education includes monitoring of symptoms and/or lung function, a written asthma action plan, and regular review by a health professional.516 (For children 5 years and younger, see Section 11, p.185). A written (i.e., documented) asthma action plan may be printed, digital, or pictorial, to suit the patient’s needs and literacy. A written asthma action plan helps patients to recognize and respond appropriately to worsening asthma. It should include specific instructions for the patient about changes to reliever and/or maintenance medications, when and how to use OCS if needed (Box 9-2, p.162) and when and how to access medical care. The criteria for initiating an increase in maintenance medication will vary from patient to patient. In studies that evaluated an increase in maintenance ICS-containing treatment, this was usually initiated when there was a clinically important change from the patient’s usual level of asthma control, for example, if asthma symptoms were interfering with normal activities, or PEF had fallen by >20% for more than 2 days.521 For patients prescribed an anti-inflammatory reliever (as-needed combination ICS-formoterol or ICS-SABA), this provides a small extra dose of ICS as well as a rapid-acting bronchodilator without delay whenever the reliever is used, as the first step in the patient’s action plan; this approach reduces the risk of progressing to severe exacerbation and need for oral corticosteroids. In the case of as-needed ICS-formoterol, both the ICS and the formoterol appear to contribute to the reduction in severe exacerbations compared with using a SABA reliever.399 See Box 4-8 (p.84) for more details about as-needed ICS-formoterol, including medications and dosages. A specific action plan template is available for patients prescribed maintenance-and-reliever therapy with ICS-formoterol;313 it can also be modified for patients prescribed as-needed-only ICS-formoterol. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 162 Box 9-2. Self-management of worsening asthma in adults and adolescents with a written asthma action plan Medication Short-term change (1–2 weeks) for worsening asthma Evidence level Increase usual reliever: Low-dose ICS-formoterol† Increase frequency of as-needed low-dose ICS-formoterol (for patients prescribed this as-needed only, or with maintenance ICS-formoterol).† See Box 4-8 (p.84) for details of medications and doses. A Short-acting beta2 agonist (SABA) Increase frequency of SABA use For pMDI, add spacer A A Combination ICS-SABA Increase frequency of as-needed ICS-SABA# (see below and p.88) B Increase usual maintenance treatment: Maintenance-and-reliever ICS-formoterol (MART)† Continue usual maintenance dose of ICS-formoterol and increase ICS-formoterol reliever doses as needed.† See Box 4-8 (p.84) for details. A Maintenance ICS with SABA as reliever Consider quadrupling ICS dose. B Maintenance ICS-formoterol with SABA as reliever† Consider quadrupling maintenance ICS-formoterol.† B Maintenance ICS plus other LABA with SABA as reliever Step up to higher dose formulation of ICS plus other LABA, if available In adults, consider adding a separate ICS inhaler to quadruple ICS dose. B D Add oral corticosteroids (OCS) and contact doctor; review before ceasing OCS (prednisone or prednisolone) Add OCS for severe exacerbations (e.g., PEF or FEV1 <60% personal best or predicted), or patient not responding to treatment over 48 hours. Once started, morning dosing is preferable. Adults: prednisolone 40–50 mg/day, usually for 5–7 days. A D Tapering is not needed if OCS are prescribed for less than 2 weeks. B See list of abbreviations (p.11). or equivalent dose of prednisone. † ICS-formoterol as-needed for relief of symptoms (‘AIR-only’), or as part of maintenance-and-reliever therapy (MART) with low-dose combination budesonide-formoterol or beclometasone (BDP)-formoterol. See Box 4-8 (p.84) for details of medications and doses. The maximum recommended total dose of budesonide-formoterol in a single day for adults and adolescents gives 72 mcg formoterol (54 mcg delivered dose); GINA suggests that. for BDP-formoterol. the maximum total metered dose should be the same (maximum total 12 inhalations in a day). # Combination budesonide-salbutamol (albuterol) 2 puffs of 100/100 mcg (delivered dose 80/90 mcg) maximum 6 times in a day.343 See text below for more details about action plan options in adults, adolescents and children. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 163 Treatment options for written asthma action plans – relievers Inhaled combination ICS-formoterol reliever In adults and adolescents, use of as-needed combination low-dose ICS-formoterol for symptom relief (without maintenance treatment) reduced the risk of severe exacerbations requiring OCS or requiring emergency department visit or hospitalization by 65% compared with SABA-only treatment.183 It also reduced the risk of needing an emergency department visit or hospitalization by 37% compared with daily ICS plus as-needed SABA.183 After a day of even small increased doses of ICS-formoterol, the risk of severe exacerbation in the following 3 weeks was reduced compared with using the same doses of SABA alone.128 Details of the evidence are found on p.79 and p.81. In adults, adolescents and children 6–11 years, maintenance-and-reliever therapy (MART) with very low- or low-dose ICS-formoterol reduced the risk of severe exacerbations compared with the same or higher dose of ICS or ICS-LABA, with similar symptom control.224,226,410 Details about the evidence are found on p.98 and p.99. Information about medications and doses for use of as-needed ICS-formoterol is summarized in Box 4-8 (p,84). For adults and adolescents, the evidence is with use of budesonide-formoterol 200/6 mcg metered dose (160/4.5 mcg delivered dose) by dry-powder inhaler and, for children aged 6–11 prescribed MART, budesonide-formoterol 100/6 mcg metered dose (80/4.5 mcg delivered dose) by dry-powder inhaler. Patients prescribed ICS-formoterol as their reliever (with or without maintenance ICS-formoterol) should take 1 inhalation of their ICS-formoterol reliever whenever needed for symptom relief; for formulations with 3 mcg [2.25 delivered dose] of formoterol per inhalation, 2 inhalations should be taken whenever needed for symptom relief. If necessary, an extra dose can be taken a few minutes later. Additional doses are taken when symptoms recur, even if this is within 4 hours, but the maximum total recommended dose in any single day for adults and adolescents (as-needed plus maintenance doses, if used) is 12 inhalations for budesonide-formoterol (total 72 mcg formoterol [54 mcg delivered dose]). Based on extensive evidence for efficacy and safety of budesonide-formoterol up to this total maximum dose in any day, GINA suggests that the same maximum total dose in a single day should also apply to beclometasone-formoterol. For children, budesonide-formoterol can, if needed, be used up to a total (as-needed and maintenance doses, if used) of 8 inhalations in any day. This is the maximum total of as-needed doses and maintenance doses, if used. See Box 4-8 (p.84) for specific details. If the patient is rapidly worsening, or has failed to respond to an increase in as-needed doses of ICS-formoterol over 2–3 days, they should contact their healthcare provider or seek medical assistance. Inhaled combination ICS-SABA reliever For adults prescribed as-needed combination ICS-SABA reliever with maintenance ICS-containing therapy, the recommended dose is 2 inhalations of budesonide-salbutamol [albuterol] 100/100 mcg metered dose (80/90 mcg delivered dose) as needed, a maximum of 6 times in a day. Overall, in patients on Step 3–5 therapy, this reduced the risk of severe exacerbations by 26% compared with using a SABA reliever, with the greatest benefit seen in patients taking maintenance low-dose ICS-LABA or medium-dose ICS.343 There is only one study to date about use of as-needed combination ICS-SABA alone, i.e., without maintenance ICS or ICS-LABA (see p.98).324 If the patient is rapidly worsening, or needs repeated doses of as-needed ICS-SABA reliever over 1–2 days, they should contact their healthcare provider or seek medical assistance. Inhaled SABA reliever For patients prescribed a SABA bronchodilator reliever, repeated dosing provides temporary relief until the cause of the worsening symptoms passes or increased ICS-containing treatment has had time to take effect. However, use of SABA reliever is less effective in preventing progression to severe exacerbation requiring OCS than use of low-dose ICS formoterol reliever, either with224 or without301,302 daily maintenance ICS-containing treatment, or than combination ICS-SABA reliever (see Section 4, p.67). The need for repeated doses of SABA over more than 1–2 days signals the need to review, and possibly increase, ICS containing treatment if this has not already been done. This is particularly important if there has been a lack of response. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 164 Treatment options for written asthma action plans – maintenance medications Maintenance-and-reliever therapy (MART) with combination low-dose ICS-formoterol In adults and adolescents, the combination of a rapid-onset LABA (formoterol) and low-dose ICS (budesonide or beclometasone) in a single inhaler as both the maintenance and the reliever medication was effective in improving asthma symptom control,318 and it reduced exacerbations requiring OCS, and hospitalizations,224,718-721 compared with the same or higher dose of ICS or ICS-LABA with as-needed SABA reliever (Evidence A). This regimen was also effective in reducing exacerbations in children aged 4–11 years (Evidence B).410 For adults and adolescents prescribed MART, the recommended maximum total dose of formoterol in 24 hours with budesonide-formoterol is 72 mcg (delivered dose 54 mcg), with extensive evidence from large studies of its safety and efficacy up to this frequency in a single day (as above). Based on this evidence, GINA suggests that the same maximum total dose in a single day should apply to beclometasone-formoterol (See Box 4-8, p.84). This approach should not be attempted with other combination ICS-LABA medications with a slower-onset LABA (e.g., ICS-salmeterol), or that lack the dose response and safety profile that is required for a maintenance-and-reliever regimen. The benefit of the MART regimen in reducing the risk of severe exacerbations requiring OCS appears to be due to the increase in doses of both the ICS and the formoterol at a very early stage of worsening asthma.126-128 In an action plan for patients prescribed maintenance-and-reliever therapy with ICS-formoterol, the maintenance dose does not normally need to be increased. Instead, the patient increases their as-needed doses of ICS-formoterol. More details of medications and doses for different age-groups are available in Box 4-8, p.84. Examples of action plans customized for MART are available online.313,314 Other ICS and ICS-LABA maintenance treatment regimens plus as-needed SABA In a systematic review, self-management studies in which the ICS dose was at least doubled were associated with improved asthma outcomes and reduced healthcare utilization (Evidence A).521 In placebo-controlled trials, temporarily doubling the dose of ICS was not effective (Evidence A);722 however, the delay before increasing the ICS dose (mean 5–7 days)719,720 may have contributed. Some studies in adults721 and young children723 have reported that higher ICS doses might help prevent worsening asthma progressing to a severe exacerbation. In a randomized controlled trial in primary care with patients aged ≥16 years, those who quadrupled their ICS dose (to average of 2000 mcg/day beclometasone dipropionate (BDP) equivalent) after their PEF fell were significantly less likely to require OCS.724 In an open-label primary care randomized controlled trial of adult and adolescent patients using ICS with or without LABA, early quadrupling of ICS dose (to average 3200 mcg/day BDP equivalent) was associated with a modest reduction in prescribing of OCS.725 However, a double-blind placebo-controlled study in children 5–11 years with high adherence to low-dose ICS found no difference in the rate of severe exacerbations requiring OCS if maintenance ICS was quintupled (to 1600 mcg BDP equivalent) versus continuing maintenance low-dose therapy.726 Given the shape of the ICS dose-response curve, little benefit may be seen from increasing maintenance ICS when background adherence is high, as in this study. In addition, in several of the studies evaluating ICS increases,719,720,726 a pre-specified level of deterioration in symptoms (± lung function) had to be reached before the extra ICS could be started. This may help to explain the greater reduction in severe exacerbations seen with maintenance-and-reliever therapy with ICS-formoterol, where there is no lag before the doses of both ICS and formoterol are increased. In adult with an acute deterioration, high-dose ICS for 7–14 days (500–1600 mcg BDP-HFA standard-particle equivalent) had an equivalent benefit to a short course of OCS (Evidence A).721 For adults taking combination ICS-LABA with as-needed SABA, the ICS dose may be increased by adding a separate ICS inhaler (Evidence D).721,725 Leukotriene receptor antagonists If patients are using a leukotriene receptor antagonist (LTRA) as their only controller, there are no specific studies about how to manage worsening asthma. Clinicians’ judgment should be used (Evidence D). For ongoing treatment, the patient should be switched to an ICS-containing controller to reduce the risk of further exacerbations.347 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 165 Oral corticosteroids For most patients, the written asthma action plan should provide instructions for when and how to commence OCS. Typically, a short course of OCS is used (e.g., for adults, 40–50 mg/day usually for 5–7 days, Evidence B)721 for patients who: • Fail to respond to an increase in reliever and ICS-containing maintenance medication for 2–3 days • Deteriorate rapidly or who have a PEF or FEV1 <60% of their personal best or predicted value • Have worsening asthma and a history of sudden severe exacerbations. For children 6–11 years, the recommended dose of prednisone is 1–2 mg/kg/day to a maximum of 40 mg/day (Evidence B), usually for 3–5 days. Patients should be advised about common side-effects, including sleep disturbance, increased appetite, reflux, and mood changes.727 Patients should contact their doctor if they start taking OCS (Evidence D). Even occasional short courses of OCS are associated with significant short-term and cumulative long-term adverse effects,225,571 with a pronounced dose response. For all patients, therefore, asthma management should be optimized to reduce the risk of further exacerbations requiring OCS (Box X). This includes optimizing ICS-containing therapy (with a switch for adults and adolescents to Track 1 with ICS-formoterol if available), treating modifiable risk factors and comorbidities, using relevant non-pharmacologic strategies, and providing education and skills training including a written asthma action plan (see Section 5, p.108 for details). Box 9-3. Optimizing asthma treatment to minimize need for OCS Optimize asthma treatment to minimize cumulative adverse effects of OCS use • OCS can be life-saving during severe asthma exacerbations, but there is increasing awareness of the risks of repeated courses. • In adults, short-term adverse effects of OCS include sleep disturbance, increased appetite, reflux, mood changes,727 sepsis, pneumonia, and thromboembolism.571 • In adults, even 4–5 lifetime courses of OCS are associated with a significantly increased dose-dependent risk of diabetes, cataract, heart failure, osteoporosis and several other conditions.225 • The need for OCS can be reduced by optimizing asthma therapy, including ICS-containing medications, treating modifiable risk factors, using relevant non-pharmacological strategies, and providing education and skills training, including inhaler technique and adherence. Refer patients for expert advice if needed (Box 3-8, p.66). • Make sure that all patients are receiving ICS-containing therapy. For adults and adolescents, GINA Track 1 with ICS-formoterol as anti-inflammatory reliever reduces the risk of severe exacerbations requiring OCS compared with using a SABA reliever (see Box 4-6, p.77). • All patients should have a written asthma action plan, showing them how to increase their inhaled medications and when to contact medical care. See p.11 for abbreviations. Reviewing response Patients should see their doctor immediately or go to an acute care unit if their asthma continues to deteriorate despite following their written asthma action plan, or if their asthma suddenly worsens. Follow up after a self-managed exacerbation After a self-managed exacerbation, patients should see their primary care healthcare provider for a semi-urgent review (e.g., within 1–2 weeks, but preferably before ceasing oral corticosteroids if prescribed), for assessment of symptom control and additional risk factors for exacerbations (Box 2-2, p.37), and to identify the potential cause of the exacerbation. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 166 This visit provides an opportunity for additional asthma education by a trained asthma educator or trained lay healthcare worker. The written asthma action plan should be reviewed to see if it met the patient’s needs. Maintenance asthma treatment can generally be reduced to previous levels 2–4 weeks after the exacerbation (Evidence D), unless the history suggests that the exacerbation occurred on a background of long-term poorly controlled asthma. In this situation, provided inhaler technique and adherence have been checked, a step up in treatment may be indicated (Box 4-6, p.77). Patients with more than 1–2 exacerbations per year despite Step 4–5 therapy (or Step 4 therapy in children 6–11 years), or with several emergency department visits, should be referred to a specialist center, if available, for assessment and strategies to reduce their risk of future exacerbations and their risk of exposure to OCS. See decision tree for difficult-to-treat and severe asthma in Section 8 (p.139). PRIMARY CARE MANAGEMENT OF ASTHMA EXACERBATIONS (ADULTS, ADOLESCENTS, CHILDREN 6–11 YEARS) Assessing exacerbation severity A brief focused history and relevant physical examination should be conducted concurrently with the prompt initiation of therapy, and findings documented in the notes. If the patient shows signs of a severe or life-threatening exacerbation, treatment with SABA, controlled oxygen and systemic corticosteroids should be initiated while arranging for the patient’s urgent transfer to an acute care facility where monitoring and expertise are more readily available. Milder exacerbations can usually be treated in a primary care setting, depending on resources and expertise. History The history should include: • Timing of onset and cause (if known) of the present exacerbation • Severity of asthma symptoms, including any limiting exercise or disturbing sleep • Any symptoms of anaphylaxis • Any risk factors for asthma-related death (Box 9-1, p.160) • All current reliever and maintenance medications, including doses and devices prescribed, adherence pattern, any recent dose changes, and response to current therapy. Physical examination The physical examination should assess: • Signs of exacerbation severity (Box 9-4, p.167) and vital signs (e.g., level of consciousness, temperature, pulse rate, respiratory rate, blood pressure, ability to complete sentences, use of accessory muscles, wheeze). • Complicating factors (e.g., anaphylaxis, pneumonia, pneumothorax) • Signs of alternative conditions that could explain acute breathlessness (e.g., cardiac failure, inducible laryngeal obstruction, inhaled foreign body or pulmonary embolism). Objective measurements Pulse oximetry: Saturation levels <90% in children or adults signal the need for aggressive therapy. Under conditions of hypoxemia, oxygen saturation may be over-estimated by pulse oximetry in people with dark skin color.728 PEF in patients older than 5 years (Box 9-4, p.167) COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 167 Box 9-4. Management of asthma exacerbations in primary care (adults, adolescents, children 6–11 years) See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 168 Treating exacerbations in primary care The main initial therapies (Box 9-4, p.167) include repetitive administration of rapid-acting inhaled bronchodilators, early introduction of systemic corticosteroids, and controlled flow oxygen supplementation.716 The aim is to rapidly relieve airflow obstruction and hypoxemia, address the underlying inflammatory pathophysiology, and prevent relapse. Infection control procedures should be followed. Inhaled short-acting beta2 agonists Currently, inhaled salbutamol (albuterol) is the usual bronchodilator for acute asthma management. For mild to moderate exacerbations, repeated administration of inhaled SABA (up to 4–10 puffs every 20 minutes for the first hour) is an effective and efficient way to achieve rapid reversal of airflow limitation (Evidence A).729 After the first hour, the dose of SABA required varies from 4–10 puffs every 3–4 hours up to 6–10 puffs every 1–2 hours, or more often. No additional SABA is needed if there is a good response to initial treatment (e.g., PEF >60–80% of predicted or personal best for 3–4 hours). Delivery of SABA via a pMDI and spacer or a DPI leads to a similar improvement in lung function as delivery via nebulizer (Evidence A); 729,730 however, patients with acute severe asthma were not included in these studies. The most cost-effective route of delivery is pMDI and spacer,731 provided the patient can use this device. Because of static charge, some spacers require pre-washing with detergent before use. The manufacturer’s advice should be followed. Combination ICS-formoterol in management of acute asthma exacerbations Combination ICS-formoterol (budesonide-formoterol or beclometasone-formoterol) is now widely used as an anti-inflammatory reliever as part of routine asthma management in adults and adolescents, because it reduces the risk of severe exacerbations and exposure to OCS, compared with use of a SABA reliever (GINA Track 1, p.62). Up to a maximum total of 12 inhalations of budesonide-formoterol 200/6 mcg (160/4.5 mcg delivered dose) can be taken in a single day if needed (total of as-needed and maintenance doses, if used), based on evidence from large studies of its efficacy and safety up to this level of use.224,226 Given this extensive evidence, GINA suggests that the same maximum total use in a single day should apply to beclometasone-formoterol (see Box 4-8, p.84 for details of medications and doses). In emergency departments, a randomized controlled trial in adult and adolescent patients with average FEV1 42–45% predicted compared the effect of 2 doses of budesonide-formoterol 400/12 mcg (delivered dose 320/9 mcg) versus 8 doses of salbutamol (albuterol) 100 mcg (delivered dose 90 mcg), with these doses repeated again after 5 minutes; all patients received OCS. Lung function was similar over 3 hours, but pulse rate was higher in the SABA group.732 A meta-analysis of earlier RCTs found that the efficacy and safety of formoterol itself was similar to that of salbutamol (albuterol) in management of acute asthma.733 Formoterol is no longer used for this purpose, but there is no evidence that budesonide-formoterol would be less effective in management of asthma exacerbations.733 More studies are needed. There are no published data on use of combination ICS-SABA in an emergency department setting. Controlled oxygen therapy (if available) Oxygen therapy should be titrated against pulse oximetry (if available) to maintain oxygen saturation at 93–95% (94–98% for children 6–11 years); note the potential for overestimation of oxygen saturation in people with dark skin color . In hospitalized asthma patients, controlled or titrated oxygen therapy is associated with lower mortality and better outcomes than high concentration (100%) oxygen therapy (Evidence A).734-737 Oxygen should not be withheld if oximetry is not available, but the patient should be monitored for deterioration, somnolence or fatigue because of the risk of hypercapnia and respiratory failure.734-737 If supplemental oxygen is administered, oxygen saturation should be maintained no higher than 96% in adults.738 Systemic corticosteroids OCS should be given promptly, especially if the patient is deteriorating, or had already increased their reliever and maintenance ICS-containing medications before presenting (Evidence B). The recommended dose of prednisolone for adults is 1 mg/kg/day or equivalent up to a maximum of 50 mg/day, and 1–2 mg/kg/day for children 6–11 years up to a COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 169 maximum of 40 mg/day). OCS should usually be continued for 5–7 days in adults739,740 and 3–5 days in children (Evidence B).741 Patients should be advised about common short-term side-effects, including sleep disturbance, increased appetite, reflux and mood changes.727 In adults, the risk of sepsis and thromboembolism is also increased after a short course of OCS.571 While OCS are life-saving for acute severe asthma, use of 4–5 lifetime courses in adults is associated with a dose-dependent increased risk of long-term adverse effects such as osteoporosis, fractures, diabetes, heart failure and cataract.225 This emphasizes the importance of optimizing asthma management after any severe exacerbation to reduce the risk of further exacerbations (see Section 4, p.67). Maintenance ICS-containing medication Patients already prescribed maintenance ICS-containing medication should be provided with advice about increasing the dose for the next 2–4 weeks, as summarized in Box 9-2 (p.162). Patients not currently taking controller medication should be commenced on ICS-containing therapy, as SABA-only treatment of asthma is no longer recommended. An exacerbation requiring medical care indicates that the patient is at increased risk of future exacerbations (Box 2-2, p.37). Antibiotics (not recommended) Evidence does not support routine use of antibiotics in the treatment of acute asthma exacerbations unless there is strong evidence of lung infection (e.g., fever and purulent sputum or radiographic evidence of pneumonia).742 Reviewing response During treatment, patients should be closely monitored, and treatment titrated according to their response. Patients who present with signs of a severe or life-threatening exacerbation (Box 9-4, p.167), who fail to respond to treatment, or who continue to deteriorate should be transferred immediately to an acute care facility. Patients with little or slow response to SABA treatment should be closely monitored. For many patients, lung function can be monitored after SABA therapy is initiated. Additional treatment should continue until PEF or FEV1 reaches a plateau or (ideally) returns to the patient’s previous best. A decision can then be made whether to send the patient home or transfer them to an acute care facility. Follow up Discharge medications should include regular maintenance ICS-containing treatment (see Box 4-8, p.84 and Box 9-5, p.170), as-needed reliever medication (low-dose ICS-formoterol, ICS-SABA or SABA) and a short course of OCS. SABA-only treatment is not recommended. Inhaler technique and adherence should be reviewed before discharge. Patients should be advised to use their reliever inhaler only as-needed, rather than routinely. A follow-up appointment should be arranged for about 2–7 days later, depending on the clinical and social context. At the review visit the healthcare provider should assess whether the flare-up has resolved, and whether OCS can be ceased. They should assess the patient’s level of symptom control and risk factors; explore the potential cause of the exacerbation; and review the written asthma action plan (or provide one if the patient does not already have one). Maintenance ICS-containing treatment can generally be stepped back to pre-exacerbation levels 2–4 weeks after the exacerbation. However, if the exacerbation was preceded by symptoms suggestive of chronically poorly controlled asthma, and inhaler technique and adherence are good, a step up in treatment (Box 4-6, p.77) may be indicated. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 170 Box 9-5. Discharge management after acute care for asthma Medications Inhaled corticosteroid (ICS)-containing therapy Initiate ICS-containing treatment, if not already being taken. For adults/adolescents, maintenance-and-reliever therapy with ICS-formoterol (MART) is preferred as it reduces risk of future exacerbations compared with using a SABA reliever (see Box 4-6, p.77). For adults/adolescents, start MART at Step 4 on discharge (Box 4-5, p.76, Box 4-6, p.77 and Box 4-8, p.84). If prescribing an ICS regimen with SABA reliever, step maintenance dose up for 2–4 weeks (Box 9-2, p.162). Emphasize good adherence and check and correct inhaler technique.
Oral corticosteroids (OCS) To reduce the risk of relapse, prescribe at least a 5–7 day course of OCS for adults (prednisolone or equivalent 40–50 mg/day) and 3–5 days for children (1–2 mg/kg/day to a maximum of 40 mg/day) (Evidence A).743 Review progress before ceasing OCS. If the OCS is dexamethasone, treatment is only for total 1–2 days,744 but if there is failure of resolution, or relapse of symptoms, consider switching to prednisolone.
Reliever medication: as-needed rather than regular Switch patients to as-needed rather than regular reliever medication use, and monitor symptomatic and objective improvement. Regular use of SABA for even 1–2 weeks leads to beta-receptor down-regulation, increased airway hyperresponsiveness and increased eosinophilic inflammation, with reduced bronchodilator response,311,662,745 and regular SABA can mask worsening asthma. Ipratropium bromide, if used in the ED or hospital, may be quickly discontinued as it is unlikely to provide ongoing benefit. Patients prescribed ICS–formoterol as their reliever should return to this on/before discharge if SABA was substituted in ED or hospital.
Risk factors and triggers that contributed to the exacerbation Identify factors that may have contributed to the exacerbation, and implement strategies to reduce modifiable risk factors (Box 3-5, p.55). These may include irritant or allergen exposure, viral respiratory infections, inadequate long-term ICS treatment, problems with adherence, and/or lack of a written asthma action plan. Handwashing, masks and social/physical distancing may reduce the risk of acquiring viral respiratory infections, including influenza. Self-management skills and written asthma action plan • Review inhaler technique and correct if needed (Box 5-2, p.110).
• Provide a written asthma action plan (Box 9-2, p.162) or review the patient’s existing plan, either at discharge or as soon as possible afterwards. Patients discharged from the ED with an action plan and PEF meter have better outcomes than patients discharged without these resources.746 For patients prescribed ICS-formoterol reliever, use an action plan template customized for this treatment.313,314 Review technique with PEF meter if used.
• Evaluate the patient’s response as the exacerbation was developing. If it was inadequate, review the action plan and provide further written guidance to assist if asthma worsens again.746,747 • Review the patient’s use of medications before and during the exacerbation. Was ICS-containing treatment increased promptly and by how much? Was ICS-formoterol reliever (if prescribed) increased appropriately in response to symptoms? If OCS were indicated, were they taken; did the patient experience adverse effects? If the patient is provided with a prescription for OCS to be on hand for subsequent exacerbations, beware of inappropriate use, as even 4–5 lifetime courses of OCS in adults increase the risk of serious adverse effects.225 Follow up communication and appointment • Inform the patient’s usual healthcare provider about their ED presentation/admission, instructions given on discharge, and any treatment changes.
• Make a follow-up appointment within 2–7 days of discharge (1–2 days for children) to ensure that treatment is continued. The patient should be followed to ensure that asthma symptoms return to well controlled, and that their lung function returns to their personal best (if known). Refer for expert advice if the patient required ICU treatment, or if they already had one or more other exacerbations in the last 12 months; see Box 3-8, p.66.
See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 171 EMERGENCY DEPARTMENT MANAGEMENT OF EXACERBATIONS (ADULTS, ADOLESCENTS, CHILDREN 6–11 YEARS) Severe exacerbations of asthma are life-threatening medical emergencies, which are most safely managed in an acute care setting e.g., emergency department (Box 9-6, p.171). Infection control procedures should be followed. Management of asthma in the intensive care unit is beyond the scope of this report and readers are referred to a comprehensive review.748 Box 9-6. Management of asthma exacerbations in acute care facility (e.g., emergency department) See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 172 Assessment History A brief history and physical examination should be conducted concurrently with the prompt initiation of therapy. Include: • Time of onset and cause (if known) of the present exacerbation • Severity of asthma symptoms, including any limiting exercise or disturbing sleep • Any symptoms of anaphylaxis • Risk factors for asthma-related death (Box 9-1, p.160) • All current reliever and maintenance medications, including doses and devices prescribed, adherence pattern, any recent dose changes, and response to current therapy. Physical examination The physical examination should assess: • Signs of exacerbation severity (Box 9-6, p.171), including vital signs (e.g., level of consciousness, temperature, pulse rate, respiratory rate, blood pressure, ability to complete sentences, use of accessory muscles) • Complicating factors (e.g., anaphylaxis, pneumonia, atelectasis, pneumothorax or pneumomediastinum) • Signs of alternative conditions that could explain acute breathlessness (e.g., cardiac failure, inducible laryngeal obstruction, inhaled foreign body or pulmonary embolism). Objective assessments Objective assessments are also needed as the physical examination alone may not indicate the severity of the exacerbation.749,750 However, patients, and not their laboratory values, should be the focus of treatment. Measurement of lung function: this is strongly recommended. If possible, and without unduly delaying treatment, PEF or FEV1 should be recorded before treatment is initiated, although spirometry may not be possible in children with acute asthma. Lung function should be monitored at one hour and at intervals until a clear response to treatment has occurred or a plateau is reached. Oxygen saturation: this should be closely monitored, preferably by pulse oximetry. In children, oxygen saturation is normally ≥95% when breathing room air at sea level, and saturation <92% is a predictor of the need for hospitalization (Evidence C).751 Saturation levels <90% in children or adults signal the need for aggressive therapy. Subject to clinical urgency, saturation should be assessed before oxygen is commenced, or 5 minutes after oxygen is removed or when saturation stabilizes. Of concern, under conditions of hypoxemia, oxygen saturation may be over-estimated by pulse oximeters in people with dark skin color.728 Arterial blood gas measurements are not routinely required:752 They should be considered for patients with PEF or FEV1 <50% predicted,753 or for those who do not respond to initial treatment or are deteriorating. Supplemental controlled oxygen should be continued while blood gases are obtained. During an asthma exacerbation PaCO2 is often below normal (<40 mmHg). Fatigue and somnolence suggest that pCO2 may be increasing and airway intervention may be needed. PaO2<60 mmHg (8 kPa) and normal or increased PaCO2 (especially >45 mmHg, 6 kPa) indicate respiratory failure. Chest X-ray is not routinely recommended: In adults, chest X-ray should be considered if a complicating or alternative cardiopulmonary process is suspected (especially in older patients), or for patients who are not responding to treatment where a pneumothorax may be difficult to diagnose clinically.754 Similarly, in children, routine check X-ray is not recommended unless there are physical signs suggestive of pneumothorax, parenchymal disease or an inhaled foreign body. Features associated with positive chest X-ray findings in children include fever, no family history of asthma, and localized lung examination findings.755 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 173 Treatment in acute care settings such as the emergency department The following treatments are usually administered concurrently to achieve rapid improvement.756 Oxygen To achieve arterial oxygen saturation of 93–95% (94–98% for children 6–11 years), oxygen should be administered by nasal cannulae or mask. Note the potential for overestimation of saturation in people with dark skin color.728 In severe exacerbations, controlled low flow oxygen therapy using pulse oximetry to maintain saturation at 93–95% is associated with better physiological outcomes than with high concentration (100%) oxygen therapy (Evidence B).734-736 However, oxygen therapy should not be withheld if pulse oximetry is not available (Evidence D). Once the patient has stabilized, consider weaning them off oxygen using oximetry to guide the need for ongoing oxygen therapy. Inhaled short-acting beta2 agonists Currently, inhaled salbutamol (albuterol) is the usual bronchodilator in acute asthma management. The most cost-effective and efficient delivery is by pMDI with a spacer (Evidence A).729,731 Evidence for pMDI and spacer is less robust in severe and near-fatal asthma. Systematic reviews of intermittent versus continuous SABA in acute asthma, which mostly used nebulized SABA, provide conflicting results. Use of nebulizers can disseminate aerosols and potentially contribute to spread of respiratory viral infections.757 Current evidence does not support the routine use of intravenous beta2 agonist in most patients with severe asthma exacerbations (Evidence A).758 Combination ICS-formoterol as an alternative to high dose SABA Compared with SABA, similar efficacy and safety have been reported from emergency department studies with formoterol,733 and in one study with budesonide-formoterol.732 The later study showed that high-dose budesonide-formoterol had similar efficacy and safety profile to high dose SABA.732 In this study, patients received 2 doses of budesonide-formoterol 400/12 mcg (delivered dose 320/9 mcg) or 8 doses of salbutamol (albuterol) 100 mcg (delivered dose 90 mcg), repeated once after 5 minutes; all patients received OCS.732 While more studies are needed, meta-analysis of data from earlier studies comparing high-dose formoterol with high dose salbutamol (albuterol) for treatment of acute asthma in the ED setting suggest that budesonide-formoterol would also be effective.733 Formoterol alone is no longer used for this purpose. Epinephrine (for anaphylaxis) Intramuscular epinephrine (adrenaline) is indicated in addition to standard therapy for acute asthma associated with anaphylaxis and angioedema. It is not routinely indicated for other asthma exacerbations. Systemic corticosteroids Systemic corticosteroids speed resolution of exacerbations and prevent relapse, and in acute care settings should be utilized in all but the mildest exacerbations in adults, adolescents and children 6–11 years.759,760 (Evidence A). Where possible, systemic corticosteroids should be administered to the patient within 1 hour of presentation;759 some studies showed similar benefit with high-dose ICS.761 Use of systemic corticosteroids is particularly important in the emergency department if: • Initial SABA treatment fails to achieve lasting improvement in symptoms • The exacerbation developed while the patient was taking OCS • The patient has a history of previous exacerbations requiring OCS. Route of delivery: oral administration is as effective as intravenous. The oral route is preferred because it is quicker, less invasive and less expensive.762,763 For children, a liquid formulation is preferred to tablets. OCS require at least 4 hours to produce a clinical improvement. Intravenous corticosteroids can be administered when patients are too dyspneic to COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 174 swallow; if the patient is vomiting; or when patients require non-invasive ventilation or intubation. Evidence does not demonstrate a benefit of intramuscular corticosteroids over oral corticosteroids.760 Dosage: daily doses of OCS equivalent to 50 mg prednisolone as a single morning dose, or 200 mg hydrocortisone in divided doses, are typically used for adults. For children, a prednisolone dose of 1–2 mg/kg up to a maximum of 40 mg/day is suggested.764 Duration: 5- and 7-day courses of prednisone or prednisolone in adults have been found to be as effective as 10- and 14-day courses respectively (Evidence B),739,740 and in most children, a 3–5-day course is usually considered sufficient. Evidence from studies in which all patients were taking maintenance ICS after discharge suggests that there is no benefit in tapering the dose of OCS, either in the short term765 or over several weeks766 (Evidence B). In adults, a small number of studies examined oral dexamethasone 12–16 mg given once daily for 1–2 days in children and adults; the relapse rate was similar to that with prednisolone for 3–5 days, and adverse events rates were similar.744,767,768 In children, a systematic review found no difference in relapse rate with oral dexamethasone 0.3 mg/kg or 0.6 mg/kg once daily for 1–2 days versus oral prednisone/prednisolone for 3–5 days; adherence was better, and there was a substantially lower risk of vomiting with dexamethasone.769 Oral dexamethasone should not be continued beyond 2 days because of concerns about metabolic side-effects. If there is a failure of resolution, or relapse of symptoms, consideration should be given to switching to prednisolone. Inhaled corticosteroids Within the emergency department: high-dose ICS given within the first hour after presentation reduces the need for hospitalization in patients not receiving systemic corticosteroids (Evidence A).761 When added to systemic corticosteroids, evidence is conflicting in adults.770 In children, administration of ICS with or without concomitant systemic corticosteroids within the first hours of attendance to the emergency department might reduce the risk of hospital admission and need for systemic corticosteroids (Evidence B).771 Overall, add-on ICS are well tolerated; however, cost may be a significant factor, and the agent, dose and duration of treatment with ICS in the management of asthma in the emergency department remain unclear. Patients admitted to hospital for an asthma exacerbation should continue on, or be prescribed, ICS-containing therapy. On discharge home: patients should be prescribed ongoing ICS-containing treatment since the occurrence of a severe exacerbation is a risk factor for future exacerbations (Evidence B) (Box 2-2, p.37), and ICS-containing medications significantly reduce the risk of asthma-related death or hospitalization (Evidence A).329 SABA-only treatment of asthma is no longer recommended. For short-term outcomes such as relapse requiring admission, symptoms, and quality of life, a systematic review found no significant differences when ICS were added to systemic corticosteroids after discharge.772 There was some evidence, however, that post-discharge ICS were as effective as systemic corticosteroids for milder exacerbations, but the confidence limits were wide (Evidence B).772 Cost may be a significant factor for patients in the use of high-dose ICS, and further studies are required to establish their role.772 After an ED presentation or hospitalization, the preferred ongoing treatment is maintenance-and-reliver therapy (MART) with ICS-formoterol. In patients with a history of ≥1 severe exacerbations, MART reduces the risk of another severe exacerbation in the next 12 months by 32% compared with same dose ICS or ICS-LABA plus as-needed SABA, and by 23% compared with higher dose ICS-LABA plus as-needed SABA.224 See Box 4-8 (p.84) for medications and doses. Other treatments Ipratropium bromide For adults and children with moderate-severe exacerbations, treatment in the emergency department with both SABA and ipratropium, a short-acting anticholinergic, was associated with fewer hospitalizations (Evidence A for adults;773 Evidence B for adolescents/children774) and greater improvement in PEF and FEV1 compared with SABA alone (Evidence A, adults/adolescents).773-775 For children hospitalized for acute asthma, no benefits were seen from adding ipratropium to SABA, including no reduction in length of stay, but the risk of nausea and tremor was reduced.774 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 175 Aminophylline and theophylline (not recommended) Intravenous aminophylline and theophylline should not be used in the management of asthma exacerbations, in view of their poor efficacy and safety profile, and the greater effectiveness and relative safety of SABA.776 Nausea and/or vomiting are more common with aminophylline.774,776 The use of intravenous aminophylline is associated with severe and potentially fatal side-effects, particularly in patients already treated with sustained-release theophylline. In adults with severe asthma exacerbations, add-on treatment with aminophylline does not improve outcomes compared with SABA alone.776 Magnesium Intravenous magnesium sulfate is not recommended for routine use in asthma exacerbations; however, when administered as a single 2 g infusion over 20 minutes, it reduces hospital admissions in some patients, including adults with FEV1 <25–30% predicted at presentation; adults and children who fail to respond to initial treatment and have persistent hypoxemia; and children whose FEV1 fails to reach 60% predicted after 1 hour of care (Evidence A).777-779 Randomized, controlled trials that excluded patients with more severe asthma showed no benefit with the addition of intravenous or nebulized magnesium compared with placebo in the routine care of asthma exacerbations in adults and adolescents780-782 or children781,783 (Evidence B). Helium oxygen therapy A systematic review of studies comparing helium–oxygen with air–oxygen suggests there is no role for this intervention in routine care (Evidence B), but it may be considered for patients who do not respond to standard therapy; however, availability, cost and technical issues should be considered.784 Leukotriene receptor antagonists (LTRAs) There is limited evidence to support the use of oral or intravenous LTRAs in acute asthma. Small studies have demonstrated improvement in lung function,785,786 but the clinical role and safety of these agents requires more study. Antibiotics (not recommended) Evidence does not support the routine use of antibiotics in the treatment of acute asthma exacerbations unless there is strong evidence of lung infection (e.g., fever or purulent sputum or radiographic evidence of pneumonia).742 Non-invasive ventilation (NIV) The evidence regarding the role of NIV in asthma is weak. A systematic review identified five studies in adults involving 206 patients with acute severe asthma treated with NIV or placebo.787 Two studies found no difference in need for endotracheal intubation but one study identified fewer admissions in the NIV group. No deaths were reported in either study. Given the small size of the studies, no recommendation is offered. If NIV is tried, the patient should be monitored closely (Evidence D). It should not be attempted in agitated patients, and patients should not be sedated to receive NIV (Evidence D). Sedatives (MUST BE AVOIDED) Sedation should be strictly avoided during exacerbations of asthma because of the respiratory depressant effect of anxiolytic and hypnotic drugs. An association between the use of these drugs and avoidable asthma deaths has been reported.788,789 Reviewing response Clinical status and oxygen saturation should be re-assessed frequently, with further treatment titrated according to the patient’s response (Box 9-6, p.171). Lung function should be measured after one hour, i.e., after the first three bronchodilator treatments, and patients who deteriorate despite intensive bronchodilator and corticosteroid treatment should be re-evaluated for transfer to the intensive care unit. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 176 Criteria for hospitalization versus discharge from the emergency department From retrospective analyses, clinical status (including the ability to lie flat) and lung function 1 hour after commencement of treatment are more reliable predictors of the need for hospitalization than the patient’s status on arrival.790,791 Spirometric criteria that have been proposed for hospital admission or discharge from the emergency department:792 • If pre-treatment FEV1 or PEF is <25% predicted or personal best, or post-treatment FEV1 or PEF is <40% predicted or personal best, hospitalization is recommended. • If post-treatment lung function is 40–60% predicted, discharge may be possible after considering the patient’s risk factors (Box 9-1, p.160) and availability of follow-up care. • If post-treatment lung function is >60% predicted or personal best, discharge is recommended after considering risk factors and availability of follow-up care. Other factors associated with increased likelihood of need for admission include:793-795 • Female sex, older age and non-white race • Use of more than eight beta2 agonist puffs in the previous 24 hours • Severity of the exacerbation (e.g., need for resuscitation or rapid medical intervention on arrival, respiratory rate >22 breaths/minute, oxygen saturation <95%, final PEF <50% predicted) • Past history of severe exacerbations (e.g., intubations, asthma admissions) • Previous unscheduled office and emergency department visits requiring use of OCS. Overall, these risk factors should be considered by clinicians when making decisions on admission/discharge for patients with asthma managed in the acute care setting. The patient’s social circumstances should also be considered. DISCHARGE PLANNING AND FOLLOW-UP Prior to discharge from the emergency department or hospital to home, arrangements should be made for a follow-up appointment within 2–7 days (1–2 days for children), and strategies to improve asthma management including medications, inhaler skills and written asthma action plan, should be addressed (Box 9-5, p.170).418 All patients should be prescribed ongoing ICS-containing treatment to reduce the risk of further exacerbations. For adults and adolescents, the preferred regimen after discharge is maintenance-and-reliever therapy (MART) with the anti-inflammatory reliever ICS-formoterol, because this will reduce the risk of future severe exacerbations and reduce the need for OCS compared with a regimen with a SABA reliever. In the context of a recent ED visit or hospitalization, it would be appropriate to commence treatment with ICS-formoterol in adults and adolescents at Step 4. For medications and doses, see Box 4-8 (p.84), The maintenance dose can be stepped down later, once the patient has fully recovered and asthma has remained stable for 2–3 months (see Box 4-13, p.102). Follow up after emergency department presentation or hospitalization for asthma Following discharge, the patient should be reviewed by their healthcare provider regularly over subsequent weeks until good symptom control is achieved and personal best lung function is reached or surpassed. Incentives such as free transport and telephone reminders improve primary care follow up but have shown no effect on long-term outcomes.418 At follow-up, again ensure that the patient’s treatment has been optimized to reduce the risk of future exacerbations. Consider switching to GINA Track 1 with the anti-inflammatory reliever ICS-formoterol, if not already prescribed. See Box 4-8 (p.84) for medications and doses. Check and correct inhaler technique and adherence. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 177 Patients discharged following an emergency department presentation or hospitalization for asthma should be especially targeted for an asthma education program, if one is available. Patients who were hospitalized may be particularly receptive to information and advice about their illness. Healthcare providers should take the opportunity to review: • The patient’s understanding of the cause of their asthma exacerbation • Modifiable risk factors for exacerbations (including, where relevant, smoking) (Box 3-5, p.55) • The patient’s understanding of the purposes and correct uses of medications, including ICS-containing maintenance treatment and anti-inflammatory reliever, if prescribed • The actions the patient needs to take to respond to worsening symptoms or peak flows. After an emergency department presentation, comprehensive intervention programs that include optimization of asthma treatment, inhaler technique, and elements of self-management education (self-monitoring, written action plan and regular review)193 are cost effective and have shown significant improvement in asthma outcomes (Evidence B).418 Referral for expert advice should be considered for patients who have been hospitalized for asthma, or who have had several presentations to an acute care setting despite having a primary care provider. Follow-up by a specialist is associated with fewer subsequent emergency department visits or hospitalizations and better asthma control.418 Optimize asthma treatment to minimize the use of OCS OCS can be life-saving during severe asthma exacerbations, but there is increasing awareness of the risks of repeated courses. In adults, short-term adverse effects of OCS include sleep disturbance, increased appetite, reflux, mood changes,727 sepsis, pneumonia, and thromboembolism.571 In adults, even 4–5 lifetime courses of OCS are associated with a significantly increased dose-dependent risk of diabetes, cataract, heart failure, osteoporosis and several other conditions.225 The need for OCS can be reduced by optimizing inhaled therapy, including attention to inhaler technique and adherence. For adults and adolescents, GINA Track 1 with ICS-formoterol as anti-inflammatory reliever reduces the risk of severe exacerbations requiring OCS compared with using a SABA reliever (see Box 4-6, p.77). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 178 10. Diagnosis of asthma in children 5 years and younger KEY POINTS Recurrent wheezing occurs in a large proportion of children 5 years and younger, typically with viral upper respiratory tract infections. It is difficult to discern when this is the initial presentation of asthma. Previous classifications of wheezing phenotypes (episodic wheeze and multiple-trigger wheeze; or transient wheeze, persistent wheeze and late-onset wheeze) do not appear to identify stable phenotypes, and their clinical usefulness is uncertain. However, emerging research suggest that more clinically relevant phenotypes will be described and phenotype-directed therapy possible. A diagnosis of asthma in young children with a history of wheezing is more likely if they have: • Wheezing or coughing that occurs with exercise, laughing or crying, or in the absence of an apparent respiratory infection • A history of other allergic disease (eczema or allergic rhinitis), allergen sensitization or asthma in first-degree relatives • Clinical improvement during 2–3 months of low-dose inhaled corticosteroid (ICS) treatment plus as-needed short-acting beta2 agonist (SABA) reliever, and worsening after cessation. ASTHMA AND WHEEZING IN YOUNG CHILDREN Asthma is the most common chronic disease of childhood and the leading cause of childhood morbidity from chronic disease as measured by school absences, emergency department visits and hospitalizations.796 Asthma often begins in early childhood; in up to half of people with asthma, symptoms commence during childhood.797 Onset of asthma is earlier in males than females.207,798,799 No intervention has yet been shown to prevent the development of asthma or modify its long-term natural course. Atopy is present in the majority of children with asthma who are over 3 years old, and allergen-specific sensitization (and particularly multiple early-life sensitizations) is one of the most important risk factors for the development of asthma.800 Viral-induced wheezing Recurrent wheezing occurs in a large proportion of children aged 5 years or younger. It is typically associated with upper respiratory tract infections (URTI), which occur in this age group around 6–8 times per year.801 Some viral infections (respiratory syncytial virus and rhinovirus) are associated with recurrent wheeze throughout childhood.802 Wheezing in this age group is a highly heterogeneous condition, and not all wheezing indicates asthma. A large proportion of wheezing episodes in young children is virally induced whether the child has asthma or not. Therefore, deciding when wheezing with a respiratory infection is truly an isolated event or represents a recurrent clinical presentation of childhood asthma may be difficult.207,803 In children aged under 1 year, bronchiolitis may present with wheeze. It is usually accompanied by other chest signs such as crackles on auscultation. Wheezing phenotypes In the past, two main classifications of wheezing (called ‘wheezing phenotypes’) were proposed: • Symptom-based classification:804 this was based on whether the child had only episodic wheeze (wheezing during discrete time periods, often in association with URTI, with symptoms absent between episodes) or multiple-trigger wheeze (episodic wheezing with symptoms also occurring between these episodes, e.g., during sleep or with triggers such as activity, laughing, or crying). • Time trend-based classification: this system was initially based on retrospective analysis of data from a cohort study207. It included transient wheeze (symptoms began and ended before the age of 3 years); persistent wheeze COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 179 (symptoms began before the age of 3 years and continued beyond the age of 6 years), and late-onset wheeze (symptoms began after the age of 3 years). These general patterns have been confirmed in subsequent studies using unsupervised statistical approaches.805,806 However, prospective allocation of individual children to these phenotypes has been challenging in ‘real-life’ clinical situations, and the clinical usefulness of these, and other, classification and asthma prediction systems remain a subject of active investigation. For example, one study conducted in a research setting with high medication adherence found that daily ICS treatment reduced exacerbations in preschool children characterized as ‘sensitization with indoor pet exposure’ or ‘multiple sensitization with eczema’, but not among those characterized as ‘minimal sensitization’ or ‘sensitization with tobacco smoke exposure’.807 CLINICAL DIAGNOSIS OF ASTHMA It may be challenging to make a confident diagnosis of asthma in children 5 years and younger, because episodic respiratory symptoms such as wheezing and cough are also common in children without asthma, particularly in those aged 0–2 years,349,350 and it is not possible to routinely assess airflow limitation or bronchodilator responsiveness in this age group. A probability-based approach, based on the pattern of symptoms during and between viral respiratory infections,808 may be helpful for discussion with parents/caregivers (Box 10-1, Box 10-2 and Box 10-3, p.179). This allows individual decisions to be made about whether to give a trial of controller treatment. It is important to make decisions for each child individually, to avoid either over- or under-treatment. Box 10-1. Probability of asthma diagnosis in children 5 years and younger COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 180 Box 10-2. Features suggesting a diagnosis of asthma in children 5 years and younger Feature Characteristics suggesting asthma Cough • Recurrent or persistent non-productive cough that may be worse at night or accompanied by wheezing and breathing difficulties • Cough occurring with exercise, laughing, crying or exposure to tobacco smoke, particularly in the absence of an apparent respiratory infection Wheezing • Recurrent wheezing, including during sleep or with triggers such as activity, laughing, crying or exposure to tobacco smoke or air pollution Difficult or heavy breathing or shortness of breath • Occurring with exercise, laughing, or crying Reduced activity • Not running, playing or laughing at the same intensity as other children; tires earlier during walks (wants to be carried) Past or family history • Other allergic disease (atopic dermatitis or allergic rhinitis, food allergy). Asthma in first-degree relative(s) Therapeutic trial with low-dose ICS (Box 11-2, p.190) plus as-needed SABA • Clinical improvement during 2–3 months of low-dose ICS treatment and worsening when treatment is stopped See list of abbreviations (p.11). Box 10-3. Questions that can be used to elicit features suggestive of asthma • Does your child have wheezing? Wheezing is a high-pitched noise which comes from the chest and not the throat. Use of a video questionnaire,809 or asking a parent/caregiver to record an episode on a smartphone if available can help to confirm the presence of wheeze and differentiate from upper airway abnormalities. • Does your child wake up at night because of coughing, wheezing, or difficult breathing, heavy breathing, or breathlessness? • Does your child have to stop running, or play less hard, because of coughing, wheezing or difficult breathing, heavy breathing, or shortness of breath? • Does your child cough, wheeze or get difficult breathing, heavy breathing, or shortness of breath when laughing, crying, playing with animals, or when exposed to strong smells or smoke? Additional features may help to elicit features that support the diagnosis of asthma or allergic asthma: • Has your child ever had eczema, or been diagnosed with allergy to foods? • Has anyone in your family had asthma, hay fever, food allergy, eczema, or any other disease with breathing problems? COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 181 Symptoms suggestive of asthma in children 5 years and younger An asthma diagnosis in children 5 years and younger can often be based on: • Symptom patterns (recurrent episodes of wheeze, cough, breathlessness (typically manifested by activity limitation), and nocturnal symptoms or awakenings) • Presence of risk factors for development of asthma, such as family history of atopy, allergic sensitization, allergy or asthma, or a personal history of food allergy or atopic dermatitis • Therapeutic response to controller treatment • Exclusion of alternate diagnoses.
Box 10-1 shows the estimated probability of an asthma diagnosis in children aged 5 years or younger who have viral-induced cough, wheeze or heavy breathing, based on the pattern of symptoms.810,811 Many young children wheeze with viral infections; it may be difficult to decide when a child should be given controller treatment. The frequency and severity of wheezing episodes and the temporal pattern of symptoms (only with viral colds or also in response to other triggers) should be considered. Any controller treatment should be viewed as a treatment trial, with follow up scheduled after 2–3 months to review the response. Review is also important since the pattern of symptoms tends to change over time in a large proportion of children. A diagnosis of asthma in young children is therefore based largely on recurrent symptom patterns combined with a careful clinical assessment of family history and physical findings with careful consideration of the differential diagnostic possibilities. A positive family history of allergic disorders, or the presence of atopy or allergic sensitization provide additional predictive support, as early allergic sensitization increases the likelihood that a wheezing child will develop persistent asthma.800 Wheeze Wheeze is the most common and specific symptom associated with asthma in children 5 years and younger. Wheezing occurs in several different patterns, but a wheeze that occurs recurrently, during sleep, or with triggers such as activity, laughing, or crying, is consistent with a diagnosis of asthma. Clinician confirmation is important, as parents/caregivers may describe any noisy breathing as ‘wheezing’.812 Some cultures do not have a word for wheeze. Wheezing may be interpreted differently based on: • Who observes it (e.g., parent/caregiver versus the healthcare provider) • The environmental context (e.g., high income countries versus areas with a high prevalence of parasites that involve the lung) • The cultural context (e.g., the relative importance of certain symptoms can differ between cultures, as can the diagnosis and treatment of respiratory tract diseases in general).
Cough Cough due to asthma is generally non-productive, recurrent and/or persistent, and is usually accompanied by wheezing episodes and breathing difficulties. Allergic rhinitis may be associated with cough in the absence of asthma. A nocturnal cough (when the child is asleep) or a cough that occurs with exercise, laughing or crying, in the absence of an apparent respiratory infection, supports a diagnosis of asthma. The common cold and other respiratory illnesses including pertussis are also associated with coughing. Prolonged cough in infancy, and cough without cold symptoms, are associated with later parent/caregiver-reported physician-diagnosed asthma, independent of infant wheeze. Characteristics of cough in infancy may be early markers of asthma susceptibility, particularly among children with maternal asthma.813 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 182 Breathlessness Parents/caregivers may also use terms such as ‘difficult breathing’, ‘heavy breathing’, or ‘shortness of breath’. Breathlessness that occurs during exercise and is recurrent increases the likelihood of the diagnosis of asthma. In infants and toddlers, crying and laughing are equivalent to exercise in older children. Activity and social behavior Physical activity is an important trigger of asthma symptoms in young children. Young children with poorly controlled asthma often abstain from strenuous play or exercise to avoid symptoms, but many parents/caregivers are unaware of such changes in their children’s lifestyle. Engaging in play is important for a child’s normal social and physical development. For this reason, careful review of the child’s daily activities, including their willingness to walk and play, is important when assessing a potential asthma diagnosis in a young child. Parents/caregivers may report irritability, tiredness and mood changes in their child as the main problems when asthma is not well controlled. TESTS TO ASSIST IN DIAGNOSIS While no tests specifically and definitively diagnose asthma with certainty, in children 5 years and younger, the following are useful adjuncts. Therapeutic trial A trial of treatment for at least 2–3 months with as-needed SABA and regular low-dose ICS may provide some guidance about the diagnosis of asthma (Evidence D). Response should be evaluated by symptom control (daytime and night-time), and the frequency of wheezing episodes and exacerbations. Marked clinical improvement during treatment, and deterioration when treatment is stopped, support a diagnosis of asthma. Due to the variable nature of asthma in young children, a therapeutic trial may need to be repeated to confirm the diagnosis. Tests for allergic sensitization Sensitization to allergens can be assessed using either skin prick testing or allergen-specific immunoglobulin E. Allergic sensitization is present in the majority of children with asthma once they are over 3 years of age; however, absence of sensitization to common aeroallergens does not rule out a diagnosis of asthma. Allergic sensitization is the best predictor for development of persistent asthma.814 Chest X-ray Radiographs are rarely indicated; however, if there is doubt about the diagnosis of asthma in a wheezing or coughing child, a plain chest X-ray may help to exclude structural abnormalities (e.g., congenital lobar emphysema, vascular ring) chronic infections such as tuberculosis, an inhaled foreign body, or other diagnoses. Other imaging investigations may be appropriate, depending on the condition being considered. Lung function testing Due to the inability of most children 5 years and younger to perform reproducible expiratory maneuvers, lung function testing, bronchial provocation testing, and other physiological tests do not have a major role in the diagnosis of asthma at this age. However, by 5 years of age, many children are capable of performing reproducible spirometry if coached by an experienced technician and with visual incentives. Exhaled nitric oxide Measurement of fractional concentration of exhaled nitric oxide (FeNO) is not widely available for most children in this age group and currently remains primarily a research tool. FeNO can be measured in young children with tidal breathing, and normal reference values have been published for children aged 1–5 years.815 In preschool children with recurrent coughing and wheezing, an elevated FeNO recorded 4 weeks from any URTI predicted physician-diagnosed asthma at school age,816 and increased the odds for wheezing, asthma and ICS use by school age, independent of clinical history and presence of specific IgE.817 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 183 RISK PROFILES A number of risk profile tools aimed at identifying which wheezing children aged 5 years and younger are at high risk of developing persistent asthma symptoms have been evaluated for use in clinical practice. However, these tools have shown limited performance for clinical practice. Only three prediction tools have been externally validated (Asthma Predictive Index818 from Tucson, USA, Prevention and Incidence of Asthma and Mite Allergy (PIAMA) index803 from the Netherlands, and Leicester tool819 from the UK), and a systematic review has shown that these tools have poor predictive accuracy, with variation in sensitivity and positive predictive value.820 Larger predictive studies using more advanced statistical methods, and with objective measurements for asthma diagnosis, are probably needed to propose a practical tool in clinical care to predict persistent asthma in recurrent wheezers in infancy and preschool age. The role of these tools is to help identify children at greater risk of developing persistent asthma symptoms, not as criteria for the diagnosis of asthma in young children. Each tool demonstrates different performance characteristics with varying criteria used to identify risk.821 DIFFERENTIAL DIAGNOSIS A definite diagnosis of asthma in this young age group is challenging but has important clinical consequences. It is particularly important in this age group to consider and exclude alternative causes that can lead to symptoms of wheeze, cough, and breathlessness before confirming an asthma diagnosis (Box 10-4).822 Box 10-4. Common differential diagnoses of asthma in children 5 years and younger Condition Typical features Recurrent viral respiratory tract infections Mainly cough, runny congested nose for <10 days; no symptoms between infections Gastroesophageal reflux Cough when feeding; recurrent chest infections; vomits easily especially after large feeds; poor response to asthma medications Foreign body aspiration Episode of abrupt, severe cough and/or stridor during eating or play; recurrent chest infections and cough; focal lung signs Pertussis Protracted paroxysms of coughing, often with stridor and vomiting Persistent bacterial bronchitis Persistent wet cough; poor response to asthma medications Tracheomalacia Noisy breathing when crying or eating, or during upper airway infections (noisy inspiration if extrathoracic or expiration if intrathoracic); harsh cough; inspiratory or expiratory retraction; symptoms often present since birth; poor response to asthma medications Tuberculosis Persistent noisy respirations and cough; fever unresponsive to normal antibiotics; enlarged lymph nodes; poor response to bronchodilators or inhaled corticosteroids; contact with someone who has tuberculosis Congenital heart disease Cardiac murmur; cyanosis when eating; failure to thrive; tachycardia; tachypnea or hepatomegaly; poor response to asthma medications Cystic fibrosis Cough starting shortly after birth; recurrent chest infections; failure to thrive (malabsorption); loose greasy bulky stools COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 184 Primary ciliary dyskinesia Cough and recurrent chest infections; neonatal respiratory distress, chronic ear infections and persistent nasal discharge from birth; poor response to asthma medications; situs inversus occurs in about 50% of children with this condition Vascular ring Persistently noisy breathing; poor response to asthma medications Bronchopulmonary dysplasia Infant born prematurely; very low birth weight; needed prolonged mechanical ventilation or supplemental oxygen; difficulty with breathing present from birth Immune deficiency Recurrent fever and infections (including non-respiratory); failure to thrive See list of abbreviations (p.11). Box 10-5. Key indications for referral of a child 5 years or younger for expert advice Any of the following features in a child 5 years or younger suggest an alternative diagnosis and indicate the need for further investigations: • Failure to thrive • Neonatal or very early onset of symptoms (especially if associated with failure to thrive) • Vomiting associated with respiratory symptoms • Continuous wheezing • Failure to respond to asthma medications (inhaled ICS, oral steroids or SABA) • No association of symptoms with typical triggers, such as viral URTI • Focal lung or cardiovascular signs, or finger clubbing • Hypoxemia outside context of viral illness.
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 185 11. Assessment and management of asthma in children 5 years and younger KEY POINTS • The goals of asthma management in young children are similar to those in older patients: o To achieve best possible control of symptoms and maintain normal activity levels o To minimize the risk of asthma flare-ups, impaired lung development and medication side-effects.
• Wheezing episodes in young children should be treated initially with inhaled short-acting beta2 agonist (SABA), regardless of whether the diagnosis of asthma has been made. However, for initial episodes of wheeze in children <1 year in the setting of infectious bronchiolitis, SABAs are generally ineffective.
• A trial of low-dose inhaled corticosteroid (ICS) treatment should be given if the symptom pattern suggests asthma, alternative diagnoses have been excluded and respiratory symptoms are uncontrolled and/or wheezing episodes are frequent or severe.
• Response to treatment should be reviewed before deciding whether to continue it. If the response is absent or incomplete, reconsider alternative diagnoses.
• The choice of inhaler device should be based on the child’s age and capability. The preferred device is a pressurized metered-dose inhaler (pMDI) and spacer, with face mask for <3 years and mouthpiece for most children aged 3–5 years. Children should be switched from a face mask to mouthpiece as soon as they are able to demonstrate good technique.
• Review the need for asthma treatment frequently, as asthma-like symptoms remit in many young children. Advise parents/caregivers that asthma symptoms will often return later in life.
GOAL OF ASTHMA MANAGEMENT As with other age groups, the goal of asthma management in young children is to achieve the best possible long-term asthma outcomes for the child: • To achieve and maintain good long-term control of symptoms and maintain normal activity levels • To minimize future risk; that is to reduce the risk of flare-ups, maintain lung function and lung development as close to normal as possible, and minimize medication side-effects.
Maintaining normal activity levels is particularly important in young children because engaging in play is important for their normal social and physical development. Avoiding flare-ups is important not only because of the health concerns, but also because of the disruption they cause to social and educational progress. It is important to also elicit the goals of the parent/caregiver, as these may differ from conventional medical goals. The long-term goals of asthma management are achieved through a partnership between the parent/caregiver and the health professional team, with a cycle of: • Assess (diagnosis, symptom control, risk factors, inhaler technique, adherence, parent preference) • Adjust treatment (medications, non-pharmacological strategies, and treatment of modifiable risk factors) • Review response including medication effectiveness and side-effects.
COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 186 This is carried out in combination with education of parent/caregiver, and child (depending on the child’s age): • Skills training for effective use of inhaler devices and encouragement of good adherence • Monitoring of symptoms by parent/caregiver • A written personalized asthma action plan.
ASSESSMENT OF ASTHMA What does ‘asthma control’ mean? Asthma control means the extent to which the manifestations of asthma are controlled, with or without treatment.38,84 It has two components (Box 11-1, p.188): the frequency and severity of symptoms (symptom control), and how asthma may affect them in the future, for example exacerbations in the next 12 months (future risk). In young children, as in older patients, both symptom control and future risk should be monitored (Evidence D). The rationale for this is described on p.41.
Assessing asthma symptom control Defining satisfactory symptom control in children 5 years and younger depends on information derived from family members and careers, who may be unaware either of how often the child has experienced asthma symptoms, or that their respiratory symptoms represent uncontrolled asthma. Few objective measures to assess symptom control have been validated for children <4 years. The Childhood Asthma Control Test can be used for children aged 4–11 years.142 The Test for Respiratory and Asthma Control in Kids (TRACK) is a validated questionnaire for parent/caregiver completion for preschool-aged children with symptoms consistent with asthma; it includes both symptom control and courses of systemic corticosteroids in the previous year.146 However, children with no interval symptoms can still be at risk of exacerbations. Box 6-5 shows a working schema for assessing asthma control in children ≤5 years, based on current expert opinion. It incorporates assessment of symptoms; the child’s level of activity and their need for reliever/rescue treatment; and assessment of risk factors for adverse outcomes (Evidence D). There are no validated tools for assessing symptom control over longer periods than 1–4 weeks, but ask the parent/caregiver whether the child’s recent status is usual for them. REMISSION OF CHILDHOOD WHEEZING AND ASTHMA Remission of asthma has been investigated extensively in the past, most commonly remission of childhood asthma off treatment. Definitions and criteria vary, but they commonly refer to either clinical remission (e.g., no asthma symptoms or exacerbations for a specific period) or complete (or pathophysiological) remission (e.g., also including normal lung function, airway responsiveness and/or inflammatory markers). There has been interest in remission off treatment, and remission on treatment, for example with biologic therapy for severe asthma.204-206 The concept of clinical remission on treatment is consistent with the long-term goal of asthma management promoted by GINA (see p.50), to achieve the best possible asthma outcomes for the patient. This includes control of symptoms (long-term, not just in recent days/weeks), unimpaired physical activity, improved or stable optimized lung function, prevention of exacerbations (particularly those requiring OCS), avoidance of maintenance OCS, prevention of asthma deaths, and avoidance of adverse effects of asthma medications. Reported rates of remission off treatment from studies in children with wheezing or asthma vary depending on the populations, definitions, and length of follow-up. For example, in one study, 59% of wheezing preschool children had no wheezing at 6 years,207 whereas in another study, only 15% of children with persistent wheezing at/after 9 years had no wheezing at 26 years.208 Clinical remission is more frequent than pathophysiological remission at all ages.209,210 The most important predictors of asthma remission during school years in children with childhood wheezing are fewer, milder or decreasing frequency of symptomatic episodes,211-214 good or improving lung function, and less airway COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 187 hyperresponsiveness.210 Risk factors for persistence of childhood asthma include atopy, parental asthma/allergy, later onset of symptoms, wheezing without colds, and maternal smoking or tobacco smoke exposure. Remission is not cure: asthma often recurs later in life, and children whose asthma has remitted have an increased risk of accelerated lung decline in adulthood, independent from, but synergistic with, tobacco smoking; and they may develop persistent airflow limitation, although this is less likely than for those whose asthma has persisted.215 This suggests the importance of monitoring lung function in people with remission of asthma symptoms. To date, there is no evidence that interventions in childhood increase the likelihood of remission of asthma or reduce the risk of recurrence. However, treatment of asthma in childhood with ICS substantially reduces the burden of asthma on the child and family, reduces absence from school and social events, reduces the risk of exacerbations and hospitalizations, and allows the child to participate in normal physical activity. Parents/caregivers often ask if their child will grow out of their asthma, and will not need treatment in the future. Current consensus advice for discussions like these includes: • If the child has no reported symptoms, check for evidence of ongoing disease activity, e.g., wheezing; child avoiding physical activity; lung function if available • Use language such as ‘asthma has gone quiet for the present’ to help avoid misunderstandings. If you use the term ‘remission’ with parents/caregivers, explain the medical meaning, because it is often interpreted as meaning a permanent cure • Advise parents/caregivers that even if the child’s symptoms resolve completely, their asthma may recur later • Emphasize the benefits of taking controller treatment for the child’s current health, their risk of asthma attacks, and their ability to participate in school and sporting activities, avoiding claims about effect of therapy on future asthma outcomes.
Research needs: clinical questions about remission off treatment in children focus on risk factors for asthma persistence and recurrence (including clinical, pathological ,and genetic factors), the effect of risk reduction strategies on the likelihood of remission, whether monitoring after remission to allow early identification of asthma recurrence improves outcomes, and whether progression to persistent airflow limitation can be prevented. Clinical questions about remission on treatment (e.g., in children with severe asthma treated with biologic therapy) include whether inhaled anti-inflammatory therapy can be down-titrated. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 188 Box 11-1. GINA assessment of asthma control in children 5 years and younger A. Recent symptom control (also ask about whole period since last visit) Level of asthma symptom control In the past 4 weeks, has the child had: Well controlled Partly controlled Uncontrolled Daytime asthma symptoms for more than a few minutes, Yes No more than once a week? None of these 1–2 of these 3–4 of these Any activity limitation due to asthma? (Runs/plays less Yes No than other children, tires easily during walks/playing?) SABA reliever medication needed more than once a week? Yes No Any night waking or night coughing due to asthma? Yes No B. Future risk for poor asthma outcomes Risk factors for asthma exacerbations within the next few months • Uncontrolled asthma symptoms • One or more severe exacerbations (ED attendance, hospitalization, or course of OCS) in previous year • The start of the child’s usual ‘flare-up’ season (especially if autumn/fall) • Exposures: tobacco smoke; indoor or outdoor air pollution; indoor allergens (e.g., house dust mite, cockroach, pets, mold), especially in combination with viral infection823 • Major psychological or socio-economic problems for child or family • Poor adherence with ICS medication, or incorrect inhaler technique • Outdoor pollution (NO2 and particles)101 Risk factors for persistent airflow limitation • Severe asthma with several hospitalizations • History of bronchiolitis Risk factors for medication side-effects • Systemic: Frequent courses of OCS, high-dose and/or potent ICS (for low ICS doses, see Box 11-3, p.191) • Local: moderate-to high-dose or potent ICS; incorrect inhaler technique; failure to protect skin or eyes when using ICS by nebulizer or spacer with face mask See list of abbreviations (p.11). Excludes reliever taken before exercise. Before stepping up treatment, ensure that the child’s symptoms are due to asthma, and that the child has good inhaler technique and good adherence to existing treatment. Assessing future risk of adverse outcomes The relationship between symptom control and future risk of adverse outcomes, such as exacerbations (Box 6-5, p.182), has not been sufficiently studied in young children. Although exacerbations may occur in children after months of apparently good symptom control, the risk is greater if current symptom control is poor. Preschool children at high risk of asthma (based on modified API) who were treated with daily low-dose ICS experienced fewer days with asthma symptoms and a reduced risk of exacerbations than those receiving placebo.824 The future risk of harm due to excessive doses of inhaled or systemic corticosteroids must also be avoided. This can be minimized by ensuring that the prescribed treatment is appropriate and reduced to the lowest dose that maintains satisfactory symptom control and minimizes exacerbations. The child’s height should be measured and recorded at least yearly, as growth velocity may be lower in the first 1–2 years of ICS treatment,141 and poorly controlled asthma can affect COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 189 growth.140 The minimum effective dose of ICS to maintain good asthma control should be used. If decreased growth velocity is seen, other factors should be considered, including poorly controlled asthma, frequent use of oral corticosteroids (OCS), and poor nutrition, and referral should be considered. If ICS is delivered through a face-mask or nebulizer, the skin on the nose and around the mouth should be cleaned shortly after inhalation to avoid local side-effects such as steroid rash (reddening and atrophy). MEDICATIONS FOR SYMPTOM CONTROL AND RISK REDUCTION Choosing medications for children 5 years and younger Good control of asthma can be achieved in the overwhelming majority of young children with a pharmacological intervention strategy.825 This should be developed in a partnership between the family/career and the healthcare provider. As with older children and adults, medications comprise only one component of asthma management in young children; other key components include education, skills training for inhaler devices and adherence, non-pharmacological strategies including environmental control where appropriate, regular monitoring, and clinical review (see later sections in this chapter). When recommending treatment for a young child, both general and individual questions apply (Box 3-4, p.54): • What is the ‘preferred’ medication option at each treatment step to control asthma symptoms and minimize future risk? These decisions are based on data for efficacy, effectiveness and safety from clinical trials, and on observational data. Studies suggest that consideration of factors such as allergic sensitization and/or peripheral blood count may help to better identify which children are more likely to have a short-term response to ICS.826 However, further studies are needed to assess the applicability of these findings in a wider range of settings, particularly in areas where blood eosinophilia may reflect helminth infection rather than asthma or atopy. • How does this individual child differ from other children with asthma, in terms of: - Response to previous treatment - Patient characteristics that contribute to symptoms or risk of flare-ups: e.g., clinical phenotype, risk factors for flare-ups, comorbidities including allergic rhinitis, environmental exposures - Preferences of the parent/caregiver (goals, beliefs and concerns about medications) - Practical issues (cost, inhaler technique and adherence)? The following treatment recommendations for children of 5 years of age or younger are based on the available evidence and on expert opinion. Although the evidence is expanding it is still rather limited as most clinical trials in this age group have not characterized participants with respect to their symptom pattern, and different studies have used different outcomes and different definitions of exacerbations. A stepwise treatment approach is recommended (Box 11-2, p.190), based on symptom patterns, risk of exacerbations and side-effects, and response to initial treatment. Generally, treatment includes the daily, long-term use of low-dose ICS treatment to keep asthma well controlled (see Box 11-3 for doses), and reliever medications for as-needed symptom relief. The choice of inhaler device is also an important consideration (Box 11-4, p.191). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 190 Box 11-2. Personalized management of asthma in children 5 years and younger See list of abbreviations (p.11). For ICS doses in children, see Box 11-3 (p.191) †If prescribing LTRA, advise parent/caregiver about risk of neuropsychiatric adverse effects. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 191 Box 11-3. Low daily doses of inhaled corticosteroids for children 5 years and younger This is not a table of equivalence, but instead, suggestions for ‘low’ total daily doses for the ICS treatment recommendations for children aged 5 years and younger in Box 11-2 (p.190), based on available studies and product information. Data on comparative potency are not readily available, particularly for children. This table does NOT imply potency equivalence. For example, if you switch a child’s treatment from a ‘low’ dose of one ICS to a ‘low’ dose of another ICS, this may represent a decrease (or increase) in potency. The child’s asthma may become unstable (or they may be at increased risk of adverse effects). Children should be monitored to ensure stability after any change of treatment. Doses and potency may also differ by country, depending on local products, inhaler devices, regulatory labelling and clinical guidelines. The doses listed here are the lowest approved doses for which safety and effectiveness have been adequately studied in this age group. Low-dose ICS provides most of the clinical benefit for most children with asthma. Higher doses are associated with an increased risk of local and systemic side-effects, which must be balanced against potential benefits. Inhaled corticosteroid Low total daily dose in mcg (age-group with adequate safety and effectiveness data) BDP (pMDI, standard particle, HFA) 100 (ages 5 years and older) BDP (pMDI, extrafine particle, HFA) 50 (ages 5 years and older) Budesonide nebulized 500 (ages 1 year and older) Fluticasone propionate (pMDI, standard particle, HFA) 50 (ages 4 years and older) Fluticasone furoate (DPI) Not sufficiently studied in children 5 years and younger Mometasone furoate (pMDI, standard particle, HFA) 100 (ages 5 years and older) Ciclesonide (pMDI, extrafine particle, HFA) Not sufficiently studied in children 5 years and younger BDP : beclometasone dipropionate. For other abbreviations see p.11. In children, pMDI should always be used with a spacer Box 11-4. Choosing an inhaler device for children 5 years and younger Age Preferred device Alternate device 0–3 years Pressurized metered-dose inhaler plus dedicated spacer with face mask Nebulizer with face mask 4–5 years Pressurized metered-dose inhaler plus dedicated spacer with mouthpiece Pressurized metered-dose inhaler plus dedicated spacer with face mask or nebulizer with mouthpiece or face mask See list of abbreviations (p.11). If nebulizer is used, follow infection control procedures, as respiratory viruses can be dispersed by up to 1 meter. See p.109 and Box 5-1 (p.109) for other factors to consider in choice of an inhaler device. Which children should be prescribed regular controller treatment? Intermittent or episodic wheezing of any severity may represent an isolated viral-induced wheezing episode, an episode of seasonal or allergen-induced asthma, or unrecognized uncontrolled asthma. The initial treatment of wheezing is identical for all of these – a SABA every 4–6 hours as needed until symptoms disappear, usually within 1 to 7 days. Further treatment of the acute wheezing episodes themselves is described below (see Acute asthma exacerbations in children 5 years and younger, p.200). However, uncertainty surrounds the addition of other medications in these children, especially when the nature of the episode is unclear. In general, the following principles apply: COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 192 • If the history and symptom pattern suggest a diagnosis of asthma (Box 10-2, p.180; Box 10-3, p.180) and respiratory symptoms are uncontrolled (Box 11-1, p.188) and/or wheezing episodes are frequent (e.g., three or more episodes in a season), regular controller treatment (usually maintenance low-dose ICS) should be initiated (Step 2, Box 11-2, p.190) and the response evaluated (Evidence D). Regular ICS treatment may also be indicated in a child with less frequent, but more severe episodes of viral- induced wheeze (Evidence D). • If the diagnosis of asthma is in doubt, and inhaled SABA therapy or courses of antibiotics need to be repeated frequently, e.g., more than every 6–8 weeks, a trial of regular ICS treatment should be considered to confirm whether the symptoms are due to asthma (Evidence D). Referral for specialist opinion should also be considered at this stage. It is important to discuss the decision to prescribe controller treatment and the choice of treatment with the child’s parents or caregivers. They should be aware of both the relative benefits and risks of the treatments, and the importance of maintaining normal activity levels for their child’s normal physical and social development. Although effects of ICS on growth velocity are seen in pre-pubertal children in the first 1–2 years of treatment, this is not progressive or cumulative, and the one study that examined long-term outcomes showed a difference of only 0.7% in adult height.141,827 Poorly controlled asthma itself adversely affects adult height.140 Treatment steps to control asthma symptoms and minimize future risk for children 5 years and younger Asthma treatment in young children follows a stepwise approach (Box 11-2), with medication adjusted up or down to achieve good symptom control and minimize future risk of exacerbations and medication side-effects. The need for controller treatment should be re-assessed regularly. Before considering a step-up of controller treatment If symptom control is poor and/or exacerbations persist despite 3 months of adequate controller therapy, check the following before any step up in treatment is considered: • Confirm that the symptoms are due to asthma rather than a concomitant or alternative condition (Box 10-4, p.183). Refer for expert assessment if the diagnosis is in doubt. • Check and correct inhaler technique. • Confirm good adherence with the prescribed dose. • Consider trial of one of the other treatment options for that step, as many children may respond to one of the options. • Enquire about risk factors such as allergen or tobacco smoke exposure (Box 11-1, p.188). ASTHMA TREATMENT STEPS FOR CHILDREN AGED 5 YEARS AND YOUNGER Step 1: Preferred option: as-needed inhaled short-acting beta2 agonist (SABA) All children who experience wheezing episodes should be provided with inhaled SABA for relief of symptoms (Evidence D), although it is not effective in all children. See Box 11-4 (p.191) for choice of inhaler device. Use of SABA for the relief of symptoms on average more than twice a week over a 1-month period indicates the need for a trial of low-dose ICS treatment. Initial episodes of wheeze in children <1 year often occur in the setting of infectious bronchiolitis, and this should be managed according to local bronchiolitis guidelines. SABAs are generally ineffective for bronchiolitis.828 Other options Oral bronchodilator therapy is not recommended due to its slower onset of action and higher rate of side-effects, compared with inhaled SABA (Evidence D). For children with intermittent viral-induced wheeze and no interval symptoms, particularly those with underlying atopy (positive for modified API) in whom inhaled SABA medication is not sufficient, intermittent high-dose ICS may be considered723,829,830 (see Management of worsening asthma and exacerbations, p.158), but because of the risk of side-effects, this should only be considered if the physician is confident that the treatment will be used appropriately. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 193 Step 2: Preferred option: regular daily low-dose ICS plus as-needed SABA Regular daily, low-dose ICS (Box 11-3, p.191) is recommended as the preferred initial treatment to control asthma in children 5 years and younger (Evidence A).831-833 This initial treatment should be given for at least 3 months to establish its effectiveness in achieving good asthma control. Other options In young children with persistent asthma, regular treatment with a leukotriene receptor antagonist (LTRA) modestly reduces symptoms and need for oral corticosteroids compared with placebo.834 However, for young children with recurrent viral- induced wheezing, a review concluded that neither regular nor intermittent LTRA reduces OCS requiring exacerbations (Evidence A).835 A further systematic review found that in preschool children with asthma or recurrent wheezing, daily ICS was more effective in improving symptom control and reducing exacerbations than regular LTRA monotherapy.836 Parents/caregivers should be counselled about the potential adverse effects of montelukast on sleep and behavior, and health professionals should consider the benefits and risks of side effects before prescribing.295 For preschool children with asthma characterized by frequent viral-induced wheezing and interval asthma symptoms, as-needed (prn)837 or episodic ICS838 may be considered, but a trial of regular daily low-dose ICS should be undertaken first. The effect on exacerbation risk seems similar for regular daily low-dose and episodic high-dose ICS.833 See also Initial home management of asthma exacerbations (p.197). If good asthma control is not achieved with a given therapy, trials of the alternative Step 2 therapies are recommended prior to moving to Step 3.826 Step 3: Double the ‘low’ daily ICS dose plus as-needed SABA. Consider specialist referral If 3 months of initial therapy with a low-dose ICS fails to control symptoms, or if exacerbations continue to occur, check the following before any step up in treatment is considered: • Confirm that the symptoms are due to asthma rather than a concomitant or alternative condition (Box 10-4, p.183). • Check and correct inhaler technique. Consider alternative delivery systems if indicated. • Confirm good adherence with the prescribed dose. • Enquire about risk factors, such as exposure to allergens or tobacco smoke (Box 11-1, p.188). Preferred option: medium-dose ICS (double the ‘low’ daily dose) Doubling the initial low dose of ICS may be the best option (Evidence C). Assess response after 3 months. The child should be referred for expert assessment if symptom control remains poor and/or flare-ups persist, or if side-effects of treatment are observed or suspected. Other options Addition of a LTRA to low-dose ICS may be considered, based on data from older children (Evidence D). The relative cost of different treatment options in some countries may be relevant to controller choices for children. Note the concern about potential neuropsychiatric adverse effects with montelukast.295 Not recommended There are insufficient data about the efficacy and safety of ICS in combination with a long-acting beta2 agonist (LABA) in children <4 years old to recommend their use. A short-term (8 week) placebo-controlled study did not show any significant difference in symptoms between combination fluticasone propionate-salmeterol versus fluticasone propionate alone; no additional safety signals were noted in the group receiving LABA.839 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 194 Step 4: Continue controller treatment and refer for expert assessment If the Step 3 approach of doubling the initial dose of ICS fails to achieve and maintain good asthma control, carefully reassess inhaler technique and medication adherence as these are common problems in this age group. In addition, reassess and address control of environmental factors where relevant, and reconsider the asthma diagnosis. Other options The best treatment for this population has not been established. If the diagnosis of asthma has been confirmed, options to consider, with specialist advice, are: • Further increase the dose of ICS for a few weeks until the control of the child’s asthma improves (Evidence D). Monitor for side-effects. • Add LTRA (data based on studies in older children, Evidence D). Benefits, and risks of side effects, should be considered, as described previously; inform the parent/caregiver about the potential risk of neuropsychiatric adverse effects.572 • Add long-acting beta agonist (LABA) in combination with ICS; data based on studies in children ≥4 years of age. • Add a low dose of oral corticosteroid (for a few weeks only) until asthma control improves (Evidence D); monitor for side-effects. • Add intermittent high-dose ICS at onset of respiratory illnesses to the regular daily ICS if exacerbations are the main problem (Evidence D). The need for additional controller treatment should be re-evaluated at each visit and maintained for as short a period as possible, with consideration of potential risks and benefits. Treatment goals and their feasibility should be reconsidered and discussed with the child’s family/career. REVIEWING RESPONSE AND ADJUSTING TREATMENT Assessment at every visit should include asthma symptom control and risk factors (Box 11-1, p.188), and side-effects. The child’s height should be measured every year, or more often. Asthma-like symptoms remit in a substantial proportion of children of 5 years or younger,840-842 so the need for continued controller treatment should be regularly assessed (e.g., every 3–6 months) (Evidence D). If therapy is stepped-down or discontinued, schedule a follow-up visit 3–6 weeks later to check whether symptoms have recurred, as therapy may need to be stepped-up or reinstituted (Evidence D). Marked seasonal variations may be seen in symptoms and exacerbations in this age-group. For children with seasonal symptoms whose daily long-term controller treatment is to be discontinued (e.g., 4 weeks after their season ends), the parent/caregiver should be provided with a written asthma action plan detailing specific signs of worsening asthma, the medications that should be initiated to treat it, and when and how to contact medical care. CHOICE OF INHALER DEVICE Inhaled therapy constitutes the cornerstone of asthma treatment in children 5 years and younger. General information about inhaler devices, and the issues that should be considered, are found in Section 5 (p.108) and in Box 5-1 (p.109). These include, first, choosing the right medication(s) for the child to control symptoms, allow normal activity, and reduce the risk of severe exacerbations; considering which delivery device is available; whether they can use it correctly after training; and, if more than one type of inhaler device is available, their relative environmental impact. For children aged 5 years and younger, the preferred delivery system is a pressurized metered-dose inhaler (pMDI) with a valved spacer (Box 11-4, p.191), with or without a face mask, depending on the child’s age (Evidence A).843 This recommendation is based on studies with beta2 agonists. The spacer device should have documented efficacy in young children. The dose delivered may vary considerably between spacers, so consider this if changing from one spacer to another. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 195 The only possible inhalation technique in young children is tidal breathing. The optimal number of breaths required to empty the spacer depends on the child’s tidal volume, and the dead space and volume of the spacer. Generally, 5–10 breaths will be sufficient per actuation. The way a spacer is used can markedly affect the amount of drug delivered: • Spacer size may affect the amount of drug available for inhalation in a complex way depending on the drug prescribed and the pMDI used. Young children can use spacers of all sizes, but theoretically a lower volume spacer (<350 mL) is advantageous in very young children. • A single pMDI actuation should be delivered at a time, with the inhaler shaken in between. Multiple actuations into the spacer before inhalation may markedly reduce the amount of drug inhaled. • Delay between actuating the pMDI into the spacer and inhalation may reduce the amount of drug available. This varies between spacers, but to maximize drug delivery, inhalation should start as soon as possible after actuation. If a healthcare provider or a carer is giving the medication to the child, they should actuate the pMDI only when the child is ready and the spacer is in the child’s mouth. • If a face mask is used it must be fitted tightly around the child’s mouth and nose, to avoid loss of drug. • Ensure that the valve is moving while the child is breathing through the spacer. • Static charge may accumulate on some plastic spacers, attracting drug particles and reducing lung delivery. This charge can be reduced by washing the spacer with detergent (without rinsing) and allowing it to air dry, but it may re-accumulate over time. Spacers made of anti-static materials or metals are less subject to this problem. If a patient or healthcare provider carries a new plastic spacer for emergency use, it should be regularly washed with detergent (e.g., monthly) to reduce static charge. • Nebulizers, the only viable alternative delivery systems in children, are reserved for the minority of children who cannot be taught effective use of a spacer device. If a nebulizer is used for delivery of ICS, it should be used with a mouthpiece to avoid the medication reaching the eyes. If a nebulizer is used, follow local infection control procedures. ASTHMA SELF-MANAGEMENT EDUCATION FOR CARERS OF YOUNG CHILDREN Asthma self-management education should be provided to family members and carers of wheezy children 5 years and younger when wheeze is suspected to be caused by asthma. An educational program should contain: • A basic explanation about asthma and the factors that influence it • Training about correct inhalation technique • Information on the importance of the child’s adherence to the prescribed medication regimen • A written asthma action plan. Crucial factors for a successful asthma education program include a partnership between patient/carer and healthcare providers, with a high level of agreement regarding the goals of treatment for the child, and intensive follow-up (Evidence D).39 Written asthma action plans Asthma action plans should be provided for the family/carers of all children with asthma, including those aged 5 years and younger (Evidence D). Action plans, developed through collaboration between an asthma educator, the healthcare provider and the family, have been shown to be of value in older children,844 although they have not been extensively studied in children of 5 years and younger. A written asthma action plan includes: • A description of how the parent or caregiver can recognize when symptom control is deteriorating • The medications to administer • When and how to obtain medical care, including telephone numbers of services available for emergencies (e.g., doctors’ offices, emergency departments and hospitals, ambulance services and emergency pharmacies). Details of treatments that can be initiated at home are provided in Section 12. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 196 12. Management of worsening asthma and exacerbations in children 5 years and younger KEY POINTS Symptoms of exacerbation in young children • Early symptoms of exacerbations in young children may include increased symptoms; increased coughing, especially at night; lethargy or reduced exercise tolerance; impaired daily activities including feeding; and a poor response to reliever medication. Home management in a written asthma action plan • Give a written asthma action plan to parents/caregivers of young children with asthma so they can recognize an impending severe attack, start treatment, and identify when urgent hospital treatment is required. • Initial treatment at home is with inhaled short-acting beta2 agonist (SABA), with review after 1 hour or earlier. • Parents/caregivers should seek urgent medical care if the child is acutely distressed, lethargic, fails to respond to initial bronchodilator therapy, or is worsening, especially in children <1 year of age. • Medical attention should be sought on the same day if inhaled SABA is needed more often than 3-hourly or for more than 24 hours. • There is no compelling evidence to support parent/caregiver-initiated oral corticosteroids. Management of exacerbations in primary care or acute care facility • Assess severity of the exacerbation while initiating treatment with SABA (2–6 puffs every 20 minutes for first hour) and oxygen (to maintain saturation 94–98%). • Recommend immediate transfer to hospital if there is no response to inhaled SABA within 1–2 hours; if the child is unable to speak or drink, has a respiratory rate >40/minute or is cyanosed, if resources are lacking in the home, or if oxygen saturation is <92% on room air. • Consider oral prednisone/prednisolone 1–2 mg/kg/day for children attending an Emergency Department (ED) or admitted to hospital, up to a maximum of 20 mg/day for children aged 0–2 years, and 30 mg/day for children aged 3–5 years, for up to 5 days; or dexamethasone 0.6 mg/kg/day for 2 days. If there is failure of resolution, or relapse of symptoms with dexamethasone, consideration should be given to switching to prednisolone. • Be aware that oxygen saturation by pulse oximetry may be overestimated in people with dark skin color. Arrange early follow-up after an exacerbation • Children who have experienced an asthma exacerbation are at risk of further exacerbations. Arrange follow-up within 1–2 days of an exacerbation and again 1–2 months later to plan ongoing asthma management. DIAGNOSIS OF EXACERBATIONS A flare-up or exacerbation of asthma in children 5 years and younger is defined as an acute or sub-acute deterioration in symptom control that is sufficient to cause distress or risk to health, and necessitates a visit to a healthcare provider or requires treatment with systemic corticosteroids. In pediatric literature, the term ‘episode’ is commonly used, but understanding of this term by parents/caregivers is not known. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 197 Early symptoms of an exacerbation may include any of the following: • Onset of symptoms of respiratory tract infection • An acute or sub-acute increase in wheeze and shortness of breath • An increase in coughing, especially while the child is asleep • Lethargy or reduced exercise tolerance • Impairment of daily activities, including feeding • A poor response to reliever medication. In a study of children aged 2–5 years, the combination of increased daytime cough, daytime wheeze, and night-time beta2 agonist use was a strong predictor at a group level of an imminent exacerbation (1 day later). This combination predicted around 70% of exacerbations, with a low false positive rate of 14%. In contrast, no individual symptom was predictive of an imminent asthma exacerbation.845 Upper respiratory symptoms frequently precede the onset of an asthma exacerbation, indicating the important role of viral URTI in precipitating exacerbations in many, although not all, children with asthma. In a randomized controlled trial of acetaminophen versus ibuprofen, given for pain or fever in children with mild persistent asthma, there was no evidence of a difference in the subsequent risk of flare-ups or poor symptom control.826 INITIAL HOME MANAGEMENT OF ASTHMA EXACERBATIONS Initial management includes an action plan to enable the child’s family members and carers to recognize worsening asthma and initiate treatment, recognize when it is severe, identify when urgent hospital treatment is necessary, and provide recommendations for follow up (Evidence D). The action plan should include specific information about medications and dosages and when and how to access medical care. Need for urgent medical attention Parents/caregivers should be advised to seek medical attention immediately if: • The child is acutely distressed • The child’s symptoms are not relieved promptly by inhaled bronchodilator • The period of relief after doses of SABA becomes progressively shorter • A child younger than 1 year requires repeated inhaled SABA over several hours. Initial treatment at home Inhaled SABA via a mask or spacer, and review response The parent/caregiver should initiate treatment with two puffs of inhaled SABA (200 mcg salbutamol [albuterol] or equivalent), given one puff at a time via a spacer device with or without a facemask (Evidence D). This may be repeated a further two times at 20-minute intervals, if needed. The child should be observed by the family/carer and, if improving, maintained in a restful and reassuring atmosphere for an hour or more. Medical attention should be sought urgently if any of the features listed above apply; or on the same day if more than 6 puffs of inhaled SABA are required for symptom relief within the first 2 hours, or if the child has not recovered after 24 hours. Family/carer-initiated corticosteroids Evidence to support the initiation of oral corticosteroid (OCS) treatment by family/carers in the home management of asthma exacerbations in children is weak,846-850 despite this practice in some regions. Preemptive episodic high-dose nebulized ICS may reduce exacerbations in children with intermittent viral triggered wheezing.833 However, because of the high potential for side-effects, especially if the treatment is continued inappropriately or is given frequently, family-administered high-dose ICS should be considered only where the healthcare provider is confident that the medications will be used appropriately, and the child is closely monitored for side-effects. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 198 Leukotriene receptor antagonists In children aged 2–5 years with intermittent viral wheezing, one study found that a short course of an oral LTRA (for 7– 20 days, commenced at the start of an URTI or the first sign of asthma symptoms) reduced symptoms, healthcare utilization and time off work for the carer.851 In contrast another study found no significant effect with LTRA, compared with placebo, on episode-free days (primary outcome), OCS use, healthcare utilization, quality of life or hospitalization in children with or without a positive Asthma Predictive Index (API). However, activity limitation and a symptom trouble score were significantly improved, particularly in children with a positive API.852 Parents/caregivers should be counseled about the risk of adverse effects on sleep, behavior and mental health with montelukast.295 PRIMARY CARE OR HOSPITAL MANAGEMENT OF ACUTE ASTHMA EXACERBATIONS IN CHILDREN 5 YEARS OR YOUNGER Assessment of exacerbation severity Conduct a brief history and examination concurrently with the initiation of therapy (Box 12-1, p.199). The presence of any of the features of a severe exacerbation listed in Box 12-2 are an indication of the need for urgent treatment and immediate transfer to hospital (Evidence D). Oxygen saturation from pulse oximetry of <92% on presentation (before oxygen or bronchodilator treatment) is associated with high morbidity and likely need for hospitalization; saturation of 92– 95% is also associated with higher risk.751 Note that oxygen saturation by pulse oximetry may be overestimated in people with dark skin color.728 Agitation, drowsiness and confusion are features of cerebral hypoxemia. A quiet chest on auscultation indicates minimal ventilation, insufficient to produce a wheeze. Several clinical scoring systems such as PRAM (Preschool Respiratory Assessment Measure) and PASS (Pediatric Asthma Severity Score) have been developed for assessing the severity of acute asthma exacerbations in children.853 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 199 Box 12-1. Management of acute asthma or wheezing in children 5 years and younger See list of abbreviations (p.11). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 200 Box 12-2. Initial assessment of acute asthma exacerbations in children 5 years and younger Symptoms Mild Severe Altered consciousness No Agitated, confused or drowsy Oximetry on presentation (SaO2) >95% <92% Speech† Sentences Words Pulse rate <100 beats/minute >180 beats/minute (0–3 years) >150 beats/minute (4–5 years) Respiratory rate ≤40/minute >40/minute Central cyanosis Absent Likely to be present Wheeze intensity Variable Chest may be quiet See list of abbreviations (p.11). Any of these features indicates a severe asthma exacerbation. Oximetry before treatment with oxygen or bronchodilator. Note potential for overestimation of oxygen saturation with pulse oximetry in people with dark skin color.728 † The child’s developmental stage and usual capability must be considered.
Indications for immediate transfer to hospital Children with features of a severe exacerbation that fail to resolve within 1–2 hours despite repeated dosing with inhaled SABA must be referred to hospital for observation and further treatment (Evidence D; Box 12-3). Other indications are respiratory arrest or impending arrest; lack of supervision in the home or doctor’s office; and recurrence of signs of a severe exacerbation within 48 hours (particularly if treatment with OCS has already been given). In addition, early medical attention should be sought for children with a history of severe life-threatening exacerbations, and those aged less than 2 years, as the risk of dehydration and respiratory fatigue is increased (Box 12-4, p.201). Box 12-3. Indications for immediate transfer to hospital for children 5 years and younger Immediate transfer to hospital is indicated if a child ≤5 years with asthma has ANY of the following: • At initial or subsequent assessment: o Child is unable to speak or drink o Cyanosis o Respiratory rate >40 per minute o Oxygen saturation <92% when breathing room air (note potential for overestimation of oxygen saturation with pulse oximetry in people with dark skin color.
o Silent chest on auscultation • Lack of response to initial bronchodilator treatment: o Lack of response to 6 puffs of inhaled salbutamol [albuterol] (2 separate puffs, repeated 3 times) over 1–2 hours o Persisting tachypnea despite three administrations of inhaled SABA, even if the child shows other clinical signs of improvement • Social environment that limits delivery of acute treatment, or parent/caregiver unable to manage acute asthma at home.
During transfer to hospital, continue to give inhaled SABA, oxygen (if available) to maintain saturation 94–98%, and give systemic corticosteroids (see Box 12-1, p.199) See list of abbreviations (p.11). Normal respiratory rates: <60 breaths/minute in children 0–2 months; <50 breaths/minute in children 2– 12 months; <40 breaths/minute in children 1–5 years. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 201 Box 12-4. Initial emergency department management of asthma exacerbations in children 5 years and younger Therapy Dose and administration Supplemental oxygen Delivered by face nasal prongs or mask, as indicated to maintain oxygen saturation at 94–98% Short-acting beta2 agonist (SABA) 2–6 puffs of salbutamol [albuterol] by spacer, or 2.5 mg by nebulizer, every 20 minutes for first hour, then reassess severity. If symptoms persist or recur, give an additional 2–3 puffs per hour. Admit to hospital if >10 puffs required in 3–4 hours. Systemic corticosteroids Give initial dose of oral prednisolone (1–2 mg/kg up to a maximum 20 mg for children <2 years old; 30 mg for children 2–5 years) OR, intravenous methylprednisolone 1 mg/kg 6-hourly on day 1 Additional options in the first hour of treatment Ipratropium bromide Consider adding 1–2 puffs of ipratropium bromide by pMDI and spacer For children with moderate-severe exacerbations with a poor response to initial SABA, give nebulized ipratropium bromide 250 mcg every 20 minutes for 1 hour only Magnesium sulfate Consider nebulized isotonic magnesium sulfate (150 mg) 3 doses in the first hour of treatment for children aged ≥2 years with severe exacerbation (Box 12-2, p.200) See list of abbreviations (p.11). If inhalation is not possible an intravenous bolus of terbutaline 2 mcg/kg may be given over 5 minutes, followed by continuous infusion of 5 mcg/kg/hour (Evidence C).854 The child should be closely monitored, and the dose should be adjusted according to clinical improvement and side-effects. See below for additional and ongoing treatment, including maintenance ICS. If a nebulizer is used, follow infection control procedures. Emergency treatment and initial pharmacotherapy Oxygen Treat hypoxemia urgently with oxygen by face mask to achieve and maintain percutaneous oxygen saturation 94–98% (Evidence A). Note the potential for overestimation of oxygen saturation in people with dark skin color. To avoid hypoxemia during changes in treatment, children who are acutely distressed should be treated immediately with oxygen and SABA (2.5 mg of salbutamol or equivalent diluted in 3 mL of sterile normal saline) delivered by an oxygen-driven nebulizer (if available). This treatment should not be delayed, and may be given before the full assessment is completed. Transient hypoxemia due to ventilation/perfusion mismatch may occur during treatment with SABAs. Inhaled bronchodilator therapy The initial dose of inhaled SABA may be given by a pMDI with spacer and mask or mouthpiece or an air-driven nebulizer; or, if oxygen saturation is low, by an oxygen-driven nebulizer (as described above). For most children, pMDI plus spacer is favored as it is more efficient than a nebulizer for bronchodilator delivery (Evidence A),855 and nebulizers can spread infectious particles. The initial dose of SABA is two puffs of salbutamol (100 mcg per puff) or equivalent, except in acute, severe asthma when six puffs should be given. When a nebulizer is used, a dose of 2.5 mg salbutamol solution is recommended, and infection control procedures should be followed. The frequency of dosing depends on the response observed over 1–2 hours (see below). For children with moderate-severe exacerbations and a poor response to initial SABA, nebulized ipratropium bromide may be added every 20 minutes for 1 hour only.855 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 202 Magnesium sulfate The role of magnesium sulfate is not established for children 5 years and younger, because there are few studies in this age group. Nebulized isotonic magnesium sulfate may be considered as an adjuvant to standard treatment with nebulized salbutamol and ipratropium in the first hour of treatment for children ≥2 years old with acute severe asthma (e.g., oxygen saturation <92%, Box 6-10, p.201), particularly those with symptoms lasting <6 hours.856 Intravenous magnesium sulfate in a single dose of 40–50 mg/kg (maximum 2 g) by slow infusion (20–60 minutes) has also been used.857 Assessment of response and additional bronchodilator treatment Children with a severe asthma exacerbation must be observed for at least 1 hour after initiation of treatment, at which time further treatment can be planned: • If symptoms persist after initial bronchodilator: a further 2–6 puffs of salbutamol (depending on severity) may be given 20 minutes after the first dose and repeated at 20-minute intervals for an hour. Consider adding 1–2 puffs of ipratropium. Failure to respond at 1 hour, or earlier deterioration, should prompt urgent admission to hospital, addition of nebulized ipratropium, and a short course of oral corticosteroids (Evidence D). • If symptoms have improved by 1 hour but recur within 3–4 hours: the child may be given more frequent doses of bronchodilator (2–3 puffs each hour), and oral corticosteroids should be given. The child may need to remain in the emergency department, or, if at home, should be observed by the family/carer and have ready access to emergency care. Children who fail to respond to 10 puffs of inhaled SABA within a 3–4 hour period should be referred immediately to hospital (Evidence D). • If symptoms resolve rapidly after initial bronchodilator and do not recur for 1–2 hours: no further treatment may be required. Further SABA may be given as needed up to every 3–4 hours (up to a total of 10 puffs/24 hours). If symptoms persist beyond 1 day, other treatments including inhaled and/or oral corticosteroids are indicated (Evidence D), as outlined below. Additional treatment When treatment in addition to SABA is required for an exacerbation, the options available for children aged 5 years and under include ICS, a short course of oral corticosteroid, and/or LTRA (see p.197). However, the clinical benefit of these interventions – particularly on endpoints such as hospitalizations and longer-term outcomes – has not been impressive. Parents/caregivers should be informed about the potential for neuropsychiatric adverse effects associated with LTRA. Maintain current controller treatment (if prescribed) Children who have been prescribed maintenance therapy with ICS, LTRA or both should continue to take the prescribed dose during and after an exacerbation (Evidence D). As above, parents/caregivers should be informed about the potential neuropsychiatric adverse effects associated with LTRA. Inhaled corticosteroids For children not previously on ICS, an initial dose of ICS twice the low daily dose indicated in Box 11-3 (p.191) may be given and continued for a few weeks or months (Evidence D). Some studies have used high-dose ICS (1600 mcg/day, preferably divided into four doses over the day and given for 5–10 days) as this may reduce the need for OCS.723,829,830,858,859 Addition of ICS to standard care (including OCS) does not reduce risk of hospitalization but reduces length of stay and acute asthma scores in children in the emergency department.860 However, the potential for side-effects with high-dose ICS should be considered, especially if used repeatedly, and the child should be monitored closely. For those children already on ICS, doubling the dose was not effective in a small study of mild-moderate exacerbations in children aged 6–14 years,861 nor was quintupling the dose in children aged 5–11 years with good adherence. This approach should be reserved mainly for individual cases, and should always involve regular follow-up and monitoring of adverse effects (Evidence D). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 203 Oral corticosteroids For children with severe exacerbations, a dose of OCS equivalent to prednisolone 1–2 mg/kg/day, with a maximum of 20 mg/day for children under 2 years of age and 30 mg/day for children aged 2–5 years, is currently recommended (Evidence A),862 although several studies have failed to show any benefits when given earlier (e.g., by parents or caregivers) during periods of worsening wheeze managed in an outpatient setting (Evidence D).846-849,863,864 A meta-analysis demonstrated a reduced risk of hospitalization when oral corticosteroids were administered in the emergency department, but no clear benefit in risk of hospitalization when given in the outpatient setting.865 A course of 3–5 days is sufficient in most children of this age, and can be stopped without tapering (Evidence D), but the child must be reviewed after discharge (as below) to confirm they are recovering. In children discharged from the emergency department, an intramuscular corticosteroid may be an alternative to a course of OCS for preventing relapse,760 but the risk of long-term adverse effects must be considered. There is insufficient evidence to recommend intramuscular over oral corticosteroids.760 Regardless of treatment, the severity of the child’s symptoms must be carefully monitored. The sooner therapy is started in relation to the onset of symptoms, the more likely it is that the impending exacerbation may be clinically attenuated or prevented. DISCHARGE AND FOLLOW-UP AFTER AN EXACERBATION Before discharge, the condition of the child should be stable (e.g., out of bed and able to eat and drink without problems). Children who have recently had an asthma exacerbation are at risk of further exacerbations and require follow up. The purpose is to ensure complete recovery, to establish the cause of the exacerbation, and, when necessary, to establish appropriate maintenance treatment and adherence (Evidence D). Prior to discharge from the emergency department or hospital, family/carers should receive the following advice and information (all are Evidence D): • Instruction on recognition of signs of recurrence and worsening of asthma. The factors that precipitated the exacerbation should be identified, and strategies for future avoidance of these factors implemented. • A written, individualized action plan, including details of accessible emergency services • Careful review of inhaler technique • SABAs should be used on an as-needed basis to avoid masking worsening asthma, but the daily requirement should be recorded to ensure it is being decreased over time to pre-exacerbation levels. • Confirm that ICS has been initiated where appropriate (at twice the low initial dose in Box 11-3 (p.191) for the first month after discharge, then adjusted as needed) or continued, for those previously prescribed controller medication. • A supply of SABA and, where applicable, the remainder of the course of oral corticosteroid, ICS or LTRA • A follow-up appointment within 1–2 days and another within 1–2 months, depending on the clinical, social and practical context of the exacerbation. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 204 13. Primary prevention of asthma KEY POINTS The development and persistence of asthma are driven by gene–environment interactions. For children, a ‘window of opportunity’ to prevent asthma exists in utero and in early life, but intervention studies are limited. With regard to allergen avoidance strategies aimed at preventing asthma in children: • Strategies directed at a single allergen have not been effective in reducing the incidence of asthma • Multifaceted strategies may be effective, but the essential components have not been identified. Current recommendations for preventing asthma in children, based on high-quality evidence or consensus are: • Avoid exposure to environmental tobacco smoke during pregnancy and the first year of life. • Encourage vaginal delivery where possible. • Where possible, avoid use of broad-spectrum antibiotics during the first year of life. Breast-feeding is advised, not for prevention of allergy and asthma, but for its other positive health benefits). In patients with adult-onset asthma, always ask about occupational or domestic exposures, as these exposures may explain 5–20% of new cases of asthma. In adults and adolescents, the early identification and elimination of occupational sensitizers and the removal of sensitized patients from any further exposure are important aspects of the prevention and management of occupational asthma. FACTORS ASSOCIATED WITH INCREASED OR DECREASED RISK OF ASTHMA IN CHILDREN Asthma is a heterogeneous disease whose inception and persistence are driven by gene–environment interactions that are not yet fully understood. The most important of these interactions may occur in early life and even in utero. There is consensus that a ‘window of opportunity’ exists during pregnancy and early in life when environmental factors may influence asthma development. Multiple environmental factors, both biological and sociological, may be important in the development of asthma. Data from studies investigating the role of environmental risk factors for the development of asthma support further research on prevention strategies focusing on nutrition, allergens (both inhaled and ingested), pollutants (particularly environmental tobacco smoke), microbes, and psychosocial factors. ‘Primary prevention’ refers to preventing the onset of disease. Nutrition of mother and baby Maternal diet A large body of research investigating the development of allergy and asthma in children has focused on the mother’s diet during pregnancy. Current evidence does not clearly demonstrate that ingestion of any specific foods during pregnancy increases the risk for asthma. However, a study of a pre-birth cohort observed that maternal intake of foods commonly considered allergenic (peanut and milk) was associated with a decrease in allergy and asthma in the offspring.866 Similar data have been shown in a very large Danish National birth cohort, with an association between ingestion of peanuts, tree nuts and/or fish during pregnancy and a decreased risk of asthma in the offspring.867,868 Epidemiological studies and randomized controlled trials on maternal dietary intake of fish or long-chain polyunsaturated fatty acids during pregnancy showed no consistent effects on the risk of wheeze, asthma or atopy in the child.869-872 Dietary changes during pregnancy are therefore not recommended for prevention of allergies or asthma. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 205 Maternal obesity and weight gain during pregnancy Data suggest that maternal obesity and weight gain during pregnancy pose an increased risk for asthma in children. A meta-analysis873 showed that maternal obesity in pregnancy was associated with higher odds of ever asthma or wheeze or current asthma or wheeze; each 1 kg/m2 increase in maternal body-mass index (BMI) was associated with a 2% to 3% increase in the odd of childhood asthma. High gestational weight gain was associated with higher odds of ever asthma or wheeze. However, no recommendations can be made at present, as unguided weight loss in pregnancy should not be encouraged. Breastfeeding Despite the existence of many studies reporting a beneficial effect of breastfeeding on asthma prevention, results are conflicting,874 and caution should be taken in advising families that breastfeeding will prevent asthma. Breastfeeding decreases wheezing episodes in early life; however, it may not prevent development of persistent asthma (Evidence D). Regardless of any effect on development of asthma, breastfeeding should be encouraged for all of its other positive benefits (Evidence A). Timing of introduction of solids Beginning in the 1990s, many national pediatric agencies and societies recommended delay of introduction of solid food, especially for children at a high risk for developing allergy. However, meta-analyses have found no evidence that this practice reduces the risk of allergic disease (including asthma).875 Early introduction of peanuts may prevent peanut allergy in high-risk infants.875 Dietary supplements for mothers and/or babies Vitamin D Intake of vitamin D may be through diet, dietary supplementation or sunlight. A systematic review of cohort, case control and cross-sectional studies concluded that maternal dietary intake of vitamin D, and of vitamin E was associated with lower risk of wheezing illnesses in children.876 This was not confirmed in two randomized controlled trials (RCTs) of vitamin D supplementation in pregnancy, which compared standard-dose with high-dose vitamin D; however, a significant effect was not disproven.877,878 When the results from these two trials were combined, there was a 25% reduction of risk of asthma/recurrent wheeze at ages 0–3 years.879 The effect was greatest among women who maintained 25(OH)vitamin D levels of at least 30 ng/mL from the time of study entry through delivery, suggesting that sufficient levels of Vitamin D during early pregnancy may be important in decreasing risk for early life wheezing episodes,879 although in both trials, no effects of vitamin D supplementation on the development of asthma and recurrent wheeze were evident at the age of 6 years.880 Secondary analysis of the VDAART study878 suggested that earlier supplementation may be more effective in reducing the risk of asthma.881 Fish oil and long-chain polyunsaturated fatty acids Systematic reviews of cohort studies about maternal dietary intake of fish or seafood during pregnancy869,882 and of RCTs on maternal dietary intake of fish or long-chained polyunsaturated fatty acids during pregnancy869 showed no consistent effects on the risk of wheeze, asthma or atopy in the child. One study demonstrated decreased wheeze/asthma in preschool children at high risk for asthma when mothers were given a high-dose fish oil supplement in the third trimester;883 however, ‘fish oil’ is not well defined, and the optimal dosing regimen has not been established. Probiotics A meta-analysis provided insufficient evidence to recommend probiotics for the prevention of allergic disease (asthma, rhinitis, eczema or food allergy).884 Inhalant allergens Sensitization to indoor, inhaled aeroallergens is generally more important than sensitization to outdoor allergens for the presence of, and/or development of, asthma. While there appears to be a linear relationship between exposure and COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 206 sensitization to house dust mite,885,886 the relationship for animal allergen appears to be more complex.874 Some studies have found that exposure to pet allergens is associated with increased risk of sensitization to these allergens,887,888 and of asthma and wheezing.889,890 By contrast, other studies have demonstrated a decreased risk of developing allergy with exposure to pets.891,892 Analyses of data from large populations of school-age children from birth cohorts in Europe have found no association between pets in the homes early in life and higher or lower prevalence of asthma in children.893,894 For children at risk of asthma, dampness, visible mold and mold odor in the home environment are associated with increased risk of developing asthma.895 Overall, there are insufficient data to recommend efforts to either reduce or increase prenatal or early-life exposure to common sensitizing allergens, including pets, for the prevention of allergies and asthma. Birth cohort studies provide some evidence for consideration. A meta-analysis found that studies of interventions focused on reducing exposure to a single allergen did not significantly affect asthma development, but that multifaceted interventions such as in the Isle of Wight study,896 the Canadian Asthma Primary Prevention Study,897 and the Prevention of Asthma in Children study898 were associated with lower risk of asthma diagnosis in children younger than 5 years.899 Two multifaceted studies that followed children beyond 5 years of age demonstrated a significant protective effect both before and after the age of 5 years.896,900 The Isle of Wight study has shown a continuing positive benefit for early-life intervention through to 18 years of age;901 however, it remains unclear which components of the intervention contributed to the effects reported, and the precise mechanism of these effects. Treatment with grass pollen sublingual allergen immunotherapy (SLIT) for 3 years did not reduce the incidence of asthma diagnosis (primary outcome) in a large randomized double-blind placebo-controlled trial in children aged 5–12 years with grass-allergic rhinoconjunctivitis, but asthma symptoms and asthma medication use were reduced. 902 At present, there is insufficient evidence to make a recommendation for SLIT in children with grass allergic rhinoconjunctivitis for the purpose of asthma prevention. More studies are needed. Pollutants Maternal smoking during pregnancy is the most direct route of prenatal environmental tobacco smoke exposure.903 A meta-analysis concluded that prenatal smoking had its strongest effect on young children, whereas postnatal maternal smoking appeared only to affect asthma development in older children.904 Exposure to outdoor pollutants, such as living near a main road, is associated with increased risk of asthma.905,906 A 2019 study suggested that up to 4 million new pediatric asthma cases (13% of the global incidence) may be attributable to exposure to traffic-related air pollution.907 Prenatal NO2, SO2, and PM10 exposures are associated with an increased risk of asthma in childhood,908 but it is difficult to separate effects of prenatal and postnatal exposure. Microbial effects The ‘hygiene hypothesis’, and the more recently coined ‘microflora hypothesis’ and ‘biodiversity hypothesis’,909 suggest that human interaction with microbiota may be beneficial in preventing asthma. For example, there is a lower risk of asthma among children raised on farms with exposure to stables and consumption of raw farm milk than among children of non-farmers.910 The risk of asthma is also reduced in children whose bedrooms have high levels of bacterial-derived lipopolysaccharide endotoxin.911,912 Similarly, children in homes with ≥2 dogs or cats are less likely to be allergic than those in homes without dogs or cats.892 Exposure of an infant to the mother’s vaginal microflora through vaginal delivery may also be beneficial; the prevalence of asthma is higher in children born by cesarean section than those born vaginally.913,914 This may relate to differences in the infant gut microbiota according to their mode of delivery.915 Respiratory syncytial virus (RSV) infection in infancy is associated with recurrent wheeze at age 5 years.802 Preventative treatment of premature infants with monthly injections of palivizumab, a monoclonal antibody prescribed for prophylaxis of severe RSV infection, was associated with a reduction in recurrent wheezing in the first year of life.916 However, although the risk of parent-reported asthma with infrequent wheeze was reduced at 6 years, there was no impact on doctor-diagnosed asthma or lung function.917 The long-term effect of RSV-specific monoclonal antibodies in the prevention of asthma remains uncertain.918 Studies of RSV vaccination of pregnant women919 and healthy infants920 suggest a reduction COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 207 in RSV infection requiring medical attention in the first year of life. However, it has not yet been established whether these interventions will lead to a reduced risk of further wheezing episodes, or will prevent development of asthma. Medications and other factors Antibiotic use during pregnancy and in infants and toddlers has been associated with the development of asthma later in life,921 although not all studies have shown this association.922 Intake of the analgesic, paracetamol (acetaminophen), may be associated with an increased risk of asthma in both children and adults,923 although exposure during infancy may be confounded by use of paracetamol for respiratory tract infections.923 Frequent use of paracetamol by pregnant women has been associated with increased risk of asthma in their children.924 Maternal folic acid supplementation during pregnancy at higher than recommended doses may be associated with a small increase in the risk of childhood asthma in offspring.925 However, this small risk is far outweighed by the well-established role of folate supplementation in reducing the risk of clinically important neural tube defects. Women should therefore be advised and encouraged to follow recommendations by local health authorities on folic acid supplementation during pregnancy. There is no evidence that vaccinations increase a child’s risk of developing asthma. Psychosocial factors The social environment to which children are exposed may also contribute to the development and severity of asthma. Maternal distress during pregnancy926 or during the child’s early years927 has been associated with an increased risk of the child developing asthma. Obesity A meta-analysis of 18 studies found that being either overweight or obese was a risk factor for childhood asthma and wheeze, particularly in girls.528 In adults, there is evidence suggesting that obesity affects the risk of asthma, but that asthma does not affect the risk of obesity.928,929 ADVICE ABOUT PRIMARY PREVENTION OF ASTHMA Based on the results of cohort and observational studies,930 and a GRADE-based analysis conducted for the Allergic Rhinitis and its Impact on Asthma (ARIA) guidelines,874 parents/caregivers enquiring about how to reduce the risk of their children developing asthma can be provided with the advice summarized in Box 13-1. Possibly the most important factor is the need to provide a positive, supportive environment for discussion that decreases stress, and which encourages families to make choices with which they feel comfortable. Box 13-1. Advice about primary prevention of asthma in children 5 years and younger Parents/caregivers enquiring about how to reduce the risk of their child developing asthma can be given the following advice: • Children should not be exposed to environmental tobacco smoke during pregnancy or after birth. • Identification and correction of Vitamin D insufficiency in women with asthma who are pregnant, or planning pregnancy, may reduce the risk of early life wheezing episodes. • Where possible, vaginal delivery should be encouraged. • Where possible, the use of broad-spectrum antibiotics during the first year of life should be discouraged. • Breastfeeding is advised, not for prevention of allergy or asthma, but for its other positive health benefits. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 208 PREVENTION OF OCCUPATIONAL ASTHMA IN ADULTS An estimated 5–20% of new cases of adult-onset asthma can be attributed to occupational exposure.62 Asthma may be induced or (more commonly) aggravated by exposure to allergens or other sensitizing agents at work, or sometimes from a single, massive exposure. Occupational rhinitis may precede asthma by up to a year. Early diagnosis is essential, as persistent exposure is associated with worse outcomes.62,63 Asthma acquired in the workplace is frequently missed. The occurrence of adult-onset asthma requires a systematic inquiry about work history and exposures, including hobbies. An essential screening question is to ask patients whether their symptoms improve when they are away from work (weekends or vacation).64 It is important to confirm the diagnosis of occupational asthma objectively as it may lead to the patient changing their occupation, which may have legal and socioeconomic implications. Specialist referral is usually necessary, and frequent PEF monitoring at and away from work is often used to help confirm the diagnosis. The early identification and elimination of occupational sensitizers and the removal of sensitized patients from any further exposure are important aspects of the management of occupational asthma (Evidence A). Attempts to reduce occupational exposure have been successful, especially in industrial settings.62 For example, cost-effective minimization of latex sensitization can be achieved by using non-powdered low-allergen gloves instead of powdered latex gloves.62 Patients with suspected or confirmed occupational asthma should be referred for expert assessment and advice, if this is available, because of the economic and legal implications of the diagnosis (Evidence A). There is more information about occupational asthma in specific guidelines.62 COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 209 14. Implementing asthma management strategies into health systems KEY POINTS • To improve asthma care and patient outcomes, evidence-based recommendations must not only be developed, but also disseminated and implemented at a national and local level, and integrated into clinical practice. • Recommendations for implementing asthma care strategies are based on many successful programs worldwide. • Implementation requires an evidence-based strategy involving professional groups and stakeholders, and should take into account local cultural and socioeconomic conditions. • Cost-effectiveness of implementation programs should be assessed so a decision can be made to pursue or modify them. • Local adaptation and implementation of asthma care strategies is aided by the use of tools developed for this purpose. INTRODUCTION Due to the exponential increase in medical research publications, practical syntheses are needed to guide policy makers and healthcare professionals in delivering evidence-based care. When asthma care is consistent with evidence-based recommendations, outcomes improve.219,931,932 This Strategy Report is a resource document for healthcare professionals, intended to set out the main goals of asthma treatment and the actions required to ensure their fulfilment, as well as to facilitate the achievement of standards for quality asthma care. These objectives can only be realized through local implementation in each country, region and healthcare organization. The use of rigorous methodologies such as GRADE9 for the development of clinical practice recommendations, and of ADAPTE933 and similar approaches for assisting the adaptation of recommendations for local country and regional conditions, has assisted in reducing biased opinion as the basis for asthma programs worldwide. Adaptation of clinical practice recommendations to local conditions using the GRADE method is costly, and often requires expertise that is not available locally; in addition, regular revision is required to remain abreast of developments, including drug availability and new evidence, and this is not easily achieved.934 Further, there is generally very limited high quality evidence addressing the many decision nodes in comprehensive clinical practice guidelines, particularly in developing countries. The GINA annual report is not a formal guideline but an evidence-based strategy, updated yearly from a review of the evidence published in the last 18 months. Each year’s report is an update on the entire strategy, so it does not use individual PICOT questions and GRADE, but the review process includes systematic reviews using these methodologies. (See section on methodology at www.ginasthma.org). As with other evidence-based clinical recommendations, the GINA strategy must be adapted to the local context for implementation in clinical practice. ADAPTING AND IMPLEMENTING ASTHMA CLINICAL PRACTICE GUIDELINES Implementation of asthma management strategies may be carried out at a national, regional or local level.935 Ideally, implementation should be a multidisciplinary effort involving many stakeholders, and using cost-effective methods of knowledge translation.935-937 Each implementation initiative needs to consider the nature of the local health system and its resources, including human resources, infrastructure, and available treatments (Box 14-1). Moreover, goals and implementation strategies will need to vary from country to country and within countries, based on economics, culture and the physical and social environment. Priority should be given to high-impact interventions. Specific steps need to be followed before clinical practice recommendations can be embedded into local clinical practice and become the standard of care, particularly in low resource settings. The individual steps are summarized in Box 14-2. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 210 Box 14-1. Approach to implementation of the Global Strategy for Asthma Management and Prevention Box 14-2. Essential elements required to implement a health-related strategy Steps in implementing an asthma strategy into a health system 1. Develop a multidisciplinary working group. 2. Assess the current status of asthma care delivery, outcomes e.g., exacerbations, admissions, deaths, care gaps and current needs. 3. Select the material to be implemented, agree on main goals, identify key recommendations for diagnosis and treatment, and adapt them to the local context or environment. In treatment recommendations, consider environmental issues (planetary health) in addition to patient health 4. Identify barriers to, and facilitators of, implementation. 5. Select an implementation framework and its component strategies. 6. Develop a step-by-step implementation plan: • Select target populations and evaluable outcomes, and specify data coding requirements (if relevant). • Identify local resources to support implementation. • Set timelines. • Distribute tasks to members. • Evaluate outcomes. 7. Continually review progress and results to determine if the strategy requires modification. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 211 Barriers and facilitators Many barriers to, and facilitators of, implementation procedures have been described.937-940 Some of the barriers to implementation of evidence-based asthma management relate to the delivery of care, while others occur at individual or community level (see Box 14-3). Cultural and economic barriers can particularly affect the application of recommendations. Box 14-3. Examples of barriers to the implementation of evidence-based recommendations Healthcare providers Patients • Insufficient knowledge of recommendations • Lack of agreement with recommendations or expectation that they will be effective • Resistance to change • External barriers (organizational, health policies, financial constraints) • Lack of time and resources • Medico-legal issues • Lack of accurate coding (diagnosis, exacerbations, emergency department and hospital admissions, and deaths) • Low health literacy • Insufficient understanding of asthma and its management • Lack of agreement with recommendations • Cultural and economic barriers • Peer influence • Attitudes, beliefs, preferences, fears and misconceptions Examples of high-impact implementation interventions Ideally, interventions should be applied at the level of both the patient and the healthcare provider and, where relevant, the health system. Studies of the most effective means of medical education show that it may be difficult to change clinical practice. Examples of highly effective implementation interventions are shown in Box 14-4. Box 14-4. Examples of high-impact implementation interventions in asthma management • Free inhaled corticosteroids (ICS) for patients with a recent hospital admission and/or severe asthma941 • Early treatment with ICS, guided self-management, reduction in exposure to tobacco smoke, improved access to asthma education219 • Checklist memory aid for primary care, prompting assessment of asthma control and treatment strategies942 • Use of individualized written asthma action plans as part of self-management education521 • An evidence-based care process model for acute and chronic pediatric asthma management, implemented at multiple hospitals943 Evaluation of the implementation process An important part of the implementation process is to establish a means of evaluating the effectiveness of the program and any improvements in quality of care. Evaluation involves surveillance of traditional epidemiological parameters, such as morbidity and mortality, as well as specific audits of both process and outcome within different sectors of the healthcare system. Each country should determine its own minimum sets of data to audit health outcomes. How can GINA help with implementation? The GINA Strategy Report provides an annually updated summary of evidence relevant to asthma diagnosis, management and prevention that may be used in the formulation and adaptation of local guidelines; where evidence is lacking, the report provides approaches for consideration. The GINA Dissemination Group assists in the dissemination of the recommendations in the Strategy Report. GINA can be contacted via the website (www.ginasthma.org/contact-us). COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 212 Glossary of asthma medication classes For more details about medications, see full 2023 GINA report (www.ginasthma.org) and Product Information from manufacturers. Always check local eligibility criteria. MEDICATIONS for MAINTENANCE TREATMENT Inhaled corticosteroids (ICS) Medications Beclometasone, budesonide, ciclesonide, fluticasone propionate, fluticasone furoate, mometasone, triamcinolone. Delivery pMDI or DPI Action and use ICS-containing medications are the most effective anti-inflammatory medications for asthma. ICS reduce symptoms, increase lung function, reduce airway hyperresponsiveness, improve quality of life, and reduce the risk of exacerbations, asthma-related hospitalizations and death. ICS differ in their potency and bioavailability, but most of the benefit is seen at low doses (see Box 4-2, p.71) for low, medium and high doses of different ICS). Adherence with ICS alone (i.e., not in combination with a bronchodilator) is usually very poor. Adverse effects Most patients do not experience side-effects. Local side-effects include oropharyngeal candidiasis and dysphonia; these can be reduced by use of a spacer with pMDIs, and rinsing with water and spitting out after inhalation. Long-term high doses increase the risk of systemic side-effects such as osteoporosis, cataract and glaucoma. Concomitant treatment with cytochrome P450 inhibitors such as ketoconazole, ritonavir, itraconazole, erythromycin and clarithromycin may increase the risk of ICS adverse effects such as adrenal suppression. ICS in combination with a long-acting beta2 agonist bronchodilator (ICS-LABA) Medications Beclometasone-formoterol, budesonide-formoterol, fluticasone furoate-vilanterol, fluticasone propionate formoterol, fluticasone propionate-salmeterol, mometasone-formoterol and mometasone-indacaterol. Delivery pMDI or DPI Action and use When a low-dose of ICS alone fails to achieve good control of asthma, the addition of LABA to maintenance ICS improves symptoms, lung function and reduces exacerbations in more patients, more rapidly, than doubling the dose of ICS. Two regimens are available: low-dose combination beclometasone or budesonide with low-dose formoterol for both maintenance-and-reliever treatment (MART, GINA Track 1), and maintenance ICS-LABA with SABA or ICS-SABA as reliever (Track 2). MART with low-dose ICS-formoterol reliever is preferred as it reduces exacerbations compared with conventional maintenance therapy with SABA as reliever, and is a simpler regimen. For as-needed-only use of ICS-formoterol in mild asthma, see section on anti-inflammatory relievers below; and for ICS-LABA-LAMA, see section on add-on medications. See box 4-2, p.71 for low, medium and high doses of ICS in combination with LABA. See Box 4-8, p.84 for medications and doses for anti-inflammatory reliever therapy with ICS-formoterol. Adverse effects The LABA component may be associated with tachycardia, headache or cramps. LABA is safe for asthma when used in combination with ICS. LABA or LAMA should not be used without ICS in asthma (or in patients with asthma+COPD) due to increased risk of serious adverse outcomes. Concomitant treatment with cytochrome P450 inhibitors such as ketoconazole, ritonavir, itraconazole, erythromycin and clarithromycin may increase the risk of ICS adverse COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 213 effects such as adrenal suppression. Leukotriene modifiers (leukotriene receptor antagonists, LTRA) Medications Montelukast, pranlukast, zafirlukast, zileuton. Delivery Tablets Action and use Target one part of the inflammatory pathway in asthma. Sometimes used as an option for maintenance therapy, mainly only in children. When used alone: less effective than low-dose ICS. When added to ICS: less effective than ICS-LABA. Adverse effects Few in placebo-controlled studies except elevated liver function tests with zileuton and zafirlukast. There are concerns in adults and children about risk of serious behavioral and mood changes, including suicidal ideation, associated with montelukast; this should be discussed with patients/parents/caregivers. ADD-ON MAINTENANCE MEDICATIONS Long-acting muscarinic antagonists (LAMA) (check your local eligibility criteria) Medications Tiotropium, ≥6 years, by mist inhaler, added to separate ICS-LABA. Combination ICS-LABA-LAMA inhalers for adults ≥18 years: beclometasone-formoterol-glycopyrronium; fluticasone furoate-vilanterol-umeclidinium; mometasone-indacaterol-glycopyrronium. Delivery pMDI or DPI or mist inhaler Action and use An add-on option at Step 5 (or at Step 4, non-preferred because of weaker evidence for benefit) in combination or separate inhalers for patients with uncontrolled asthma despite ICS-LABA. Modestly improves lung function but not symptoms or quality of life; small reduction in exacerbations. For patients with exacerbations, ensure that ICS is increased to at least medium dose before considering need for add-on LAMA. Adverse effects Uncommon, but include dry mouth, urinary retention. Anti-IgE (check your local eligibility criteria) Medications Omalizumab, ≥6 years Delivery Syringe or pen for subcutaneous injection Action and use An add-on option for patients with severe allergic asthma uncontrolled on high-dose ICS-LABA. May also be indicated for nasal polyps and chronic spontaneous (idiopathic) urticaria. Self-administration may be an option. Adverse effects Reactions at the site of injection are common but minor. Anaphylaxis is rare. Anti-IL5 and anti-IL5Rα (check your local eligibility criteria) Medications Anti-IL5: mepolizumab (≥6 years, SC injection) or reslizumab (≥18 years, intravenous infusion). Anti-IL5 receptor benralizumab (≥12 years, SC injection). Delivery Depends on the specific medication, as above Action and use Add-on options for patients with severe eosinophilic asthma uncontrolled on high-dose ICS-COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 214 LABA. Maintenance OCS dose can be significantly reduced with benralizumab and mepolizumab. Mepolizumab may also be indicated for eosinophilic granulomatosis with polyangiitis (EGPA), hypereosinophilic syndrome or chronic rhinosinusitis with nasal polyposis. For mepolizumab and benralizumab, self-administration may be an option. Adverse effects Headache, and reactions at injection site are common but minor. Anti-IL4Rα (check your local eligibility criteria) Medications Anti-interleukin 4 receptor alpha: dupilumab, ≥6 years Delivery Syringe or pen for subcutaneous injection Action and use An add-on option for patients with severe eosinophilic or Type 2 asthma uncontrolled on high-dose ICS-LABA, or patients requiring maintenance OCS. Not advised for patients with current or historical blood eosinophils ≥1500/mL. May also be indicated for treatment of skin conditions including moderate-severe atopic dermatitis, chronic rhinosinusitis with nasal polyps, and eosinophilic esophagitis. Self-administration may be an option. Adverse effects Reactions at injection site are common but minor. Transient blood eosinophilia occurs in 4–13% of patients. Rarely, cases of eosinophilic granulomatosis with polyangiitis (EGPA) may be unmasked following reduction/cessation of OCS treatment on dupilumab. Anti-TSLP (check your local eligibility criteria) Medications Tezepelumab, SC injection, ≥12 years Delivery Syringe or pen for subcutaneous injection Action and use An add-on option for patients with severe asthma uncontrolled on high-dose ICS-LABA. In patients taking maintenance OCS, no significant reduction in OCS dose compared with placebo. Adverse effects Injection-site reactions; anaphylaxis is rare; adverse events generally similar between active and placebo groups. Systemic corticosteroids Medications include prednisone, prednisolone, methylprednisolone, hydrocortisone tablets, dexamethasone. Delivery Given by tablets or suspension or by IM or IV injection Action and use Short-term treatment (usually 5–7 days in adults) is important in the treatment of severe acute exacerbations, with main effects seen after 4–6 hours. For acute severe exacerbations, oral corticosteroid (OCS) therapy is preferred to IM or IV therapy and is effective in preventing short-term relapse. Tapering is required if OCS given for more than 2 weeks. Patients should be reviewed after any exacerbation, to optimize their inhaled treatment to reduce the risk of future exacerbations requiring OCS. As a last resort, long-term treatment with OCS may be required for some patients with severe asthma, but serious side-effects are problematic. Patients for whom this is considered should be referred for specialist review if available, to have treatment optimized and phenotype assessed. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 215 Adverse effects Short courses: adverse effects include sepsis, thromboembolism, sleep disturbance, reflux, appetite increase, hyperglycemia, mood changes. Even 4–5 lifetime courses increase cumulative risk of long-term adverse effects e.g., diabetes, osteoporosis, cataract, glaucoma, heart failure. Maintenance use: consider only as last resort, because of significant adverse effects e.g., cataract, glaucoma, hypertension, diabetes, adrenal suppression osteoporosis. Assess for these risks and treat appropriately. ANTI-INFLAMMATORY RELIEVER MEDICATIONS Low-dose combination ICS-formoterol Medications Beclometasone-formoterol or budesonide-formoterol. Delivery pMDI or DPI Action and use This is the anti-inflammatory reliever inhaler for GINA Track 1, for patients prescribed maintenance-and-reliever therapy (MART) with maintenance ICS-formoterol in Steps 3-5, or for patients prescribed as-needed-only ICS-formoterol in Steps 1-2. In both settings, it reduces the risk of severe exacerbations compared with using SABA as reliever, with similar symptom control. In patients with mild asthma, as-needed-only ICS-formoterol reduces emergency visits/hospitalizations by 65% compared with SABA alone, and by 37% compared with daily ICS plus as-needed SABA. See Box 4-8, p.84 for details of medications and doses for AIR-only and MART. Low-dose ICS-formoterol can be taken before exercise to reduce exercise-induced bronchoconstriction, and it can be taken before or during allergen exposure to reduce allergic responses. Recommended maximum doses in any day For adults and adolescents, the maximum total number of inhalations in a single day (maintenance plus reliever doses) for budesonide-formoterol gives 72 mcg metered dose (delivered dose 54 mcg) of the formoterol component. Since the safety and efficacy of budesonide-formoterol up to this maximum total daily use has been established from large studies (>50,000 patients), GINA suggests that the same maximum total daily dose should also apply for beclometasone-formoterol. For children 6–11 years prescribed MART with budesonide-formoterol, the maximum total dose recommended in a single day gives 48 mcg metered dose (delivered dose 36 mcg) of the formoterol component. See Box 4-7, p.78 for details of medications and doses for different age-groups. Adverse effects As for ICS-formoterol above. Low-dose combination ICS-SABA Medications Budesonide-salbutamol (also described as albuterol-budesonide); beclometasone-salbutamol. Delivery pMDI or DPI Action and use Anti-inflammatory reliever option (instead of SABA) for GINA Track 2. Budesonide-salbutamol 100/100 mcg (delivered dose 80/90 mcg) taken 2 inhalations as needed for symptom relief on top of maintenance ICS or ICS-LABA reduced the risk of severe exacerbations in adults compared with SABA reliever; most of the benefit was seen in Step 3. ICS-SABA cannot be used for maintenance-and-reliever therapy. No evidence for as-needed-only use of budesonide-salbutamol in Steps 1–2. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 216 Recommended maximum doses in any day: Maximum 6 doses, each of 2 inhalations, in any day. Adverse effects As for ICS and SABA. SHORT-ACTING BRONCHODILATOR RELIEVER MEDICATIONS Short-acting inhaled beta2 agonist bronchodilators (SABA) Medications e.g., salbutamol (albuterol), terbutaline. Delivery Administered by pMDI, DPI or, rarely, as solution for nebulization or injection Action and use Inhaled SABAs provide quick relief of asthma symptoms and bronchoconstriction, and for pre-treatment before exercise. SABAs should be used only as-needed (not regularly) and at the lowest dose and frequency required. SABA-only treatment is not recommended because of the risk of severe exacerbations and asthma-related death. Currently, inhaled SABAs are the most commonly used bronchodilator for acute exacerbations requiring urgent primary care visit or ED presentation. Adverse effects Tremor and tachycardia are commonly reported with initial use of SABA. Tolerance develops rapidly with even 1–2 weeks of regular use, with increased airway hyperresponsiveness, reduced bronchodilator effect, and increased airway inflammation. Excess use, or poor response indicate poor asthma control and risk of exacerbations. Dispensing of 3 or more 200-dose canisters per year is associated with increased risk of exacerbations, and dispensing of 12 or more canisters per year is associated with markedly increased risk of death. Short-acting antimuscarinics (anticholinergics) Medications e.g., ipratropium bromide, oxitropium bromide. May be in combination with SABA. Delivery pMDI or DPI. Action and use As-needed use: ipratropium is a less effective reliever medication than SABA, with slower onset of action. Short-term use in severe acute asthma, where adding ipratropium to SABA reduces the risk of hospital admission. Adverse effects Dryness of the mouth or a bitter taste. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 217 References 1. Asher I, Bissell K, Chiang CY, et al. Calling time on asthma deaths in tropical regions-how much longer must people wait for essential medicines? Lancet Respir Med 2019; 7: 13-15. 2. Meghji J, Mortimer K, Agusti A, et al. 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Riedler J, Braun-Fahrlander C, Eder W, et al. Exposure to farming in early life and development of asthma and allergy: a cross-sectional survey. Lancet 2001; 358: 1129-1133. 911. Braun-Fahrlander C, Riedler J, Herz U, et al. Environmental exposure to endotoxin and its relation to asthma in school-age children. N Engl J Med 2002; 347: 869-877. 912. Karvonen AM, Hyvarinen A, Gehring U, et al. Exposure to microbial agents in house dust and wheezing, atopic dermatitis and atopic sensitization in early childhood: a birth cohort study in rural areas. Clin Exp Allergy 2012; 42: 1246-1256. 913. Huang L, Chen Q, Zhao Y, et al. Is elective cesarean section associated with a higher risk of asthma? A meta-analysis. J Asthma 2015; 52: 16-25. 914. Keag OE, Norman JE, Stock SJ. Long-term risks and benefits associated with cesarean delivery for mother, baby, and subsequent pregnancies: Systematic review and meta-analysis. PLoS Med 2018; 15: e1002494. 915. Azad MB, Konya T, Maughan H, et al. Gut microbiota of healthy Canadian infants: profiles by mode of delivery and infant diet at 4 months. CMAJ 2013; 185: 385-394. 916. Blanken MO, Rovers MM, Molenaar JM, et al. Respiratory syncytial virus and recurrent wheeze in healthy preterm infants. N Engl J Med 2013; 368: 1791-1799. 917. Scheltema NM, Nibbelke EE, Pouw J, et al. Respiratory syncytial virus prevention and asthma in healthy preterm infants: a randomised controlled trial. Lancet Respir Med 2018; 6: 257-264. 918. Quinn LA, Shields MD, Sinha I, et al. Respiratory syncytial virus prophylaxis for prevention of recurrent childhood wheeze and asthma: a systematic review. Syst Rev 2020; 9: 269. 919. Kampmann B, Madhi SA, Munjal I, et al. Bivalent prefusion F vaccine in pregnancy to prevent RSV illness in infants. N Engl J Med 2023; 388: 1451-1464. 920. Hammitt LL, Dagan R, Yuan Y, et al. Nirsevimab for prevention of RSV in healthy late preterm and term infants. N Engl J Med 2022; 386: 837-846. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 260 921. Baron R, Taye M, der Vaart IB, et al. The relationship of prenatal antibiotic exposure and infant antibiotic administration with childhood allergies: a systematic review. BMC Pediatr 2020; 20: 312. 922. Celedon JC, Fuhlbrigge A, Rifas-Shiman S, et al. Antibiotic use in the first year of life and asthma in early childhood. Clin Exp Allergy 2004; 34: 1011-1016. 923. Cheelo M, Lodge CJ, Dharmage SC, et al. Paracetamol exposure in pregnancy and early childhood and development of childhood asthma: a systematic review and meta-analysis. Arch Dis Child 2015; 100: 81-89. 924. Eyers S, Weatherall M, Jefferies S, et al. Paracetamol in pregnancy and the risk of wheezing in offspring: a systematic review and meta-analysis. Clin Exp Allergy 2011; 41: 482-489. 925. Yang F, Zhu J, Wang Z, et al. Relationship between maternal folic acid supplementation during pregnancy and risk of childhood asthma: Systematic review and dose-response meta-analysis. Front Pediatr 2022; 10: 1000532. 926. Flanigan C, Sheikh A, DunnGalvin A, et al. Prenatal maternal psychosocial stress and offspring's asthma and allergic disease: A systematic review and meta-analysis. Clin Exp Allergy 2018; 48: 403-414. 927. Kozyrskyj AL, Mai XM, McGrath P, et al. Continued exposure to maternal distress in early life is associated with an increased risk of childhood asthma. Am J Respir Crit Care Med 2008; 177: 142-147. 928. Xu S, Gilliland FD, Conti DV. Elucidation of causal direction between asthma and obesity: a bi-directional Mendelian randomization study. Int J Epidemiol 2019; 48: 899-907. 929. Sun YQ, Brumpton BM, Langhammer A, et al. Adiposity and asthma in adults: a bidirectional Mendelian randomisation analysis of The HUNT Study. Thorax 2020; 75: 202-208. 930. Beasley R, Semprini A, Mitchell EA. Risk factors for asthma: is prevention possible? Lancet 2015; 386: 1075-1085. 931. Burgers J, Eccles M. Clinical guidelines as a tool for implementing change in patient care. Oxford: Butterworth-Heinemann; 2005. 932. Woolf SH, Grol R, Hutchinson A, et al. Clinical guidelines: potential benefits, limitations, and harms of clinical guidelines. BMJ 1999; 318: 527-530. 933. The ADAPTE Collaboration. The ADAPTE process: resource toolkit for guideline adaptation. Version 2.0: Guideline International Network; 2009. Available from: 934. Brouwers MC, Kho ME, Browman GP, et al. AGREE II: advancing guideline development, reporting and evaluation in health care. CMAJ 2010; 182: E839-842. 935. Boulet LP, FitzGerald JM, Levy ML, et al. A guide to the translation of the Global Initiative for Asthma (GINA) strategy into improved care. Eur Respir J 2012; 39: 1220-1229. 936. Davis DA, Taylor-Vaisey A. Translating guidelines into practice. A systematic review of theoretic concepts, practical experience and research evidence in the adoption of clinical practice guidelines. CMAJ 1997; 157: 408-416. 937. Harrison MB, Legare F, Graham ID, et al. Adapting clinical practice guidelines to local context and assessing barriers to their use. CMAJ 2010; 182: E78-84. 938. Partridge MR. Translating research into practice: how are guidelines implemented? Eur Respir J Suppl 2003; 39: 23s-29s. 939. Baiardini I, Braido F, Bonini M, et al. Why do doctors and patients not follow guidelines? Curr Opin Allergy Clin Immunol 2009; 9: 228-233. 940. Boulet LP, Becker A, Bowie D, et al. Implementing practice guidelines: a workshop on guidelines dissemination and implementation with a focus on asthma and COPD. Can Respir J 2006; 13 Suppl A: 5-47. 941. Franco R, Santos AC, do Nascimento HF, et al. Cost-effectiveness analysis of a state funded programme for control of severe asthma. BMC Public Health 2007; 7: 82. 942. Renzi PM, Ghezzo H, Goulet S, et al. Paper stamp checklist tool enhances asthma guidelines knowledge and implementation by primary care physicians. Can Respir J 2006; 13: 193-197. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE 261 943. Nkoy F, Fassl B, Stone B, et al. Improving pediatric asthma care and outcomes across multiple hospitals. Pediatrics 2015; 136: e1602-1610. COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE Visit the GINA website at www.ginasthma.org ©2024 Global Initiative for Asthma COPYRIGHTED MATERIAL - DO NOT COPY OR DISTRIBUTE |
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2: Math Review
GSU-TM-Introductory Physics II (1112)
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2.5: Finding Angle Measurements
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2.5: Finding Angle Measurements
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2.4: The Rectangular Coordinate Systems and Graphs
2.6: Parallel and Perpendicular Lines
Page ID 83868
Henry Africk
CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Angles
1. Example 2.5.4
1. Solution
2. Example 2.5.5/02:_Math_Review/2.05:_Finding_Angle_Measurements#Example_.5C(.5CPageIndex.7B5.7D.5C))
1. Solution/02:_Math_Review/2.05:_Finding_Angle_Measurements#Solution_2)
3. Try It Now 2/02:_Math_Review/2.05:_Finding_Angle_Measurements#Try_It_Now_2)
Finding Angle Measurements
Example 2.5.8
Solution
Example 2.5.9
Solution
Try It Now 2
Supplementary and Complementary
Example 2.5.10
Solution
Example 2.5.11
Solution
Example 2.5.12
Solution
Example 2.5.13
Solution
Try It Now 3
Summary
Try It Now Answers
Theorem 2.5.1: Parallel Postulate
Theorem 2.5.1 The "Z" Theorem
Example 2.5.1
Theorem 2.5.2: The "F" Theorem
Example 2.5.2
Theorem 2.5.3:The "C" Theorem
Example 2.5.3
Example 2.5.4
Example 2.5.5
Example 2.5.6
SUMMARY
Historical Note
Problems
Angles
Lines, line segments, points, and rays are the building blocks of other figures. For example, two rays with a common endpoint make up an angle. The common endpoint of the angle is called the vertex.
The angle ABC is shown below. This angle can also be called ∠ABC, ∠CBA or simply ∠B. When you are naming angles, be careful to include the vertex (here, point B) as the middle letter.
The image below shows a few angles on a plane. Notice that the label of each angle is written “point-vertex-point,” and the geometric notation is in the form ∠ABC.
Sometimes angles are very narrow; sometimes they are very wide. When people talk about the “size” of an angle, they are referring to the arc between the two rays. The length of the rays has nothing to do with the size of the angle itself. Drawings of angles will often include an arc (as shown above) to help the reader identify the correct ‘side’ of the angle.
Think about an analog clock face. The minute and hour hands are both fixed at a point in the middle of the clock. As time passes, the hands rotate around the fixed point, making larger and smaller angles as they go. The length of the hands does not impact the angle that is made by the hands.
An angle is measured in degrees, represented by the symbol º. A circle is defined as having 360º. (In skateboarding and basketball, “doing a 360” refers to jumping and doing one complete body rotation.
Aright angle is any degree that measures exactly 90º. This represents exactly one-quarter of the way around a circle. Rectangles contain exactly four right angles. A corner mark is often used to denote a right angle, as shown in right angle DCB below.
Angles that are between 0º and 90º (smaller than right angles) are called acute angles. Angles that are between 90º and 180º (larger than right angles and less than 180º) are called obtuse angles. And an angle that measures exactly 180º is called a straight angle because it forms a straight line.
Figure 2.5.5: Examples of Angles
Example 2.5.4
Label each angle below as acute, right, or obtuse.
Solution
You can start by identifying any right angles.
∠GFI is a right angle, as indicated by the corner mark at vertex F.
Acute angles will be smaller than ∠GFI (or less than 90º). This means that ∠DAB and ∠MLN are acute.
∠TQS is larger than ∠GFI, so it is an obtuse angle.
Answer: ∠DAB and ∠MLN are acute angles. ∠GFI is a right angle. ∠TQS is an obtuse angle.
Example 2.5.5
Identify each point, ray, and angle in the picture below.
Solution
Begin by identifying each point in the figure. There are 4: E, F, G, and J.
Now find rays. A ray begins at one point, and then continues through another point towards infinity (indicated by an arrow). Three rays start at point J: JE→, JF→, and JG→. But also notice that a ray could start at point F and go through J and G, and another could start at point G and go through J and F. These rays can be represented by GF→ and FG→.
Finally, look for angles. ∠EJG is obtuse, ∠EJF is acute, and ∠FJG is straight. (Don’t forget those straight angles!)
Answer: Points: E, F, G,J
Rays: JE→, JG→, JF→, GF→, FG→
Angles: ∠EJG, ∠EJF, ∠FJG
Try It Now 2
Identify the acute angles in the given image:
Finding Angle Measurements
Understanding how parallel and perpendicular lines relate can help you figure out the measurements of some unknown angles. To start, all you need to remember is that perpendicular lines intersect at a 90º angle and that a straight angle measures 180º.
The measure of an angle such as ∠A is written as m∠A. Look at the example below. How can you find the measurements of the unmarked angles?
Example 2.5.8
Find the measurement of ∠IJF.
Solution
Only one angle, ∠HJM, is marked in the image. Notice that it is a right angle, so it measures 90º. ∠HJM is formed by the intersection of lines IM↔ and HF↔. Since IM↔ is a line, ∠IJM is a straight angle measuring 180º.
You can use this information to find the measurement of ∠HJI :
m∠HJM + m∠HJI = m∠IJM
90º + m∠HJI = 180º
m∠HJI = 90º
Now use the same logic to find the measurement of ∠IJF. ∠IJF is formed by the intersection of lines IM↔ and HF↔. Since HF↔ is a line, ∠HJF will be a straight angle measuring 180º.
You know that ∠HJI measures 90º. Use this information to find the measurement of ∠IJF:
m∠HJM + m∠IJF = m∠HJF
90º + m∠IJF = 180º
m∠IJF = 90º
Answer: m∠IJF = 90º
In this example, you may have noticed that angles ∠HJI, ∠IJF, and ∠HJM are all right angles. (If you were asked to find the measurement of ∠FJM, you would find that angle to be 90º, too.) This is what happens when two lines are perpendicular—the four angles created by the intersection are all right angles.
Not all intersections happen at right angles, though. In the example below, notice how you can use the same technique as shown above (using straight angles) to find the measurement of a missing angle.
Example 2.5.9
Find the measurement of ∠DAC.
Solution
This image shows the line BC↔ and the ray AD→ intersecting at point A. The measurement of ∠BAD is 135º. You can use straight angles to find the measurement of ∠DAC.
∠BAC is a straight angle, so it measures 180º.
Use this information to find the measurement of ∠DAC.
m∠BAD + m∠DAC = m∠BAC
135º + m∠DAC = 180º
m∠DAC = 45º
Answer: m∠DAC = 45º
Try It Now 2
Find the measurement of ∠CAD.
Supplementary and Complementary
In the example above, m∠BAC and m∠DAC add up to 180º. Two angles whose measures add up to 180º are called supplementary angles. There’s also a term for two angles whose measurements add up to 90º, they are called complementary angles.
One way to remember the difference between the two terms is that “corner” and “complementary” each begin with c (a 90º angle looks like a corner), while straight and “supplementary” each begin with s (a straight angle measures 180º).
If you can identify supplementary or complementary angles within a problem, finding missing angle measurements is often simply a matter of adding or subtracting.
Example 2.5.10
Two angles are supplementary. If one of the angles measures 48º, what is the measurement of the other angle?
Solution
Two supplementary angles make up a straight angle, so the measurements of the two angles will be 180º.
m∠A + m∠B = 180º
You know the measurement of one angle. To find the measurement of the second angle, subtract 48º from 180º.
48º+ m∠B = 180º
m∠B = 180º - 48º
m∠B = 132º
Answer: The measurement of the other angle is 132º
Example 2.5.11
Find the measurement of ∠AXZ.
Solution
This image shows two intersecting lines, AB↔ and YZ↔. They intersect at point X, forming four angles. Angles ∠AXY and ∠AXZ are supplementary because together they make up the straight angle ∠YXZ.
Use this information to find the measurement of ∠AXZ.
m∠AXY + m∠AXZ = m∠YXZ
30º + m∠AXZ = 180º
m∠AXZ = 150º
Answer: m∠AXZ = 150º
Example 2.5.12
Find the measurement of ∠BAC.
Solution
This image shows the line CF↔ and the rays AB↔ and AD↔, all intersecting at point A. Angle ∠BAD is a right angle. Angles ∠BAC and ∠CAD are complementary because together they create ∠BAD.
Use this information to find the measurement of ∠BAC .
m∠BAC + m∠CAD = m∠BAD
m∠BAC + 50º = 90º
m∠BAC = 40º
Answer: m∠BAC = 40º
Example 2.5.13
Find the measurement of ∠CAD.
Solution
You know the measurements of two angles here: ∠CAB and ∠DAE. You also know that m∠BAE = 180º.
Use this information to find the measurement of ∠CAD.
m∠BAC + m∠CAD + m∠DAE = m∠BAE
25º + m∠CAD + 75º = 180º
m∠CAD + 100º = 180º
m∠CAD = 80º
Answer: m∠CAD = 80º
Try It Now 3
Which pair of angles is complementary?
A) ∠PKO and ∠MKN
B) ∠PKO and ∠PKM
C) ∠LKP and ∠LKN
D) ∠LKM and ∠MKN
Summary
Parallel lines do not intersect, while perpendicular lines cross at a 90º angle. Two angles whose measurements add up to 180º are said to be supplementary, and two angles whose measurements add up to 90º are said to be complementary. For most pairs of intersecting lines, all you need is the measurement of one angle to find the measurements of all other angles formed by the intersection.
Try It Now Answers
C) FH || EG; both EG and FH are marked with >> on each line, and those markings mean they are parallel.
137º; ∠BAD is a straight angle measuring 180º. Since ∠BAC measures 43º, the measure of ∠CAD must be 180º – 43º = 137º.
D) ∠LKM and ∠MKN; the measurements of two complementary angles will add up to 90º. ∠LKP is a right angle, so ∠LKN must be a right angle as well. ∠LKM + ∠MKN = ∠LKN, so ∠LKM and ∠MKN are complementary.
Two lines are parallelif they do not meet, no matter how far they are extended. The symbol for parallel is ||. In Figure 2.5.1, AB↔||CD↔. The arrow marks are used to indicate the lines are parallel.
Figure 2.5.1: AB↔ and CD^ are parallel.They do not meet no matter how far they are extended.
Figure 2.5.1: EF↔ and GH↔ are not parallel. They meet at point P.
We make the following assumption about parallel lines, called theparallel postulate.
Theorem 2.5.1: Parallel Postulate
The probabilities assigned to events by a distribution function on a sample space are given by
Through a point not on a given line one and only one line can be drawn parallel to the given line. So in Figure 2.5.3, there is exactly one line that can be drawn through C that is parallel to AB←.
Figure 2.5.3: There is exactly one line that can be drawn through C parallel to AB→.
Figure 2.5.4: EF↔ is a transversal.
A transversalis a line that intersects two other lines at two distinct points. In Figure 2.5.4, EF↔ is a transversal. ∠x and ∠x′ are called alternate interior angles of lines AB↔ and CD↔. The word "alternate," here, means that the angles are on different sides of the transversal, one angle formed with AB↔ and the other formed with CD↔. The word "interior" means that they are between the two lines. Notice that they form the letter "Z." (Figure 2.5.5). ∠y and ∠y′ are also alternate interior angles. They also form a "Z" though It is stretched out and backwards. Viewed from the side, the letter "Z" may also look like an "N."
Figure 2.5.5: Alternate interior form the letters "Z" or "N". The letter may be stretched out or backwards.
Alternate interior angles are important because of the following theorem:
Theorem 2.5.1 The "Z" Theorem
If two lines are parallel then their alternate interior angles are equal, If the alternate interior angles of two lines are equal then the lines must 'oe parallel,
In Figure 2.5.6, AB↔ must be parallel to CD↔ because the alternate interior angles are both 30∘. Notice that the other pair of alternate interior angles, ∠y and ∠y′, are also equal. They are both 150∘. In Figure 2.5.7, the lines are not parallel and none of the alternate interior angles are equal.
Figure 2.5.6: The lines are parallel and their alternate interior angles are equal.
Figure 2.5.7: The lines are not parallel and their alternate interior angles are not equal.
The Proof of Theorem 2.5.1 is complicated and will be deferred to the appendix.
Example 2.5.1
Find x,y and z:
Solution
AB↔||CD↔ since the arrows indicate parallel lines. x∘=40∘ because alternate interior angles of parallel lines are equal. y∘=z∘=180∘−40∘=140∘.
Answer: x=40,y=140,z=140.
Corresponding angles of two lines are two angles which are on the same side of the two lines and the same side of the transversal, In Figure 2.5.8, ∠w and ∠w′ are corresponding angles of lines AB↔ and CD↔. They form the letter "F." ∠x and ∠x′, ∠y and ∠y′, and ∠z and ∠z′ are other pairs of corresponding angles of AB↔ and CD↔. They all form the letter "F", though it might be a backwards or upside down "F" (Figure 2.5.9).
Figure 2.5.8: Four pairs of corresponding angles are illustrated.
Figure 2.5.9: Corresponding angles form the letter "F," though it may be a backwards or upside down "F."
Corresponding angles are important because of the following theorem:
Theorem 2.5.2: The "F" Theorem
If two lines are parallel then their corresponding angles are equal. If the corresponding angles of two lines are equal then the lines must be parallel.
Example 2.5.2
Find x:
Solution
The arrow indicate AB↔||CD↔. Therefore x∘=110∘ because x∘ and 110∘ are the measures of corresponding angles of the parallel lines AB↔ and CD↔.
Answer: x=110.
Figure PageIndex10: Each pair of corresponding angles is equal.
Notice that we can now find all the other angles in Example PageIndex2. Each one is either supplementary to one of the 110∘ angles or forms equal vertical angles with one of them (Figure PageIndex10). Thereforeall the corresponding angles are equal, Also each pair of alternate interior angles is equal. It is not hard to see that if just one pair of corresponding angles or one pair of alternate interior angles are equal then so are all other pairs of corresponding and alternate interior angles.
Proof of Theorem 2.5.2: The corresponding angles will be equal if the alternate interior angles are equal and vice versa. Therefore Theorem 2.5.2 follows directly from Therorem 2.5.1.
In Figure 2.5.11, ∠x and ∠x′ are called interior angles on the same side of the transversal.(In some textbooks, interior angles on the same sdie of the transversal are called cointerior angles.) ∠y and ∠y′ are also interior angles on the same side of the transversal, Notice that each pair of angles forms the letter "C." Compare Figure 2.5.11 with Figure 10 and also with Example 2.5.1, The following theorem is then apparent:
Figure 2.5.11: Interior angles on the same side of the transversal form the letter "C". It may also be a backwards "C."
Theorem 2.5.3:The "C" Theorem
If two lines are parallel then the interior angles on the same side of the transversal are supplementary (they add uP to 180∘). If the interior angles of two lines on the same side of the transversal are supplementary then the lines must be parallel.
Example 2.5.3
Find x and the marked angles:
Solution
The lines are parallel so by Theorem 2.5.3 the two labelled angles must be supplementary.
(2.5.1)x+2x+30=180 3x+30=180 3x=180−30 3x=150 x=50
∠CHG=x=50∘
∠AGH=2x+30=2(50)+30=100+30=130∘.
Check:
Answer: x=50, ∠CHG=50∘, ∠AGHa=130∘.
Example 2.5.4
Find x and the marked angles:
Solution
∠BEF=3x+40∘ because vertical angles are equal. ∠BEF and ∠DFE are interior angles on the same side of the transversal, and therefore are supplementary because the lines are parallel.
(2.5.2)3x+40+2x+50=180 5x+90=180 5x=180−90 5x=90 x=18
∠AEC=3x+40=3(18)+40=54+40=94∘
∠DFE=2x+50=2(18)+50=36+50=86∘
Check:
Answer: x=18, ∠AEG=94∘, ∠DFE=86∘.
Example 2.5.5
List all pairs of alternate interior angles in the diagram, (The single arrow indicates AB↔ is parallel to CD↔ and the double arrow indicates AD↔ is parallel to BC↔.
Solution
We see if a letter Z or N can be formed using the line segments in the diagram (Figure 2.5.12),
Answer: ∠DCA and ∠CAB are alternate interior angles of lines AB↔ and CD↔. ∠DAC and ∠ACB are alternate interior angles of lines AD↔ and BC↔
Example 2.5.6
A telescope is pointed at a star 70∘ above the horizon, What angle x∘ must the mirror BD make with the horizontal so that the star can be seen in the eyepiece E?
Solution
x∘=∠BCE because they are alternate interior angles of parallel lines AB↔ and CE↔. ∠DCF=∠BCE=x∘ because the angle of incidence is equal to the angle of reflection. Therefore
(2.5.3)x+70+x=180 2x+70=180 2x=110 x=55
Answer: 55∘
SUMMARY
Alternate interior angles of paralle lines are equal. They form the letter "Z."
Corresponding angles of parallel lines are equal. They form the letter "F."
Interior angles on the same sides of the transversal of parallel lines are supplementary. They form the letter "G."
Historical Note
The parallel postulate given earlier in this section is the equivalent of the fifth postulate of Euclid's Elements. Euclid was correct in assuming it as a postulate rather than trying to prove it as a theorem, However this did not become clear to the mathematical world until the nineteenth century, 2200 years later, In the interim, scores of prominent mathematicians attempted unsuccessfully to give a satisfactory proof of the parallel postulate. They felt that it was not as self-evident as a postulate should be, and that it required some formal justification,
In 1826, N, I, Lobachevsky, a Russian mathematician, presented a system of geometry based on the assumption that through a given point more than one straight line can be drawn parallel to a given line (Figure 2.5.13). In 1854, the German mathematician Georg Bernhard Riemann proFosed a system of geometriJ in which there are no parallel lines at all, A gecmetry in which the parallel postulate has been replaced by some other postulate is called a non-Euclidean geometry. The existence of these geometries shows that the parallel postulate need not necessarily be true. Indeed Einstein used the geometry of Riemann as the basis for his theory of relativity.
Figure PageIndex13: In the geometry of Lobachevsky, more than one line can be drawn through C parallel to AB.
Of course our original parallel postulate makes the most sense for ordinary applications, and we use it throughout this book, However, for applications where great distances are involved, such as in astronomy, it may well be that a non-Euclidean geometry gives a better approximation of physical reality.
Problems
For each of the following, state the theorem(s) used in obtaining your answer (for example, "alternate interior angles of parallel lines are equal"). Lines marked with the same arrow are assumed to be parallel,
1 - 2. Find x,y, and z:
2.
3 - 4. Find t, u, v, w, x, y, and z:
4.
5 - 10. Find x:
6.
8.
10.
11 - 18. Find x and the marked angles:
12.
14.
16.
18.
19 - 26. For each of the following, list all pairs of alternate interior angles and corresponding angles, If there are none, then list all pairs of interior angles on the same side of the transversal. Indicate the parallel lines which form each pair of angles.
20.
22.
24.
26.
A telescope is pointed at a star 50∘ above the horizon. What angle x∘ must the mirror BD make wiht the horizontal so that the star can be seen in the eyepiece E?
A periscope is used by sailors in a submarine to see objects on the surface of the water, If ∠ECF=90∘, what angle x∘ does the mirror BD make with the horizontal?
This page titled 2.5: Finding Angle Measurements is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Henry Africk (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform.
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2.4: The Rectangular Coordinate Systems and Graphs
2.6: Parallel and Perpendicular Lines
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1328 | https://www.doubtnut.com/qna/642562996 | Determine f(0) so that the function f(x) defined by f(x)=(4x−1)3sinx4log(1+x23) becomes continuous at x=0
To determine f(0) such that the function f(x)=(4x−1)3sinx4log(1+x23) is continuous at x=0, we need to find the limit of f(x) as x approaches 0. For the function to be continuous at x=0, we must have: limx→0f(x)=f(0) Step 1: Evaluate the limit We start by evaluating the limit: limx→0f(x)=limx→0(4x−1)3sinx4log(1+x23) Step 2: Expand 4x−1 Using the expansion for 4x: 4x=exln4≈1+xln4+(xln4)22+… Thus, 4x−1≈xln4 So, (4x−1)3≈(xln4)3=x3(ln4)3 Step 3: Expand sinx Using the Taylor series expansion for sinx: sinx≈x−x36+… Therefore, sin(x4)≈x4−16(x4)3+…=x4−x3384+… Step 4: Expand log(1+x23) Using the Taylor series expansion for log(1+x): log(1+x)≈x−x22+… Thus, log(1+x23)≈x23−12(x23)2+…≈x23 Step 5: Substitute expansions into f(x) Now substituting these expansions back into f(x): f(x)≈x3(ln4)3x4⋅x23=x3(ln4)3x312=12(ln4)3 Step 6: Take the limit as x→0 Taking the limit as x approaches 0: limx→0f(x)=12(ln4)3 Step 7: Set f(0) For continuity at x=0: f(0)=limx→0f(x)=12(ln4)3 Thus, we conclude: f(0)=12(ln4)3
To determine f(0) such that the function
f(x)=(4x−1)3sinx4log(1+x23)
is continuous at x=0, we need to find the limit of f(x) as x approaches 0. For the function to be continuous at x=0, we must have:
limx→0f(x)=f(0)
Step 1: Evaluate the limit
We start by evaluating the limit:
limx→0f(x)=limx→0(4x−1)3sinx4log(1+x23)
Step 2: Expand 4x−1
Using the expansion for 4x:
4x=exln4≈1+xln4+(xln4)22+…
Thus,
4x−1≈xln4
So,
(4x−1)3≈(xln4)3=x3(ln4)3
Step 3: Expand sinx
Using the Taylor series expansion for sinx:
sinx≈x−x36+…
Therefore,
sin(x4)≈x4−16(x4)3+…=x4−x3384+…
Step 4: Expand log(1+x23)
Using the Taylor series expansion for log(1+x):
log(1+x)≈x−x22+…
Thus,
log(1+x23)≈x23−12(x23)2+…≈x23
Step 5: Substitute expansions into f(x)
Now substituting these expansions back into f(x):
f(x)≈x3(ln4)3x4⋅x23=x3(ln4)3x312=12(ln4)3
Step 6: Take the limit as x→0
Taking the limit as x approaches 0:
limx→0f(x)=12(ln4)3
Step 7: Set f(0)
For continuity at x=0:
f(0)=limx→0f(x)=12(ln4)3
Thus, we conclude:
f(0)=12(ln4)3
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Guide: Long Sleeve Dresses
April 8 2024 - Average reading time: 7 min
Written by: Salva Jovells
Share:
Every woman wants to dress to impress without actually looking like she tried too hard. It’s all about looking effortlessly chic while staying comfortable and feeling confident. One of the easiest ways to accomplish just that is by slipping into an elegant, long sleeve dress. Whether for a busy workday, an afternoon occasion, or an evening event, a long sleeve dress is a quick way to instantly elevate your look.
Burgundy Boat Neck Long Dress
Of course, dresses come in various silhouettes, fabrics, styles, and other variations. Not all dresses are for everyone. Like with all clothing, certain styles work best for certain body types and certain cuts can feel more comfortable to wear. Having said this, long sleeve dresses are the ideal choice for many women and for multiple reasons.
Burgundy Long-Sleeve Dress
We know that not all women are comfortable showing off their arms, especially for certain occasions or on busy days. During a hectic workday, the last thing you want to have to worry about is feeling self-conscious about raising your arms or moving around in a sleeveless dress. A long sleeve midi dress will spare you from all the unnecessary stress. On the other hand, during formal events, you want to be able to enjoy the occasion and the company you’re with. You don’t want to feel self-conscious and you shouldn’t have to concern yourself with how your arms will look in photos or if your strapless dress will slip down. A black long sleeve dress is the perfect choice to enjoy the occasion and look timelessly elegant while doing so.
| |
| |
| Black Halter Maxi Dress | | Green Dusty V-Neck Dress |
Long sleeve dresses are the obvious choice for the fall and winter seasons when the temperature drops and you want to cover up. A long sleeve shift dress or a long sleeve maxi dress paired with chic knee-high boots is not only fashionable but will keep you feeling warm and cozy too. However, long sleeve dresses are not only meant for the short, fall days and long winter nights.
Green Long Sleeve Dress
Even during the spring season or the warmer summer months, there’s a time and place for long sleeve dresses. The changeable spring weather makes it ideal to wear a long sleeve midi dress that can take you from day to night. Whether the temperatures lean towards cool or cold, you’ll be comfortable from sun up to sun down. On the other hand, we all know that during the summer months, while it may be warm outside, it’s usually frigid indoors. When most places set their air conditioning to maximum, a long sleeve knee-length dress is the smart choice, allowing for air circulation while keeping you adequately covered.
Lavender Long Sleeve V-Neck Dress
Of course, you can also wear a long sleeve dress for work. This time, reach for a sheath or shift dress reaching at least to your knees and you've got a perfect, appropriate work outfit.
Blue Sheath Dress with Long Sleeves, Pink Shift Dress
When it comes to styling, long sleeve dresses are a great canvas for accessories. Whatever silhouette the dress may have, whatever length it may be, and whatever shape its neckline is, you can pair it with a plethora of beautiful jewelry and eye-catching accessories. Go for clean and simple with a burgundy long sleeve evening dress and statement earrings. If you’re wearing a knee-length long sleeve dress with a deep v-neckline, fasten a necklace or a fun choker around your neck. Opting for a long sleeve midi dress instead? Why not accentuate the sleeves with a chic cuff around both of your wrists? Whatever your personal style may be, there’s surely a long sleeve dress for you.
Types of Long Sleeve Dresses
Elegant Evening Gowns: When it comes to formal events, long sleeve dresses offer a blend of sophistication and style. Elegant evening gowns with long sleeves are not just about covering up; they are a fashion statement in themselves. Look for gowns crafted from luxurious fabrics like silk, velvet, or lace to make a lasting impression. The fit and design of these dresses play a crucial role - a well-fitted bodice with a flowing skirt can be both flattering and comfortable. Pay attention to details like embroidery, sequins, or beading for that extra touch of elegance.
Casual Day Dresses: For everyday wear, casual long sleeve dresses are a must-have in your wardrobe. Emphasizing comfort and style, these dresses are perfect for a range of activities, from running errands to casual outings with friends. Opt for breathable fabrics like cotton or linen to stay comfortable throughout the day. Styles like wrap dresses, shirt dresses, or A-line dresses with long sleeves offer both ease and a fashion-forward look.
Professional Work Attire: In a professional setting, long sleeve dresses can convey a sense of authority and poise. Sheath and shift dresses with long sleeves are particularly suitable for the workplace. These styles are known for their sleek and straightforward design, making them an excellent choice for a business environment. Look for dresses in solid colors or subtle patterns, and pair them with a blazer or cardigan for a polished look.
Bohemian Styles: For those who love a more laid-back and artistic style, boho-chic long sleeve dresses are an ideal choice. These dresses often feature relaxed fits, unique prints, and flowy fabrics, embodying a free-spirited aesthetic. Long sleeves with details like bell shapes, ruffles, or lace inserts add to the bohemian charm. These dresses are perfect for festivals, outdoor events, or just a casual day out.
Cocktail Dresses: Long sleeve cocktail dresses strike the perfect balance between sophistication and allure for evening social events. These dresses are often shorter in length and come in a variety of styles, from bodycon to fit-and-flare. Look for features like sheer sleeves, cutouts, or off-shoulder designs to add a modern twist. Fabrics like satin, chiffon, or embellished tulle can elevate the overall look, making you the center of attention at any social gathering.
Green Boat Neck Maxi Dress
Summer Dresses Full Sleeves
Lightweight and Airy Fabrics: The key to staying comfortable in summer long sleeve dresses is choosing the right fabric. Opt for lightweight and breathable materials like cotton, linen, or chiffon. These fabrics allow for air circulation, keeping you cool even on the hottest days. Long sleeve dresses made from such materials are not only practical but can also be incredibly stylish, offering a perfect blend of comfort and fashion.
Vibrant Colors and Patterns: Summer is the time to embrace bright colors and fun patterns. Full sleeve dresses in vivid hues like turquoise, coral, or sunshine yellow can instantly uplift your mood and style. Floral prints, abstract patterns, or even bold stripes add a playful touch to your summer wardrobe. These lively designs are great for daytime outings, garden parties, or casual brunches.
Beach to Bar Styles: Versatility is key when choosing summer dresses. Look for long sleeve dresses that can easily transition from a day at the beach to a night out. Flowy maxi dresses or relaxed shirt dresses are perfect for this. Pair them with sandals for a beach look, and switch to heels and a clutch for an evening bar setting. Such adaptable styles make packing for summer vacations easier and more efficient.
Layering for Cooler Evenings: Even in summer, evenings can get a bit cooler. Layering your full sleeve dress is a stylish way to stay warm. A light denim jacket or a sheer kimono can be the perfect addition to your outfit. These layers can be easily carried around during the day and thrown on when the temperature drops, ensuring you’re prepared for any change in weather.
Accessorizing for Summer: The right accessories can elevate your summer long sleeve dress. Pair your dress with a wide-brimmed hat and oversized sunglasses for a chic, sun-protected look. When it comes to jewelry, opt for light and airy pieces like thin chain necklaces or bangle bracelets. A crossbody bag or a tote can complete the look, making you ready for any summer adventure.
Incorporating these elements into your summer wardrobe will ensure that your long sleeve dresses are not only fashionable but also functional for the warm season.
Body Type and Long Sleeve Dress Selection
Selecting the right long sleeve dress for your body type can make all the difference in how you look and feel. A flattering fit enhances your natural features, boosts confidence, and ensures comfort. Here’s a guide to help you choose the perfect long sleeve dress for various body types:
1. Pear-Shaped Body:
Objective: Balance the broader hips with the upper body.
Recommended Styles: A-line dresses or dresses with embellished or detailed tops draw attention upwards. Long sleeve dresses with a fit-and-flare silhouette can beautifully accentuate the waist and balance the hips.
Fabrics & Patterns: Opt for lighter colors or bolder patterns on the top half of the dress to create an illusion of balance.
2. Apple-Shaped Body:
Objective: Create a more defined waistline.
Recommended Styles: Empire line dresses or styles that flare out slightly from under the bust can be very flattering. Wrap dresses with long sleeves are excellent for creating a more cinched-in waist effect.
Fabrics & Patterns: Go for dresses with a darker color around the midsection, combined with brighter or lighter sleeves to draw attention to the arms and shoulders.
3. Hourglass Body Type:
Objective: Accentuate the naturally well-proportioned body structure.
Recommended Styles: Bodycon long sleeve dresses or belted styles that highlight the waist are ideal. Wrap dresses also work well by following the natural body line.
Fabrics & Patterns: You can experiment with both solid colors and patterns, as hourglass figures are versatile in terms of styling.
4. Rectangle Body Type:
Objective: Create curves or add dimension to the silhouette.
Recommended Styles: Peplum dresses or long sleeve dresses with ruching at the waist can add shape. Dresses with a belt or cinched waist can also create an illusion of curves.
Fabrics & Patterns: Play with bold prints and textures to add fullness and dimension to your figure.
5. Inverted Triangle Body Type:
Objective: Balance broader shoulders with the lower part of the body.
Recommended Styles: Dresses that are simple and unembellished at the top but have volume at the bottom, like a long-sleeve A-line dress, are ideal.
Fabrics & Patterns: Choose darker colors for the top part of the dress and brighter or more detailed designs for the bottom half.
General Tips:
Sleeve Style: The type of sleeve can also impact the overall look. Flared or bell sleeves can add volume to the arms, while fitted sleeves offer a sleeker appearance.
Length Matters: The length of the dress can influence the overall effect. Midi or knee-length dresses often work well for most body types, offering a balanced look.
Remember, these are just guidelines. The most important factor in selecting a long sleeve dress is how it makes you feel. Confidence is the best accessory, so choose a dress that you feel great wearing!
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1330 | https://www.youtube.com/watch?v=ed8iFz1jErU | The Difference of Powers Formula
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We go over the difference of powers formula.
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Transcript:
hello in this video I'm going to show you the difference of powers formula so the difference of powers formula so this is a powerful formula so it's a to the N minus B to the N okay and that's equal to a minus B time the finite sum as K runs from 0 to n -1 of a to the N - 1 - K time B to the K and that would be the difference of powers formula okay so uh for example we can have uh a 5 minus B the 5 and then you can apply uh this formula to work out um the rest and yeah that's it I hope it's been helpful take care |
1331 | https://math.stackexchange.com/questions/940171/how-to-prove-that-sin90v-cosv | trigonometry - how to prove that $\sin(90+v) = \cos(v)$ - Mathematics Stack Exchange
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how to prove that sin(90+v)=cos(v)
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I need to prove that sin(90+v)=cos v and that cos(90+v)=−sin v
So I did the following steps to prove these statements
sin(90+v)=sin(90−(−v))=cos(−v)=cos(v)
cos(90+v)=cos(90−(−v))=sin(−v)=−sin(v)
Is this correct?
Thanks!!
trigonometry
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edited Sep 21, 2014 at 13:14
Sejad DulicSejad Dulic
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Sejad DulicSejad Dulic
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You do not need to go through v=−(−v). Just do it directly.Claude Leibovici –Claude Leibovici 2014-09-21 13:11:34 +00:00 Commented Sep 21, 2014 at 13:11
Yes, that is absolutely correct! If you already know the identities sin(90−x)=cos x and cos(90−x)=sin x, this is a great way to prove it.user795305 –user795305 2014-09-21 13:14:00 +00:00 Commented Sep 21, 2014 at 13:14
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Invariably, the proof requires the angle-sum/angle-difference identities:
sin(a±b)=sin a cos b±sin b cos a
cos(a±b)=cos a cos b∓sin a sin b
Your work ultimately appeals to the angle-difference identities, where one angle is 90∘.
I'll base the following on the angle-sum identities.
From the first, we have sin(90+x)=sin(90)cos(x)+sin x cos(90)=cos x+0=cos x
For the second, we have cos(90+x)=cos(90)cos(x)−sin(90)sin(x)=0−sin(x)=−sin x
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edited Sep 21, 2014 at 13:20
answered Sep 21, 2014 at 13:11
amWhyamWhy
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Use the following identities:
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)cos(a+b)=cos(a)cos(b)−sin(a)sin(b)
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answered Sep 21, 2014 at 13:12
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1332 | https://my.clevelandclinic.org/health/diseases/21771-gingivostomatitis | Gingivostomatitis: Types, Symptoms & Treatment
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Gingivostomatitis
Gingivostomatitis is an oral infection that causes painful sores. Certain viruses and bacteria cause it, and poor oral hygiene makes you more likely to get it. Treatment usually involves antivirals or antibiotics — and you can avoid triggers to reduce your risk of flare-ups.
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ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentPreventionOutlook / PrognosisLiving WithAdditional Common Questions
ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentPreventionOutlook / PrognosisLiving WithAdditional Common Questions
Overview
What is gingivostomatitis?
Gingivostomatitis is a painful infection that can cause blisters on your lips and canker sores in your mouth. Certain viruses and bacteria cause it, and poor oral hygiene can make it worse. Gingivostomatitis is most common in young children, but anyone can get it. Gingivostomatitis isn’t the same as gingivitis, the earliest stage of gum disease.
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Gingivostomatitis usually doesn’t cause serious health issues, especially when you get prompt treatment. But it can be very painful. It’s also contagious. You can pass it to another person through things like kissing or sharing eating utensils.
How common is gingivostomatitis?
Gingivostomatitis is quite common, partly because the condition can come back (recur) repeatedly in the form of flare-ups. Once initially infected, approximately 40% of children will develop recurring mouth sores.
Symptoms and Causes
What are the symptoms of gingivostomatitis?
Gingivostomatitis symptoms can vary, and they may be mild or severe, including:
Bad breath (halitosis).
Dry mouth (xerostomia).
Fever.
Headaches.
Loss of appetite due to mouth pain.
Painful sores on your lips, gums, tongue or inner cheeks.
Red, tender or swollen gums.
Swollen lymph nodes in your neck.
The above symptoms can affect both children and adults.
What causes gingivostomatitis?
Gingivostomatitis can develop due to certain viruses or bacteria, including:
Herpes simplex virus type 1 (HSV-1). The most common cause of gingivostomatitis, HSV-1 is the same virus that causes cold sores. When HSV-1 causes gingivostomatitis, providers call it “herpes simplex gingivostomatitis” or “herpetic gingivostomatitis.” (Primary herpetic gingivostomatitis refers to the first time you encountered the virus. Secondary herpetic gingivostomatitis refers to each flare-up, or reactivation, of the virus.)
Streptococcus. This bacteria commonly causes strep throat or blood infections, but it can also result in gingivostomatitis.
Actinomyces. This bacteria naturally occurs in your mouth, but it can lead to gingivostomatitis if it enters your bloodstream. This can be a side effect of dental trauma or oral surgery.
Coxsackieviruses. Coxsackieviruses are viruses that usually spread through unwashed hands or other surfaces contaminated with feces (poop). Coxsackieviruses also commonly cause hand, foot and mouth disease.
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Gingivostomatitis risk factors
Poor oral hygiene, like not brushing or flossing enough, is the main risk factor for gingivostomatitis.
If you have herpetic gingivostomatitis, you’re more likely to experience flare-ups when exposed to:
Fever.
Trauma.
Stress.
UV light.
Diagnosis and Tests
How is gingivostomatitis diagnosed?
Healthcare providers can usually diagnose gingivostomatitis during a physical examination. Your provider will also ask about symptoms.
They may also recommend a swab culture or biopsy of the affected area to confirm which type of bacteria or virus caused the infection. They’ll send the sample to a pathologist for testing.
Management and Treatment
How is gingivostomatitis treated?
Gingivostomatitis treatment may include antibiotics or antivirals to get rid of the infection and ease your symptoms. In some cases, your healthcare provider will need to clean the affected areas.
To further ease gingivostomatitis symptoms:
Take over-the-counter (OTC) pain relievers, like acetaminophen (Tylenol®) or ibuprofen (Advil®).
Rinse your mouth with an antibacterial mouthwash twice a day.
Gently swish with warm salt water a few times daily to soothe your mouth.
Avoid eating hot, spicy or salty foods.
Brush twice a day and floss once a day.
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Prevention
Can gingivostomatitis be prevented?
To reduce your risk for gingivostomatitis:
Practice good oral hygiene.
Routinely clean any oral appliances like dentures or retainers.
Visit your dentist regularly for checkups and cleanings.
Outlook / Prognosis
What’s the outlook for gingivostomatitis?
It depends on the severity. Some people have mild discomfort; others have severe pain. Most gingivostomatitis-related mouth sores heal in about two to three weeks.
After mouth sores appear, you’ll be contagious with gingivostomatitis for about seven days. You should be fever-free for at least 24 hours before having close contact with anyone else.
Living With
When should I see my healthcare provider?
Any time you develop mouth sores along with a fever, you should make an appointment with your healthcare provider. If your symptoms worsen or don’t respond to treatment within three weeks, you should ask your provider about next steps.
What questions should I ask my doctor?
If you have gingivostomatitis, here are some questions you might want to ask your healthcare provider:
Why did I get gingivostomatitis?
What kind of treatment do I need?
What precautions should I take?
How can I reduce my risk for flare-ups?
Additional Common Questions
Is gingivostomatitis an STI?
No, it’s not a sexually transmitted infection. But HSV-1 can cause it. HSV-1 commonly causes oral herpes, but it can cause genital herpes in some cases.
A note from Cleveland Clinic
Gingivostomatitis can cause frustrating and embarrassing symptoms. It’s painful, and it makes routine tasks like eating and speaking uncomfortable. The good news is that gingivostomatitis is treatable. If you develop symptoms, don’t wait. Call your healthcare provider right away. They can find ways to ease symptoms and help you feel better faster.
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1333 | https://www.xconvert.com/unit-converter/cubic-feet-per-minute-to-gallons-per-second | Cubic feet per minute to Gallons per second | Convert ft3/min To gal/s Online - XConvert
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Cubic feet per minute (ft 3/min) to Gallons per second (gal/s) conversion
Volume Flow Rate
From Cubic feet per minute
ft 3/min
To Gallons per second
gal/s
Swap units Copy result
Cubic feet per minute to Gallons per second conversion table
| Cubic feet per minute (ft 3/min) | Gallons per second (gal/s) |
--- |
| 0 | 0 |
| 1 | 0.1246752604167 |
| 2 | 0.2493505208333 |
| 3 | 0.37402578125 |
| 4 | 0.4987010416667 |
| 5 | 0.6233763020833 |
| 6 | 0.7480515625 |
| 7 | 0.8727268229167 |
| 8 | 0.9974020833333 |
| 9 | 1.12207734375 |
| 10 | 1.2467526041667 |
| 20 | 2.4935052083333 |
| 30 | 3.7402578125 |
| 40 | 4.9870104166667 |
| 50 | 6.2337630208333 |
| 60 | 7.480515625 |
| 70 | 8.7272682291667 |
| 80 | 9.9740208333333 |
| 90 | 11.2207734375 |
| 100 | 12.467526041667 |
| 1000 | 124.67526041667 |
How to convert cubic feet per minute to gallons per second?
Converting Cubic Feet per Minute (CFM) to Gallons per Second (GPS) involves two main steps.
Convert CFM to Cubic Feet per Second (CFS): Since there are 60 seconds in a minute, divide the number of cubic feet by 60 to convert to cubic feet per second.
Convert Cubic Feet to Gallons: One cubic foot is equivalent to 7.48052 gallons.
Using these steps, let's convert 1 CFM to GPS:
Step 1: Convert CFM to CFS
1 CFM / 60 = 0.0166667 cubic feet per second (CFS)
Step 2: Convert CFS to GPS
0.0166667 CFS × 7.48052 gallons per cubic foot = 0.1246753 gallons per second (GPS)
So, 1 CFM is approximately 0.1247 GPS.
Real-World Examples for Other Quantities of CFM:
10 CFM:
Convert to CFS: 10 / 60 = 0.1667 CFS
Convert to GPS: 0.1667 CFS 7.48052 gallons per cubic foot = 1.24675 GPS
So, 10 CFM is approximately 1.2467 GPS.
50 CFM:
Convert to CFS: 50 / 60 = 0.8333 CFS
Convert to GPS: 0.8333 CFS 7.48052 gallons per cubic foot = 6.23375 GPS
So, 50 CFM is approximately 6.2337 GPS.
100 CFM:
Convert to CFS: 100 / 60 = 1.6667 CFS
Convert to GPS: 1.6667 CFS 7.48052 gallons per cubic foot = 12.4675 GPS
So, 100 CFM is approximately 12.4675 GPS.
Additional Real-world Scenarios:
Air Conditioning systems: Large HVAC units often have air flow rates around 400 to 600 CFM. Using our conversion, 600 CFM would be approximately 74.805 GPS (600 / 60 7.48052).
Bathroom Exhaust Fans: These typically run around 50 to 110 CFM. So, a 100 CFM exhaust fan would approximately translate to 12.4675 GPS.
Industrial Blowers: Systems used in industrial settings can have air flow rates in the thousands. For instance, a blower with 2000 CFM would convert to approximately 249.35 GPS (2000 / 60 7.48052).
These conversions and examples demonstrate how CFM can be translated into GPS, helping to understand the fluid dynamics in a variety of systems and settings.
See below section for step by step unit conversion with formulas and explanations. Please refer to the table below for a list of all the Gallons per second to other unit conversions.
What is cubic feet per minute?
What is Cubic feet per minute?
Cubic feet per minute (CFM) is a unit of measurement that expresses the volume of a substance (usually air or gas) flowing per minute. It's commonly used to measure airflow in ventilation, HVAC systems, and other industrial processes. Understanding CFM helps in selecting appropriate equipment and ensuring efficient system performance.
Understanding Cubic Feet per Minute (CFM)
Definition
CFM defines the amount of cubic feet that passes through a specific area in one minute. It is a standard unit for measuring volume flow rate in the United States.
How it is formed?
CFM is derived from the units of volume (cubic feet, f t 3 ft^3) and time (minutes, min). Therefore, 1 CFM means one cubic foot of a substance passes a specific point every minute.
Formula
The relationship between volume, time, and CFM can be expressed as:
C F M=V o l u m e(f t 3)T i m e(m i n u t e s)CFM = \frac{Volume (ft^3)}{Time (minutes)}
Real-World Applications and Examples
HVAC Systems
Home Ventilation: A typical bathroom exhaust fan might have a CFM rating of 50-100, depending on the bathroom's size. This ensures adequate removal of moisture and odors.
Air Conditioners: The CFM rating of a central air conditioning system is crucial for proper cooling. For instance, a 2.5-ton AC unit might require around 1000 CFM to effectively cool a space.
Furnaces: Furnaces use CFM to ensure proper airflow across the heat exchanger, maintaining efficiency and preventing overheating.
Industrial Applications
Pneumatic Tools: Air compressors powering pneumatic tools (like nail guns or impact wrenches) are often rated by CFM delivered at a certain pressure (PSI). For example, a heavy-duty impact wrench might require 5 CFM at 90 PSI.
Spray Painting: Air compressors used for spray painting need a specific CFM to atomize the paint properly. An automotive paint job may require a compressor delivering 10-15 CFM at 40 PSI.
Dust Collection: Dust collection systems in woodworking shops use CFM to extract sawdust and debris from the air, maintaining a clean and safe working environment. A small shop might use a system with 600-800 CFM.
Other Examples
Computer Cooling: Fans used to cool computer components (CPUs, GPUs) are rated in CFM to indicate how much air they can move across the heat sink.
Leaf Blowers: Leaf blowers are often specified by CFM, indicating their ability to move leaves and debris.
Interesting Facts
Standard Conditions
When comparing CFM values, it's important to note the conditions under which they were measured. Standard conditions for airflow are typically at a specific temperature and pressure (e.g., Standard Temperature and Pressure, or STP).
Conversion to Other Units
CFM can be converted to other volume flow rate units, such as cubic meters per hour (m 3/h m^3/h) or liters per second (L/s), using appropriate conversion factors.
1 CFM ≈ 1.699 m 3/h m^3/h
1 CFM ≈ 0.472 L/s
Relationship to Velocity
CFM is related to air velocity and the cross-sectional area of the flow. The formula linking these is:
C F M=A r e a(f t 2)×V e l o c i t y(f t/m i n)CFM = Area (ft^2) \times Velocity (ft/min)
This relationship is crucial in designing ductwork and ventilation systems to ensure proper airflow. You can find more about this relationship on engineering websites such as How to measure air volume flow or air velocity?
What is Gallons per Second (GPS)?
Gallons per second (GPS) is a measurement unit that tells you how many gallons of a liquid are moving past a certain point every second. It's a rate, showing volume over time. It is commonly used in the US to measure high volume flow rates.
How is GPS Formed?
GPS is formed by dividing a volume measured in gallons by a time measured in seconds.
G P S=V o l u m e(G a l l o n s)T i m e(S e c o n d s)GPS = \frac{Volume (Gallons)}{Time (Seconds)}
For example, if 10 gallons of water flow out of a pipe in 2 seconds, the flow rate is 5 gallons per second.
Conversions and Relationships
GPS can be converted to other common flow rate units:
1 Gallon ≈ 0.00378541 Cubic Meters
1 GPS ≈ 0.00378541 m 3/s m^3/s
1 GPS ≈ 3.78541 Liters/second
Real-World Applications and Examples
Firefighting: Fire hoses and sprinkler systems are often rated in GPS to indicate their water delivery capacity. A typical fire hydrant might deliver 500-1000 GPS.
Pumping Stations: Large pumping stations, such as those used in water treatment plants or flood control, can have flow rates measured in thousands of GPS.
Industrial Processes: Many industrial processes, such as chemical manufacturing or oil refining, involve the movement of large volumes of fluids, and GPS is used to measure flow rates in these processes.
River Flow: While not a direct measurement, river discharge rates can be expressed in terms relatable to GPS (e.g., converting cubic feet per second to GPS for easier understanding).
The average flow rate of the Mississippi River is around 600,000 cubic feet per second, which is approximately 4.5 million GPS.
Pool filling: Average garden hose has 5-10 gallons per minute. This means it will take around 30 minutes to fill a 150 gallon pool. This is 0.08 - 0.17 GPS.
Historical Context and Interesting Facts
While no single person is specifically associated with the "invention" of GPS as a unit, its use is tied to the development of fluid mechanics and hydraulics. Understanding flow rates became crucial with the rise of industrialization and the need to efficiently manage and transport fluids.
The measurement of flow rates dates back to ancient civilizations that developed aqueducts and irrigation systems. However, the standardization of units like GPS is a more recent development, driven by the need for precise measurements in engineering and scientific applications.
Complete Cubic feet per minute conversion table
Enter # of Cubic feet per minute
| Convert 1 ft 3/min to other units | Result |
--- |
| Cubic feet per minute to Cubic Millimeters per second (ft 3/min to mm 3/s) | 471947.19998024 |
| Cubic feet per minute to Cubic Centimeters per second (ft 3/min to cm 3/s) | 471.94719998024 |
| Cubic feet per minute to Cubic Decimeters per second (ft 3/min to dm 3/s) | 0.4719471999802 |
| Cubic feet per minute to Cubic Decimeters per minute (ft 3/min to dm 3/min) | 28.316831998815 |
| Cubic feet per minute to Cubic Decimeters per hour (ft 3/min to dm 3/h) | 1699.0099199289 |
| Cubic feet per minute to Cubic Decimeters per day (ft 3/min to dm 3/d) | 40776.238078293 |
| Cubic feet per minute to Cubic Decimeters per year (ft 3/min to dm 3/a) | 14893520.958096 |
| Cubic feet per minute to Millilitres per second (ft 3/min to ml/s) | 471.94719998024 |
| Cubic feet per minute to Centilitres per second (ft 3/min to cl/s) | 47.194719998024 |
| Cubic feet per minute to Decilitres per second (ft 3/min to dl/s) | 4.7194719998024 |
| Cubic feet per minute to Litres per second (ft 3/min to l/s) | 0.4719471999802 |
| Cubic feet per minute to Litres per minute (ft 3/min to l/min) | 28.316831998815 |
| Cubic feet per minute to Litres per hour (ft 3/min to l/h) | 1699.0099199289 |
| Cubic feet per minute to Litres per day (ft 3/min to l/d) | 40776.238078293 |
| Cubic feet per minute to Litres per year (ft 3/min to l/a) | 14893520.958096 |
| Cubic feet per minute to Kilolitres per second (ft 3/min to kl/s) | 0.0004719471999802 |
| Cubic feet per minute to Kilolitres per minute (ft 3/min to kl/min) | 0.02831683199881 |
| Cubic feet per minute to Kilolitres per hour (ft 3/min to kl/h) | 1.6990099199289 |
| Cubic feet per minute to Cubic meters per second (ft 3/min to m 3/s) | 0.0004719471999802 |
| Cubic feet per minute to Cubic meters per minute (ft 3/min to m 3/min) | 0.02831683199881 |
| Cubic feet per minute to Cubic meters per hour (ft 3/min to m 3/h) | 1.6990099199289 |
| Cubic feet per minute to Cubic meters per day (ft 3/min to m 3/d) | 40.776238078293 |
| Cubic feet per minute to Cubic meters per year (ft 3/min to m 3/a) | 14893.520958096 |
| Cubic feet per minute to Cubic kilometers per second (ft 3/min to km 3/s) | 4.7194719998024e-13 |
| Cubic feet per minute to Teaspoons per second (ft 3/min to tsp/s) | 95.7506 |
| Cubic feet per minute to Tablespoons per second (ft 3/min to Tbs/s) | 31.916866666667 |
| Cubic feet per minute to Cubic inches per second (ft 3/min to in 3/s) | 28.800117906793 |
| Cubic feet per minute to Cubic inches per minute (ft 3/min to in 3/min) | 1728.0070744076 |
| Cubic feet per minute to Cubic inches per hour (ft 3/min to in 3/h) | 103680.42446446 |
| Cubic feet per minute to Fluid Ounces per second (ft 3/min to fl-oz/s) | 15.958433333333 |
| Cubic feet per minute to Fluid Ounces per minute (ft 3/min to fl-oz/min) | 957.506 |
| Cubic feet per minute to Fluid Ounces per hour (ft 3/min to fl-oz/h) | 57450.36 |
| Cubic feet per minute to Cups per second (ft 3/min to cup/s) | 1.9948041666667 |
| Cubic feet per minute to Pints per second (ft 3/min to pnt/s) | 0.9974020833333 |
| Cubic feet per minute to Pints per minute (ft 3/min to pnt/min) | 59.844125 |
| Cubic feet per minute to Pints per hour (ft 3/min to pnt/h) | 3590.6475 |
| Cubic feet per minute to Quarts per second (ft 3/min to qt/s) | 0.4987010416667 |
| Cubic feet per minute to Gallons per second (ft 3/min to gal/s) | 0.1246752604167 |
| Cubic feet per minute to Gallons per minute (ft 3/min to gal/min) | 7.480515625 |
| Cubic feet per minute to Gallons per hour (ft 3/min to gal/h) | 448.8309375 |
| Cubic feet per minute to Cubic feet per second (ft 3/min to ft 3/s) | 0.01666666666667 |
| Cubic feet per minute to Cubic feet per hour (ft 3/min to ft 3/h) | 60 |
| Cubic feet per minute to Cubic yards per second (ft 3/min to yd 3/s) | 0.0006172830432927 |
| Cubic feet per minute to Cubic yards per minute (ft 3/min to yd 3/min) | 0.03703698259756 |
| Cubic feet per minute to Cubic yards per hour (ft 3/min to yd 3/h) | 2.2222189558537 |
Volume flow rate conversions
Cubic feet per minute to Cubic Millimeters per second (ft 3/min to mm 3/s)
Cubic feet per minute to Cubic Centimeters per second (ft 3/min to cm 3/s)
Cubic feet per minute to Cubic Decimeters per second (ft 3/min to dm 3/s)
Cubic feet per minute to Cubic Decimeters per minute (ft 3/min to dm 3/min)
Cubic feet per minute to Cubic Decimeters per hour (ft 3/min to dm 3/h)
Cubic feet per minute to Cubic Decimeters per day (ft 3/min to dm 3/d)
Cubic feet per minute to Cubic Decimeters per year (ft 3/min to dm 3/a)
Cubic feet per minute to Millilitres per second (ft 3/min to ml/s)
Cubic feet per minute to Centilitres per second (ft 3/min to cl/s)
Cubic feet per minute to Decilitres per second (ft 3/min to dl/s)
Cubic feet per minute to Litres per second (ft 3/min to l/s)
Cubic feet per minute to Litres per minute (ft 3/min to l/min)
Cubic feet per minute to Litres per hour (ft 3/min to l/h)
Cubic feet per minute to Litres per day (ft 3/min to l/d)
Cubic feet per minute to Litres per year (ft 3/min to l/a)
Cubic feet per minute to Kilolitres per second (ft 3/min to kl/s)
Cubic feet per minute to Kilolitres per minute (ft 3/min to kl/min)
Cubic feet per minute to Kilolitres per hour (ft 3/min to kl/h)
Cubic feet per minute to Cubic meters per second (ft 3/min to m 3/s)
Cubic feet per minute to Cubic meters per minute (ft 3/min to m 3/min)
Cubic feet per minute to Cubic meters per hour (ft 3/min to m 3/h)
Cubic feet per minute to Cubic meters per day (ft 3/min to m 3/d)
Cubic feet per minute to Cubic meters per year (ft 3/min to m 3/a)
Cubic feet per minute to Cubic kilometers per second (ft 3/min to km 3/s)
Cubic feet per minute to Teaspoons per second (ft 3/min to tsp/s)
Cubic feet per minute to Tablespoons per second (ft 3/min to Tbs/s)
Cubic feet per minute to Cubic inches per second (ft 3/min to in 3/s)
Cubic feet per minute to Cubic inches per minute (ft 3/min to in 3/min)
Cubic feet per minute to Cubic inches per hour (ft 3/min to in 3/h)
Cubic feet per minute to Fluid Ounces per second (ft 3/min to fl-oz/s)
Cubic feet per minute to Fluid Ounces per minute (ft 3/min to fl-oz/min)
Cubic feet per minute to Fluid Ounces per hour (ft 3/min to fl-oz/h)
Cubic feet per minute to Cups per second (ft 3/min to cup/s)
Cubic feet per minute to Pints per second (ft 3/min to pnt/s)
Cubic feet per minute to Pints per minute (ft 3/min to pnt/min)
Cubic feet per minute to Pints per hour (ft 3/min to pnt/h)
Cubic feet per minute to Quarts per second (ft 3/min to qt/s)
Cubic feet per minute to Gallons per second (ft 3/min to gal/s)
Cubic feet per minute to Gallons per minute (ft 3/min to gal/min)
Cubic feet per minute to Gallons per hour (ft 3/min to gal/h)
Cubic feet per minute to Cubic feet per second (ft 3/min to ft 3/s)
Cubic feet per minute to Cubic feet per hour (ft 3/min to ft 3/h)
Cubic feet per minute to Cubic yards per second (ft 3/min to yd 3/s)
Cubic feet per minute to Cubic yards per minute (ft 3/min to yd 3/min)
Cubic feet per minute to Cubic yards per hour (ft 3/min to yd 3/h) |
1334 | https://www.gauthmath.com/solution/Expand-and-simplify-cos-sin-2--1695634841730054 | Solved: Expand and simplify: (cos θ -sin θ )^2 [Math]
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Expand and simplify: (cos θ -sin θ )^2
Expert Verified Solution
97%(551 rated)
Answer
$$1 - 2\cos \theta \sin \theta$$1−2 cos θ sin θ
Explanation
Apply the square of a binomial formula $$(a - b)^{2} = a^{2} - 2ab + b^{2}$$(a−b)2=a 2−2 ab+b 2 to the expression $$(\cos \theta - \sin \theta)^{2}$$(cos θ−sin θ)2
Calculate the square of each term: $$\cos^{2} \theta$$cos 2 θ and $$\sin^{2} \theta$$sin 2 θ
Multiply the terms: $$2 \cdot \cos \theta \cdot \sin \theta$$2⋅cos θ⋅sin θ
Combine the results to get the expanded form: $$\cos^{2} \theta - 2\cos \theta \sin \theta + \sin^{2} \theta$$cos 2 θ−2 cos θ sin θ+sin 2 θ
Recognize that $$\cos^{2} \theta + \sin^{2} \theta = 1$$cos 2 θ+sin 2 θ=1 (Pythagorean identity).
Substitute $$\cos^{2} \theta + \sin^{2} \theta$$cos 2 θ+sin 2 θ with 1 to simplify the expression: $$1 - 2\cos \theta \sin \theta$$1−2 cos θ sin θ
So, the simplified form of $$(\cos \theta - \sin \theta)^{2}$$(cos θ−sin θ)2 is $$1 - 2\cos \theta \sin \theta$$1−2 cos θ sin θ
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a Solve, in the interval 0< θ <180 ° , icosec θ =2cot θ ⅱ 2cot 2 θ =7cos ec θ -8 b Solve, in the interval 0 ≤ slant θ ≤ slant 360 ° , i sec 2 θ -15 ° =cosec 135 ° ⅱ sec 2 θ +tan θ =3 c Solve, in the interval 0 ≤ slant x ≤ slant 2 π , icos ecx+frac π 15=- square root of 2 ⅱ sec 2x= 4/3
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1335 | https://www.medscape.com/viewarticle/456334_2 | Calcific Aortic Stenosis in the Elderly: A Brief Overview - Page 2
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The American Journal of Geriatric Cardiology
Calcific Aortic Stenosis in the Elderly: A Brief Overview
Neil Sawhney, MD, Alborz Hassankhani, MD, PhD, Barry H. Greenberg, MD
Disclosures;)
Am J Geriatr Cardiol.2003;12(3)
5
In This Article
Abstract and Introduction
Pathophysiology
Diagnosis
Management
Conclusion
References
Pathophysiology
The most common etiologies of aortic stenosis are rheumatic heart disease, congenital bicuspid valve, and senile calcific aortic valve disease. In industrialized countries, rheumatic aortic valvular disease has become very uncommon. Furthermore, it usually causes multivalvular disease, and only rarely causes isolated aortic stenosis. On the other hand, bicuspid aortic valve is the most common congenital abnormality (excluding mitral valve prolapse) -- occurring in almost 2% of all births. It is estimated that one third of patients with bicuspid valves will develop aortic stenosis. However, since the vast majority of these patients develop symptoms in their fourth to sixth decades of life, congenital bicuspid valve is not the major cause of aortic stenosis in the elderly.
In patients greater than age 65, more than 90% of aortic stenosis involves trileaflet valves which have developed heavy calcification. In these instances the calcium deposits are thought to alter the structure and motion of the valve leaflets and thus prevent the normal opening of the aortic cusps. The cause of this calcification is unknown. It is thought that mechanical stress on the valve over time causes structural changes predisposing it to calcification. Interestingly, diabetes, hypertension, hypercholesterolemia, and end-stage renal disease have all been shown to be risk factors for the development of aortic stenosis, leading to speculation that aortic stenosis may be a form of atherosclerotic disease and thus may be prevented by aggressive lipid lowering.[1,6] The available data suggests that the statins could have the potential to attenuate progression of calcific aortic stenosis.
The pathophysiologic changes caused by a calcified aortic valve are the same as those caused by other forms of aortic stenosis and are related to the pressure gradient that develops between the left ventricle and aorta. Usually, when the aortic valve opens in systole, there is no gradient between the two chambers. Even in early aortic stenosis there is no gradient at rest. However, with increased blood flow, as occurs during exercise, a gradient may develop. As the stenosis progresses, a resting gradient develops across the aortic valve. The left ventricle responds by developing hypertrophy to overcome the stenosis. The increased wall thickness helps normalize wall stress and maintain cardiac output in the normal range. Since there is little hemodynamic consequence to mild and moderate aortic stenosis, the patient can tolerate this condition for years before the onset of symptoms. At some point, however, the left ventricle cannot generate an adequate pressure to maintain flow -- thus resulting in symptoms. These symptoms appear first during exercise. As the degree of stenosis becomes worse, symptoms may occur at rest. In addition, the left ventricle stiffens as it develops hypertrophy. This leads to decreased compliance and subsequent diastolic filling abnormalities. The pulmonary capillary bed is usually protected from these diastolic filling abnormalities by a forceful left atrial contraction, which ensures filling of the stiff ventricle. In normal subjects, the left atrial "kick" contributes in the range of 20% to the stroke volume. Since this percentage increases substantially in aortic stenosis, the loss of the atrial kick with atrial fibrillation can significantly compromise cardiac output.
Severe aortic stenosis is usually defined as an aortic valve area (AVA) of 1.0 cm 2 because at this degree of stenosis the peak resting gradient often approaches 50 mm Hg. Moreover, the gradient increases substantially with exercise, often causing symptoms. However, it is difficult to define critical aortic stenosis in terms of gradients because late in the course of the disease the left ventricle may begin to fail. When this occurs, the resultant decrease in left ventricular contraction may decrease the transvalvular gradient despite significant obstruction. Furthermore, it is important to judge the AVA in the context of the patient's size. For instance, an AVA of 1.0 cm 2 may be quite adequate for a 110-pound grandmother, but would certainly not suffice for a 6 foot, 350-pound professional athlete. Given these limitations, critical aortic stenosis is usually defined as an AVA of 0.70 cm 2 or less. As discussed below, the distinction between severe and critical aortic stenosis may be of little clinical consequence.
5
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Am J Geriatr Cardiol.2003;12(3)©2003 Le Jacq Communications, Inc.
Cite this: Calcific Aortic Stenosis in the Elderly: A Brief Overview-Medscape-May 01,2003.
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Abstract and Introduction
Pathophysiology
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Management
Conclusion
References
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Table. Differences Between Calcific Aortic Stenosis (AS) in the Elderly, Bicuspid AS, and Benign Flow Murmurs
References
Authors and Disclosures
Authors and Disclosures
Neil Sawhney, MD; Alborz Hassankhani, MD, PhD; Barry H. Greenberg, MD, Division of Cardiology, Department of Medicine, University of California San Diego, UCSD Medical Center, San Diego, CA
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2001 /viewarticle/tavr-beats-surveillance-asymptomatic-aortic-stenosis-2024a1000kji newsTAVR Beats Surveillance for Asymptomatic Aortic Stenosis
2001 /viewarticle/sleep-disorders-linked-aortic-stenosis-2025a10006fx newsSleep Disorders Linked to Aortic Stenosis
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1336 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3132917/ | WATERHOUSE-FRIDERICHSEN SYNDROME IN AN ADULT PATIENT WITH MENINGOCOCCAL MENINGITIS - PMC
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Indian J Dermatol
. 2011 May-Jun;56(3):326–328. doi: 10.4103/0019-5154.82496
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WATERHOUSE-FRIDERICHSEN SYNDROME IN AN ADULT PATIENT WITH MENINGOCOCCAL MENINGITIS
Alka Sonavane
Alka Sonavane
1 From the Department of Microbiology, Lokmanya Tilak Municipal Medical College and General Hospital, Sion, Mumbai, India
Find articles by Alka Sonavane
1, Vasant Baradkar
Vasant Baradkar
1 From the Department of Microbiology, Lokmanya Tilak Municipal Medical College and General Hospital, Sion, Mumbai, India
Find articles by Vasant Baradkar
1,✉, Parul Salunkhe
Parul Salunkhe
1 From the Department of Microbiology, Lokmanya Tilak Municipal Medical College and General Hospital, Sion, Mumbai, India
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1, Simit Kumar
Simit Kumar
1 From the Department of Microbiology, Lokmanya Tilak Municipal Medical College and General Hospital, Sion, Mumbai, India
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1
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1 From the Department of Microbiology, Lokmanya Tilak Municipal Medical College and General Hospital, Sion, Mumbai, India
✉
Address for correspondence: Dr. Vasant P. Baradkar, Department of Microbiology, L.T.M.M.C and L.T.M.G.H, Sion, Mumbai - 400 022, India. E-mail: vasantbaradkar@yahoo.com
Received 2009 Nov; Accepted 2010 Jul.
© Indian Journal of Dermatology
This is an open-access article distributed under the terms of the Creative Commons Attribution-Noncommercial-Share Alike 3.0 Unported, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
PMC Copyright notice
PMCID: PMC3132917 PMID: 21772601
Abstract
Waterhouse-Friderichsen syndrome is one of the fatal complications of meningococcal infection. Here we report a fatal case of this syndrome due to Neisseria meningitidis in a 29-year-old male patient who was admitted with high-grade fever and chills and vomiting since 7 days, a skin rash over the abdomen and trunk, and altered sensorium since 2 days. On examination, the signs of meningitis were present along with the hemorrhagic rash. The diagnosis of adrenal hemorrhage was confirmed by computerized tomographic scan findings. The patient was started on intravenous ceftriaxone, and the cerebrospinal fluid was processed for bacterial culture, which yielded growth of N meningitidis. The patient's condition deteriorated; he developed purpura along with a fall in platelet count, and died due to shock. This case is being reported as such a complication is comparatively rare in this antibiotic era, especially in adults, and starting steroids like dexamethasone prior to antibacterial therapy may be useful to diminish the inflammation brought about by bacterial cell death and thus help in reducing the otherwise high mortality in these cases.
Keywords:Adult, meningococcal meningitis, Waterhouse-Friderichsen syndrome
Introduction
Despite the advent of new microbials, meningococcal infection remains a leading cause of morbidity and mortality.[1–4] Based on the sequence of pathophysiological events and the agent and host factors, a wide spectrum of presentations may be seen. Patients with invasive meningococcal disease can present with meningitis alone or meningitis with shock.[1–4]
The Waterhouse Friderichsen syndrome is a severe complication of meningococcal infection.[1–3] The patient presents with meningococcal disease along with shock due to adrenal hemorrhage.[1–4] Here we report a fatal case of Waterhouse-Friedrichsen syndrome in an adult male patient with meningococcal disease.
Case Report
A 29-year-old male patient was admitted with a history of high-grade fever with chills and vomiting of 7 days’ duration, along with skin rash over the abdomen and trunk for the last 2 days. After admission he developed a hemorrhagic rash. There was no significant finding noted in the family history or the past history. On examination, his general condition was unstable and cyanosis was present. He was febrile, with a pulse rate of 90/min and systolic BP of 70 mm Hg. The neurological examination showed that the Kerning and Brudzinski signs were positive. A rash was observed over the whole of the body, but it was predominantly over the abdomen and trunk [Figure 1]. The cerebrospinal fluid (CSF) tapped was collected under aseptic precautions and was processed according to standard bacteriological procedures. Routine microscopy of the CSF showed a cell count of 9400/mm 3, with 86% polymorphs and 14% lymphocytes. The CSF protein was raised to 309 mg% and the sugar was reduced to 20 mg%. The Gram stained smear of the CSF showed pus cells along with gram-negative diplococci, some of which were intracellular and some extracellular [Figure 2]. The CSF was cultured on chocolate agar, blood agar, and MacConkey agar and the media were incubated at 37°C in humid conditions in a candle jar. After overnight incubation, tiny translucent colonies were observed on chocolate agar and blood agar, which was later identified by standard laboratory procedures to be that of Neisseria meningitidis. CT scan confirmed adrenal hemorrhage, which is supposed to be a diagnostic factor for Waterhouse-Friderichsen syndrome. Antibiotic sensitivity tests showed that the isolate was sensitive to all the antibiotics tested, i.e., ceftriaxone, chloramphenicol, penicillin, and trimethoprim + sulfamethoxazole. The patient meanwhile was started on ceftriaxone and steroids, but he went into shock and expired on the third day after admission.
Figure 1.
Open in a new tab
Purpuric, hemorrhagic skin lesions associated with the Waterhouse-Friedrichsen syndrome
Figure 2.
Open in a new tab
Gram stained smear showing pus cells along with gram-negative diplococci
This case is being reported as Waterhouse-Friderichsen syndrome is comparatively rare in this antibiotic era, especially in adults. If early diagnosis and antibacterial treatment, along with steroids, is not administered then there is high associated mortality.
Discussion
Asia has been the focus of meningococcal meningitis. Many outbreaks of meningococcal meningitis were documented during 1966 and 1985 in Delhi and adjoining areas. in early 2005, a spurt in the number of cases of meningococcemia and meningitis due to N meningitidis was reported from India. Upto June 2005, 429 cases were reported. The majority of the cases, and all the deaths, occurred in young adults between 16–30 years of age.[2,5]
Skin hemorrhages are hallmark of invasive meningococcal disease. Microscopically, these lesions are characterized by endothelial damage, which leads to hemorrhages, and microthrombi in small vessels. This is consistent with a generalized Sanarelli-Shwartzman reaction. The lesions are the result of endotoxins- and cytokine-primed vasculitis that is mediated by the upregulation of adhesion molecules on endothelium and degranulation activated neutrophils. Massive adrenal hemorrhage (Waterhouse-Friderichsen syndrome) is an uncommon, but usually fatal, consequence of overwhelming sepsis. Although disseminated intravascular coagulation is a generalized phenomenon, it may affect any organ and when it causes adrenal hemorrhages the condition is called Waterhouse-Friderichsen syndrome. Gram's staining of the CSF is still considered an important method for rapid diagnosis, but culture from CSF or skin lesion biopsy is the gold standard.[1–6] In some cases, blood culture may yield the organism. In the present case only CSF was processed, which showed growth of N meningitidis. In India, there is a paucity of studies on antimicrobial susceptibility of N meningitidis. Most of the studies showed that the isolates were sensitive to penicillin, ampicillin, and ceftriaxone; in two-thirds of the cases the isolates were resistant to ciprofloxacin.[2,5] Our isolate was sensitive to all the drugs tested, but the patient expired despite adequate antibiotic therapy, probably due to the adrenal hemorrhage leading to shock.
Waterhouse-Friderichsen syndrome, first reported in 1911 by Rubert Waterhouse, is characterized by fever, rash, purpura, coagulopathy, and shock. It has been suggested by many authors that this syndrome is more common than what literature reports indicate, many cases probably being missed due to lack of familiarity with the condition. The majority of the diseases caused by meningococci occur in children under 2 years of age, but it can occur at any age. Though Waterhouse-Friderichsen syndrome is common with N meningitidis, other organisms are also associated with this syndrome, including Streptococcus pneumoniae, β-hemolytic streptococcus group A, Staphylococcus aureus, Neisseria gonorrhoeae, Escherichia coli, Hemophilus influenzae, Klebsiella sp, and Pasturella sp. Although the condition is predominantly associated with meningococcal infection and with sepsis due to other organisms, there are also noninfectious causes of the Waterhouse-Friderichsen syndrome, such as anticoagulant treatment, antiphospholipid syndrome, trauma, and postoperative adrenal hemorrhages.[7,8]
Though N meningitidis is susceptible to the commonly used antibiotics, the mortality in Waterhouse-Friderichsen syndrome is approximately 20%, rising to 50% if the patient is in shock as happened in the present case.
If there are any signs and symptoms suggestive of Waterhouse-Friderichsen syndrome then 100 mg of hydrocortisone should also be given to the patient. If possible it should be given intravenously, but if a vein is not accessible the intramuscular route will suffice. Studied have shown that even prophylactically administered steroid can be life saving. Starting steroids like dexamethasone prior to antibacterial therapy may be useful to diminish the meningococcal inflammation brought about by bacterial cell death. Thus early diagnosis and proper treatment is necessary to prevent death from this otherwise fatal condition.
Footnotes
Source of Support: Nil
Conflict of Interest: Nil.
References
1.Raja NS, Parasakthi N, Puthucheary SD, Kamarulzaman A. Invasive meningococcal disease in the University of Malaya Medical Centre, Kuala Lumpur, Malaysia. J Postgrad Med. 2006;52:23–9. [PubMed] [Google Scholar]
2.Manchanda V, Gupta S, Bhalla P. Meningococcal disease: History, epidemiology, pathogenesis, clinical manifestation, diagnosis, antimicrobial susceptibility and prevention. Ind J Med Microbiol. 2008;24:7–19. doi: 10.4103/0255-0857.19888. [DOI] [PubMed] [Google Scholar]
3.Adem PV, Montgomery CP, Housain A, Koogler TK, Aragelvovich V, Humilier M, et al. Staphylococcus aureus sepsis and Waterhouse Friedrichsen syndrome in children. N Eng J Med. 2005;353:1245–51. doi: 10.1056/NEJMoa044194. [DOI] [PubMed] [Google Scholar]
4.Waterhouse R. A case of suprarenal apoplexy. Lancet. 1911;1:577–8. [Google Scholar]
5.Basu RN, Prasad R, Ichpujani RL. Meningococcal meningitis in Delhi and other areas. Commun Dis Bull. 1985;2:1. [Google Scholar]
6.Van Deuren M, Bran dtzaeg P, Van der Meer JW. Update on meningococcal disease with emphasis on pathogenesis and clinical management. Clin Microbiol Rev. 2003;13:144–66. doi: 10.1128/cmr.13.1.144-166.2000. [DOI] [PMC free article] [PubMed] [Google Scholar]
7.Hamilton D, Harris MD, Foweraker J, Greshman GA. Waterhouse Friedrichsen syndrome as a result of nonmenigococcal infection. J Clin Pathol. 2004;57:208–9. doi: 10.1136/jcp.2003.9936. [DOI] [PMC free article] [PubMed] [Google Scholar]
8.Karkouris PC, Page KR, Varello MA. Waterhouse- Friedrichsen syndrome after infection with Group A streptococcus. Mayo Clin Proc. 2001;76:1167–70. doi: 10.4065/76.11.1167. [DOI] [PubMed] [Google Scholar]
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1337 | https://www.cs.yale.edu/homes/aspnes/pinewiki/MathematicalLogic.html | | | |
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| | MathematicalLogic [FrontPage] [TitleIndex] [WordIndex] |
Note: You are looking at a static copy of the former PineWiki site, used for class notes by James Aspnes from 2003 to 2012. Many mathematical formulas are broken, and there are likely to be other bugs as well. These will most likely not be fixed. You may be able to find more up-to-date versions of some of these notes at
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Mathematical logic is the discipline that mathematicians invented in the late nineteenth and early twentieth centuries so they could stop talking nonsense. It's the most powerful tool we have for reasoning about things that we can't really comprehend, which makes it a perfect tool for ComputerScience.
The basic picture
Axioms, models, and inference rules
Consistency
What can go wrong
The language of logic
Standard axiom systems and models
Propositional logic
Operations on propositions
Precedence
Truth tables
Tautologies and logical equivalence
Inverses, converses, and contrapositives
Normal forms
Predicate logic
Variables and predicates
Quantifiers
Universal quantifier
Existential quantifier
Negation and quantifiers
Restricting the scope of a quantifier
Nested quantifiers
Examples
Functions
Equality
Uniqueness
Models
Examples
Proofs
Inference Rules
More on proofs vs implication
The Deduction Theorem
Inference rules for equality
Inference rules for quantified statements
Proof techniques
Example: The natural numbers
The Peano axioms
A simple proof
Defining addition
Other useful properties of addition
A scary induction proof involving even numbers
Defining more operations
1. The basic picture
| | | | | |
--- ---
| | | Abstract model | | |
| herds of sheep piles of rocks collections of fingers | | natural numbers: 0, 1, 2, ... | ⇒ | ∀x ∃y y = x+1 |
We want to model something we see in reality with something we can fit in our heads. Ideally we drop most of the features of the real thing that we don't care about and keep the parts that we do care about. But there is a second problem: if our model is very big (and the natural numbers are very very big), how do we know what we can say about them?
1.1. Axioms, models, and inference rules
One approach is to true to come up with a list of axioms that are true statements about the model and a list of inference rules that let us derive new true statements from the axioms. The axioms and inference rules together generate a theory that consists of all statements that can be constructed from the axioms by applying the inference rules. The rules of the game are that we can't claim that some statement is true unless it's a theorem: something we can derive as part of the theory.
Simple example: All fish are green (axiom). George Washington is a fish (axiom). From "all X are Y" and "Z is X", we can derive "Z is Y" (inference rule). Thus George Washington is green (theorem). Since we can't do anything else with our two axioms and one inference rule, these three statements together form our entire theory about George Washington, fish, greenness, etc.
Theories are attempts to describe models. A model is typically a collection of objects and relations between them. For a given theory, there may be many models that are consistent with it: for example, a model that includes both green fishy George Washington and MC 10,000-foot Abraham Lincoln is consistent with the theory above, because it doesn't say anything about Abraham Lincoln.
1.2. Consistency
A theory is consistent if it can't prove both P and not-P for any P. Consistency is incredibly important, since all the logics people actually use can prove anything starting from P and not-P.
1.3. What can go wrong
If we throw in too many axioms, you can get an inconsistency: "All fish are green; all sharks are not green; all sharks are fish; George Washington is a shark" gets us into trouble pretty fast.
If we don't throw in enough axioms, we underconstrain the model. For example, the Peano axioms for the natural numbers (see example below) say (among other things) that there is a number 0 and that any number x has a successor S(x) (think of S(x) as x+1). If we stop there, we might have a model that contains only 0, with S(0)=0. If we add in 0≠S(x) for any x, then we can get stuck at S(0)=1=S(1). If we add yet another axiom that says S(x)=S(y) if and only if x=y, then we get all the ordinary natural numbers 0, S(0)=1, S(1)=2, etc., but we could also get some extras: say 0', S(0') = 1', S(1') = 0'. Characterizing the "correct" natural numbers historically took a lot of work to get right, even though we all know what we mean when we talk about them. The situation is of course worse when we are dealing with objects that we don't really understand; here the most we can hope for is to try out some axioms and see if anything strange happens.
Better yet is to use some canned axioms somebody else has already debugged for us. In this respect the core of mathematics acts like a system library—it's a collection of useful structures and objects that are known to work, and (if we are lucky) may even do exactly what we expect.
1.4. The language of logic
The basis of mathematical logic is propositional logic, which was essentially invented by Aristotle. Here the model is a collection of statements that are either true or false. There is no ability to refer to actual things; though we might include the statement "George Washington is a fish", from the point of view of propositional logic that is an indivisible atomic chunk of truth or falsehood that says nothing in particular about George Washington or fish. If we treat it as an axiom we can prove the truth of more complicated statements like "George Washington is a fish or 2+2=5" (true since the first part is true), but we can't really deduce much else. Still, this is a starting point.
If we want to talk about things and their properties, we must upgrade to predicate logic. Predicate logic adds both constants (stand-ins for objects in the model like "George Washington") and predicates (stand-ins for properties like "is a fish"). It also lets use quantify over variables and make universal statements like "For all x, if x is a fish then x is green." As a bonus, we usually get functions ("f(x) = the number of books George Washington owns about x") and equality ("George Washington = 12" implies "George Washington + 5 = 17"). This is enough machinery to define and do pretty much all of modern mathematics.
We will discuss both of these logics in more detail below.
1.5. Standard axiom systems and models
Rather than define our own axiom systems and models from scratch, it helps to use ones that have already proven to be (a) consistent and (b) useful. Almost all mathematics fits in one of the following models:
The natural numbers ℕ. These are defined using the Peano axioms, and if all you want to do is count, add, and multiply, you don't need much else. (If you want to subtract, things get messy.)
The integers ℤ. Like the naturals, only now we can subtract. Division is still a problem.
The rational numbers ℚ. Now we can divide. But what about √2?
The real numbers ℝ. Now we have √2. But what about √(-1)?
The complex numbers ℂ. Now we are pretty much done. But what if we want to talk about more than one complex number at a time?
The universe of sets. These are defined using the axioms of SetTheory, and produce a rich collection of sets that include, among other things, structures equivalent to the natural numbers, the real numbers, collections of same, sets so big that we can't even begin to imagine what they look like, and even bigger sets so big that we can't use the axioms to prove whether they exist or not. Fortunately, in ComputerScience we can mostly stop with finite sets, which makes life less confusing.
Various alternatives to set theory, like lambda_calculus, category_theory or second-order_arithmetic. We won't talk about these much, since they generally don't let you do anything you can't do already with sets.
In practice, the usual way to do things is to start with sets and then define everything else in terms of sets: e.g., 0 is the empty set, 1 is a particular set with 1 element, 2 a set with 2 elements, etc., and from here we work our way up to the fancier numbers. The idea is that if we trust our axioms for sets to be consistent, then the things we construct on top of them should also be consistent (although if we are not careful in our definitions they may not be exactly the things we think they are).
2. Propositional logic
Propositional logic is the simplest form of logic. Here the only statements that are considered are propositions, which contain no variables. Because propositions contain no variables, they are either always true or always false.
Examples of propositions:
2+2=4. (Always true).
2+2=5. (Always false).
Examples of non-propositions:
x+2=4. (May be true, may not be true; it depends on the value of x.)
x⋅0=0. (Always true, but it's still not a proposition because of the variable.)
x⋅0=1. (Always false.)
As the last two examples show, it is not enough for a statement to be always true—whether a statement is a proposition or not is a structural property. But if a statement doesn't contain any variables (or other undefined terms), it is a proposition, and as a side-effect of being a proposition it's always true or always false.
2.1. Operations on propositions
Propositions by themselves are pretty boring. So boring, in fact, that logicians quickly stop talking about actual statements and instead haul out placeholder names for propositions like p, q, or r. But we can build slightly more interesting propositions by combining propositions together using various logical connectives, like:
Negation
: The negation of p is written as ¬p, or sometimes -p or p with a line over it. It has the property that it is false when p is true, and true when p is false.
Or
: The or of two propositions p and q is written as p ∨ q, (or maybe p \/ q in ASCII) and is true as long as at least one, or possibly both, of p and q is true.1 This is not always the same as what "or" means in English; in English, "or" often is used for exclusive or which is not true if both p and q are true. For example, if someone says "You will give me all your money or I will stab you with this table knife", you would be justifiably upset if you turn over all your money and still get stabbed. But a logician would not be at all surprised, because the standard "or" in propositional logic is an inclusive or that allows for both outcomes.
Exclusive or
: If you want to exclude the possibility that both p and q are true, you can use exclusive or instead. This is written as p⊕q, and is true precisely when exactly one of p or q is true. Exclusive or is not used in classical logic much, but is important for many computing applications, since it corresponds to addition modulo 2 (see ModularArithmetic) and has nice reversibility properties (e.g. p⊕(p⊕q) always has the same truth-value as q).
And
: The and of p and q is written as p ∧ q, (or p /\ q) and is true only when both p and q are true.2 This is pretty much the same as in English, where "I like to eat ice cream and I own a private Caribbean island" is not a true statement when made by most people even though most people like to eat ice cream. The only complication in translating English expressions into logical ands is that logicians can't tell the difference between "and" and "but": the statement "2+2=4 but 3+3=6" becomes simply "(2+2=4) ∧ (3+3=6)."
Implication
: This is the most important connective for proofs. An implication represents an "if...then" claim. If p implies q, then we write p → q, p ⇒ q, or p ⇒ q, depending on our typographic convention and the availability of arrow symbols in our favorite font. In English, p ⇒ q" is usually rendered as "If p, then q", as in "If you step on your own head, it will hurt." The meaning of p ⇒ q is that q is true whenever p is true, and the proposition p ⇒ q is true provided (a) p is false (in which case all bets are off), or (b) q is true. In fact, the only way for p ⇒ q to be false is for p to be true but q to be false; because of this, p ⇒ q can be rewritten as ¬p ∨ q. So, for example, the statements "If 2+2=5, then I'm the Pope", "If I'm the Pope, then 2+2=4", and "If 2+2=4, then 3+3=6", are all true, provided the if...then is interpreted as implication. Normal English usage does not always match this pattern; instead, if...then in normal speech is often interpreted as the much stronger biconditional (see below).
Biconditional
: Suppose that p ⇒ q and q ⇒ p, so that either both p and q are true or both p and q are false. In this case, we write p ⇔ q, p ⇔ q, or p ⇔ q, and say that p holds if and only if q holds. The truth of p ⇔ q is still just a function of the truth or falsehood of p and q; though there doesn't seem any connection between the two sides of the statement, "2+2=5 if and only if I am the Pope" is still true (provided it is not uttered by the Pope). The only way for p ⇔ q to be false is for one side to be true and one side to be false.
The result of applying any of these operations is called a compound proposition.
Here's what all of this looks like when typeset nicely. Note that in some cases there is more than one way to write a compound expression. Which you choose is a matter of personal preference, but you should try to be consistent.
2.1.1. Precedence
When reading a logical expression, the order of precedence in the absence of parentheses is negation, and, or, implication, biconditional. So (¬p ∨ q ∧ r ⇒ s ⇔ t) is interpreted as ((((¬p) ∨ (q ∧ r)) ⇒ s) ⇔ t). Or and and are associative, so (p ∨ q ∨ r) is the same as ((p ∨ q) ∨ r) and as (p ∨ (q ∨ r)), and similarly (p ∧ q ∧ r) is the same as ((p ∧ q) ∧ r) and as (p ∧ (q ∧ r)).
There does not seem to be a standard convention for the precedence of XOR, since logicians don't use it much. There are plausible arguments for putting XOR in between AND and OR, but it's probably safest just to use parentheses.
Implication is not associative, although the convention is that it binds "to the right", so that a ⇒ b ⇒ c is read as a ⇒ (b ⇒ c); few people ever remember this, so it is usually safest to put in the parentheses. I personally have no idea what p ⇔ q ⇔ r means, so any expression like this should be written with parentheses as either (p ⇔ q) ⇔ r or p ⇔ (q ⇔ r).
2.2. Truth tables
To define logical operations formally, we give a truth table. This gives, for any combination of truth values (true or false, which as computer scientists we often write as 1 or 0) of the inputs, the truth value of the output. So truth tables are to logic what addition tables or multiplication tables are to arithmetic.
Here is a truth table for negation:
| | |
--- |
| p | ¬p |
| 0 | 1 |
| 1 | 0 |
And here is a truth table for the rest of the logical operators:
| | | | | | | |
--- --- ---
| p | q | p∨q | p⊕q | p∧q | p⇒q | p⇔q |
| 0 | 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 | 0 | 1 | 0 |
| 1 | 0 | 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 1 | 0 | 1 | 1 | 1 |
See also RosenBook §§1.1–1.2 or BiggsBook §§3.1–3.3.
We can think of each row of a truth table as a model for propositonal logic, since the only things we can describe in propositional logic are whether particular propositions are true or not. Constructing a truth table corresponds to generating all possible models. This can be useful if we want to figure out when a particular proposition is true.
2.3. Tautologies and logical equivalence
A compound proposition that is true no matter what the truth-values of the propositions it contains is called a tautology. For example, p ⇒ p, p ∨ ¬p, and ¬(p ∧ ¬p) are all tautologies, as can be verified by constructing truth tables. If a compound proposition is always false, it's a contradiction. The negation of a tautology is a contradiction and vice versa.
The most useful class of tautologies are logical equivalences. This is a tautology of the form X ⇔ Y, where X and Y are compound propositions. In this case, X and Y are said to be logically equivalent and we can substitute one for the other in more complex propositions. We write X ≡ Y if X and Y are logically equivalent.
The nice thing about logical equivalence is that is does the same thing for Boolean formulas that equality does for algebraic formulas: if we know (for example), that p∨¬p is equivalent to 1, and q∨1 is equivalent to q, we can grind q∨p∨¬p ≡ q∨1 ≡ 1 without having to do anything particularly clever. (We will need cleverness later when we prove things where the consequence isn't logically equivalent to the premise.)
To prove a logical equivalence, one either constructs a truth table to show that X ⇔ Y is a tautology, or transforms X to Y using previously-known logical equivalences. Some examples:
p ∧ ¬p ≡ 0: Construct a truth table
| | | | |
--- --- |
| p | ¬p | p ∧ ¬p | 0 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 |
and observe that the last two columns are always equal.
p ∨ p ≡ p:
| | |
--- |
| p | p∨p |
| 0 | 0 |
| 1 | 1 |
p ⇒ q ≡ ¬p ∨ q: Again construct a truth table
| | | | |
--- --- |
| p | q | p ⇒ q | ¬p ∨ q |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 |
¬(p ∨ q) ≡ ¬p ∧ ¬q: (one of De Morgan's laws; the other is ¬(p ∧ q) ≡ ¬p ∨ ¬q).
| | | | | | | |
--- --- ---
| p | q | p ∨ q | ¬(p ∨ q) | ¬p | ¬q | ¬p ∧ ¬q |
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
p∨(q∧r) ≡ (p∨q)∧(p∨r) (one of the distributive laws; the other is p∧(q∨r) ≡ (p∧q)∨(p∧r)).
| | | | | | | | |
--- --- --- --- |
| p | q | r | q∧r | p∨(q∧r) | p∨q | p∨r | (p∨q)∧(p∨r) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
(p ⇒ r) ∨ (q ⇒ r) ≡ (p ∧ q) ⇒ r. Now things are getting messy, so building a full truth table may take awhile. But we have take a shortcut by using logical equivalences that we've already proved (plus associativity of ∨):
(p ⇒ r) ∨ (q ⇒ r) ≡ (¬p ∨ r) ∨ (¬q ∨ r) [Using p ⇒ q ≡ ¬p ∨ q twice.] ≡ ¬p ∨ ¬q ∨ r ∨ r [Associativity and commutativity of ∨.] ≡ ¬p ∨ ¬q ∨ r [p ≡ p ∨ p.] ≡ ¬(p ∧ q) ∨ r [De Morgan's law.] ≡ (p ∧ q) ⇒ r [p ⇒ q ≡ ¬p ∨ q.]
This last equivalence is a little surprising. It shows, for example, that if somebody says "It is either the case that if you study you will graduate from Yale with distinction, or that if you join the right secret society you will graduate from Yale with distinction", then this statement (assuming we treat the or as ∨) is logically equivalent to "If you study and join the right secret society, then you will graduate from Yale with distinction." It is easy to get tangled up in trying to parse the first of these two propositions; translating to logical notation and simplifying using logical equivalence is a good way to simplify it.
Over the years, logicians have given names to many logical equivalences. Only a few of these are actually useful to remember. These include de Morgan's laws: ¬(p ∨ q) ≡ (¬p ∧ ¬q) and ¬(p ∧ q) ≡ (¬p ∨ ¬q) (see De Morgan's law for more versions of these), commutativity of AND and OR, the equivalence of p ⇒ q and ¬p ∨ q, and the contraposition rule p ⇒ q ≡ ¬q ⇒ ¬p (see below for more about contraposition). Anything else you need you can probably derive from these or by writing out a truth table.
2.4. Inverses, converses, and contrapositives
The inverse of p ⇒ q is ¬p ⇒ ¬q. So the inverse of "If you take CPSC 202, you will surely die" is "If you do not take CPSC 202, you will not surely die." There is often no connection between the truth of an implication and the truth of its inverse: "If I am Barack Obama then I am a Democrat" does not have the same truth-value as "If I am not Barack Obama then I am not a Democrat", at least according to current polling numbers.
The converse of p ⇒ q is q ⇒ p. E.g. the converse of "If I am Barack Obama then I am a Democrat" is "If I am a Democrat then I am Barack Obama." The converse of a statement is always logically equivalent to the inverse. Often in proving a biconditional (e.g., "I am Barack Obama if and only if I am a Democrat"), one proceeds by proving first the implication in one direction and then either the inverse or the converse.
The contrapositive of p ⇒ q is ¬q ⇒ ¬p; it is logically equivalent to the original implication ("If I am not a Democrat then I am not Barack Obama"). A proof by contraposition proves p implies q by assuming ¬q and then proving ¬p; it is similar but not indentical to an IndirectProof, which assumes ¬p and derives a contradiction.
2.5. Normal forms
A compound proposition is in conjuctive normal form (CNF for short) if it is obtained by ANDing together ORs of one or more variables or their negations (an OR of one variable is just the variable itself). So for example P, (P∨Q)∧R, (P∨Q)∧(Q∨R)∧(¬P), and (P∨Q)∧(P∨¬R)∧(¬P∨Q∨S∨T∨¬U) are in CNF, but (P∨Q)∧(P∨¬R)∧(¬P∧Q),(P∨Q)∧(P⇒R)∧(¬P∨Q), and (P∨(Q∧R))∧(P∨¬R)∧(¬P∨Q) are not. Using the equivalence P⇒Q ≡ ¬P∨Q, De Morgan's laws, and the distributive law, it is possible to rewrite any compound proposition in CNF.
CNF formulas are particularly useful because they support resolution (see Inference rules below). Using the tautology (P∨Q)∧(¬P∨R) ⇒ Q∨R, we can construct proofs from CNF formulas by looking for occurrences of some simple proposition and its negation and resolving them, which generates a new clause we can add to the list. For example, we can compute (P∨Q)∧(P∨¬R)∧(¬P∨Q)∧(¬Q∨R) ⊢ (P∨Q)∧(P∨¬R)∧(¬P∨Q)∧(¬Q∨R)∧Q ⊢ (P∨Q)∧(P∨¬R)∧(¬P∨Q)∧(¬Q∨R)∧Q∧R ⊢ (P∨Q)∧(P∨¬R)∧(¬P∨Q)∧(¬Q∨R)∧Q∧R∧P ⊢ P. This style of proof is called a resolution proof. Because of its simplicity it is particularly well-suited for mechanical theorem provers. Such proofs can also encode traditional proofs based on modus ponens: the inference P∧(P⇒Q) ⊢ Q can be rewritten as resolution by expanding ⇒ to get P∧(¬P∨Q) ⊢ Q.
Similarly, a compound proposition is in disjunctive normal form (DNF) if it consists of an OR of ANDs, e.g. (P∧Q)∨(P∧¬R)∨(¬P∧Q). Just as any compound proposition can be transformed into CNF, it can similarly be transformed into DNF.
Note that conjunctive normal forms are not unique; for example, P∨Q and (P∧¬Q)∨(P∧Q)∨(¬P∧Q) are both in conjunctive normal form and are logically equivalent to each other. So while CNF can be handy as a way of reducing the hairiness of a formula (by eliminating nested parentheses or negation of non-variables, for example), it doesn't necessarily let us see immediately if two formulas are really the same.
3. Predicate logic
Using only propositional logic, we can express a simple version of a famous argument:
Socrates is a man.
If Socrates is a man, then Socrates is mortal.
Therefore, Socrates is mortal.
This is an application of the inference rule called modus ponens, which says that from p and p ⇒ q you can deduce q. The first two statements are axioms (meaning we are given them as true without proof), and the last is the conclusion of the argument.
What if we encounter Socrates's infinitely more logical cousin Spocrates? We'd like to argue
Spocrates is a man.
If Spocrates is a man, then Spocrates is mortal.
Therefore, Spocrates is mortal.
Unfortunately, the second step depends on knowing that humanity implies mortality for everybody, not just Socrates. If we are unlucky in our choice of axioms, we may not know this. What we would like is a general way to say that humanity implies mortality for everybody, but with just propositional logic, we can't write this fact down.
3.1. Variables and predicates
The solution is to extend our language to allow formulas that involve variables. So we might let x, y, z, etc. stand for any element of our universe of discourse or domain—essentially whatever things we happen to be talking about at the moment. We can now write statements like:
"x is human."
"x is the parent of y."
"x + 2 = x2."
These are not propositions because they have variables in them. Instead, they are predicates; statements whose truth-value depends on what concrete object takes the place of the variable. Predicates are often abbreviated by single capital letters followed by a list of arguments, the variables that appear in the predicate, e.g.:
H(x) = "x is human."
P(x,y) = "x is the parent of y."
Q(x) = "x + 2 = x2."
We can also fill in specific values for the variables, e.g. H(Spocrates) = "Spocrates is human." If we fill in specific values for all the variables, we have a proposition again, and can talk about that proposition being true (e.g. H(2) and H(-1) are true) or false (H(0) is false).
3.2. Quantifiers
What we really want though is to be able to say when H or P or Q is true for many different values of their arguments. This means we have to be able to talk about the truth or falsehood of statements that include variables. To do this, we bind the variables using quantifiers, which state whether the claim we are making applies to all values of the variable (universal quantification), or whether it may only apply to some (existential quantification).
3.2.1. Universal quantifier
The universal quantifier ∀ (pronounced "for all") says that a statement must be true for all values of a variable within some universe of allowed values (which is often implicit). For example, "all humans are mortal" could be written ∀x: Human(x) ⇒ Mortal(x) and "if x is positive then x+1 is positive" could be written ∀x: x > 0 ⇒ x+1 > 0.
If you want to make the universe explicit, use set membership notation, e.g. ∀x∈Z: x > 0 ⇒ x+1 > 0. This is logically equivalent to writing ∀x: x∈Z ⇒ (x > 0 ⇒ x+1 > 0) or to writing ∀x: (x∈Z ∧ x > 0) ⇒ x+1 > 0, but the short form makes it more clear that the intent of x∈Z is to restrict the range of x.3
The statement ∀x: P(x) is equivalent to a very large AND; for example, ∀x∈N: P(x) could be rewritten (if you had an infinite amount of paper) as P(0) ∧ P(1) ∧ P(2) ∧ P(3) ∧ ... . Normal first-order logic doesn't allow infinite expressions like this, but it may help in visualizing what ∀x: P(x) actually means. Another way of thinking about it is to imagine that x is supplied by some adversary and you are responsible for showing that P(x) is true; in this sense, the universal quantifier chooses the worst case value of x.
3.2.2. Existential quantifier
The existential quantifier ∃ (pronounced "there exists") says that a statement must be true for at least one value of the variable. So "some human is mortal" becomes ∃x: Human(x) ∧ Mortal(x). Note that we use AND rather than implication here; the statement ∃x: Human(x) ⇒ Mortal(x) makes the much weaker claim that "there is some thing x, such that if x is human, then x is mortal", which is true in any universe that contains an immortal purple penguin—since it isn't human, Human(penguin) ⇒ Mortal(penguin) is true.
As with ∀, ∃ can be limited to an explicit universe with set membership notation, e.g. ∃x∈Z: x = x2. This is equivalent to writing ∃x: x∈Z ∧ x = x2.
The formula ∃x: P(x) is equivalent to a very large OR, so that ∃x∈N: P(x) could be rewritten as P(0) ∨ P(1) ∨ P(2) ∨ P(3) ∨ ... . Again, you can't generally write an expression like this if there are infinitely many terms, but it gets the idea across.
3.2.3. Negation and quantifiers
¬∀x: P(x) ≡ ∃x: ¬P(x).
¬∃x: P(x) ≡ ∀x: ¬P(x).
These are essentially the quantifier version of De Morgan's laws: the first says that if you want to show that not all humans are mortal, it's equivalent to finding some human that is not mortal. The second says that to show that no human is mortal, you have to show that all humans are not mortal.
3.2.4. Restricting the scope of a quantifier
Sometimes we want to limit the universe over which we quantify to some restricted set, e.g. all positive integers or all baseball teams. We can do this explicitly using implication, e.g.
∀x: x > 0 ⇒ x-1 ≥ 0
or we can abbreviate by including the restriction in the quantifier expression itself
∀x>0: x-1 ≥ 0.
Similarly
∃x: x > 0 ∧ x2 = 81
can be written as
∃x>0: x2 = 81.
Note that constraints on ∃ get expressed using AND rather than implication.
The use of set membership notation to restrict a quantifier is a special case of this. Suppose we want to say that 79 is not a perfect square, by which we mean that there is no integer whose square is 79. If we are otherwise talking about real numbers (two of which happen to be square roots of 79), we can exclude the numbers we don't want by writing
¬∃x∈ℤ x2 = 79
which is interpreted as
¬∃x (x∈ℤ ∧ x2 = 79)
or, equivalently
∀x x∈ℤ ⇒ x2 ≠ 79.
(Here ℤ = { ..., -2, -1, 0, 1, 2, ... } is the standard set of integers. See SetTheory for more about ∈.)
3.2.5. Nested quantifiers
It is possible to nest quantifiers, meaning that the statement bound by a quantifier itself contains quantifiers. For example, the statement "there is no largest prime number" could be written as
¬∃x: (Prime(x) ∧ ∀y: y > x ⇒ ¬Prime(y))
i.e., "there does not exist an x that is prime and any y greater than x is not prime." Or in a shorter (though not strictly equivalent) form:
∀x ∃y: y > x ∧ Prime(y)
"for any x there is a bigger y that is prime."
To read a statement like this, treat it as a game between the ∀ player and the ∃ player. Because the ∀ comes first in this statement, the for-all player gets to pick any x it likes. The ∃ player then picks a y to make the rest of the statement true. The statement as a whole is true if the ∃ player always wins the game. So if you are trying to make a statement true, you should think of the universal quantifier as the enemy (the adversary in algorithm analysis) who says "nya-nya: try to make this work, bucko!", while the existential quantifier is the friend who supplies the one working response.
Naturally, such games can go on for more than two steps, or allow the same player more than one move in a row. For example
∀x ∀y ∃z x2 + y2 = z2
is a kind of two-person challenge version of the Pythagorean_theorem where the universal player gets to pick x and y and the existential player has to respond with a winning z. (Whether the statement itself is true or false depends on the range of the quantifiers; it's false, for example, if x, y, and z are all natural numbers or rationals but true if they are all real or complex. Note that the universal player only needs to find one bad (x,y) pair to make it false.)
One thing to note about nested quantifiers is that we can switch the order of two universal quantifiers or two existential quantifiers, but we can swap a universal quantifier for an existential quantifier or vice versa. So for example ∀x∀y (x = y ⇒ x+1 = y+1) is logically equivalent to ∀y∀x (x = y ⇒ y+1 = x+1), but ∀x∃y y < x is not logically equivalent to ∃y∀x y < x. This is obvious if you think about it in terms of playing games: if I get to choose two things in a row, it doesn't really matter which order I choose them in, but if I choose something and then you respond it might make a big difference if we make you go first instead.
One measure of the complexity of a mathematical statement is how many layers of quantifiers it has, where a layer is a sequence of all-universal or all-existential quantifiers. Here's a standard mathematical definition that involves three layers of quantifiers, which is about the limit for most humans:
Now that we know how to read nested quantifiers, it's easy to see what the right-hand side means:
The adversary picks ε, which must be greater than 0.
We pick N.
The adversary picks x, which must be greater than N.
We win if f(x) is within ε of y.
So, for example, a proof of
would follow exactly this game plan:
Choose some ε > 0.
Let N > 1/ε. (Note that we can make our choice depend on previous choices.)
Choose any x > N.
Then x > N > 1/ε > 0, so 1/x < 1/N < ε ⇒ |1/x - 0| < ε. QED!
3.2.6. Examples
Here are some more examples of translating English into statements in predicate logic:
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| English | Logic | Other variations |
| All crows are black. | ∀x Crow(x) ⇒ Black(x) | Equivalent to ¬∃x Crow(x) ∧ ¬Black(x) or ∀x ¬Black(x) ⇒ ¬Crow(x). The latter is the core of a classic "paradox of induction" in philosophy: if seeing a black crow makes me think it's more likely that all crows are black, shouldn't seeing a logically equivalent non-black non-crow (e.g. a banana yellow AMC_Gremlin) also make me think all non-black objects are non-crows, i.e., that all crows are black? (I think this mostly illustrates that logical equivalence only works for true/false and not for probabilities.) |
| Some cows are brown. | ∃x Cow(x) ∧ Brown(x) | |
| No cows are blue. | ¬∃x Cow(x) ∧ Blue(x) | ≡ ∀x ¬(Cow(x) ∧ Blue(x) ≡ ∀x (¬Cow(x) ∨ ¬Blue(x)) ≡ ∀x Cow(x) ⇒ ¬Blue(x) ≡ ∀x Blue(x) ⇒ ¬Cow(x). (But see Paul_Bunyan.) |
| All that glitters is not gold. | ¬∀x Glitters(x) ⇒ Gold(x) | ≡ ∃x Glitters(x) ∧ ¬Gold(x). Note that the English syntax is a bit ambiguous: a literal translation might look like ∀x Glitters(x) ⇒ ¬Gold(x), which is not logically equivalent. This is why predicate logic is often safer than natural language. |
| No shirt, no service. | ∀x ¬Shirt(x) ⇒ ¬Served(x) | |
| Every event has a cause. | ∀x∃y Causes(y,x) | |
| Every even number greater than 2 can be expressed as the sum of two primes. | ∀x (Even(x) ∧ x > 2) ⇒ (∃p ∃q Prime(p) ∧ Prime(q) ∧ x = p+q) | Note: the truth value of this statement is currently unknown. See Goldbach's conjecture. |
3.3. Functions
A function symbol looks like a predicate but instead of computing a truth value it returns a new object. So for example the successor function S in the Peano axioms for the natural numbers returns x+1 when applied as S(x). Sometimes when there is only a single argument we omit the parentheses, e.g. Sx = S(x), SSSx = S(S(S(x))).
3.4. Equality
Often we include a special equality predicate =, written x=y. The interpretation of x=y is that x and y are the same element of the domain. It satisfies the reflexivity axiom ∀x x=x and the substitution axiom schema
∀x∀y (x=y ⇒ (Px ⇔ Py))
where P is any predicate; this immediately gives a substitution rule that says x = y, P(x) ⊢ P(y). It's likely that almost every proof you ever wrote down in high school algebra consisted only of many applications of the substitution rule.
Example
: We'll prove ∀x∀y (x=y ⇒ y=x) from the above axioms (this property is known as symmetry). Apply substitution to the predicate Px ≡ y=x to get ∀x∀y (x=y ⇒ (y=x ⇔ x=x)). Use reflexivity to rewrite this as ∀x∀y (x=y ⇒ (y=x ⇔ T)) or ∀x∀y (x=y ⇒ y=x) as claimed.
Exercise
: Prove ∀x∀y∀z (x=y ∧ y=z ⇒ x=z). (This property is known as transitivity.)
3.4.1. Uniqueness
An occasionally useful abbreviation is ∃!x P(x), which stands for "there exists a unique x such that P(x)." This is short for
(∃x P(x)) ∧ (∀x∀y P(x) ∧ P(y) ⇒ x = y).
An example is ∃!x x+1 = 12. To prove this we'd have to show not only that there is some x for which x+1 = 12 (11 comes to mind), but that if we have any two values x and y such that x+1 = 12 and y+1 = 12, then x=y (this is not hard to do). So the exclamation point encodes quite a bit of extra work, which is why we usually hope that ∃x x+1 = 12 is good enough.
3.5. Models
In propositional logic, we can build truth tables that describe all possible settings of the truth-values of the literals. In predicate logic, the analogous concept to an assignment of truth-values is a structure. A structure consists of a set of objects or elements (see SetTheory), together with a description of which elements fill in for the constant symbols, which predicates hold for which elements, and what the value of each function symbol is when applied to each possible list of arguments (note that this depends on knowing what constant, predicate, and function symbols are available—this information is called the signature of the structure). A structure is a model of a particular theory (set of statements), if each statement in the theory is true in the model.
In general we can't hope to find all possible models of a given theory. But models are useful for two purposes: if we can find some model of a particular theory, then the existence of this model demonstrates that the theory is consistent; and if we can find a model of the theory in which some additional statement S doesn't hold, then we can demonstrate that there is no way to prove S from the theory (i.e. it is not the case that T ⊢ S, where T is the list of axioms that define the theory).
3.5.1. Examples
Consider the axiom ¬∃x. This axiom has exactly one model (it's empty).
Now consider the axiom ∃!x. This axiom also has exactly one model (with one element).
We can enforce exactly k elements with one rather long axiom, e.g. for k = 3 do ∃x1∃x2∃x3∀y y = x1 ∨ y = x2 ∨ y = x3. In the absence of any special symbols, a structure of 3 undifferentiated elements is the unique model of this axiom.
Suppose we add a predicate P and consider the axiom ∃x Px. Now we have many models: take any nonempty model you like, and let P be true of at least one of its elements. If we take a model with two elements a and b, with Pa and ¬Pb, we get that ∃x Px is not enough to prove ∀x Px, since the later statement isn't true in this model.
Now let's bring in a function symbol S and constant symbol 0. Consider a stripped-down version of the Peano axioms that consists of just the axiom ∀x∀y Sx = Sy ⇒ x = y. Both the natural numbers ℕ and the integers ℤ are a model for this axiom, as is the set ℤm of integers mod m for any m (see ModularArithmetic); in each case each element has a unique predecessor, which is what the axiom demands. If we throw in the first Peano axiom ∀x Sx ≠ 0, we eliminate ℤ and ℤm because in each of these models 0 is a successor of some element. But we don't eliminate a model that consists of two copies of ℕ sitting next to each other (only one of which contains the "real" 0), or even a model that consists of one copy of ℕ (to make 0 happy) plus any number of copies of ℤ and ℤm.
A practical example: The family tree of the kings of France is a model of the theory containing the two axioms ∀x∀y∀z Parent(x,y) ∧ Parent(y,z) ⇒ GrandParent(x,z) and ∀x∀y Parent(x,y) ⇒ ¬Parent(y,x). But this set of axioms could use some work, since it still allows for the possibility that Parent(x,y) and GrandParent(y,x) are both true.
4. Proofs
A proof is a way to derive statements from other statements. It starts with axioms (statements that are assumed in the current context always to be true), theorems or lemmas (statements that were proved already), and premises (P), and uses inference rules to derive Q. The axioms, theorems, and premises are in a sense the starting position of a game whose rules are given by the inference rules. The goal of the game is to apply the inference rules until Q pops out. We refer to anything that isn't proved in the proof itself (ie., an axiom, theorem, lemma, or premise) as a hypothesis; the result Q is the conclusion.
When a proof exists of Q from some premises P1, P2, ..., we say that Q is deducible or provable from P1, P2, ..., which is written as
P1,P2, ... ⊢ Q.
Note that axioms are usually not included in list on the left-hand side. So we can write just
⊢ Q
if we can prove Q directly from our axioms without any additional premises.
The turnstile symbol ⊢ has the specific meaning that we can derive the conclusion Q by applying inference rules to the premises. This is not quite the same thing as saying P ⇒ Q. If our inference rules are particularly weak, it may be that P ⇒ Q is true but we can't prove Q starting with P. Conversely, if our inference rules are too strong (maybe they can prove anything, even things that aren't true) we might have P ⊢ Q but P ⇒ Q is false. But most of the time we will use inference rules that are just right, meaning that P ⊢ Q implies P ⇒ Q (soundness) and P ⇒ Q implies P ⊢ Q (completeness). The distinction between ⊢ and ⇒ is then whether we want to talk about the existence of a proof (the first case) or about the logical relation between two statements (the second). As a consequence, you probably won't see ⊢ again after you stop reading this page.
4.1. Inference Rules
The main source of inference rules is tautologies of the form P ⇒ Q; given such a tautology, there is a corresponding inference rule that allows us to assert Q once we have P (either because it's an axiom/theorem/premise or because we proved it already while doing the proof). The most important inference rule is modus ponens, based on the tautology (p ∧ (p ⇒ q)) ⇒ q; this lets us, for example, write the following famous argument:4
If it doesn't fit, you must acquit. [Axiom]
It doesn't fit. [Premise]
You must acquit. [Modus ponens applied to 1+2]
There are many named inference rules in classical propositional logic. We'll list some of them below. You don't need to remember the names of anything except modus ponens, and most of the rules are pretty much straightforward applications of modus ponens plus some convenient tautology that can be proved by truth tables or stock logical equivalences. (For example, the "addition" rule below is just the result of applying modus ponens to p and the tautology p ⇒ (p ∨ q).)
Inference rules are often written by putting the premises above a horizontal line and the conclusion below. In text, the horizontal line is often replaced by the symbol ⊢, which means exactly the same thing. Premises are listed on the left-hand side separated by commas, and the conclusion is placed on the right. We can then write
Addition
: p ⊢ p ∨ q.
Simplification
: p ∧ q ⊢ p.
Conjunction
: p, q ⊢ p ∧ q.
Modus ponens
: p, p ⇒ q ⊢ q.
Modus tollens
: ¬q, p ⇒ q ⊢ ¬p.
Hypothetical syllogism
: p ⇒ q, q ⇒ r ⊢ p ⇒ r.
Disjunctive syllogism
: p ∨ q, ¬p ⊢ q.
Resolution
: p ∨ q, ¬p ∨ r ⊢ q ∨ r.
Of these rules, Addition, Simplification, and Conjunction are mostly used to pack and unpack pieces of arguments. Modus ponens (and its reversed cousin modus tollens) let us apply implications. You only need to remember modus ponens if you can remember the contraposition rule (p ⇒ q) ≡ (¬q ⇒ ¬p). Hypothetical syllogism just says that implication is transitive; it lets you paste together implications if the conclusion of one matches the premise of the other. Disjunctive syllogism is again a disguised version of modus ponens (via the logical equivalence (p ∨ q) ≡ (¬p ⇒ q)); you don't need to remember it if you can remember this equivalence. Resolution is almost never used by humans but is very popular with computer theorem provers.
An argument is valid if the conclusion is true whenever the hypotheses are true. Any proof constructed using the inference rules is valid. It does not necessarily follow that the conclusion is true; it could be that one or more of the hypotheses is false:
If you give a mouse a cookie, he's going to ask for a glass of milk. [Axiom]
If he asks for a glass of milk, he will want a straw. [Axiom]
You gave a mouse a cookie. [Premise]
He asks for a glass of milk. [Modus ponens applied to 1 and 3.]
He will want a straw. [Modus ponens applied to 2 and 4.]
Will the mouse want a straw? No: Mice can't ask for glasses of milk, so axiom 1 is false.
4.2. More on proofs vs implication
Recall that P ⊢ Q means there is a proof of Q by applying inference rules to the axioms and P, while P ⇒ Q says that Q holds whenever P does. These are not the same thing: provability (⊢) is outside the theory (it's a statement about whether a proof exists or not) while implication (⇒) is inside (it's a logical connective for making compound propositions). But most of the time they mean almost the same thing.
For example, suppose that we can prove P ⇒ Q directly from whatever axioms we are currently assuming, i.e., that
⊢ P ⇒ Q.
Since we can always ignore extra premises, we get
P ⊢ P ⇒ Q
and thus
P ⊢ P, P ⇒ Q
and finally
P ⊢ Q
by applying modus ponens to the right-hand side.
So we can go from
⊢ P ⇒ Q
to
P ⊢ Q.
This means that provability is in a sense weaker than implication: it holds (assuming modus ponens) whenever implication does. But we usually don't use this fact much, since P ⇒ Q is a much more useful statement than P ⊢ Q. Can we go the other way?
4.2.1. The Deduction Theorem
Yes, using the Deduction Theorem.
Often we want to package the result of a proof as a Theorem (a proven statement that is an end in itself) or Lemma (a proven statement that is intended mostly to be used in other proofs). Typically a proof shows that if certain premises P1, P2, ... Pn hold, then some conclusion Q holds (with various axioms or previously-established theorems assumed to be true from context). To use this result later, it is useful to be able to package it as an implication P1 ∧ P2 ∧ ... Pn ⇒ Q. In other words, we want to go from
P1, P2, ..., Pn ⊢ Q
to
⊢ (P1 ∧ P2 ∧ ... ∧ Pn) ⇒ Q.
The statement that we can do this, for a given set of axioms and inference rules, is the
Deduction Theorem
: If there is a proof of Q from premises P1, P2, ..., Pn, then there is a proof of P1 ∧ P2 ∧ ... ∧ Pn ⇒ Q.
The actual proof of the theorem depends on the particular set of axioms and inference rules we start with, but the basic idea is that there exists a mechanical procedure for extracting a proof of the implication from the proof of Q assuming P1 etc.
Caveat: In predicate logic, the deduction theorem only applies if none of the premises contain any free variables (i.e. a variable that isn't bound by a containing quantifier). Usually you won't run into this, but see e.g. Deduction theorem for some bad cases that arise without this restriction.
4.3. Inference rules for equality
The equality predicate is special, in that it allows for the substitution rule
x = y, P(x) ⊢ P(y).
This can also be represented as an axiom schema: ∀x ∀y ((x = y ∧ P(x)) ⇒ P(y)).
4.4. Inference rules for quantified statements
Universal generalization
: If P ⊢ Q(x) and x does not appear in P, then P ⊢ ∀x Q(x). Typical use: We want to show that there is no biggest natural number, i.e. that ∀x∈ℕ ∃y∈ℕ y > x. Let x be any element of ℕ. Let y = x+1. Then y > x. (Note: there is also an instance of existential generalization here.)
Universal instantiation
: ∀x Q(x) ⊢ Q(c). Typical use: Given that all humans are mortal, it follows that Spocrates is mortal.
Existential generalization
: Q(c) ⊢ ∃x Q(x). Typical use: We are asked to prove that there exists an even prime number. Consider 2: it's an even prime number. QED.
Existential instantiation
: ∃x Q(x) ⊢ Q(c) for some particular c. (The real rule is more complicated and says ((∀x (Q(x) ⇒ P)) ∧ ∃y Q(y)) ⇒ P; but the intent in both cases is that once you have proven that at least one c satisfying Q exists, you can give it a name and treat it as an object you actually have in your hands.) Typical use: Suppose we are told that there exists a prime number greater than 7. Let p be some such prime number greater than 7.
5. Proof techniques
A proof technique is a template for how to go about proving particular classes of statements: this template guides you in the choice of inference rules (or other proof techniques) to write the actual proof. This doesn't substitute entirely for creativity (there is no efficient mechanical procedure for generating even short proofs unless P=NP), but it can give you some hints for how to get started.
In the table below, we are trying to prove A ⇒ B for particular statements A and B. The techniques are mostly classified by the structure of B. Before applying each technique, it may help to expand any definitions that appear in A or B.5
If you want to prove A ⇔ B, the usual approach is to prove A ⇒ B and A ⇐ B seperately; proving A ⇒ B and ¬A ⇒ ¬B also works (because of contraposition).
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| Strategy | When to use it | What to assume | What to conclude | What to do/why it works |
| Direct proof | Try it first | A | B | Apply inference rules to work forward from A and backward from B; when you meet in the middle, pretend that you were working forward from A all along. |
| Contraposition | B = ¬Q | ¬B | ¬A | Apply any other technique to show ¬B ⇒ ¬A and then apply the contraposition rule. Sometimes called an indirect proof although the term indirect proof is often used instead for proofs by contradiction (see below). |
| Contradiction | When B = ¬Q, or when you are stuck trying the other techniques. | A ∧ ¬B | False | Apply previous methods to prove both P and ¬P for some P. Note: this can be a little dangerous, because you are assuming something that is (probably) not true, and it can be hard to detect as you prove further false statements whether the reason they are false is that they follow from your false assumption, or because you made a mistake. Direct or contraposition proofs are preferred because they don't have this problem. |
| Construction | B = ∃x P(x) | A | P(c) for some specific object c. | Pick a likely-looking c and prove that P(c) holds. |
| Counterexample | B = ¬∀x P(x) | A | ¬P(c) for some specific object c. | Pick a likely-looking c and show that ¬P(c) holds. This is identical to a proof by construction, except that we are proving ∃x ¬P(x), which is equivalent to ¬∀x P(x). |
| Choose | B = ∀x (P(x) ⇒ Q(x)) | A and P(c) where c is chosen arbitrarily. | Q(c) | Choose some c and assume A and P(c). Prove Q(c). Note: c is a placeholder here. If P(c) is "c is even" you can write "Let c be even" but you can't write "Let c = 12", since in the latter case you are assuming extra facts about c. |
| Instantiation | A = ∀x P(x) | A | B | Pick some particular c and prove that P(c) ⇒ B. Here you can get away with saying "Let c = 12." (If c=12 makes B true). |
| Proof by elimination | B = C ∨ D | A ∧ ¬C | D | The reason this works is that A ∧ ¬C ⇒ D is equivalent to ¬(A ∧ ¬C) ⇒ D ≡ ¬A ∨ C ∨ D ≡ A ⇒ (C ∨ D). Of course, it works equally well if you start with A ∧ ¬D and prove C. |
| Case analysis | A = C ∨ D | C, D | B | Here you write two separate proofs: one that assumes C and proves B, and one that assumes D and proves B. A special case is when D = ¬C. You can also consider more cases, as long as A implies at least one of the cases holds. |
| Induction | B = ∀x∈ℕ P(x) | A | P(0) and ∀x∈ℕ (P(x) ⇒ P(x+1)). | If P(0) holds, and P(x) implies P(x+1) for all x, then for any specific natural number n we can consider constructing a sequence of proofs P(0) ⇒ P(1) ⇒ P(2) ⇒ ... ⇒ P(n). (This is actually a defining property of the natural numbers; see the example below.) |
Some others that are mentioned in SolowBook: Direct Uniqueness, Indirect Uniqueness, various max/min arguments. We will also cover InductionProofs in more detail later.
6. Example: The natural numbers
Here we give an example of how we can encode simple mathematics using predicate logic, and then prove theorems about the resulting structure. Our goal is to represent the natural numbers: 0, 1, 2, etc.6 The method is to use the Peano axioms, a modern version of similar axioms developed by Giuseppe Peano.
6.1. The Peano axioms
Peano represents natural numbers using a special 0 constant and a function symbol S (for "successor"; think of it as +1). Repeatedly applying S to 0 generates increasingly large natural numbers: S0 = 1, SS0 = 2, SSS0 = 3, etc. (Note that 1, 2, 3, etc., are not part of the language, although we might use them sometimes as shorthand for long strings of S's.) For convenience, we don't bother writing parentheses for the function applications unless we need to do so to avoid ambiguity: read SSSS0 as S(S(S(S(0)))).
The usual interpretation of function symbols implies that 0 and its successors exist, but it doesn't guarantee that they aren't all equal to each other. The first Peano axiom prevents this:
∀x Sx ≠ 0.
In English, 0 is not the successor of any number.
This still allows for any number of nasty little models in which 0 is nobody's successor, but we still stop before getting all of the naturals. For example, let SS0 = S0; then we only have two elements in our model (0 and S0, because once we get to S0, any further applications of S keep us where we are.
To avoid this, we need to prevent S from looping back round to some number we've already produced. It can't get to 0 because of the first axiom, and to prevent it from looping back to a later number, we take advantage of the fact that they already have one successor:
∀x ∀y Sx = Sy ⇒ x = y.
If we take the contrapositive in the middle, we get x ≠ y ⇒ Sx ≠ Sy. In other words, we can't have a single number z that is the successor of two different numbers x and y.
Now we get all of ℕ, but we may get some extra elements as well. There is nothing in the first two axioms that prevents us from having something like this:
0 → S0 → SS0 → SSS0 → ...
Bogus → S Bogus → SS Bogus → SSS Bogus → ...
The hard part of coming up with the Peano axioms was to prevent the model from sneaking in extra bogus values (that still have successors and at most one predecessor each). This is (almost) done using the third Peano axiom, which in modern terms is actually an axiom schema: an axiom that holds when substituting any predicate for its placeholder predicate P:
(P(0) ∧ (∀x P(x) ⇒ P(Sx))) ⇒ ∀x P(x).
This is known as the induction schema, and says that, for any predicate P, if we can prove that P holds for 0, and we can prove that P(x) implies P(x+1), then P holds for all x in ℕ. The intuition is that even though we haven't bothered to write out a proof of, say P(1337), we know that we can generate one by starting with P(0) and modus-pwning our way out to P(1337) using P(0) ⇒ P(1), then P(1) ⇒ P(2), then P(2) ⇒ P(3), etc. Since this works for any number (eventually), there can't be some number that we missed.
In particular, this lets us throw out the bogus numbers in the bad example above. Let bogus(x) be true if x is bogus (i.e., it's equal to Bogus or one of the other values in its chain of successors). Let P(x) ≡ ¬Bogus(x). Then P(0) holds (0 is not bogus), and if P(x) holds (x is not bogus) then so does P(Sx). It follows from the induction axiom that ∀x P(x): there are no bogus numbers.7
6.2. A simple proof
Let's use the Peano axioms to prove something that we know to be true about the natural numbers we learned about in grade school but that might not be obvious from the axioms themselves. (This will give us some confidence that the axioms are not bogus.) We want to show that 0 is the only number that is not a successor:
Claim
: ∀x x ≠ 0 ⇒ ∃y x = Sy.
To find a proof of this, we start by looking at the structure of what we are trying to prove. It's a universal statement about elements of ℕ (implicitly, the ∀x is really ∀x∈ℕ, since our axioms exclude anything that isn't in ℕ), so our table of proof techniques suggests using an induction argument, which in this case means finding some predicate we can plug into the induction schema.
If we strip off the ∀x, we are left with
x ≠ 0 ⇒ ∃y x = Sy.
Here a direct proof is suggested: assuming x ≠ 0, and try to prove ∃y x = Sy. But our axioms don't tell us much about numbers that aren't 0, so it's not clear what to do with the assumption. This turns out to be a dead end.
Recalling that A ⇒ B is the same thing as ¬A ∨ B, we can rewrite our goal as
x = 0 ∨ ∃y x = Sy.
This seems like a good candidate for P (our induction hypothesis), because we do know a few things about 0. Let's see what happens if we try plugging this into the induction schema:
P(0) ≡ 0 = 0 ∨ ∃y 0 = Sy. The right-hand term looks false because of our first axiom, but the left-hand term is just the reflexive axiom for equality. P(0) is true.
∀x P(x) ⇒ P(Sx). We can drop the ∀x if we fix an arbitrary x. Expand the right-hand side P(Sx) ≡ Sx = 0 ∨ ∃y Sx = Sy. We can be pretty confident that Sx ≠ 0 (it's an axiom), so if this is true, we had better show ∃y Sx = Sy. The first thing to try for ∃ statements is instantiation: pick a good value for y. Picking y = x works.
Since we showed P(0) and ∀x P(x) ⇒ P(Sx), the induction schema tells us ∀x P(x). This finishes the proof.
Having figured the proof out, we might go back and clean up any false starts to produce a compact version. A typical mathematican might write the preceding argument as:
Claim
: ∀x x ≠ 0 ⇒ ∃y x = Sy.
Proof
: By induction on x. For x = 0, the premise fails. For Sx, let y = x. ∎
(A black rectangle—known as a tombstone—is often used to mark the end of a proof.)
A really lazy mathematician would write:
Claim
: ∀x x ≠ 0 ⇒ ∃y x = Sy.
Proof
: Induction on x. ∎
Though laziness is generally a virtue, you probably shouldn't be quite this lazy when writing up homework assignments.
6.3. Defining addition
Because of our restricted language, we do not yet have the ability to state valuable facts like 1+1=2 (which we would have to write as S0 + S0 = SS0). Let's fix this, by adding a two-argument function symbol + which we will define using the axioms
x+0 = x.
x+Sy = S(x+y).
(We are omitting some ∀ quantifiers, since unbounded variables are implicitly universally quantified.)
This definition is essentially a recursive program for computing x+y using only successor, and there are some programming languages (e.g. Haskell) that will allow you to define addition using almost exactly this notation. If the definition works for all inputs to +, we say that + is well-defined. Not working would include giving different answers depending on which parts of the definitions we applied first, or giving no answer for some particular inputs. These bad outcomes correspond to writing a buggy program. Though we can in principle prove that this particular definition is well-defined (using induction on y), we won't bother. Instead, we will try to prove things about our new concept of addition that will, among other things, tell us that the definition gives the correct answers.
We start with a lemma, which is Greek for a result that is not especially useful by itself but is handy for proving other results.8
Lemma
: 0+x = x.
Proof
: By induction on x. When x = 0, we have 0+0=0, which is true from the first case of the definition. Now suppose 0+x = x and consider what happens with Sx. We want to show 0+Sx = Sx. Rewrite 0+Sx as S(0+x) [second case of the definition], and use the induction hypothesis to show S(0+x) = S(x). ∎
(We could do a lot of QED-ish jumping around in the end zone there, but it is more refined—and lazier—to leave off the end of the proof once it's clear we've satisifed all of our obligations.)
Here's another lemma, which looks equally useless:
Lemma
: x+Sy = Sx+y.
Proof
: By induction on y. If y = 0, then x+S0 = S(x+0) = Sx = Sx+0. Now suppose the result holds for y and show x+SSy = Sx+Sy. We have x+SSy = S(x+Sy) = S(Sx+y) [ind. hyp.] = Sx+Sy. ∎
Now we can prove a theorem: this is a result that we think of as useful in its own right. (In programming terms, it's something we export from a module instead of hiding inside it as an internal procedure.)
Theorem
: x+y = y+x. (Commutativity of addition.)
Proof
: By induction on x. If x = 0, then 0+y = y+0 (see previous lemma). Now suppose x+y = y+x, and we want to show Sx+y = y+Sx. But y+Sx = S(y+x) [axiom] = S(x+y) [induction hypothesis] = x+Sy [axiom] = Sx+y [lemma]. ∎
This sort of definition-lemma-lemma-theorem structure is typical of written mathematical proofs. Breaking things down into small pieces (just like breaking big subroutines into small subroutines) makes debugging easier, since you can check if some intermediate lemma is true or false without having to look through the entire argument at once.
Question: How do you know which lemmas to prove? Answer: As when writing code, you start by trying to prove the main theorem, and whenever you come across something you need and can't prove immediately, you fork it off as a lemma. Conclusion: The preceding notes were not originally written in order.
6.3.1. Other useful properties of addition
So far we have shown that x+y = y+x, also known as commutativity of addition. Another familiar property is associativity of addition: x+(y+z) = (x+y)+z. This is easily proven by induction (try it!)
We don't have subtraction in ℕ (what's 3-5?)9 The closest we can get is cancellation:
x+y = x+z ⇒ y = z.
We can define ≤ for ℕ directly from addition: Let x ≤ y ≡ ∃z x+z = y. Then we can easily prove each of the following (possibly using our previous results about addition having commutativity, associativity, and cancellation).
0 ≤ x.
x ≤ Sx.
x ≤ y ∧ y ≤ z ⇒ x ≤ z.
a ≤ b ∧ c ≤ d ⇒ a+c ≤ b+d.
x ≤ y ∧ y ≤ x ⇒ x = y.
(The actual proofs will be left as an exercise for the reader.)
6.4. A scary induction proof involving even numbers
Let's define the predicate Even(x) ≡ ∃y x = y+y. (The use of ≡ here signals that Even(x) is syntactic sugar, and we should think of any occurrence of Even(x) as expanding to ∃y x = y+y.)
It's pretty easy to see that 0 = 0+0 is even. Can we show that S0 is not even?
Lemma
: ¬Even(S0).
Proof
: Expand the claim as ¬∃y S0 = y+y ≡ ∀y S0 ≠ y+y. Since we are working over ℕ, it's tempting to try to prove the ∀y bit using induction. But it's not clear why S0 ≠ y+y would tell us anything about S0 ≠ Sy+Sy. So instead we do a case analysis, using our earlier observation that every number is either 0 or Sz for some z.
Case 1
: y = 0. Then S0 ≠ 0+0 since 0+0 = 0 (by the definition of +) and 0 ≠ S0 (by the first axiom).
Case 2
: y = Sz. Then y+y = Sz+Sz = S(Sz+z) = S(z+Sz) = SS(z+z).10 Suppose S0 = SS(z+z) [Note: "Suppose" usually means we are starting a proof by contradiction]. Then 0 = S(z+z) [second axiom], violating ∀x 0 ≠ Sx [first axiom]. So S0 ≠ SS(z+z) = y+y.
Since we have S0 ≠ y+y in either case, it follows that S0 is not even. ∎
Maybe we can generalize this lemma! If we recall the pattern of non-even numbers we may have learned long ago, each of them (1, 3, 5, 7, ...) happens to be the successor of some even number (0, 2, 4, 6, ...). So maybe it holds that:
Theorem
: Even(x) ⇒ ¬Even(Sx).
Proof
: Expanding the definitions gives (∃y x = y+y) ⇒ (¬∃z Sx = z+z). This is an implication at top-level, which calls for a direct proof. The assumption we make is ∃y x = y+y. Let's pick some particular y that makes this true (in fact, there is only one, but we don't need this). Then we can rewrite the right-hand side as ¬∃z S(y+y) = z+z. There doesn't seem to be any obvious way to show this (remember that we haven't invented subtraction or division yet, and we probably don't want to). We are rescued by showing the stronger statement ∀y ¬∃z S(y+y) = z+z: this is something we can prove by induction (on y, since that's the variable inside the non-disguised universal quantifier). Our previous lemma gives the base case ¬∃z S(0+0) = z+z, so we just need to show that ¬∃z S(y+y) = z+z implies ¬∃z S(Sy+Sy) = z+z. Suppose that S(Sy+Sy) = z+z for some z ["suppose" = proof by contradiction again: we are going to drive this assumption into a ditch]. Rewrite S(Sy+Sy) to get SSS(y+y) = z+z. Now consider two cases:
Case 1
: z = 0. Then SSS(y+y) = 0+0 = 0, contradicting our first axiom.
Case 2
: z = Sw. Then SSS(y+y) = Sw+Sw = SS(w+w). Applying the second axiom twice gives S(y+y) = w+w. But this contradicts the induction hypothesis.
Since both cases fail, our assumption must have been false. It follows that S(Sy+Sy) is not even, and the induction goes through. ∎
6.5. Defining more operations
Let's define multiplication (⋅) by the axioms:
0⋅y = 0.
Sx⋅y = y + x⋅y.
Some properties of multiplication:
x⋅0 = 0.
1⋅x = x.
x⋅1 = x.
x⋅y = y⋅x.
x⋅(y⋅z) = (x⋅y)⋅z.
x ≠ 0 ∧ x⋅y = x⋅z ⇒ y = z.
x⋅(y+z) = x⋅y + x⋅z.
x ≤ y ⇒ z⋅x ≤ z⋅y.
z ≠ 0 ∧ z⋅x ≤ z⋅y ⇒ x ≤ y.
(Note we are using 1 as an abbreviation for S0.)
The first few of these are all proved pretty much the same way as for addition. Note that we can't divide in ℕ any more than we can subtract, which is why we have to be content with multiplicative cancellation.
Exercise: Show that the Even(x) predicate, defined previously as ∃y y = x+x, is equivalent to Even'(x) ≡ ∃y x = 2⋅y, where 2 = SS0. Does this definition make it easier or harder to prove ¬Even'(S0)?
The symbol is a stylized V, intended to represent the Latin word vel, meaning "or." (Thanks to Noel McDermott for pointing this out.) (1)
This symbol is a stylized A, short for the latin word atque, meaning "and." (2)
Programmers will recognize this as a form of SyntacticSugar. (3)
OK, maybe not as famous as it once was. (4)
These strategies are largely drawn from SolowBook, particularly the summary table on pages 156-159 (which is the source of the order and organization of the table and the names of most of the techniques). RosenBook describes proof strategies of various sorts in §§1.5–1.7 and BiggsBook describes various proof techniques in Chapters 1, 3, and 4; both descriptions are a bit less systematic, but also include a variety of specific techniques that are worth looking at. (5)
Some people define the natural numbers as starting at 1. Those people are generally either (a) wrong, (b) number theorists, (c) extremely conservative, or (d) citizens of the United Kingdom of Great Britain and Northern Ireland. As computer scientists, we will count from 0 as the gods intended. (6)
There is a complication here. Peano's original axioms were formulated in terms of second-order logic, which allows quantification over all possible predicates (you can write things like ∀P P(x) ⇒ P(Sx)). So the bogus predicate we defined is implicitly included in that for-all. In first-order logic, which is what everybody uses now, you can't quantify over predicates, and in particular if there is no predicate that distinguishes bogus numbers from legitimate ones, the induction axiom won't kick them out. This means that the Peano axioms (in their modern form) actually do allow bogus numbers to sneak in somewhere around infinity. But they have to be very polite bogus numbers that never do anything different from ordinary numbers. This is probably not a problem except for philosophers. (Similar problems show up for any model with infinitely many elements, due to something called the Löwenheim-Skolem theorem.) (7)
It really means fork. (8)
This actually came up on a subtraction test I got in the first grade from the terrifying Mrs Garrison at Mountain Park Elementary School in Berkeley Heights, New Jersey. She insisted that -2 was not the correct answer, and that we should have recognized it as a trick question. She also made us black out the arrow the left of the zero on the number-line stickers we had all been given to put on the top of our desks. Mrs Garrison was, on the whole, a fine teacher, but she did not believe in New Math. (9)
What justifies that middle step? (10)
2014-06-17 11:58 |
1338 | https://pregatirematematicaolimpiadejuniori.wordpress.com/wp-content/uploads/2016/07/incenter-excenter-lemma-evan-chen.pdf | The Incenter/Excenter Lemma Evan Chen evanchen@mit.edu March 5, 2015 In this short note, we’ll be considering the following very useful lemma.
Lemma. Let ABC be a triangle with incenter I, A-excenter IA, and denote by L the midpoint of arc BC. Show that L is the center of a circle through I, IA, B, C.
A B C I L IA Proof. This is just angle chasing. Let A = ∠BAC, B = ∠CBA, C = ∠ACB, and note that A, I, L are collinear (as L is on the angle bisector). We are going to show that LB = LI, the other cases being similar.
First, notice that ∠LBI = ∠LBC + ∠CBI = ∠LAC + ∠CBI = ∠IAC + ∠CBI = 1 2A + 1 2B.
However, ∠BIL = ∠BAI + ∠ABI = 1 2A + 1 2B.
Hence, △BIL is isosceles. So LB = LI. The rest of the proof proceeds along these lines.
Now, let’s see where this lemma has come up before. . .
1 Evan Chen 3 Intermediate Examples 1 Mild Embarrassments Problem 1 (USAMO 1988). Triangle ABC has incenter I. Consider the triangle whose vertices are the circumcenters of △IAB, △IBC, △ICA. Show that its circumcenter coincides with the circumcenter of △ABC.
Problem 2 (CGMO 2012). The incircle of a triangle ABC is tangent to sides AB and AC at D and E respectively, and O is the circumcenter of triangle BCI. Prove that ∠ODB = ∠OEC.
Problem 3 (CHMMC Spring 2012). In triangle ABC, the angle bisector of ∠A meets the perpendicular bisector of BC at point D.
The angle bisector of ∠B meets the perpendicular bisector of AC at point E. Let F be the intersection of the perpendicular bisectors of BC and AC. Find DF, given that ∠ADF = 5◦, ∠BEF = 10◦and AC = 3.
Problem 4 (Nine-Point Circle). Let ABC be an acute triangle with orthocenter H.
Let D, E, F be the feet of the altitudes from A, B, C to the opposite sides. Show that the midpoint of AH lies on the circumcircle of △DEF.
2 Some Short-Answer Problems Problem 5 (HMMT 2011). Let ABCD be a cyclic quadrilateral, and suppose that BC = CD = 2. Let I be the incenter of triangle ABD. If AI = 2 as well, find the minimum value of the length of diagonal BD.
Problem 6 (HMMT 2013). Let triangle ABC satisfy 2BC = AB + AC and have incenter I and circumcircle ω.
Let D be the intersection of AI and ω (with A, D distinct). Prove that I is the midpoint of AD.
Problem 7 (Online Math Open 2014/F19). In triangle ABC, AB = 3, AC = 5, and BC = 7. Let E be the reflection of A over BC, and let line BE meet the circumcircle of ABC again at D. Let I be the incenter of △ABD. Compute cos ∠AEI.
Problem 8 (NIMO 2012). Let ABXC be a cyclic quadrilateral such that ∠XAB = ∠XAC. Let I be the incenter of triangle ABC and by D the foot of I on BC. Given AI = 25, ID = 7, and BC = 14, find XI.
3 Intermediate Examples Problem 9. Let ABC be an acute triangle such that ∠A = 60◦. Prove that IH = IO, where I, H, O are the incenter, orthocenter, and circumcenter.
Problem 10 (IMO 2006). Let ABC be a triangle with incenter I. A point P in the interior of the triangle satisfies ∠PBA + ∠PCA = ∠PBC + ∠PCB.
Show that AP ≥AI, and that equality holds if and only if P = I.
Problem 11 (APMO 2007). In triangle ABC, we have AB > AC and ∠A = 60◦. Let I and H denote the incenter and orthocenter of the triangle. Show that 2∠AHI = 3∠B.
2 Evan Chen 5 Bonus Problems Problem 12 (ELMO 2013, Evan Chen). Triangle ABC is inscribed in circle ω. A circle with chord BC intersects segments AB and AC again at S and R, respectively. Segments BR and CS meet at L, and rays LR and LS intersect ω at D and E, respectively. The internal angle bisector of ∠BDE meets line ER at K. Prove that if BE = BR, then ∠ELK = 1 2∠BCD.
Problem 13 (Online Math Open 2012/F27). Let ABC be a triangle with circumcircle ω. Let the bisector of ∠ABC meet segment AC at D and circle ω at M ̸= B. The circumcircle of △BDC meets line AB at E ̸= B, and CE meets ω at P ̸= C. The bisector of ∠PMC meets segment AC at Q ̸= C. Given that PQ = MC, determine the degree measure of ∠ABC.
4 Harder Tasks Problem 14 (Iran 2001). Let ABC be a triangle with incenter I and A-excenter IA.
Let M be the midpoint of arc BC not containing A, and let N denote the midpoint of arc MBA. Lines NI and NIA intersect the circumcircle of ABC at S and T. Prove that the lines ST, BC and AI are concurrent.
Problem 15 (Online Math Open 2014/F26). Let ABC be a triangle with AB = 26, AC = 28, BC = 30. Let X, Y , Z be the midpoints of arcs BC, CA, AB (not containing the opposite vertices) respectively on the circumcircle of ABC. Let P be the midpoint of arc BC containing point A. Suppose lines BP and XZ meet at M , while lines CP and XY meet at N. Find the square of the distance from X to MN.
Problem 16 (Euler). Let ABC be a triangle with incenter I and circumcenter O. Show that IO2 = R(R −2r), where R and r are the circumradius and inradius of △ABC, respectively.
Problem 17 (IMO 2010). Let I be the incenter of a triangle ABC and let Γ be its circumcircle. Let the line AI intersect Γ again at D. Let E be a point on the arc BDC and F a point on the side BC such that ∠BAF = ∠CAE < 1 2∠BAC.
Finally, let G be the midpoint of IF. Prove that DG and EI intersect on Γ.
5 Bonus Problems Problem 18 (Russia 2014). Let ABC be a triangle with AB > BC and circumcircle Ω. Points M, N lie on the sides AB, BC respectively, such that AM = CN. Lines MN and AC meet at K. Let P be the incenter of the triangle AMK, and let Q be the K-excenter of the triangle CNK. If R is midpoint of arc ABC of Ωthen prove that RP = RQ.
Problem 19. Let ABC be a triangle with circumcircle Ω, and let D be any point on BC. We draw a curvilinear incircle tangent to AD at L, to BC at K and internally tangent to Ω. Show that the incenter of triangle ABC lies on KL.
3 Evan Chen 6 Hints to the Problems 6 Hints to the Problems 1. Tautological.
2. Who is O?
3. Point F is the circumcenter of △ABC. Who are D and E?
4. What is the incenter of △DEF? What is the D-excenter?
5. Show that AC = 4.
6. Apply Ptolemy’s Theorem.
7. Who is C? Erase E.
8. Apply Ptolemy’s Theorem.
9. Since ∠BHC = ∠BIC = ∠BOC = 120◦, points H and O now lie on the magic circle too. So IH = IO is just an equality of certain arcs.
10. Use the angle condition to show that P also lies on the magic circle.
11. The point H lies on the magic circle. So ∠IHC = ∠180◦−∠IBC.
12. You need to do quite a bit of angle chasing. Show that R is the incenter of △CDE.
Who is B?
13. Both M and P are arc midpoints. (Why?) 14. First show that S, T, I, IA are concyclic, say by NI · NS = NM2 = NIA · NT.
15. Add the incenter I. Line MN is a tangent.
16. Add in point L, the midpoint of arc BC. By Power of a Point, it’s equivalent to prove AI · IL = 2Rr, which can be done with similar triangles.
17. Take a homothety with ratio 2 at I. This sends G to F and D to the A-excenter.
18. Construct arc midpoints on the circumcircles of both △AMJ and △CNK. Use spiral similarity at R.
19. Let the tangency point to Ωbe T, let M be the midpoint of arc BC, and let lines KL and AM meet at I. Show that M, K, T are collinear. Show that ALIT is cyclic. Prove that MI2 = MK · MT = MC2 = MI2.
4 |
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ResourcesAops Wiki 2020 AMC 8 Problems/Problem 23
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2020 AMC 8 Problems/Problem 23
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1 Problem
2 Solution 1 (Constructive Counting)
3 Version of Solution 1
4 Solution 3 (Complementary Counting)
5 Video Solution by Math-X (First understand the problem!!!)
6 Video Solution (🚀Under 3 min🚀)
7 Video Solution by Robin "The Smartest Dude" Shang
8 Video Solution by OmegaLearn
9 Video Solution by SpreadTheMathLove
10 Video Solution by WhyMath
11 Video Solutions by The Learning Royal
12 Video Solution by Interstigation
13 Video Solution by STEMbreezy
14 Video solution by TheNeuralMathAcademy
15 See also
Problem
Five different awards are to be given to three students. Each student will receive at least one award. In how many different ways can the awards be distributed?
Solution 1 (Constructive Counting)
Firstly, observe that a single student can't receive or awards because this would mean that one of the other students receives no awards. Thus, each student must receive either , , or awards. If a student receives awards, then the other two students must each receive award; if a student receives awards, then another student must also receive awards and the remaining student must receive award. We consider each of these two cases in turn.
If a student receives three awards, there are ways to choose which student this is, and ways to give that student out of the awards. Next, there are students left and awards to give out, with each student getting one award. There are clearly just ways to distribute these two awards out, giving ways to distribute the awards in this case.
In the other case, two students receive awards and one student receives award. We know there are choices for which student gets award. There are ways to do this. Then, there are ways to give the first student his two awards, leaving awards yet to be distributed. There are then ways to give the second student his awards. Finally, there is only student and award left, so there is only way to distribute this award. This results in ways to distribute the awards in this case. Adding the results of these two cases, we get .
Version of Solution 1
Upon inspection (specified in the above solution), there are two cases of the distribution of awards to the students: one student gets 3 awards and the other each get 1 award or one student gets 1 award and the other two get 2 awards.
In the first case, there are ways to choose the person who gets 3 awards. From here, there are ways to choose the 3 awards from the 5 total awards. Now, one person has choices for awards and the other has choice for the award. Thus, the total number of ways to choose awards in this case is .
In the other case, there are ways to choose the person who gets 1 award, and choices for his/her award. Then, one person has ways to have his/her awards and the other person has ways to have his/her awards. This gives ways for this case.
Adding these numbers together, we get ways to distribute the awards, or choice .
Solution 3 (Complementary Counting)
Without the restriction that each student receives at least one award, we could take each of the awards and choose one of the students to give it to. This would be ways to distribute the awards in total. Now we need to subtract the cases where at least one student doesn't receive an award. If a student doesn't receive an award, there are choices for which student that is, so ways of choosing a student to receive each of the awards; in total, .
However, if students both don't receive an award, then this case would be counted twice among the , so we need to add back in these cases. Said in other words, students not receiving an award is equivalent to student receiving awards, and there are choices for whom that student would be. To finish, the total number of ways to distribute the awards is , or .
Video Solution by Math-X (First understand the problem!!!)
~Math-X
Video Solution (🚀Under 3 min🚀)
Video Solution by Robin "The Smartest Dude" Shang
Video Solution by OmegaLearn
~ pi_is_3.14
Video Solution by SpreadTheMathLove
Video Solution by WhyMath
~WhyMath
Video Solutions by The Learning Royal
Video Solution by Interstigation
~Interstigation
Video Solution by STEMbreezy
~STEMbreezy
Video solution by TheNeuralMathAcademy
See also
2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22Followed by
Problem 24
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1340 | https://www.doubtnut.com/qna/54767 | If f(2+x)=f(−x) for all x∈R then differentiability at x=4 implies differentiability at
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1341 | https://youglish.com/pronounce/compassion/english | Compassion | 17981 pronunciations of Compassion in English
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Phonetic:
When you begin to speak English, it's essential to get used to the common sounds of the language, and the best way to do this is to check out the phonetics. Below is the UK transcription for 'compassion':
Modern IPA: kəmpáʃən
Traditional IPA: kəmˈpæʃən
3 syllables: "kuhm" + "PASH" + "uhn"
Test your pronunciation on words that have sound similarities with 'compassion':
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Sound it Out: Break down the word 'compassion' into its individual sounds "kuhm" + "pash" + "uhn". Say these sounds out loud, exaggerating them at first. Practice until you can consistently produce them clearly.
Self-Record & Review: Record yourself saying 'compassion' in sentences. Listen back to identify areas for improvement.
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1342 | https://www.unitscenter.com/convert/pressure/centimeter-of-water/milimetre-of-mercury | Convert Pressure Units: from Centimeter of water to Milimetre of mercury
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| 7 | 5.148914e+0 | Copy |
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| 9 | 6.620032e+0 | Copy |
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1343 | https://oercommons.org/courseware/lesson/78389 | Polynomials Investigation | OER Commons
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Overview:The purpose of this lesson is for students to discover the connection between the algebraic and the graphical structure of polynomial functions. This lesson leads to students being able to sketch a graph by identifying the end behavior, intercepts, and multiplicities from a given polynomial equation. It also leads to students being able to write a possible equation by determining the sign of the leading coefficient, minimum possible degree, x-intercepts and y-intercept from a given polynomial graph.Subject:Algebra, Functions Level: High School Material Type:Activity/Lab, Lesson, Lesson Plan Author:Victoria OlindeDate Added:03/20/2021
License:Creative Commons Attribution Non-Commercial Share AlikeLanguage:English Media Format:Downloadable docs, Interactive
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MCCRS.Math.Content.HSA-APR.A.1
Maryland College and Career Ready Math Standards
Grades 9-12
Learning Domain: Algebra: Arithmetic with Polynomials and Rational Functions
Standard: Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials.
Degree of Alignment: Not Rated (0 users)
MCCRS.Math.Content.HSA-APR.B.2
Maryland College and Career Ready Math Standards
Grades 9-12
Learning Domain: Algebra: Arithmetic with Polynomials and Rational Functions
Standard: Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x - a is p(a), so p(a) = 0 if and only if (x - a) is a factor of p(x).
Degree of Alignment: Not Rated (0 users)
MCCRS.Math.Content.HSA-APR.B.3
Maryland College and Career Ready Math Standards
Grades 9-12
Learning Domain: Algebra: Arithmetic with Polynomials and Rational Functions
Standard: Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.
Degree of Alignment: Not Rated (0 users)
MCCRS.Math.Content.HSA-APR.C.4
Maryland College and Career Ready Math Standards
Grades 9-12
Learning Domain: Algebra: Arithmetic with Polynomials and Rational Functions
Standard: Prove polynomial identities and use them to describe numerical relationships. For example, the polynomial identity (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 can be used to generate Pythagorean triples.
Degree of Alignment: Not Rated (0 users)
MCCRS.Math.Content.HSA-APR.C.5
Maryland College and Career Ready Math Standards
Grades 9-12
Learning Domain: Algebra: Arithmetic with Polynomials and Rational Functions
Standard: (+) Know and apply that the Binomial Theorem gives the expansion of (x + y)^n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal's Triangle. (The Binomial Theorem can be proved by mathematical induction or by a combinatorial argument.)
Degree of Alignment: Not Rated (0 users)
MCCRS.Math.Content.HSA-APR.D.6
Maryland College and Career Ready Math Standards
Grades 9-12
Learning Domain: Algebra: Arithmetic with Polynomials and Rational Functions
Standard: Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system.
Degree of Alignment: Not Rated (0 users)
MCCRS.Math.Content.HSA-APR.D.7
Maryland College and Career Ready Math Standards
Grades 9-12
Learning Domain: Algebra: Arithmetic with Polynomials and Rational Functions
Standard: (+) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.
Degree of Alignment: Not Rated (0 users)
MCCRS.Math.Content.HSF-IF.C.7
Maryland College and Career Ready Math Standards
Grades 9-12
Learning Domain: Functions: Interpreting Functions
Standard: Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases.
Degree of Alignment: Not Rated (0 users)
MCCRS.Math.Content.HSF-IF.C.7c
Maryland College and Career Ready Math Standards
Grades 9-12
Learning Domain: Functions: Interpreting Functions
Standard: Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior.
Degree of Alignment: Not Rated (0 users)
CCSS.Math.Content.HSA-APR.A.1
Common Core State Standards Math
Grades 9-12
Cluster: Perform arithmetic operations on polynomials
Standard: Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials.
Degree of Alignment: Not Rated (0 users)
CCSS.Math.Content.HSA-APR.B.2
Common Core State Standards Math
Grades 9-12
Cluster: Understand the relationship between zeros and factors of polynomial
Standard: Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x).
Degree of Alignment: Not Rated (0 users)
CCSS.Math.Content.HSA-APR.B.3
Common Core State Standards Math
Grades 9-12
Cluster: Understand the relationship between zeros and factors of polynomial
Standard: Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial.
Degree of Alignment: Not Rated (0 users)
CCSS.Math.Content.HSA-APR.C.4
Common Core State Standards Math
Grades 9-12
Cluster: Use polynomial identities to solve problems
Standard: Prove polynomial identities and use them to describe numerical relationships. For example, the polynomial identity (x^2 + y^2)^2 = (x^2 – y^2)^2 + (2xy)^2 can be used to generate Pythagorean triples.
Degree of Alignment: Not Rated (0 users)
CCSS.Math.Content.HSA-APR.C.5
Common Core State Standards Math
Grades 9-12
Cluster: Use polynomial identities to solve problems
Standard: (+) Know and apply that the Binomial Theorem gives the expansion of (x + y)^n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal’s Triangle. (The Binomial Theorem can be proved by mathematical induction or by a combinatorial argument.)
Degree of Alignment: Not Rated (0 users)
CCSS.Math.Content.HSA-APR.D.6
Common Core State Standards Math
Grades 9-12
Cluster: Rewrite rational expressions
Standard: Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system.
Degree of Alignment: Not Rated (0 users)
CCSS.Math.Content.HSA-APR.D.7
Common Core State Standards Math
Grades 9-12
Cluster: Rewrite rational expressions
Standard: (+) Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions.
Degree of Alignment: Not Rated (0 users)
CCSS.Math.Content.HSF-IF.C.7
Common Core State Standards Math
Grades 9-12
Cluster: Analyze functions using different representations
Standard: Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases.
Degree of Alignment: Not Rated (0 users)
CCSS.Math.Content.HSF-IF.C.7c
Common Core State Standards Math
Grades 9-12
Cluster: Analyze functions using different representations
Standard: Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior.
Degree of Alignment: Not Rated (0 users)
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1344 | https://en.wikipedia.org/wiki/Complex_conjugate_of_a_vector_space | Jump to content
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Contents
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1 Motivation
2 Complex conjugation functor
3 Structure of the conjugate
4 Complex conjugate of a Hilbert space
5 See also
6 References
7 Further reading
Complex conjugate of a vector space
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From Wikipedia, the free encyclopedia
Mathematics concept
In mathematics, the complex conjugate of a complex vector space is a complex vector space that has the same elements and additive group structure as but whose scalar multiplication involves conjugation of the scalars. In other words, the scalar multiplication of satisfies where is the scalar multiplication of and is the scalar multiplication of The letter stands for a vector in is a complex number, and denotes the complex conjugate of
More concretely, the complex conjugate vector space is the same underlying real vector space (same set of points, same vector addition and real scalar multiplication) with the conjugate linear complex structure (different multiplication by ).
Motivation
[edit]
If and are complex vector spaces, a function is antilinear if With the use of the conjugate vector space , an antilinear map can be regarded as an ordinary linear map of type The linearity is checked by noting: Conversely, any linear map defined on gives rise to an antilinear map on
This is the same underlying principle as in defining the opposite ring so that a right -module can be regarded as a left -module, or that of an opposite category so that a contravariant functor can be regarded as an ordinary functor of type
Complex conjugation functor
[edit]
A linear map gives rise to a corresponding linear map that has the same action as Note that preserves scalar multiplication because Thus, complex conjugation and define a functor from the category of complex vector spaces to itself.
If and are finite-dimensional and the map is described by the complex matrix with respect to the bases of and of then the map is described by the complex conjugate of with respect to the bases of and of
Structure of the conjugate
[edit]
The vector spaces and have the same dimension over the complex numbers and are therefore isomorphic as complex vector spaces. However, there is no natural isomorphism from to
The double conjugate is identical to
Complex conjugate of a Hilbert space
[edit]
Given a Hilbert space (either finite or infinite dimensional), its complex conjugate is the same vector space as its continuous dual space There is one-to-one antilinear correspondence between continuous linear functionals and vectors. In other words, any continuous linear functional on is an inner multiplication to some fixed vector, and vice versa.[citation needed]
Thus, the complex conjugate to a vector particularly in finite dimension case, may be denoted as (v-dagger, a row vector that is the conjugate transpose to a column vector ). In quantum mechanics, the conjugate to a ket vector is denoted as – a bra vector (see bra–ket notation).
See also
[edit]
Antidual space – Conjugate homogeneous additive mapPages displaying short descriptions of redirect targets
Linear complex structure – Mathematics concept
Riesz representation theorem – Theorem about the dual of a Hilbert space
conjugate bundle
References
[edit]
^ K. Schmüdgen (11 November 2013). Unbounded Operator Algebras and Representation Theory. Birkhäuser. p. 16. ISBN 978-3-0348-7469-4.
Further reading
[edit]
Budinich, P. and Trautman, A. The Spinorial Chessboard. Springer-Verlag, 1988. ISBN 0-387-19078-3. (complex conjugate vector spaces are discussed in section 3.3, pag. 26).
Retrieved from "
Categories:
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Vectors (mathematics and physics)
Hidden categories:
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Complex conjugate of a vector space
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1345 | https://www.math.kth.se/matstat/gru/sf2972/2015/gametheory.pdf | COMBINATORIAL GAME THEORY JONAS SJ¨ OSTRAND 1. What is a combinatorial game?
As opposed to classical game theory, combinatorial game theory deals exclusively with a specific type of two-player games. Informally, these games can be character-ized as follows.
(1) There are two players who alternate moves.
(2) There are no chance devices like dice or shuffled cards.
(3) There is perfect information, i.e. all possible moves and the complete history of the game are known to both players.
(4) The game will eventually come to an end, even if the players do not alternate moves.
(5) The game ends when the player in turn has no legal move and then he loses.
The last condition is called the normal play convention and is sometimes replaced by the mis` ere play convention where the player who makes the last move loses. In this course, however, we will stick to the normal play convention.
The players are typically called Left and Right.
1.1. Examples of games.
1.1.1. Domineering. A position in Domineering is a subset of the squares on a grid. Left places a domino to remove two adjacent vertical squares. Right places horizontally.
1.1.2. Hackenbush. A position in Hackenbush is a graph where one special vertex is called the ground. In each move, a player cuts an edge and removes any portion of the graph no longer connected to the ground.
In Blue-Red Hackenbush (also known as LR-Hackenbush or Hackenbush restrained) the edges are coloured blue and red, and Left may only cut blue edges, Right only red edges.
1.2. Game trees. We may visualize a game as a tree with the start position at the top and where Left follows down-left edges and Right follows down-right edges.
Here are the game trees for three Domineering positions.
∅ ∅ Date: March 2015.
1 2 JONAS SJ¨ OSTRAND ∅ ∅ 1.3. Formal definition of games. From now on, we will refer to game positions simply as games.
When developing an abstract theory for games, we have no reason to distinguish between games with isomorphic game trees (like the Domineering games and above) since those are really just different symbolic representations of the same game. In other words, it is the game tree itself that is important and not how we label the vertices or edges. Since a game tree is completely determined by its left and right subtrees, we may simply define a game as a set of left subgames together with a set of right subgames, and indeed we do: Axioms for games 1. A game is an ordered pair G = {GL|GR} of sets of games. (GL and GR are the sets of left and right options of G, respectively.) 2. Two games G and H are identical, and we write G ≡H, if and only if GL = HL and GR = HR.
3. There is no infinite sequence of games G1, G2, . . . such that Gi+1 is a (left or right) option of Gi for all i.
The third axiom guarantees that a game must eventually come to an end, even if the players do not alternate moves. In other words, the game tree has a finite depth. Note however that the sets of options does not need to be finite, so we allow games with infinitely many legal moves.
Note that we use the symbol “≡” for identical games. The reason is that later on we will define equivalence of games and use the ordinary equality sign “=” for that.
A typical left or right option of G will be denoted by GL and GR, respectively, and we will always omit the curly braces around the sets of options and write G = {GL 1 , GL 2 , . . . |GR 1 , GR 2 , . . . }.
The simplest of all games is {|} which we call the zero game and denote by 0.
Using 0 we can construct the three games 1 := {0|}, −1 := {|0}, and ∗:= {0|0}, whose names will be justified later.
In the creation story of games, the zero game was born on day 0 and 1, −1 and ∗were born at day one. The birthday of a game is defined recursively as 1 plus the maximal birthday of its options, with the zero game having birthday 0.
1.4. Conway induction.
COMBINATORIAL GAME THEORY 3 Theorem 1.1. Let P be any property that a game can have. If a game is P whenever any option is P, then all games are P.
More generally: Let P be a property that an n-tuple of games can have. Suppose that, for any n-tuple G1, . . . , Gn, if P holds when any game in the tuple is replaced by one of its options, then the tuple G1, . . . , Gn is P. Then, any n-tuple of games is P.
Proof. Suppose a game is P whenever any option is. If there is a game G1 that is not P, then it has an option G2 that is not P, which has an option G3 that is not P, and so on. This infinite sequence violates the third axiom of games.
The more general version is proved in a similar way.
□ 2. The outcome of a game We always assume optimal play and when we say that a player wins a game we have that assumption in mind.
Let us examine the outcomes of the four simplest games 0, 1, −1 and ∗: The game 0 is a win for the second player since the first player loses immediately; Left wins 1 and Right wins −1 whoever starts; and ∗is a win for the first player.
A moment’s thought reveals that any game belongs to exactly one of these four outcome classes.
If G and H belong to the same outcome class we write G ∼H, and if, whoever starts, the outcome of G is at least as good for Left as the outcome of H, we write G ≳H. Clearly, 1 ≳0 ≳−1 and 1 ≳∗≳−1 while 0 ̸≳∗̸≳0 so the Hasse diagram of outcome classes looks like this: 1 0 ∗ −1 Note that G ≳0 ⇔Left wins as second player, and G ≲0 ⇔Right wins as second player.
3. Sum and negation of games For convenience, if G is a game and H is a set of games, let G + H denote the set {G + H : H ∈H}. Also, let −H denote the set {−H : H ∈H}.
Definition 3.1 (Sum of games). For any two games G and H, we define their sum G + H as the game where G and H are played in parallel, and the player about to move must choose to make a move in either the G-component or the H-component.
Formally, G + H = {(GL + H) ∪(G + HL)|(GR + H) ∪(G + HR)}.
Theorem 3.2. Addition is commutative and associative, i.e. for any games G, H, and K, we have G + H ≡H + G, (G + H) + K ≡G + (H + K).
4 JONAS SJ¨ OSTRAND Proof. Intuitively, this is obvious. To find a formal proof is a simple exercise in Conway induction.
□ Corollary 3.3. The class Pg of partizan games forms an Abelian monoid under addition.
Definition 3.4 (Negation). The negation −G of a game G is the game where the roles of Left and Right are swapped. Formally, −G = {−GR| −GL}.
We write G −H for the game G + (−H).
Note that G −G ̸≡0 unless G ≡0, so the notation seems a bit misleading for the moment. However, things will get better in the next section.
Theorem 3.5. For any games G and H, we have −(−G) ≡G, −(G + H) ≡(−G) + (−H), G ≳H ⇔−G ≲−H.
Proof. Again, this is intuitively obvious and a formal proof is a simple exercise in Conway induction.
□ 4. Comparing games As combinatorial game theorists, we are primarily interested in outcome classes of games and not so much in the games themselves. However, the outcome class of G + H is not determined by the outcome classes of G and H; for instance, {1|} ∼1, but {1|} −1 ≁1 −1. We want to consider G and H as “equal” only if they behave similarly when added to other games, and this leads us to the following definition.
Definition 4.1. Two games G and H are said to be equivalent (or simply equal), and we write G = H, if G + K ∼H + K for any game K.
Furthermore, we write G ≥H if G + K ≳H + K for any game K.
Theorem 4.2. Equality of games is an equivalence relation and ≥is a partial order modulo equivalence. In other words, for any games G, H, and K, the following holds.
G = G (reflexivity), (1) G = H ⇒H = G (symmetry), (2) G = H = K ⇒G = K (transitivity), (3) G = H ⇒G ≥H (reflexivity), (4) G ≥H ≥G ⇒G = H (antisymmetry), (5) G ≥H ≥K ⇒G ≥K (transitivity).
(6) Proof. Easy exercise!
□ Theorem 4.3. The partial order and equality is compatible with addition and nega-tion in the sense that, for any games G, H, and K, G = H ⇒G + K = H + K, G = H ⇒−G = −H, G ≥H ⇒G + K ≥H + K, G ≥H ⇒−G ≤−H.
COMBINATORIAL GAME THEORY 5 Proof. Easy exercise!
□ Okay, so we have defined equivalence classes of games and a partial order on them which behave properly. What can we say about the algebraic structure of our creation?
Our first task is to characterize the games that compare with zero. To this end we need a lemma.
Lemma 4.4. G ≳0 and H ≳0 implies G + H ≳0.
Proof. If Right starts from G + H he must move either to some GR + H or to some G + HR. By symmetry, without loss of generality we may assume that he moves to GR + H. Since Right has no winning move from G ≳0, there must exist some left option GRL ≳0 of GR. By induction, GRL + H ≳0 so Left will win the game.
□ It turns out that the games that are equal to zero are exactly those won by the second player, the games that are ≥0 are those without a winning move for Right, and the games are ≤0 are those without a winning move for Left.
Theorem 4.5.
G ≥0 ⇔G ≳0, G ≤0 ⇔G ≲0, G = 0 ⇔G ∼0.
Proof. By symmetry, it suffices to show that G ≥0 ⇔G ≳0.
If G ≥0 then G + K ≳K for any game K, and in particular G + 0 ≳0.
For the converse, suppose G ≳0 and let us show that G + K ≳K for any game K.
Informally, if Left has a winning strategy from K, then she can transform it to a winning strategy from G + K by the additional rule that she will only play in the G component as a direct answer to a move there. In that way, she will be acting as the second player in G and can take advantage of her winning strategy there.
Formally, we may reason as follows.
If K ̸≲0, i.e. Left has a winning move from K to some KL, then she also has a winning move from G + K to G + KL since G + KL ≳KL by induction.
If K ≳0, i.e. Right has no winning move from K, then, by Lemma 4.4, he has no winning move from G + K either. The implications K ̸≲0 ⇒G + K ̸≲0 and K ≳0 ⇒G + K ≳0 together yield that G + K ≳K.
□ What if G −G is a win for the second player? Then, by the above theorem, it would be equal to zero and that would tranform our boring Monoid into an Abelian Group!
Theorem 4.6. G −G = 0 for any game G.
Proof. By the previous theorem, it suffices to show that G −G ∼0, and, by sym-metry, it suffices to show that G −G ≳0, that is, Right has no winning move from G −G.
Right’s move to GR −G is countered by GR −GR which is ≳0 by induction, so Left will win. Analogously, Right’s move to G −GL is countered by GL −GL ≳0.
The intuition behind is that the second player can always mimic the previous move in the other component.
□ 6 JONAS SJ¨ OSTRAND Corollary 4.7. The class Pg of partizan games is a partially ordered Abelian Group under addition modulo equality. The identity element is (the equivalence class of) 0.
For convenience, we introduce the notation G ▷H as a synonym for G ̸≤H, and we write G ∥H (pronounced “G is fuzzy to H”) if G ▷H ▷G, i.e. G and H are not related by the partial order ≥.
The following table summarizes the interpretations of comparisons of a game with zero.
G ≥0 ⇔Left wins as second player, G ≤0 ⇔Right wins as second player, G ▷0 ⇔Left wins as first player, G ◁0 ⇔Right wins as first player, G = 0 ⇔the second player wins, G ∥0 ⇔the first player wins, Theorem 4.8. GL ◁G ◁GR hold for any left option GL and any right option GR of any game G.
Proof. Right has a winning move from GL −G to GL −GL = 0 and a winning move from G −GR to GR −GR = 0.
□ Theorem 4.9. G ≥H if and only if all GR ▷H and all HL ◁G.
Proof. G ≥H ⇔G −H ≥0 ⇔G −H ≳0 which means that Right has no winning move from G−H. But Right’s move to GR −H is winning if and only if GR −H ≤0 and Right’s move to G −HL is winning if and only if G −HL ≤0.
□ Theorem 4.10. Let G = {GL|GR} and H = {HL|HR} be games and suppose there are bijections φL : GL →HL and φR : GR →HR with the property that φL(GL) = GL and φR(GR) = GR for any options GL and GR. Then, G = H.
Proof. For each left option GL we have GL = φL(GL) ◁H and for each right option GR we have H ◁φR(GR) = GR, so we conclude that G ≥H.
By symmetry, H ≥G.
□ Theorem 4.11. G ▷H ≥K ⇒G ▷K and G ≥H ▷K ⇒G ▷K.
Proof. Simple exercise!
□ 5. Simplifying games Theorem 5.1. The value of G is unaltered or increased when we (1) increase any GL or GR, or (2) remove some GR or add a new GL.
Proof. Let G′ be the game obtained by so modifying G. Then in the game G −G′ it is easy to check that Right has no good first move.
□ Definition 5.2. Let G a game.
A left option GL is dominated by another left option GL′ if GL′ ≥GL. Similarly, a right option GR is dominated by another right option GR′ if GR′ ≤GR.
COMBINATORIAL GAME THEORY 7 Theorem 5.3. Removing a dominated option does not alter the value of the game.
Proof. Let G be a game with a left option GL dominated by another left option GL′, and let G′ be the game obtained by removing the option GL from G. That G ≥G′ follows from Theorem 5.1. We will prove that G ≤G′ by showing that Left has no good first move in G −G′. Each left option in G −G′ has a right counterpart in the other component, except GL −G′. But this is countered by GL −GL′ which is ≤0, so Right will win.
The corresponding statement for dominated right options follows by symmetry.
□ Definition 5.4. Let G be a game. A left option GL of G is said to be reversible (through GLR) if it has a right option GLR ≤G. Similarly, a right option GR is reversible (through GLR) if it has a left option GRL ≥G.
Theorem 5.5. If G has a left option GL that is reversible through GLR, then the value of G does not change if we replace GL by all the left options of GLR.
Proof. Let G′ be the game obtained from G by replacing GL by all the left options of GLR. We will show that neither player has any good move in G −G′.
If Left starts and moves to GL −G′, Right moves to GLR −G′. Left can either move to some GLRL −G′, in which case Right moves to GLRL −GLRL and wins, or she can move to some GLR −GR, in which case Right wins since GLR ≤G ◁GR.
If Right starts and moves to some G−GLRL, Left will win since G ≥GLR ▷GLRL.
Any other first move for Right has counterparts for Left in the other component.
□ Definition 5.6. A game is in canonical form if it has no dominated or reversible options and if all its options are in canonical form.
Theorem 5.7. Every equivalence class of short games has exactly one representative in canonical form. In other words, for each short game G there is a unique game G′ in canonical form such that G = G′.
Proof. Since both ways of simplifying games (removing dominated options and by-passing reversible options) reduce the number of positions, we must eventually reach a game that cannot be simplified further. This proves existence. To prove unique-ness, we assume that G and H are two equal games in canonical form. We have to show that G ≡H. Let GL be some left option of G. Since GL ◁G = H, there must be a right option GLR ≤H or a left option HL such that GL ≤HL. The first is im-possible since GL is not reversible. Similarly, there is some GL′ such that HL ≤GL′, so GL ≤GL′. But there are no dominated options either, so GL = HL = GL′. By induction, GL ≡HL.
In that way, we see that G and H have the same set of (identical) left options, and the same is true for the right options.
□ 6. Numbers By x < y we mean that x ≤y and x ̸= y, just as expected.
Definition 6.1. A game x is said to be a number if • all options of x are numbers, and • xL < xR for any left option xL and any right option xR of x.
For instance, 0, 1 and −1 are numbers, but ∗is not. We usually denote numbers by the lower-case letters x, y, z etc.
8 JONAS SJ¨ OSTRAND Note that a game might equal a number without actually being one. Find such a game as an exercise!
Theorem 6.2. The class of numbers is closed under addition and negation.
Proof. Conway induction!
□ Theorem 6.3. If x is a number then xL < x < xR for any xL, xR.
Proof. Since xL ◁x ◁xR, it suffices to show that xL ≤x ≤xR.
Right’s move from x −xL to xR −xL is bad since xR > xL by the definition of a number, and Right’s move from x −xL to x −xLL is bad since x ▷xL > xLL by induction.
□ Theorem 6.4. Numbers are totally ordered, that is, for any numbers x and y, either x ≤y or x ≥y.
Proof. The inequality x ̸≥y implies either some xR ≤y or x ≤some yL, whence either x < xR ≤y or x ≤yL < y.
□ The following theorem states that if there is a number fuzzily in between the left and right options of a game, then the game equals the simplest such number, that is, the unique such number that has no option with the same property.
Theorem 6.5 (The simplicity theorem). Suppose G is a game and x is a number such that • GL ◁x ◁GR for any left option GL and right option GR of G, and • no option of x satisfies the same condition.
Then G = x.
Proof. We have G ≥x unless some GR ≤x (which is false) or G ≤some xL. But from G ≤xL we can deduce GL ◁G ≤xL < x ◁GR for all GL, GR, from which we have GL ◁xL ◁GR, contradicting the supposition about x. So G ≥x, similarly G ≤x, and so G = x.
□ 6.1. Dyadic rationals. We have already defined the numbers 0, 1 and −1, and we may define any integer by adding ones or minus ones, for instance 5 := 1+1+1+1+1.
Theorem 6.6. For any integer n, we have n = { n −1 | } if n is positive and n = { | n + 1 } if n is negative.
We leave the proof as an exercise.
Multiplication of a number (or, indeed, any game) by an integer can also be defined just by iterated addition, for instance 3x := x + x + x. But is it possible to divide a number by an integer?
Let us first observe that if a quotient exists it must be unique (up to equality).
Indeed, if m is a non-zero integer and mx = my = z then m(x −y) = 0 and it follows that x −y = 0. (Can you see why?) Hence, z/m is well-defined if it exists.
Also, for quotients that exist, ordinary algabraic rules apply: • If z/m and w/m exist, then the quotient (z + w)/m = z/m + w/m exists.
• If z/m exists, then, for any non-zero integer k, (kz)/m exists and equals k(z/m), and kz/km exists and equals z/m.
As an exercise, show that 1/2 exists and equals { 0 | 1 }.
Theorem 6.7. For any integer n ≥1, the fraction 1/2n exists and equals { 0 | 1/2n−1 }.
COMBINATORIAL GAME THEORY 9 Proof. Let x = 1/2n−1 and y = { 0 | x }. By induction, x = { 0 | 2x}. Clearly, 0 < y < x < x + y < 2x, so x, but no option of x, lies between y and x + y. Thus, by the simplicity theorem, x = { y | x + y } which equals 2y by the definition of addition.
□ It follows that any quotient of an integer by a power of two exists. Such fractions are called dyadic rationals.
Theorem 6.8. For any nonnegative integer n ≥0 and any odd integer k, we have (7) k 2n = ( k −1 2n k + 1 2n ) .
Proof.
k 2n = 1 2n + · · · + 1 2n = ( k −1 2n + 0 k −1 2n + 1 2n−1 ) = ( k −1 2n k + 1 2n ) .
□ The following picture shows how all numbers form an infinite tree. Each node has two children, namely the first later numbers born just to the left and right of it.
0 1 1 2 2 1 4 3 4 3 2 3 1 8 3 8 5 8 7 8 5 4 7 4 5 2 4 −1 −1 2 −2 −1 4 −3 4 −3 2 −3 −1 8 −3 8 −5 8 −7 8 −5 4 −7 4 −5 2 −4 Every number x can be found in the infinite tree, and the sign-expansion of x tells us how to reach x if we start at the top node 0 and walk along the edges down the tree. The sign-expansion is a (possibly) infinite sequence of pluses and minuses, where plus means right and minus means left. For instance, the sign-expansion of 7/4 is + + −+.
Numbers are the coolest class of games in the sense that no player wants to be the first to make a move in a number. The following theorem shows that, in order to win a game with several components, you do not have to move in a number unless there is nothing else to do.
10 JONAS SJ¨ OSTRAND Theorem 6.9 (Weak number avoidance theorem). If G is a game that is not equal to a number and x is a number, then G + x ▷0 ⇐ ⇒some GL + x ≥0.
Proof. Suppose on the contrary that Left has a winning move to some G + xL but not to any GL + x. Then all GL ◁−x < −xL ≤G ◁all GR, where the second inequality follows from the definition of a number and the last inequality follows from Theorem 4.8. But now the simplicity theorem yields that G is a number, which contradicts the assumption in the theorem.
□ 7. Blue-Red Hackenbush is always a number Lemma 7.1. In Blue-Red Hackenbush, on chopping a blue edge, the value stricly decreases; on chopping a red one it strictly increases.
Proof. Let GL be the result after chopping a blue edge e from a position G. If Right starts in the game G −GL, Left can win by mimicking Right’s move in the other component until Right chops an edge in the G-component that has no counterpart in the −GL-component. If this happens, Left simply chops the edge e from the G-component, thereby obtaining a zero game. This shows that G −GL ≥0 and thus GL < G (since an option never is equal to the game itself).
The argument for G < GR is completely analogous.
□ Now the following theorem follows by induction.
Theorem 7.2. In Blue-Red Hackenbush every position is a number.
7.1. Trees. There is a simple rule for computing the Blue-Red Hackenbush value of a tree.
For a Hackenbush tree x with the root on the ground, let 1 : x (and −1 : x) denote the tree obtained by inserting a blue (red) edge between the ground and the root.
Every tree can be written as either 1 : x or −1 : x, where x is a sum of smaller trees, so in order to compute the value of any tree we only need a method to compute the value of 1 : x given the value of x.
The moves from 1 : x are to 0 and 1 : xL for Left and to 1 : xR for Right, and the moves from −1 : x are to −1 : xL for Left and to 0 and −1 : xR for Right. Now, the function 1 : x = { 0, 1 : xL | 1 : xR} maps all numbers onto positive numbers in order of simplicity. Thus 0, the simplest number, maps to 1, the simplest positive number. Then −1 and 1 map to the simplest positive numbers to the left and right of 1, namely 1/2 and 2 respectively, and so on. In terms of sign-expansions, 1 : x is obtained by inserting a + at the beginning of the sign-expansion of x, and −1 : x is obtained by inserting a minus sign at the beginning.
8. Comparing games with numbers A game is short if it has only finitely many positions altogether. Short games are bounded in the following sense.
Theorem 8.1. For any short game G there is some integer n such that −n < G < n.
COMBINATORIAL GAME THEORY 11 Proof. Take n greater than the total number of positions of G and consider playing in G+n. Left can win this by just decreasing n by 1 each time he moves, waiting for Right to run himself down in G. Since G + n > 0, we have G > −n, and similarly G < n.
□ Definition 8.2. A stop is a symbol of the form x+ or x−, where x is a number.
The union of stops and numbers is totally ordered, larger numbers being larger and the subscript sign being used for tie-breaks, so for any numbers x < y we have x−< x < x+ < y−< y < y+.
Definition 8.3. Define the left stop L(G) and the right stop R(G) of a short game G by the rule L(G) := ( x− if G is equal to the number x, maxGL R(GL) if G is not equal to a number, R(G) := ( x+ if G is equal to the number x, minGR L(GR) if G is not equal to a number.
To see that the maximum and the mininum in the definition are well-defined, note that if a short game has no left option or no right option, by the simplicity theorem and Theorem 8.1, it must be a number.
Theorem 8.4. Let G be a short game. For any number x the following equivalences hold.
L(G) > x ⇐ ⇒G ▷x R(G) < x ⇐ ⇒G ◁x.
Proof. If G is a number, then L(G) = G−and R(G) = G+, and the statements are true. If G is not a number, we have the following equivalences.
L(G) > x ⇔some R(GL) > x (by definition of left stop) ⇔some GL ≥x (by induction) ⇔G ▷x (by the weak number avoidance theorem).
The equivalence R(G) < x ⇔G ◁x is proved analogously.
□ Corollary 8.5. Let G be a short game. Then, for any number x, the following equivalences hold.
x ≥G ⇐ ⇒x > L(G), x ∥G ⇐ ⇒L(G) > x > R(G), G ≥x ⇐ ⇒R(G) > x.
Furthermore, if G = H then L(G) = L(H) and R(G) = R(H).
Theorem 8.6. A short game G is equal to a number if and only if R(GL) < L(GR) for all GL and GR.
Furthermore, if G is equal to a number, then it equals the simplest number between all R(GL) and all L(GR).
12 JONAS SJ¨ OSTRAND Proof. First, suppose G is equal to a number x. Then, for any GR we have GR ▷x which implies that L(GR) > x according to Theorem 8.4. In the same way we can prove that R(GL) < x for any GL, and it follows that R(GL) < L(GR).
Now, suppose R(GL) < L(GR) for all GL and GR and let x be the simplest number in between. We will prove that G −x = 0 by showing that all moves from G −x are bad. Left’s move to GL −x is bad since R(GL) < x implies that GL ◁x by Theorem 8.4. Consider Left’s move to G −xR. Since x is the simplest number between all R(GL) and all L(GR) there exists a GR such that L(GR) < xR, and Right counters by moving to GR −xR for such a GR. By Theorem 8.4, GR ≤xR so Right will win the game. Since Left has no good move from G −x, we conclude that G −x ≤0. Similarly, we can show that G −x ≥0 and thus G = x.
□ Lemma 8.7. If G is a short game that is not equal to a number, then there is a left option GL such that GL −G is greater than any negative number.
Proof. Choose GL such that R(GL) = L(G) and let ε be any positive number. Write ε as a sum of two positive numbers ε = ε1+ε2. This can always be done, for instance by letting ε1 = { 0 | ε }. Now, by Corollary 8.5 we have GL + ε1 > R(GL) = L(G) > G −ε2 whence GL −G > −ε.
□ Theorem 8.8 (Translation theorem). If x is a number and G is a short game that is not equal to a number, then G + x = { GL + x | GR + x }.
Proof. Consider a left option of the form G + xL.
Since xL −x is negative, by Lemma 8.7, there is a GL such that GL −G > xL −x. Therefore, the option G+ xL is dominated by the option GL + x and can be removed. An analogous argument applies to G + xR.
□ We may add a number to a stop by the rule y+ + x := (y + x)+ and (y−+ x) := (y + x)−.
Corollary 8.9. For any short game G and dyadic rational x, we have L(G + x) = L(G) + x and R(G + x) = R(G) + x.
Proof. The case when G is equal to a number is trivial, so suppose G is not equal to a number. Then, by the translation theorem, L(G + x) = maxGL R(GL + x) which, by induction, equals maxGL R(GL) + x = L(G) + x. That R(G + x) = R(G) + x is shown analogously.
Note that we used the fact that the left and right stops respect equality of games.
□ 9. Thermography — cooling games down A game is called hot if the left stop is strictly greater than the right stop. In a hot game, the players are keen to be the first to make a move. Numbers are cold — since xL < x < xR for numbers, no player will ever want to make any move. (There are also games where the left and right stop coincide; they are sometimes called tepid games.) A hot game can be cooled down by introducing a tax on moves.
COMBINATORIAL GAME THEORY 13 Definition 9.1. If G is a short game and t a positive dyadic rational, then we define the cooled game Gt by the formula Gt := { GLt −t | GRt + t }.
unless this formula defines a number (which it will for all sufficiently large t). For the smallest values of t for which this happens, the number turns out to be independent of t, and we define Gt to be this number.
We will justify the assertions in the definition later on.
The thermograph of a short game G is a diagram plotting the left and right stops of Gt as a function of t, with t increasing vertically, and the stop values increasing to the left. (The subscript signs of the stops can be safely ignored. We will recover them later.) The following theorem tells us how to construct the thermograph of a game if we have already constructed the thermographs of its options.
Theorem 9.2. For all short games G and dyadic rationals t ≥0, we have L(Gt) = max GL R(GLt) −t =: Lt, say, and R(Gt) = min GR L(GRt) + t =: Rt, say, unless possibly Lt < Rt. In this latter case, Gt is a number x, namely the simplest number between Lu and Ru for all small enough u with Lu < Ru, and we then have L(Gt) = x−and R(Gt) = x+.
Proof. This follows from Corollary 8.9 and Theorem 8.6. For the moment, we are continuing to suppose that Gt is well-defined.
□ As an example, here are the thermographs of G = { 5 2, {4 | 2} | {−1 | −2}, {0 | − 4} } (thick lines) and of its options (thin lines): 1 3 2 0 −1 −2 −3 −4 2 1 4 The following theorem justifies the assertions in the definition of Gt and inciden-tally makes Theorem 9.2 an honest theorem.
Theorem 9.3. For any short game G, the left border of the thermograph is a line proceeding either vertically or diagonally up and right in stretches, the right bound-ary being in stretches vertical or diagonal up and left. Beyond some point, both boundaries coincide in a single vertical line — the mast. The coordinates of all corners in the diagram are dyadic rationals.
14 JONAS SJ¨ OSTRAND Proof. This requires only the observation that on subtracting t from a line which is vertical or diagonal up-and-left we obtain one correspondingly diagonal up-and-right or vertical, and that two such lines aiming towards each other must meet at a point whose coordinates can be found with a single division by 2.
□ Theorem 9.4. The left and right stops L(Gt) and R(Gt) are “just inside” the boundary of the thermograph on vertical stretches, “just outside” on diagonal stretches.
At the points of the mast above its foot, L(Gt) < R(Gt). At corners of the diagram the subscript sign is the same as for immediately smaller values of t, so the behaviour is “continuous downwards”.
Proof. These properties are preserved in the passage from the thermographs for GL and GR to that for G.
□ The height of the root of the mast is called the temperature of G and is denoted by t(G). It is equal to inf{t : Gt equals a number}.
The horizontal coordinate of the mast is called the mean value of G and is denoted by G∞or m(G). It is the number G eventually becomes when it is frozen.
For our example game above, t(G) = 5/2 and G∞= 1/2.
Lemma 9.5. For any short game G, any number x and any dyadic rational t ≥0, we have (G + x)t = Gt + x.
Proof. This follows from the translation theorem.
□ Theorem 9.6. For any games G and H and any dyadic rational t, we have (G + H)t = Gt + Ht.
Proof. Let s be the smallest of the three temperatures t(G), t(H) and t(G + H).
There is a u slightly larger than s such that, for any r ≤u, Gr = { GLr −r | GRr + r }, Hr = { HLr −r | HRr + r }, and (G + H)r = { (G + H)L r −r | (G + H)R r + r }.
If t ≤u we have (G + H)t = { (GL + H)t −t, (G + HL)t −t | (GR + H)t + t, (G + HR)t + t } = { (GLt + Ht −t, Gt + HLt −t | GRt + Ht + t, Gt + HRt + t } = Gt + Ht, where the second equality follows from induction. If t > u we have Gt = (Gu)t−u, Ht = (Hu)t−u and (G + H)t = ((G + H)u)t−u = (Gu + Hu)t−u. Since at least one of Gu, Hu and (G + H)u is a number, it follows from Lemma 9.5 that (Gu + Hu)t−u = (Gu)t−u + (Hu)t−u.
□ Theorem 9.7. For any short games G and H, (G + H)∞= G∞+ H∞ t(G + H) ≤max{t(G), t(H)} Proof. The equality follows directly from the preceding theorem.
Suppose t = t(G + H) > max{t(G), t(H)}. Then Gt and Ht would be numbers but not (G + H)t = Gt + Ht — a contradiction.
□ COMBINATORIAL GAME THEORY 15 Theorem 9.8. For any short game G, the inequalities G∞−t(G) −ε < G < G∞+ t(G) + ε hold for any dyadic rational ε > 0.
Proof. This follows from Corollary 8.5 and the fact that the borders of the thermo-graph of G has at least a 45 degree slope.
□ Theorem 9.9 (The mean value theorem). For any short game G and any integer n, the inequalities nG∞−t(G) −ε < nG < nG∞+ t(G) + ε hold for any dyadic rational ε > 0.
Proof. This follows from the above theorem since (nG)∞= nG∞and t(nG) ≤t(G) by Theorem 9.7.
□ 10. Impartial games and the game of Nim A game is called impartial if, from any of its positions, both players would have the same legal moves if they were about to play. Formally, a game is impartial if • all its options are impartial, and • its set of left options and its set of right options are equal.
An example of a game that is not impartial is chess, since white can only move white chessmen and black can only move black chessmen.
A classical example of an impartial game is the game of Nim, which is played as follows. On a table are a number of piles of sticks. In each move a player chooses one of the piles and removes one or more sticks from it. The player that removes the last stick wins.
If there is only one pile, clearly the first player wins by removing all sticks. If there are two piles things get slightly more complicated: If the piles contain the same number of sticks, the second player wins by mimicking the first player’s strategy — when the first player removes some sticks from one of the piles, the second player immediately removes the same number of sticks from the other pile. If the piles contain different numbers of sticks, the first player wins after equalising the piles.
What if there are three or more piles? In 1901, Charles Bouton found the general strategy: • For two nonnegative integers a and b, define the nim sum a⊕b as the bitwise XOR of a and b when they are written as binary numbers. For instance, if a = 14 = (1110)2 and b = 5 = (101)2 then c = (1011)2 = 11.
• If the nim sum of all piles is zero, the second player wins. If it is positive, the first player wins by a move that makes it zero.
To see that this strategy works, we must check that • if the nim sum is zero, any move makes it positive, and • if the nim sum is positive, some move makes it zero.
Suppose the nim sum is zero and consider a move that removes r sticks from a pile with a sticks. Then the new nim sum will not be zero since the binary expansion of a −r is not the same as the binary expansion of a.
For the converse, suppose the nim sum is positive, say q, and let qjqj−1 · · · q0 be the binary expansion of q. Then, some pile, of size a say, must have a one at position 16 JONAS SJ¨ OSTRAND j in its binary expansion. The move that reduces a to q ⊕a will make the new nim sum zero.
10.1. A more efficient notation. As impartial games have the same set of left and right options, our usual notation G = { GL | GR } is redundant, and we will usually identify G by its set of (left or right) options.
For instance, instead of G = { ∗, { ∗| ∗} | ∗, { ∗| ∗} } we will simply write G = {∗, {∗}}.
In an impartial game, either the first or the second player to move will win the game, independent of who is Left and who is Right. If the first player wins it is an N-game (as in the next player) and if the second player wins it is a P-game (as in the previous player). Recall that N-games are fuzzy to zero and P-games are equal to zero.
For integers n ≥0, let ∗n denote the game of Nim with a single pile of n sticks, also called a nimber. As an example, let us see how the nimber ∗3 is represented as a set.
The options of ∗3 are ∗2, ∗1, and ∗0. The options of ∗2 are ∗1 and ∗0. The game ∗1 has only one option, ∗0, and the game ∗0 has no options at all so it is the empty set ∗0 = ∅= {}. We get ∗3 = { ∗2, ∗1, ∗0 } = { {∗1, ∗0}, {∗0}, {} } = { {{∗0}, {}}, {{}}, {} } = { {{{}}, {}}, {{}}, {} }.
Note that ∗0 ≡0 and ∗1 ≡∗.
Theorem 10.1. For any nonnegative integer n, the canonical form of ∗n is ∗n = {∗0, ∗1, . . . , ∗(n −1)}.
Proof. We just have to check that there are no dominated or reversible options, and we leave that as an exercise.
□ Note that for any impartial game G, we have −G ≡G and thus G + G = 0; the choice between minus and plus does not matter for impartial games.
10.2. Grundy values and Grundy’s theorem. We define the mex (or minimum excluded value) of a finite set of nimbers to be the smallest nimber not in that set.
Theorem 10.2 (Grundy’s theorem). For any finite set G of nimbers, G = mex G as games.
Proof. Let ∗n be the smallest nimber not in G, that is, mex G = ∗n = {∗0, ∗1, . . . , ∗(n− 1)}. We must show that there are no good moves from G −∗n.
A move to G −∗k for k < n can be countered by ∗k −∗k since ∗k ∈G. A move to ∗k −∗n is countered by ∗k −∗k if k < n and by ∗n −∗n if k > n.
□ So Grundy’s theorem shows that every short impartial game is equal to a nimber!
(In fact this is true for impartial games in general if we allow infinite ordinal nimbers, but that is not important for us.) The integer n such that G = ∗n is called the Grundy value of G and is often denoted by g(G). It follows that an impartial game is a P-game if and only if its Grundy value is zero.
The optimal strategy of Nim shows us how to add nimbers: COMBINATORIAL GAME THEORY 17 Theorem 10.3. For any nonnegative integers m and n, we have ∗m + ∗n = ∗(m ⊕n).
Proof. Since m⊕n ⊕(m⊕n) = 0, a game of Nim with three piles of sizes m, n, and m ⊕n is a zero game. In other words, ∗m + ∗n + ∗(m ⊕n) = 0.
□ E-mail address: jonass@kth.se |
1346 | https://link.springer.com/article/10.1140/epjd/s10053-023-00698-2 | High-resolution X-ray emission study for Xe $$^{54+}$$ on Xe collisions | The European Physical Journal D
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The European Physical Journal D
Article
High-resolution X-ray emission study for Xe 54+ on Xe collisions
Regular Article – Atomic and Molecular Collisions
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Published: 01 July 2023
Volume 77, article number 125, (2023)
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High-resolution X-ray emission study for Xe 54+ on Xe collisions
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Marc Oliver HerdrichORCID: orcid.org/0000-0002-3939-33531,2,3,
Daniel Hengstler4,
Michael Keller4,
Jeschua Geist4,
Christian Schötz4,
Matthäus Krantz4,
Andreas FleischmannORCID: orcid.org/0000-0002-0218-50594,
Christian Enss4,
Tobias Gassner1,
Pierre-Michel HillenbrandORCID: orcid.org/0000-0003-0166-26661,5,
Alexandre GumberidzeORCID: orcid.org/0000-0002-2498-971X1,
Uwe SpillmannORCID: orcid.org/0000-0001-7281-50631,
Sergiy Trotsenko1,
Paul IndelicatoORCID: orcid.org/0000-0003-4668-89586&
…
Thomas StöhlkerORCID: orcid.org/0000-0003-0461-35601,2,3
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Abstract
We report on an experiment conducted at the ESR storage ring aiming at the study of the X-ray emission of Xe 54+ ions colliding with Xe atoms at a beam energy of 50 MeV/u. The radiation resulting from the ion–atom interaction was observed using a high-resolution spectrometer based on metallic–magnetic calorimeter technology. In order to benchmark the capabilities of these detectors for high-precision atomic physics experiments, we identified several transitions from H-like and He-like xenon and determined their energies. Furthermore, the 1 s-Lamb shift in Xe 53+ was estimated using the measured line energies. The results are compared with previous experimental studies and theoretical predictions.
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1 Introduction
X-ray spectroscopy experiments involving highly charged heavy ions have become an indispensable tool for tests of quantum electrodynamic (QED) effects in strong fields [19
")]. In the past, measurements with H-like and He-like Xe ions have been performed by several groups at different experiment facilities (see, for example, [2, 225–230 (1989).
"),3, 444–466 (2000).
"),4, 0163001 (2009).
")]). In this work, we present the results of an experiment that was conducted at the _ESR_ ion storage ring of GSI[5, 1–99 (2007).
")] utilizing a cryogenic calorimeter detector for high-precision X-ray spectroscopy. Metallic–magnetic calorimeters (MMC) like the maXs-series detectors developed in cooperation with the Kirchhoff-Institute for Physics (KIP) in Heidelberg [6.
"),7.
"),8.
")] combine several advantages over conventional energy-dispersive X-ray photon detectors. They can reach resolving powers of E/Δ E>6000 [9.
")]–comparable to crystal spectrometers–over a broad spectral acceptance range–comparable to typical semiconductor detectors. Together with an excellent linearity [10, 269–279 (2012).
")] as well as signal rise times up to τ 0≈100 ns [11, pp.151–216.
")] they are particularly well suited for high-precision X-ray studies in atomic and fundamental physics experiments dealing with heavy, highly charged ions. However, in order to achieve this extraordinary performance, a transition from conventional analog to fully digital data processing and analysis is required. Therefore, a software framework for analyzing MMC data has been developed during the course of several experiments conducted at the ion storage facilities of GSI/FAIR in the last years. In the following, the experiment and the MMC detector are briefly described in Sects.2 and 2.1, respectively. The detailed analysis of the recorded data in Sect.3 is followed by a short conclusion about the feasibility of using cryogenic microcalorimeter detectors in the context of precision spectroscopy related to fundamental tests of the standard model (see Sect.4).
2 Experiment
The experiment is described in detail in 124
"), [13, 022810 (2022).
")]. It involved an electron-cooled 132 Xe 54+ ion beam at 50 MeV/u beam energy colliding with a Xe gas-jet at the internal target of the ESR ion storage ring. In order to record X-ray photons resulting from the ion–atom interaction, a maXs-200 detector [7.
")] was positioned in front of the 60∘ port of the target chamber. The pixels of this 1×8 spectrometer array consist of a gold absorber and a Au:Er-based paramagnetic sensor in a gradiometric configuration with two pixels connected to a SQUID-based readout and amplification stage. It was mounted with its pixels in vertical orientation (to reduce Doppler broadening) at the tip of the 60 cm long cold finger of a 3 He/4 He dilution cryostat in order to achieve the required operation temperature of <30 mK. The nominal absorption efficiency at 100 keV is 45% with an active detection area of 1 mm 2 per pixel. From previous testing of an identical detector chip, an energy resolution of 62 eV full width at half maximum (FWHM) at 60 keV was reported. The distance between the detector and the interaction point amounted to 2 m resulting in a 2.5×10−7 sr solid angle per pixel. A vacuum window made of diamond was installed on the target chamber, and the X-ray entry window in front of the MMC was made of aluminum-coated Mylar foils as well as a beryllium window for the separation of the surrounding vacuum from air. During the 48 h of almost continuous recording of X-ray radiation at the ESR, a 241 Am source was positioned in front of the detector (offset from the view port into the target chamber) for simultaneous calibration.
We note that an experiment with a similar setup was performed very recently [14, 012809 (2021).
")]; however, only conventional photon detectors were used. Furthermore, a comparable experiment at even lower collision energies was conducted [15, 042825 (2020).
")] utilizing a maXs-30 type detector. Though, too few events were detected by the microcalorimeter at that time to perform a proper analysis.
2.1 Detector performance
The absolute energy calibration of the detector was performed by fitting a gaussian distribution to the peaks of the well-known lines of the 241 Am calibration source for each pixel individually. The central positions of the peaks were then compared to the corresponding energies reported in the literature. A fit using a second-order polynomial thereby takes into account the well understood nonlinearities of the detector. Additionally, unavoidable temperature fluctuations of the detector’s substrate during the measurement lead to a spectral broadening due to a linear temperature dependance of the sensor’s gain behavior. This effect could be mostly compensated by using the baseline level (before the trigger) of each recorded signal as its temperature information. Comparing the measured energies of events associated with the 59.5 keV calibration line to their corresponding sensor temperatures allowed for the calculation of a regression line which was then applied inversely to the rest of the events. An in-depth description of all utilized calibration and correction procedures can be found in [8f
")].
The instrumental energy resolution determined from the line width of the observed calibration line at 26.3 keV is found to be 42.0 eV (FWHM). However, from the characterization of the baseline noise one finds an expected resolution of 38.2 eV [124
")]. This can be partially explained by a loss of resolution power due to a remaining fraction of uncorrected temperature-dependent gain drift and signal shape (see [16")]) artifacts. The line width of the measured X-ray lines from the ion–atom interaction amounts to 77 eV at around 35 keV. This difference to the calibration lines can be partially attributed to the finite size of the gas-jet target resulting in Doppler broadening, but has not been fully understood, yet [8.
")]. Systematic uncertainties from the Doppler broadening between pixels are approximated to be below 4 eV [8.
")]. Additional uncertainties due to calibration errors are also small compared to the overall resolution (approximated mean deviation from literature values are in the order of 1 eV). Therefore, both effects are ignored in the following analysis. During the experiment only two of the eight pixels were connected, therefore limiting the total amount of events available for the analysis.
3 Experimental data and results
Because of the comparably slow collision of 50 MeV/u and the symmetrical and heavy systems chosen (Xe 54+ on Xe), one can assume an adiabatic exchange of one or more electrons between the fully ionized projectile ions and neutral target atoms [224
")]. The entire recorded spectrum can be seen in Fig.1. The most prominent features stem from K-shell transitions (see also Fig.2) within both collision partners, while L-shell transitions are not observable due to photon absorption in the diamond window and gap filled with air between target and detector. A first evaluation of the measured spectra can be found in [8.
"), 23.
")].
3.1 Target radiation
At 30 keV, a group of transitions is visible that can be attributed to radiation emitted by the target atoms. It results from subsequent relaxation of excited states into electron vacancies produced by electron transfer from the target into the projectile ions. Observed are K-α transitions from different charge states of xenon with multiple satellite features caused by screening effects. A detailed analysis of this phenomenon is found in [14, 012809 (2021).
")]. A comparison of energy limits for neutral down to H-like xenon ranges from Xe I K-α 2: 29,458.3 eV [24, 35 (2003).
")] to Ly-α 1: 31,283.9 eV [17")]. This matches the recorded spectral features and indicates that in such collisions potentially a multitude of electrons can be stripped from the target atoms.
Fig. 1
Shown is the full spectrum recorded with a maXs-200 detector during the presented beam time. It contains X-ray events (from both pixels combined) from collisions between Xe 54+ at 50 MeV/u and a Xe gas-jet target (blue curve). In red, the fitted model function containing the annotated transitions is overlaid. A preliminary version of the spectrum can be found in [124
")]
Full size image
3.2 Projectile radiation
Electrons transferred into excited states of the projectile ions through non-radiative electron capture (NRC) [254
")] also lead to the emission of characteristic X-ray radiation via subsequent cascades. As can be seen in Fig.1, the projectile radiation is significantly more intense than the radiation from the target. This can be explained by the fact that the NRC process (involving electrons from all target shells) has a much larger cross section for this collision system than the target K-shell vacancy production including electron transfer and direct Coulomb ionization responsible for the target X-rays.
Fig. 2
Displayed are the level schemes for hydrogen (left) and helium-like xenon ions (right). The observed K-α (blue, dash-dot) and K-β (red, long dash) transitions are highlighted. The respective binding energies were taken from our theoretical predictions calculated using the MDFGME code (see [18/
")] and refs. there in): It includes finite nuclear size effects (experimental charge radius from [19, 69 (2013).
")]), nuclear recoil contributions including relativistic corrections as well as first- and second-order QED corrections. For the two-electron calculations, more details can be found in [20, 639 (1987).
"), 21, 651 (1987)")]
Full size image
3.2.1 Doppler correction and comparison to theoretical predictions
Observed K-shell transitions are shifted to higher energies compared to the target radiation due to a Doppler shift at the used observation angle resulting from the relativistically moving projectiles:
E=E′⋅f=E′⋅γ(1−β cosθ′)
(1)
With a beam energy of E k i n=50.43±0.09 MeV/u (calculated from the applied electron cooler voltage of 27.70±0.05 kV at an electron beam current of 100 mA) and an observation angle of θ=60±1∘ a Doppler shift of f=1.1269±0.0064 is expected. A model of superposed normal distributed peaks for all identified transition lines is fitted to the spectrum to extract the measured line positions. The most prominent K-α- and K-β-transition energies E∗ from hydrogen-like xenon (see Table1 and Fig.3) are then used to perform a linear fit against the binding energies E i—assumed to be known from our theoretical predictions—of the participating excited states:
E∗=f⋅(E i−E f)
(2)
This not only yields a more precise value for the actual Doppler shift but also allows to estimate the common ground state binding energy E f=41,299.93±2.78 eV (uncertainty taken from the fit). The resulting Doppler correction factor of f= 1.125534 ± 9.8 ×10−5 is in good agreement with the expectation. Assuming that the beam energy is well known, this yields a corrected observation angle of 60.21±0.03∘. At a distance of 2 m this gives rise to a difference of 7.39±0.92 mm to the assumed ideal detector position at 60∘ in regards to the interaction point (a reasonable finding consistent with previous experiments). Using the spectrum itself to perform a Doppler correction eliminates these kinds of systematic errors arising from difficult to control experimental uncertainties (like the exact detector and target positions). Having multiple line positions and spectral features within the same spectrum is an immediate advantage of using high-resolution, broad range MMC detectors compared to conventional detection methods. Because of an unresolved superposition of the two transitions from the and states of ≈7 eV, the Ly-α 2 lines were excluded from the fit.
Table 1 The table contains the measured line positions of the most prominent observed X-ray transitions in hydrogen-like xenon
Full size table
Besides K-shell transitions in hydrogen-like xenon, several lines stemming from helium-like xenon ions were identified as well (see Table2). This means, that in some collisions two electrons were transferred from the target atoms to the projectiles at once. A comparison between measured line positions with theoretical predictions shows an overall good agreement within the boundaries of statistical and systematic uncertainty. The 1 s 2 p 1 P 1→1 s 2 1 S 0 line is overlapped by a second transition from the 1 s 2 p 3 P 2 state and could not be resolved.
Fig. 3
The spectrum shows spectral details of several recorded Doppler-shifted K-α lines emitted from both hydrogen- and helium-like xenon projectile ions. The satellite features of the Ly-α lines are highlighted by red circles
Full size image
Table 2 In this table, the measured line positions of K-α transitions in helium-like xenon projectile ions are listed
Full size table
3.2.2 K-α satellite peaks
There is no apparent reason to assume that the second electron always has to be captured into the ground state of the helium-like system. Thus, one expects to find K-shell transitions for two electron configurations with the second electron in a higher orbital as well. Both, the Ly-α 1 and Ly-α 2 lines seem to exhibit small satellites on their lower energy edges. The analysis of the Ly-α 1 satellite yields a plateau with a center distance of 123.0 eV in relation to the main peak and a line broadening to 100 eV. A calculation was performed using FAC [297
")] exploring all possible transition energies and branching ratios for Ly-α 1-like transitions with an additional electron in the 2 _p_-orbitals. The second charge in the L-shell partially shields the electric field of the nucleus acting upon the other electron, thus reducing its binding energy. As a result, since the K-shell binding energy is less effected by the L-electron’s shielding effect, the overall K-α-transition energy of the system is reduced. In total, two groups of lines are found. Their position coincides with the observed satellites. The transition strength weighted average line energy for the Ly-α 1 satellite yields a distance of 109.4 eV to the main peak and a width of σ=25 eV. Taking into account the measured X-ray’s line width of 77 eV, the resulting total width of the plateau is expected to be 96.8 eV which fits the observed spectral feature’s properties well. Therefore, we attribute these peaks to satellite features of the Ly-α lines. However, due to low statistics an exact determination is not possible. Although the simultaneous transfer of three or more electrons from the target to the projectile ions cannot be ruled out completely, no indication of the occurrence was observed.
3.2.3 Comparison to previous experiment findings
Fig. 4
The spectrum shows details of recorded Doppler-shifted lines stemming from K-shell transitions emitted from both hydrogen- and helium-like xenon projectile ions up the the series limit
Full size image
In the past, similar experiments were conducted by other groups measuring K-α-transition energies in different charge states of xenon. For example, a single-pass collision experiment using Xe ions as projectiles on a thin carbon foil was performed at the LISE beam line of GANIL, France[27
")]: The reported transition energies for K-α 1 (31,278 ±10 eV) and 1 _s_ 2 _p_ 3 P 1→1 s 2 1 S 0 (30,209.6 ± 3.5 eV) agree with the results presented in this work. In both experiments, occasionally more than one electron was captured from the target, leading to an unavoidable mixture of transitions from different charge states of Xe ions in the final spectra. However, in comparison to the Ge-based semiconductor detector used at LISE, the higher resolving power of the MMC detector enabled the simultaneous observation of both H-like and He-like transitions as separate lines. This not only eliminates the need to disentangle potentially overlapping transitions during the analysis, but also allows for a more direct comparison of the energies and intensities of these lines.
Another high-resolution measurement of highly charged Xe ions was performed using a different type of microcalorimeter (Si-thermistor based) detector at an electron-beam ion-trap (EBIT) at the Lawrence Livermore National Laboratory, USA[41
")]: Again, the experimental findings for the K-α 1 (31,284.9 ± 1.8 eV) as well as the y- (1 _s_ 2 _p_ 3 P 1→1 s 2 1 S 0, 30,207.1 ± 1.4 eV) and z- (1 _s_ 2 _s_ 3 S 1→1 s 2 1 S 0, 30,128.6 ± 1.3 eV) lines of the helium-like system are in good agreement with results of our current work. Though, in principle MMCs are expected to have a higher intrinsic resolution power than semiconductor-based calorimeters [30.
")], the absorber volume of the maXs-200 pixels was optimized for a higher energy range (up to 200 keV compared to 100 keV of the Si thermistor). Because of that, both, the MMC and the semiconductor-based _EBIT calorimeter system_ (ECS) used at by Thorn et al. yield a comparable energy resolution of around 35–40 eV (FWHM) in the region of interest around 35 keV. However, despite the similar energy resolutions, a higher precision was achieved at the EBIT. This can be partially explained by the fact that during the ESR measurement fewer events were recorded overall. More importantly though, the line widths of transitions stemming from the fast moving projectile ions were broadened to 77 eV as briefly discussed in Sect.2.1. This highlights the inherent property of the EBIT measurement of not requiring a Doppler correction or being effected by other relativistic effects compared to the ion–atom collision. However, performing the experiment at an ion storage ring might still be advantageous as a projectile beam of clean isotopes and a well defined charge state is available for the collision. Less overlapping transitions of different charge states are present in the resulting spectrum and even transitions from higher orbitals with quantum numbers n>2 are observable up to the continuum edge as shown in Fig.4.
3.2.4 1 s-Lamb shift of hydrogen-like xenon
By definition, the 1 s-Lamb shift of hydrogen-like xenon amounts to the difference between the real ground state binding energy and the value predicted by Dirac theory for the interaction of a single 1 s-electron with a point like nucleus (−41,346.80 eV). The resulting value of 46.87±2.78 eV is in excellent agreement with our theoretical prediction (47.094 eV, ) and calculations by V. A. Yerokhin (46.920 eV, [31, 033103 (2015).
")]). Compared to previous experimental results (54±10 eV [2, 225–230 (1989).
")]), a higher accuracy was achieved due to the better energy resolution of the MMC detector in comparison to the Ge detector (reported 270 eV FWHM in the region of interest [2, 225–230 (1989).
")]).
4 Conclusion and outlook
In the present study, the results achieved are strongly affected by the moderate counting statistics. A long distance between the detector and interaction point due to geometrical constraints of the setup and a small detector area because of missing pixels reduced the available solid angle of the observation. Additionally, many of the temperature correction procedures that were developed over the period of subsequent experiment campaigns were not yet available at the time of this experiment, thus further reducing the final energy resolution. Nevertheless, quite accurate results for the transition energies could be obtained. Overall, the findings are in a good agreement with both theoretical predictions as well as previous experimental findings. Taking into account that this was one of the first experiments utilizing a maXs-series detector for atomic physics measurements at an ion storage ring, it proves the feasibility of future beam times exploiting the excellent performance of these detectors. In particular, the recorded spectrum did contain almost no background events even without applying time coincidence filtering. Systematic uncertainties stemming from difficult to predict experiment setup parameters like the position of the gas-jet within the target chamber or even the slight shift of the detector position due to thermal contraction during the cool-down phase could be mostly eliminated by using multiple highly resolved lines within the same recorded spectrum.
Meanwhile, detectors of the maXs-series have been improved significantly. The newly designed maXs-30 [8f
")] and -100 [9.
")] detectors consist of 8×8 pixels with an increased active detection area and correct several issues that occurred over time. For example, temperature sensitive pixels have been added to allow for an intrinsic temperature correction of the obtained spectra. Accompanying the changes to the hardware, calibration and measurement procedures have been updated as well. Furthermore, the maXs-detector family will be integrated into the heterogeneous detector environment of the FAIR complex which is currently being built [32, 033001 (2019).
")]. This allows for them to be used, e.g., with particle detectors to perform coincidence measurements and further improve their accuracy by suppressing background events. This could help with a better identification of small spectral features like the discussed satellite structures found near the Ly-α lines. Such an application has been recently demonstrated in the first high-resolution measurement of K-α transitions in He-like uranium U 90+ at the electron cooler of the _CRYRING@ESR_ low energy ion storage ring of FAIR [33, 0114005 (2022).
")]. Further experiments utilizing MMC detectors for high-precision atomic physics experiments involving highly charged ions in storage rings like CRYRING@ESR are already planned.
Data Availability Statement
This manuscript has no associated data or the data will not be deposited. The data that support the findings of this study are available upon reasonable request from the authors.
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Acknowledgements
We thank all participating members of the research division for atomic, quantum and fundamental research of GSI and of the KIP that helped with the experiment setup and execution. This work was created within the SPARC collaboration and was supported in part by the ExtreMe Matter Institute EMMI at the GSI Helmholtzzentrum für Schwerionenforschung, Darmstadt.
Funding
Open Access funding enabled and organized by Projekt DEAL. We acknowledge financial support by the European Union and the federal state of Thuringia via Thüringer Aufbaubank within the ESF Project (2018 FGR 0080). The work was supported by the BMBF Grant 05P12VHFA5.
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Authors and Affiliations
GSI Helmholtz Center for Heavy Ion Research, Planckstraße 1, 64291, Darmstadt, Hesse, Germany
Marc Oliver Herdrich,Tobias Gassner,Pierre-Michel Hillenbrand,Alexandre Gumberidze,Uwe Spillmann,Sergiy Trotsenko&Thomas Stöhlker
Helmholtz-Institute Jena, Fröbelstieg 3, 07743, Jena, Thuringia, Germany
Marc Oliver Herdrich&Thomas Stöhlker
Institute for Optics and Quantum Electronics, Friedrich-Schiller-University Jena, Max-Wien-Platz 1, 07743, Jena, Thuringia, Germany
Marc Oliver Herdrich&Thomas Stöhlker
Kirchhoff-Institute for Physics, Ruprecht Karls University Heidelberg, Im Neuenheimer Feld 227, 69120, Heidelberg, Baden-Württemberg, Germany
Daniel Hengstler,Michael Keller,Jeschua Geist,Christian Schötz,Matthäus Krantz,Andreas Fleischmann&Christian Enss
Institute of Experimental Physics I, Justus-Liebig-University Giessen, Heinrich-Buff-Ring 16, 35392, Giessen, Hesse, Germany
Pierre-Michel Hillenbrand
Laboratoire Kastler Brossel, Sorbonne Université, CNRS, ENS-PSL Research University, 24 Rue Lhomond, 75005, Paris, France
Paul Indelicato
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Analysis software, data analysis and presentation of the results: MOH; Design, manufacturing and operation of microcalorimeter detectors: DH, MKe, JG, CS, MKr, AF and all other participating members of the group of CE; Experiment at ESR: P-MH, TG, AG, ST, US and all participating members of GSI; Theory: PI; Supervision, project administration and idea: Th.S. All authors have read and agreed to the published version of the manuscript.
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Herdrich, M.O., Hengstler, D., Keller, M. et al. High-resolution X-ray emission study for Xe 54+ on Xe collisions. Eur. Phys. J. D77, 125 (2023).
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P. Indelicato, Topical Review: QED tests with highly-charged ions. J. Phys. B 52, 232001 (2019).
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1347 | https://www.youtube.com/watch?v=5BFDdVqAFZE | 73939133 - Probably the Most Interesting Prime Number [Part 1]
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Today we would liek to talk about right-truncatable prime numbers and the math behind them! :D Truncatable Prime numbers are special numbers, where you can take the last digit off and still receive a prime number! Later we program a python algorithm to get ourselves a list of all possible right truncatable prime numbers and hence also the biggest one, 73939133. The maths behind this video deals with digit decompositions in base 10, modulus and prime number properties :9 Enjoy! =D
This video is part of the MegaFavNumbers project. Maths YouTubers have come together to make videos about their favourite numbers bigger than one million, which we are calling #MegaFavNumbers.
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Intro a huge thanks to my patreon supporters for making this episode possible [Music] good morning fellow mathematicians welcome back to the video huge thanks to matt parker and james crime for inviting me to their mega faith numbers playlist here on youtube if you don't know what all of this is about take a look into the description info box or pin comment you can find a link to the playlist there basically a bunch of mathematicians including me of course that's why making the video in the first place obviously are going to talk about their most favorite number over a million minus obviously seventy three million nine hundred fifty nine thousand one hundred fifty three because it's the biggest one of very special kind and this is only part one basically of a two or three part series you can already find another video out here on this channel in the info box description etc where we do some python implementation we are going to implement the most efficient algorithm i came up with in python and we are going to show that there is a finite list of those special numbers and that this one is the biggest one of its kind so i hope you are going to enjoy everything i post here if you haven't subscribed already feel free to do so also subscribe to flemboms for way more mass content now we are going to Special Property dive right in now what is the special property about this number well it's a prime number i think i swatting it no um it's a prime number which is already pretty cool i mean it looks pretty funky it's a prime number um over a million in some way but it's also extremely special because it's the biggest it's the biggest prime number of a kind namely right truncatable primes what does it mean for a number to be right truncatable well you can take off the last digit and it's going to be your prime number that's pretty cool but this is not the only property it has to satisfy it's only right truncatable if you can repeat this process up until the last digit meaning if we take another digit off it's going to be a right truncated prime if you take another digit off it's going to be a right truncatable prime and so on up until the last digit so that's extremely interesting right um if you ask yourself if there are also left truncatable primes so you take the left digit off and it's going to be a prime yes those are out there you can also generate a list of those they are going to be more left truncatable than right truncatable primes there are also truncatable primes which are basically from the left and the right truncatable up until the middle and so on it's pretty interesting but i choose the right truncatable primes because that looks pretty sexy if you ask me now we are going to dive into the mathematics behind rightful and capable primes put some restrictions on the numbers that satisfy this very property and in the next video we are going to implement everything into a python algorithm so yeah we are going to dive right in at first i would like to play around with base 10 and digits basically i would like to take a look at the digit decomposition of some random arbitrary prime number with n digits at first so we are going to have a prime number i'm Digit Decomposition going to call p with an index n this n just um basically states that we have n digits now what's the digit decomposition well it's just going to be the digit one and then we have to digit two it's just going to be a tuple up until the digit n so think about it digit one is going to be well the first one here on the left seven digit two is going to be 3 up until the nth digit we don't know yet how many there are going to be digits the n is going to be 3 in our case with this prime number up there okay now we are going to induce the property of a right truncatable prime onto this digit decomposition so what have we done we took off the last digit meaning prime number within digits turned into a prime number with n minus one digit and the most important property is that all the other digits are going to be preserved up until the n minus one digit so after taking a digit off we are going to get ourselves a digit decomposition as follows d1 d2 blah blah blah up until the n minus one and so on we are going to continue this process for it to be a right truncatable prime up until the very last digit meaning after n minus 1 iterations overall we are going to get to p1 which is just well prime number with a digit one okay that was quite easy but how does this help us it really doesn't help us right now this is why we need to take a look at further emphasizing on what the digit decomposition actually is and for this i would like to just take a look at my most favorite number as a little example let us take a look at 135 it's my most favorite number out there and how can we decompose this into something in base 10 meaning we're going to decompose it into digits corresponding to powers of 10. well at first let us write out what the digits actually are so our digits are one okay and then we have three and last digit is five okay i hope you can see where this came from but now they have to correspond to powers of ten meaning how is our one our digit one connected to what we have here okay think about the word 135 meaning a one digit one corresponds to the number 100 our number 100 is nothing but well 10 squared hope you can see where this came from we have 100 plus and then we get 30 plus 5 okay our 3 corresponds to the number 30 meaning the power of 10 is just 1. now our number 5 is just well it's just 5 basically times 1 and 1 is 10 to the 0th power for the sake of notation we are going to also put 10 to the zero power into here meaning overall what are we getting exactly i mean if we were to put this into digit notation we would get okay one is our d1 then we have three being our d2 and then our d3 is going to be five and take a look at the index it ranges up until the number of digits okay three is the number of total digits we have here and 135 now we also have powers of 10 in some way all the time and now we need to see how those powers of 10 are connected to our total number of digits and the running index we are going to sum up over okay now we have 10 squared now how is 2 connected to 3 well we are going to take the total number of digits free and we are going to subtract our running index from it okay three minus one is going to give us two does this pattern work out for the next one i mean ten to the first power is just three total number of digits minus our running index 2 that we are having here hey works out same speed here 3 minus 3 total number of digits minus our running index 3. and thus we have found out a nice formula meaning overall if we were to put this into summation notation we would get in some way a running index we are going to call it i which is bounded between okay what's the lowest index that is one and what's the highest index there would be n in some way and our n is three okay now how does it work for our digits our digits are just going to be um raised by one all the time so one two three we're just going to count over the running index then we have 10 tuta okay what is constant up here in the power it's always the total number of digits which is three in our case or for the most generalized version is going to be n and other than that we are going to subtract our running index from it so i this is how we can decompose a 135 into a digit decomposition using powers of 10 basically and our digits now we are going to expand this knowledge upon those prime numbers that we have up here meaning overall what we have here our pn is going to be the summation where our running index i goes from one to well now we have n instead of three and obviously we just have the same pattern we are going to have our d i's here and then we are going to multiply it with 10 to the okay our total number of digits is n and then minus our i okay what about our p n minus 1 our p n minus 1 is going to be well just the same thing just with n minus 1 instead of n so our i is bounded between 1 and n minus 1 overall and then we have d i times 10 to the n minus 1 minus 5 power now we are going to write a bit more stuff out and we are going to see if we can find the connection between our prime number with n digits and our truncated prime numbers because this is a really important relationship that we need to take care of i'm going to put this here at first let us write out what pn actually was pn was nothing but okay now we are going to write everything out this was the first digit times t to the n minus one power this is just the pattern that we are having here plus then we have d2 times ten to the n minus two power plus blah blah blah up until okay then we have the n minus one times two to the first power this is always going to happen the n minus one times 10 to the first power and the last digit that we are going to have is just dn okay how can we continue from this point onwards well it's not really obvious but this is how i continued you see that we have 10 as a common factor up until this nearly last term okay one term before the last term meaning we're going to factor out a 10 right here leaving us with d1 times okay then we're going to get 10 to the n minus 2 power okay we factored out the 10 and then plus d2 10 to the n minus 3 power plus blah blah blah up until okay the n minus 1 plus the n but how is this going to help exactly well now i want you guys to take a look at this right here and we're going to write it out a tiny little bit how about we write out the first and the second and maybe up until the last term so we're going to get d1 times okay 10 to the n minus 1 minus 1 this is n minus 2. this is cool right plus and then we're going to get d2 times 10 to the n minus one minus two is going to give us n minus three plus blah blah blah up until d and the last index is n minus one this is cool right what we have here is exactly our p n minus one okay i i had to look for it so we have p n minus one here meaning overall p n the n prime number in some way can be expressed as ten times p n minus one plus the last digit the n and this is going to help us a lot finding out what our digits that we can append basically to the very first digit are going to be now we are going to write it out yet again and we are going to take a look at some modulo calculations so we have that Prime Factorization pn is nothing but what is 10 exactly in its prime factorization it's 2 times 5 times pn minus 1 plus dn and now here's one key fact that comes in with prime numbers prime numbers pn are only divisible by 1 and themselves meaning if for example our dn were to word 2 then we could factor out the number two right here and thus pn would be divisible by two doesn't work out what about the n being equal to four well four is nothing but two times two we could factor out the two yet again and thus pn would be divisible by two what about um the n being equal to zero well then our our p n is divisible by two and five which wouldn't work out meaning overall if we were to take a look at modular calculations our dn right here must not be congruent to zero modulo 2 or it must not be congruent to 0 modulo 5. what this means in general is that we can restrict our dn right here the n must be element so or must not be element let's let's put it like this if we were to take a look at the modulo it must not be element of zero and then we have two four five six and eight okay if it is any of those elements it's not going to work out because pn would be divisible by some number for example two or five but this overall means that our the ends must be element of all the numbers all the other digits not in the set right here so what's not in the set okay the n could possibly be one this would work out if it were one we don't have any other factor which can be shared by um those two terms okay it could be one it could be okay three is missing it could be seven or it could be nine and if you take a look at our biggest right truncatable prime number well those are exactly the digits we have here one three nine and also seven at the beginning the first digit is a bit of an odd one out because um it doesn't really need to satisfy what we have here just because um if we were to write python script for example or just take a look at this digit decomposition it could be any or prime number that you can find because the last digit must only be a prime number meaning overall our first digit d1 this is another restriction that we can put on here d1 must be element of all the single digit prime numbers that you can find so for example two three five and seven this is what they could possibly be and this is all the information that we need to create ourselves a list of all the right truncatable primes that there are so meaning if we were to implement this into python script and this is the most efficient one i could come up with it gives you the list of all the prime numbers which are right truncatable in the matters of seconds now how does this work so at first we are going to start off with d1 we are going to get ourselves a list of all the d1 so the first digits which is a list of 2 3 5 and seven and what we are going to do is we are going to create ourselves another list which includes those other digits those the ends which we are going to append to this string basically now what this means is we are going to take the number two at first and we have four possibilities for number two we can append one three seven or nine to it so we have 21 we have 23 we have 27 and we have 29 and now we are going to let an algorithm run which checks if those numbers are prime because well we need them to be prime numbers so 21 is not prime 27 not prime so 23 and 29 are right truncatable primes just because we can take the last digit off and we are still going to get a prime number out we are going to continue this process for free and then for five and then for seven then you are going to get a list of all the two digit right truncatable prime numbers and from this point onwards we are going to continue this process now we are going to take the list of all right truncatable two-digit prime numbers and we are going to append new digits to it we are going to take 23 and we are going to append new digits to it starting with one so 231 and then 233 and so on blah blah blah up until we get all the three digit right truncatable prime numbers let the prime number algorithm check run yet again so um this one is not prime because it's divisible by 3 but 233 should be divisible by nothing so it's a prime number and thus we get this and then we are going to append those digits yet again to it and so on up until we reach the point where we have all combinations checked meaning this right here is going to be our biggest right truncatable prime number then i hope you understood everything i said here i hope this video was to your liking if it was then make sure to subscribe to the channel and to check out the other videos that come with this very video and yeah thanks again to matt park and james crime for inviting me to this special event if you have any big number video that you want to share with us then make sure to check out the playlist and to add your video to it and up until next video i wish you guys flammable day ciao you |
1348 | https://www.scienzaatscuola.it/esercizi%20pdf/moti-piani.pdf | Moti in due e tre dimensioni 1 Esercizi discussi e svolti dal prof. Gianluigi Trivia Posizione e Spostamento Esercizio 1. Un’anguria in un campo è collocata nella posizione data dalle seguenti coordinate: x = −5.0 m, y = 8.0 m e z = 0 m. Trovare il vettore posizione tramite le sue componenti e in termini di intensità e di orientamento.
Soluzione. esprimiamo il vettore posizione tramite i vettori unitari (o versori) semplicemente considerando le componenti come multipli di tali versori ⃗ r = −5.0 ⃗ i + 8.0⃗ j + 0⃗ k per determinare l’intensità del vettore, calcoliamo il suo modulo r = q (−5.0)2 + 8.02 = √ 89 = 9.4 m e l’angolo formato con l’asse orizzontale tan (π −α) = 8.0 −5.0 per cui π −α = arctan 8.0 −5.0 = 58◦ da cui α = 122◦ Esercizio 2. Il vettore posizione di un protone è inizialmente ⃗ r = 5.0 ⃗ i −3.0⃗ j + 2.0⃗ k, espresso in metri, e in seguito ⃗ r ′ = −2.0 ⃗ i + 6.0⃗ j + 2.0⃗ k. Determinare il vettore spostamento del vettore e a quale piano è parallelo.
Soluzione 3. il vettore spostamento è il vettore differenza, cioè △⃗ r = ⃗ r ′ −⃗ r = (−2.0 −5.0)⃗ i + (6.0 −3.0)⃗ j + (2.0 −2.0)⃗ k = −7.0 ⃗ i + 3.0⃗ j + 0⃗ k essendo nulla la componente lungo l’asse z, il vettore è parallelo al piano x, y.
Velocità e velocità media Esercizio 4. Un aeroplano vola per 300 km verso est dalla città A alla città B in 45 min, e quindi per 600 km a sud da B a C in 1, 50 h. (a) Quale vettore spostamento corrisponde al viaggio completo? Quali sono (b) il vettore velocità media e (c) la velocità scalare media complessiva?
Caso 1.
(a): Il vettore spostamento che rappresenta il viaggio completo è l’ipotenusa del triangolo rettangolo ABC.
AC = p 3002 + 6002 = 671 km mentre la sua direzione, riferita alla direzione sud, è α = arctan 300 600 = 26.6◦ 2 Asserzione. (b): per calcolare il vettore velocità media terremo conto del modulo e direzione del vettore spostamento e del tempo totale impiegato, vm = 671 km 2.25 h = 298 km h con direzione uguale a quella del vettore spostamento, cioè 26.6◦rispetto al sud Caso 1.
(c): la velocità scalare media deve tenere conto dei due spostamenti avvenuti e ovviamente del tempo totale impiegato vscalare m = 900 km 2.25 h = 400 km h Esercizio 5. Un treno viaggia alla velocità costante di 60.0 km/h per 40.0 min verso est, quindi per 20.0 min in direzione che forma un angolo di 50◦verso est rispetto al nord, e infine per 50.0 min verso ovest. Qual è la sua velocità vettoriale media su tutto il tragitto?
Soluzione. nei primi 40.0 min il treno percorre, viaggiando sempre a velocità costante, s1 = 60.0 km h · 2 3h = 40 km nel secondo tratto s2 = 60.0 km h · 1 3h = 20 km le componenti dello spostamento lungo le direzioni nord-sud e est-ovest sono s2x = 20 · sin 50 = 15.3 km s2y = 20 · cos 50 = 12.9 km nel terzo tratto s3 = 60.0 km h · 5 6h = 50 km Calcoliamo le componenti del vettore risultante sx = 40 −(50 −15.3) = 5.3 km sy = s2y = 12.9 lo spostamento risultante sarà, in modulo, s = q (5.3)2 + (12.9)2 = 13.9 km l’angolo è α = arctan 12.9 15.3 = 40.1◦ La velocità media è pertanto vmedia = 13.9 km 11 6 h = 7.7 km h 3 Esercizio 6. In 3.50 h un aerostato viaggia per 21.5 km verso nord e per 9.70 km verso est, e sale di 2.88 km verticalmente rispetto al punto di decollo dal suolo. Trovare il modulo della velocità media e l’angolo fra tale velocità e il piano orizzontale.
Soluzione. calcoliamo lo spostamento complessivo s = q s2 x + s2 y + s2 z = q (21.5)2 + (9.70)2 + (2.88)2 = 23.8 km la velocità media è quindi: vmedia = 23.8 km 3.50 h = 6.8 km h l’angolo è quello tra il piano, rappresentato dalla retta che contiene la risultante tra sx e sy, s1 = √ 21.52 + 9.702 = 23.6 km e il vettore spostamento (in rosso): α = arctan 2.88 23.6 = 6.96◦ Esercizio 7. Il vettore posizione di uno ione è inizialmente − → r = 5.0− → i −6.0− → j + 2.0− → k , e dopo 10 s diventa − → r = −2.0− → i + 8.0− → j −2.0− → k , essendo il metro l’unità di misura. Qual è la sua velocità media durante questo intervallo di tempo?
Soluzione. Dobbiamo calcolare il vettore spostamento come differenza tra i due vettori che individuano la posizione finale e quella iniziale rispetto all’origine del sistema di riferimento − → r f−i = (−2.0 −5.0) − → i + (8.0 + 6.0) − → j + (−2.0 −2.0) − → k = −7.0− → i + 14.0− → j −4.0− → k calcoliamo ora il modulo di tale vettore rf−i = q (−7)2 + 142 + (−4)2 = 16.2 il vettore velocità media sarà − → v = −0.7− → i + 1.4− → j −0.4− → k e il suo modulo vmedia = 16.2 10 = 1.62 m s Esercizio 8. La posizione di un elettrone è data dalla relazione − → r = 3.0t− → i −4.0t2− → j + 2.0− → k , in unità SI. (a) Qual è la − → v (t) dell’elettrone? (b) Quanto vale − → v per t = 2.0 s? (c) Quali sono in quell’istante l’intensità e la direzione di − → v ?
4 Caso 1.
(a): per calcolare la velocità in funzione del tempo, dobbiamo prendere la derivata prima del vettore posizione, espresso come si vede in funzione del tempo stesso − → v (t) = 3.0− → i −8.0t− → j m s Caso 2.
(b): calcoliamo la velocità istantanea ponendo nella legge delle velocità t = 2 s − → v (2s) = 3.0− → i −16.0− → j (c): calcoliamo l’intensità della velocità istantanea v = √ 9 + 256 = 16.3 m s la velocità è un vettore che giace nel piano x, y, con y < 0; esprimiamo l’angolo formato con l’asse delle x utilizzando la rotazione inversa, oraria, α = arctan 16 3 = −79.4◦ Accelerazione e accelerazione media Esercizio 9. Un protone inizialmente possiede una velocità − → v = 4.0− → i −2.0− → j + 3.0− → k , che diventa − → v = −2.0− → i −2.0− → j + 5.0− → k , dopo 4 s, in unità SI. Determinare l’accelerazione media, − → a , intensità, direzione e verso, durante questi 4 s.
Soluzione. ricordando la definizione di accelerazione, − → a = △− → v △t = − → v f −− → v i △t , si ha △− → v = h (−2.0 −4.0) − → i + (−2.0 + 2.0) − → j + (5.0 −3.0) − → k i m s cioè △− → v = (−6.0) − → i + (2.0) − → k m s il modulo di tale variazione è △v = √ 36 + 4 = 6.3 m s da cui si ottiene il modulo dell’accelerazione a = 6, 3 m s 4 s = 1, 6 m s2 espressa tramite i versori − → a = (−1.5) − → i + (0.5) − → k m s avendo diviso △− → v per 4.
Esercizio 10. Il moto di una particella è tale che la sua posizione in unità SI è espressa, in funzione del tempo, dall’equazione − → r = − → i + 4t2− → j + t− → k . Scrivere le espressioni della velocità e della sua accelerazione in funzione del tempo.
Soluzione. per ottenere l’espressione della velocità in funzione del tempo, calcoliamo la derivata prima della legge oraria − → v = 8t− → j + − → k mentre per l’espressione della accelerazione, calcoliamo la derivata prima della relazione che esprime la legge delle velocità − → a = 8− → j 5 Esercizio 11. La posizione − → r di una particella in moto nel piano xy è data dall’espressione − → r = 2.00t3 −5.00t − → i + 6.00 −7.00t4 − → j ove − → r è in metri e t in secondi. Calcolare la posizione − → r , la velocità − → v e l’accelerazione − → a per t = 2.00 s. Quale sarà l’orientamento di una linea tangente al percorso della particella all’istante t = 2.00 s?
Soluzione. calcoliamo la posizione, sostituendo a t il valore assegnato t = 2.00 s − → r = 2.00 · 2.003 −5.00 · 2.00 − → i + 6.00 −7.00 · 2.004 − → j = 6.00− → i −106− → j m calcoliamo ora la legge delle velocità, mediante la derivata prima − → v = 6.00t2 −5.00 − → i + −28.00t3 − → j sostituendo anche qui t = 2.00 s, si avrà − → v = (6.00 · 4.00 −5.00) − → i −28.00 · 8.00− → j = 19.0− → i −224− → j m s calcoliamo infine la legge della accelerazione − → a = 12.00t− → i −84.00t2− → j da cui sostituendo si ha − → a = 24.0− → i −336− → j m s2 calcolare l’orientamento di una linea tangente, vuol dire calcolare il coefficiente angolare di tale retta tangente, cioè il valore della derivata prima in quel punto, t = 2.00 s. Utilizzeremo pertanto il valore della velocità, calcolando il rapporto tra le sue componenti vettoriali α = arctan −224 19.0 = −85.2◦ Esercizio 12. Una slitta a vela corre su una superficie ghiacciata spinta dal vento con accelerazione costante.
Ad un tempo assegnato la sua velocità è − → v = 6.30− → i −8.42− → j m s . Dopo 3 s si arresta di colpo. Determinare la sua accelerazione media durante questi 3 s.
Soluzione. la slitta si ferma in 3 s, cioè la sua velocità assume il valore zero. Ricordando la definizione di accelerazione − → a = △− → v △t = 0− → i + 0− → j − 6.30− → i −8.42− → j 3 s m s2 = −2.10− → i + 2.81− → j m s2 Esercizio 13. Una particella parte dall’origine al tempo t = 0 con velocità iniziale di − → v = 8.0− → j m s e si muove nel piano xy con un’accelerazione costante di − → a = 4.0− → i + 2.0− → j m s2 . Trovare la coordinata y nell’istante in cui la coordinata x = 29 m. Trovare pure la velocità scalare della particella in quell’istante.
Soluzione. la legge oraria del moto uniformemente accelerato è espressa da − → s = − → v 0t + 1 2 − → a t2 essa si suddivide nelle due direzioni del moto sx = v0xt + 1 2axt2 sy = v0yt + 1 2ayt2 nel nostro caso 29 = 2.0 · t2 da cui, considerando la sola radice positiva, t = r 29 2.0 = 3.8 s 6 La coordinata y sarà pertanto sy = 8.0 · 3.8 + 1.0 · 3.82 = 45 m Per calcolare la velocità, facciamo riferimento alla alla legge delle velocità − → v = − → v 0 + − → a t da cui vx = 4.0 · 3.8 = 15.2 m s vy = 8.0 + 2.0 · 3.8 = 15.6 m s la velocità scalare è quindi v = p 15.22 + 15.62 = 22 m s Esercizio 14. Una particella parte dall’origine con velocità iniziale − → v = 3.00− → i m s . La sua velocità varia poi con un’accelerazione costante − → a = −1.00− → i −0.500− → j m s2 . Determinare posizione e velocità quando la sua coordinata x è massima.
Soluzione. scriviamo la legge oraria del moto − → s = 3.00t− → i + 1 2 −1.00− → i −0.500− → j t2 esprimiamo la legge relativa alle componenti x, y x = 3.00t −0.500t2 y = −0.250t2 essendo assimilabile all’equazione di una parabola con concavità rivolta verso il basso (coefficiente negativo di t2), il suo massimo corrisponde al vertice della parabola e quindi ad un tempo t t = −b 2a = 3.00 1.00 = 3.00 s dopo 3 s la particella è pertanto in xmax = 4.5 m, e la velocità, v = v0 + at sarà vx = 3.00 −1.00t = 3.00 −3.00 = 0 m s vy = 0 −0.500t = −1.5 m s essa sarà rappresentabile vettorialmente come − → v = −15− → j m s si troverà nella posizione − → s = 3.00 · 3.0− → i + 1 2 −1.00− → i −0.500− → j · 3.02 − → s = 9.00− → i + 1 2 −1.00− → i −0.500− → j 9.00 = 4.5− → i −2.25− → j m Esercizio 15. La velocità − → v di una particella in moto nel piano xy è data dall’espressione − → v = 6.0t −4.0t2 − → i + 8.0− → j m s . Calcolare l’accelerazione per t = 3 s; Determinare se e quando si annullano l’accelerazione e la velocità; infine determinare se e quando la velocità ha intensità 10 m s .
7 Soluzione. per ottenere l’accelerazione dobbiamo calcolare la derivata della relazione che esprime la velocità nel tempo: − → a = (6.0 −8.0t) − → i m s2 quindi − → a (t = 3) = −18.0− → i m s2 L’accelerazione si annulla quando (6.0 −8.0t) − → i m s2 = 0 per t = 0, 75 s La velocità si annulla se − → v = 6.0t −4.0t2 − → i + 8.0− → j = 0 ma, osservando che la componente lungo l’asse verticale non dipende dal tempo, se ne desume che la velocità non sarà mai nulla; Calcoliamo ora il modulo della velocità in funzione del tempo v = q (6.0t −4.0t2)2 + 64.0 la velocità assumerà il valore richiesto di 10 m s , se 10 = q (6.0t −4.0t2)2 + 64.0 cioè, svolgendo e dividendo per 4 4t4 −12t3 + 9t2 −9 = 0 risolvendo e considerando la radice reale e positiva, si ha t = 2.2 s Esercizio 16. Una particella A si sposta sulla retta y = 30 m a velocità costante − → v di modulo v = 3.0 m/s e direzione parallela all’asse x (si veda la figura). Una seconda particella B parte dall’origine, con velocità iniziale nulla e accelerazione − → a di modulo a = 0.40 m/s2, nello stesso istante in cui la particella A attraversa l’asse y.
Quale angolo θ tra − → a e il verso positivo dell’asse y potrebbe provocare una collisione tra le due particelle?
Soluzione. la particella A si muove in direzione orizzontale di moto rettilineo uniforme, mentre la particella B si muove di moto accelerato. La direzione del moto accelerato deve essere tale che le due particelle si incontrino.
Pertanto, affinché B raggiunga A, è necessario che percorra una traiettoria scomponibile in un tratto orizzontale sB x = vt = 3.0t e un tratto verticale sB y = 30 m. La distanza complessiva che deve percorrere la particella B è descritta dal segmento obliquo in figura che è ottenibile applicando il th. di Pitagora al triangolo disegnato: sB = q 302 + (3t)2 8 tale distanza deve essere percorsa con una accelerazione a = 0.4 m/s2 e partendo da ferma, per cui q 302 + (3t)2 = 1 2at2 = 0, 2t2 Risolviamo elevando al quadrato 0, 04t4 −9t2 −900 = 0 l’equazione è biquadratica e si risolve ponendo, ad es. t2 = z 0, 04z2 −9z −900 = 0 applicando la formula risolutiva, si ha, prendendo la sola soluzione positiva z = 9 + √81 + 144 0.08 = 300 risolviamo ora rispetto a t, considerando ancora la sola soluzione positiva per il tempo t = √ 300 Ora, l’angolo θ indicato in figura è legato alle due componenti dello spostamento dalla relazione tan θ = sB x sB y = 3t 30 cioè, sostituendo il valore ottenuto per t θ = arctan 3 · √ 300 30 = 60◦ Moto Dei Proiettili Esercizio 17. Una freccetta viene lanciata orizzontalmente in direzione del centro del bersaglio P in figura con v0 = 10 m/s. Dopo 0.19 s essa colpisce il punto Q sul bordo del bersaglio posto verticalmente sotto P. Calcolare la distanza PQ e la distanza dal bersaglio alla quale si trovava il lanciatore.
Soluzione. la freccetta è al momento del lancio allineata con il punto P, essendo la sua traiettoria parabolica colpirà invece il punto Q, cioè cade verso Q. Mi basta calcolare quindi la caduta verticale originata dal moto accelerato verso il basso, secondo la legge y −y0 = v0yt −1 2gt2; in questo caso la velocità iniziale è solo nella direzione orizzontale per cui v0y = 0 PQ = y0 −y = 4.9 m s2 · 0.192 s2 = 0.18 m la distanza tra il lanciatore e il bersaglio è quella che viene percorsa dalla componente orizzontale del moto della freccetta, che avviene secondo le leggi di un moto rettilineo uniforme x = v0xt = 10 m s · 0.19 s = 1.9 m 9 Esercizio 18. Un fucile è puntato orizzontalmente contro un bersaglio alla distanza di 30 m. Il proiettile colpisce il bersaglio 1.9 cm sotto il centro. Determinare il tempo di volo del proiettile e la velocità alla bocca del fucile.
Soluzione. possiamo riferirci sempre alla figura dell’esercizio precedente. La traiettoria è di tipo parabolico e quindi mirando orizzontalmente il proiettile cadrà verso il basso di 1.9 cm sotto l’azione dell’accelerazione di gravità, g. Anche qui la vy0 = 0. Quindi y −y0 = 1.9 · 10−2 m = 4.9 m s2 · t2 s2 da cui, risolvendo rispetto a t e considerando la soluzione positiva t = s 1.9 · 10−2 m 4.9 m s2 = 0.062 s per calcolare la velocità iniziale, osserviamo che essa è dovuta solamente alla componente orizzontale. Quindi il proiettile percorre 30 m in 0.062 s con un moto rettilineo uniforme. Pertanto v = 30 m 0.062 s = 482 m s Esercizio 19. Gli elettroni possono essere sottoposti a caduta libera. Se un elettrone è proiettato orizzontalmente alla velocità di 3.0 · 106 m/s, quanto cadrà percorrendo 1.0 m di distanza orizzontale?
Soluzione. il disegno riproduce schematicamente il significato di caduta, cioè la componente verticale del moto descritta da un moto uniformemente accelerato. La componente orizzontale è invece descritta da un moto rettilineo uniforme, per cui △t = △s v = 1 m 3.0 · 106 m s = 3.3 · 10−7 s in questo intervallo di tempo l’elettrone percorrerà un tratto verticale (caduta) di s = 1 2gt2 = 1 2 · 9.81 m s2 · 3.3 · 10−7 s 2 = 5.45 · 10−13 m Se la velocità orizzontale aumentasse l’elettrone percorrerebbe 1 m in un tempo inferiore e quindi la sua caduta sarebbe minore.
Esercizio 20. In un tubo a vuoto un fascio di elettroni è proiettato orizzontalmente alla velocità di 1.0·109 cm/s nella zona compresa fra due piatti orizzontali quadrati di lato 2.0 cm, ai quali è applicato un campo elettrico che accelera gli elettroni con una accelerazione costante e diretta verso il basso di 1.0 · 1017 cm/s2. Trovare (a) il tempo impiegato dagli elettroni per attraversare il campo, (b) lo spostamento verticale del raggio di elettroni nel passaggio tra i due piatti (che non tocca) e (c) la velocità del raggio all’uscita dal campo.
10 Caso 1.
(a): gli elettroni devono percorrere un tratto orizzontale (componente) di 2.0 cm. Quindi t = 2.0 cm 1.0 · 109 cm s = 2.0 · 10−9 s Caso 2.
(b): lo spostamento verticale, cioè la componente verticale del moto, rappresenta ancora la caduta del fascio di elettroni sotto l’azione dell’accelerazione prodotta dal campo (la caduta libera può considerarsi trascurabile osservando l’ordine di grandezza della accelerazione prodotta dal campo elettrico): s = 1 2 · 1.0 · 1017 cm s2 · 2.0 · 10−92 s2 = 0.2 cm Caso 3.
(c): la velocità è data dal contributo delle due componenti. La componente orizzontale, vx, rimane invariata, mentre la vy può essere calcolata attraverso la relazione del moto accelerato uniformemente v = v0 + at = 0 + 1.0 · 1017 cm s2 · 2.0 · 10−9 s = 2.0 · 108 cm s il modulo della velocità sarà v = q v2 x + v2 y = r 1.0 · 109 m s 2 + 2.0 · 108 m s 2 = 1.0 · 109 cm s esprimendola vettorialmente sarà − → v = 1.0 · 109− → i −2.0 · 108− → j cm s Esercizio 21. Una palla rotola orizzontalmente fuori dal bordo di un tavolo alto 1.20 m e cade sul pavimento alla distanza orizzontale di 1.50 m dal bordo del tavolo. Calcolare il tempo di volo della palla e la velocità all’istante in cui ha lasciato il tavolo.
Soluzione. il moto di caduta verso il basso è un moto uniformemente accelerato con accelerazione g = 9.81 m s2 ; il dislivello è di 1.20 m. Calcoliamo il tempo di percorrenza, sapendo che la componente verticale iniziale della velocità è nulla, t = s 2h g = s 2.40 m 9.81 m s2 = 0.50 s Durante questo intervallo di tempo la palla è avanzata in direzione orizzontale di 1.50 m di moto rettilineo uniforme e quindi la velocità iniziale è la velocità costante del moto v = △s △t = 1.50 m 0.5 s = 3.0 m s Esercizio 22. Un proiettile viene sparato orizzontalmente da un’arma posta a 45.0 m sopra un terreno orizzontale.
La sua velocità alla bocca dell’arma è 250 m/s. Calcolare il tempo di volo, a che distanza orizzontale raggiungerà il terreno e il modulo della componente verticale della velocità quando colpisce il terreno.
Soluzione. il tempo di volo si calcola tenendo conto del moto accelerato di caduta, con velocità iniziale 250 m/s e velocità finale nulla t = s 2h g = s 2 · 45.0 m 9.81 m s2 = 3.03 s il calcolo della distanza orizzontale si basa sul moto rettilineo uniforme della componente orizzontale del moto x = vt = 250 m s · 3.03 s = 758 m la componente verticale della velocità è quella acquisita nella caduta libera v = p 2hg = r 2 · 45.0 m · 9.81 m s2 = 29.7 m s 11 Esercizio 23. Una palla da baseball viene lanciata verso il battitore orizzontalmente a una velocità iniziale di 160 km/h. La distanza a cui si trova il battitore è 18 m. Calcolare il tempo impiegato a coprire i primi 9 m in orizzontale; I rimanenti 9 m; La caduta dovuta alla gravità nei primi 9 m in orizzontale e nei rimanenti 9 m.
Soluzione. la velocità iniziale, trasformata in m/s, è v0x = 160 3.6 = 44.4 m s , mentre la v0y = 0 (essendo la palla lanciata orizzontalmente). Il moto orizzontale può essere considerato rettilineo uniforme, da cui t = 9 m 44.4 m s = 0.2 s nei secondi 9 m il tempo rimarrà invariato, trattandosi appunto di moto rettilineo uniforme; la caduta è descrivibile mediante le leggi del moto uniformemente accelerato; nei primi 9 m si avrà, con partenza da fermo h1 = 1 2gt2 = 1 2 · 9.81 m s2 · 0.22 s2 = 0.2 m nei metri rimanenti, crescendo la velocità vy = gt = 9.81 m s2 · 0.2 s = 1.9 m/s, si percorrerà una distanza maggiore h2 = vt + 1 2gt2 = 1.9 m s · 0.2 s + 1 2 · 9.81 m s2 · 0.22 s2 = 0.6 m Esercizio 24. Un proiettile è lanciato con la velocità iniziale di 30 m/s con un alzo di 60◦rispetto al piano orizzontale. Calcolare modulo e direzione della sua velocità dopo 2.0 s e dopo 5.0 s dal lancio.
Soluzione. calcoliamo le componenti orizzontale e verticale della velocità iniziale v0x = v0 cos ϑ = 30 m s cos 60◦= 15 m s voy = v0 sin ϑ = 30 m s sin 60◦= 26 m s dopo 2.0 s la sua velocità orizzontale sarà invariata, vx(2 s) = 15 m s , mentre la componente verticale diverrà vy = v0y −gt = 26 m s −9.8 m s2 · (2.0) s = 6.4 m s il modulo della velocità sarà v = q v2 x + v2 y = q 152 + (6.4)2 = 16 m s l’angolo sarà α = arctan 6.4 15 = 23◦ sopra il piano orizzontale; dopo 5.0 s, la velocità orizzontale è sempre vx = 15 m s , quella verticale vy = v0y −gt = 26 m s −9.8 m s2 · (5.0) s = −23 m s il modulo della velocità sarà v = q v2 x + v2 y = q 152 + (−23)2 = 27.5 m s l’angolo sarà α = arctan 27.5 15 = 61◦ sotto il piano orizzontale.
Esercizio 25. Una pietra viene catapultata con la velocità iniziale di 20 m/s a un angolo di 40.0◦rispetto al piano orizzontale. Trovare i suoi spostamenti proiettati in orizzontale e in verticale dopo 1.10 s, 1.80 s, 5.00 s dopo il lancio.
12 Soluzione. calcoliamo le componenti della velocità iniziale lungo le direzioni orizzontale e verticale v0x = 20m s · cos 40.0◦= 15.3 m s v0y = 20m s · sin 40.0◦= 12.9 m s la distanza percorsa nella direzione può essere calcolata tramite le leggi del moto rettilineo uniforme x1 = v0xt = 15.3 m s · 1.1 s = 16.8 m x2 = 15.3 m s · 1.8 s = 27.5 m x3 = 15.3 m s · 5.00 s = 76.5 m lo spostamento verticale è determinato tramite la legge del moto uniformemente accelerato y1 = voyt −1 2gt2 = 12.9 m s · 1.1 s −4.9 m s2 · 1.12 s2 = 8.3 m y2 = 12.8 m s · 1.8 s −4.9 m s2 · 1.82 s2 = 7.3 m y3 = 12.8 m s · 5.00 s −4.9 m s2 · 5.002 s2 = −58 m il testo non specifica la distanza dal suolo dalla quale la pietra viene scagliata. Se fosse scagliata da terra, il terzo risultato non sarebbe fisicamente significativo, in quanto la pietra avrebbe già urtato il terreno in ricaduta.
La pietra, in tale ipotesi, urta il terreno dopo aver percorso R = v2 0 sin 2ϑ0 g = 202 sin 80◦ 9.8 = 40.2 m ciò indica che, in questo caso, anche lo spostamento orizzontale x3 non è plausibile, avendo la pietra già urtato il terreno, definendo una distanza orizzontale massima di 40.2 m; in tal caso dopo 5.00 s lo spostamento verticale è nullo.
Esercizio 26. Una palla viene lanciata dall’alto di un colle con la velocità iniziale di 15 m/s a un angolo di 20.0◦sotto il piano orizzontale. Trovare il suo spostamento proiettato sul piano orizzontale e sull’asse verticale 2.30 s dopo il lancio.
Soluzione. calcoliamo le componenti della velocità iniziale lungo le direzioni orizzontale e verticale: v0x = 15 cos 20.0◦= 14.1 m s v0y = 15 sin 20.0◦= −5.1 m s (il segno negativo di voy dipende dal fatto che la palla viene lanciata sotto il piano orizzontale) la palla ha uno spostamento in orizzontale costante nel tempo x = v0xt = 14.1 m s · 2.30 s = 32.4 m lo spostamento lungo la direzione verticale è descrivibile con un moto uniformemente accelerato (caduta libera in assenza di attriti) y = −5.1 m s · 2.30 s −1 29.8 m s2 · (2.30)2 s2 = −37.6 m Esercizio 27. Una palla viene lanciata direttamente contro un muro con la velocità iniziale di 25.0 m/s a un angolo di 40.0◦rispetto al suolo orizzontale, come indicato in figura. Il muro si trova a 22.0 m dal punto di lancio.
(a) Per quanto tempo rimane in aria la palla prima di colpire la parete? (b) Quanto più in alto del punto di lancio colpisce la parete? (c) Quali sono le componenti orizzontale e verticale della sua velocità all’istante in cui colpisce la parete? (d) In questo istante ha già superato il vertice della traiettoria?
13 Caso 1.
(a): il tempo di volo prima di colpire il muro è quello durante il quale la palla percorre i 22.0 m che la separano dal muro stesso. Calcoliamo prima le componenti della velocità v0x = 25.0 m s · cos 40.0◦= 19.2 m s v0y = 25.0 m s · sin 40.0◦= 16.1 m s È possibile calcolare il tempo attraverso la componente orizzontale, che appunto indica lo spostamento verso il muro t = s vox = 22.0 m 19.2 m s = 1.15 s Caso 2.
(b): Il punto in cui colpisce la parete al di sopra del punto di lancio, dipende dalla variazione della componente verticale della velocità, nel tempo di 1.15 s y = voyt −1 2gt2 = 16.1 m s · 1.15 s −4.9 m s2 · 1.152 s2 = 12.0 m Caso 3.
(c): calcoliamo la velocità nell’istante in cui la palla colpisce la parete. La componente orizzontale non subisce alcuna variazione, in condizioni ideali, e pertanto sarà vx = v0x = 19.2 m s la componente verticale varia invece secondo le leggi della caduta di un grave v2 y = v2 0y −2gy da cui vy = r 16.1 m s 2 −2 · 9.8 m s2 · 12 m = 4.9 m s (d): il moto complessivo è descrivibile attraverso una parabola. Il punto più in alto coincide geometricamente con il vertice di tale parabola. Dalla equazione del moto y = x tan ϑ0 −gx2 2v2 0x con ϑ0 angolo iniziale rispetto alla direzione orizzontale, si ricava l’ascissa del vertice, cioè lo spostamento in orizzontale che confronteremo con la distanza tra palla e muro. L’ascissa del vertice Vx = −b 2a = v2 0 sin 2ϑ0 2g = 25 m s 2 sin 80.0◦ 2 · 9.8 m s2 = 31.4 m essendo la distanza tra palla e parete di 22 m, dopo 1.15 s la palla è ancora in fase di salita.
La risposta si può dare anche confrontando i tempi, in quanto il punto di massimo corrisponde a quello in cui la velocità si annulla (in fase di salita la velocità ha segno positivo, che si inverte quando la palla inizia a scendere per il prevalere della gravità. Tra queste due fasi la velocità si annullerà appunto nel vertice della traiettoria parabolica). Quindi da vy = v0y −gt dovendo essere vy = 0, si ha t = v0y g = 16.1 m s 9.8 m s2 = 1.64 s anche in questo caso, il tempo è superiore al tempo necessario a colpire il muro di 1.15 s, e ciò indica ancora come la palla sia in fase di salita.
14 Esercizio 28. Dimostrare che l’altezza massima raggiunta da un proiettile è ymax = v2 0 sin2 θ0 2g .
Soluzione. Il moto di un proiettile è un moto con una traiettoria parabolica. L’equazione che caratterizza il moto verticale accelerato è data da y = (v0 sin θ0) t −1 2gt2 se indichiamo con b = v0 sin θ0 e a = −1 2g, questa equazione diviene del tipo y = at2 + bt, cioè l’equazione di una parabola passante per l’origine (il punto dal quale il proiettile è scagliato). Essendo a < 0 la parabolica ha concavità rivolta verso il basso e quindi il suo punto di massimo è rappresentato dall’ordinata del suo vertice; pertanto, essendo c = 0 yV ≡ymax = −∆ 4a = −b2 −4ac 4a = −v2 0 sin2 θ0 −0 −2g = v2 0 sin2 θ0 2g Esercizio 29. Dimostrare che lo spostamento orizzontale di un proiettile che abbia una velocità iniziale v0 e un angolo di tiro θ0 è R = v2 0 g sin 2θ0. Mostrare poi che un angolo di 45◦dà il massimo spostamento orizzontale.
Soluzione. Il massimo spostamento orizzontale è detto la gittata. La legge del moto parabolica è ottenuta scomponendo il moto nelle due componenti, orizzontale e verticale. La componente orizzontale è descritta da un moto rettilineo uniforme, mentre quella verticale da un moto uniformemente accelerato. La relazione che descrive il moto del proiettile è di secondo grado e, algebricamente, è l’equazione di una parabola. Pertanto, la massima distanza orizzontale è geometricamente il segmento che rappresenta la distanza tra le due intersezioni della parabola con l’asse x. Data l’equazione y = x tan θ0 − g 2v2 0 cos2 θ0 x2 le intersezioni con l’asse x, cioè i punti con ordinata nulla, sono x tan θ0 − g 2v2 0 cos2 θ0 x = 0 la soluzione x = 0 rappresenta il punto di partenza del proiettile, mentre l’altra x = 2v2 0 cos2 θ0 · sin θ0 cos θ0 g = v2 0 sin 2θ0 g per cui R = v2 0 sin 2θ0 g −0 = v2 0 sin 2θ0 g il valore della funzione seno quando l’angolo è uguale a 90◦, quindi 2θ0 = 90◦e θ0 = 45◦.
Dimostrare che, per un proiettile sparato da un terreno piano a un angolo ϑ0 rispetto all’orizzontale, il rapporto fra la massima altezza H e la gittata R è dato dall’espressione H/R = 1 4 tan ϑ0. Per quale angolo si ha H = R?
15 Soluzione. Soluzione La relazione che esprime il punto di massima altezza, cioè il punto in cui vy = v0 sin ϑ0 = 0, è H = v2 0 sin2 ϑ0 2g mentre la gittata, distanza tra punto di partenza e di ricaduta, cioè nel modello geometrico la distanza tra le intersezioni della parabola con l’asse orizzontale, è espressa da R = v2 0 sin 2ϑ0 g = 2v2 0 sin ϑ0 cos ϑ0 g il loro rapporto sarà pertanto H R = v2 0 sin2 ϑ0 2g 2v2 0 sin ϑ0 cos ϑ0 g = v2 0 sin2 ϑ0 2g · g 2v2 0 sin ϑ0 cos ϑ0 = 1 4 tan ϑ0 Se H = R, allora il rapporto H R = 1, cioè 1 4 tan ϑ0 = 1. Risolvendo la equazione goniometrica elementare, si ha ϑ0 = arctan 4 = 76◦ Una pietra viene proiettata verso un terrapieno di altezza h con la velocità iniziale di 42.0 m/s a un angolo di 60.0◦rispetto al suolo orizzontale (vedi figura). La pietra cade in A, 5.50 s dopo il lancio. Trovare l’altezza h del terrapieno; la velocità della pietra subito prima dell’urto col terreno e la massima altezza H sopra il suolo raggiunto dalla pietra.
Soluzione. utilizziamo la legge che descrive il moto parabolico della pietra. Nel tempo t = 5.50 s, la pietra percorre in orizzontale la distanza x = v0 cos ϑ0t = 42.0 m s · cos 60.0◦· 5.50 s = 115.5 m e in verticale y = v0 sin ϑ0t −1 2gt2 = 42.0 m s · sin 60.0◦· 5.50 s −4.9 m s2 · 5.50 m s 2 = 51.8 m Prima dell’urto con il terrapieno la velocità si otterrà sommando vettorialmente la velocità orizzontale, costante, con quella diretta verticalmente vx = v0 cos ϑ0 = 42.0 m s · cos 60.0◦= 21 m s vy = v0 sin ϑ0 −gt = 42.0 m s · sin 60.0◦−9.8 m s2 · 5.50 s = −17.5 m s la velocità sarà quindi v = q 212 + (−17.5)2 = 27.3 m s la massima altezza raggiunta è espressa da H = v2 0 sin2 ϑ0 2g = 42.0 m s 2 sin2 60.0◦ 2 · 9.8 m s2 = 67.5 m 16 Esercizio 30. La velocità di lancio di un proiettile è cinque volte maggiore della velocità che esso raggiunge alla massima altezza. Calcolare l’angolo di elevazione del lancio.
Soluzione. la velocità di lancio non è altro che la velocità iniziale v0. Inoltre, nel punto di massima altezza, la componente verticale vy = 0. Pertanto vy = v0 sin θ0 −gt = 0 cioè, la velocità nel punto di massima altezza è espressa dalla sola componente orizzontale vx = v0x = v0 cos θ0 e questa sarà un quinto della velocità iniziale; quindi v0 = q v2 0x + v2 0y = 5v0x elevando al quadrato e sommando i termini simili, si ottiene 24v2 0x = v2 oy ma il rapporto v0y v0x = tan θ0, per cui θ0 = arctan √ 24 = 78.5◦ Esercizio 31. Un fucile con una velocità alla volata di 450 m/s spara un proiettile contro un bersaglio distante 45 m. quanto più alto del bersaglio deve essere puntata la canna del fucile per riuscire a colpire il bersaglio?
Soluzione. la velocità iniziale ha una direzione orizzontale, uscendo dalla canna; l’angolo θ0 = 0. La traiettoria del proiettile è descritta dalla legge del moto parabolico: y = x tan θ0 −gx2 2v2 0x nel nostro caso, θ0 = 0 e v0x = 450 m s , per cui y = 9.8 m s2 · 452 m2 2 · 4502 m2 s2 = 0.049 m Esercizio 32. Si spara una palla da terra in aria. All’altezza di 9.1 m si osserva una velocità − → v = 7.6− → i +6.1− → j m/s.
Calcolare la massima elevazione e la distanza orizzontale complessiva percorsa. Determinare inoltre la velocità della palla nell’istante prima di cadere a terra.
Soluzione. La velocità è espressa mediante il vettore; da questo deduciamo le componenti orizzontali e verticali vx = 7.6 m s vy = 6.1 m s sappiamo che la componente orizzontale si mantiene costante. La componente verticale varia secondo le leggi del moto accelerato v2 y = v2 0 sin2 θ0 −2g (y −y0) da cui v2 0y = v2 0 sin2 θ0 = v2 y + 2g (y −y0) = 6.12 m2 s2 + 2 · 9.8 m s2 · 9.1 m = 229.3 m2 s2 la massima elevazione è data da H = v2 0 sin2 θ0 2g = 229.3 m2 s2 19.6 m s2 = 11.7 m La distanza complessiva è rappresentata dalla gittata R = v2 0 sin 2θ0 g 17 ora v2 0 sin 2θ0 = 2v2 0 sin θ0 cos θ0 = 2v0 cos θ0 · v0 sin θ0 = 2v0x · v0y = 2vx · v0y, per la costanza della componente orizzontale. Pertanto R = 2 · 7.6 m s · √ 229.3 m s 9.8 m s2 = 23.5 m Infine, la velocità prima dell’impatto corrisponde al valore assoluto massimo della componente verticale. Poiché nel punto di massima elevazione, la componente verticale della velocità è nulla, possiamo considerare la palla che cade da 11.7 m con partenza da fermo, per ottenere la velocità vy = − p 2gy = − r 2 · 9.8 m s2 · 11.7 m = −15.1 m s la velocità, espressa vettorialmente, sarà − → v = 7.6− → i −15.1− → j m s il suo modulo sarà v = q 7.62 + (−15.1)2 = 17.0 m s con θ = arctan −15.1 7.6 = 63◦ sotto il piano orizzontale.
Esercizio 33. Nel suo libro Discorsi e Dimostrazioni matematiche, Galileo afferma che “per altezze (cioè, angoli di alzo) che superano o sono minori di 45◦di uguali quantità le gittate sono uguali”. Dimostrare tale affermazione.
la gittata è espressa da R = v2 0 g sin 2ϑ0 la figura mostra l’andamento della funzione y = sin 2ϑ0 dalla quale si osserva che il valore massimo si ottiene per l’angolo di 45◦. L’alzo deve essere minore di 90◦, tiro verticale, e osservando la simmetria della curva si può ricavare che per due angoli complementari, cioè che sommati danno 90◦, il valore della funzione seno è lo stesso e ciò determinerà una uguale gittata.
Esercizio 34. Dai vulcani in eruzione vengono espulsi grossi proiettili di pietra. Se la situazione è quella rappresentata in figura, determinare a quale velocità iniziale devono essere espulsi nel punto A a un’elevazione di 35◦per cadere nel punto B ai piedi del vulcano. Determinare poi il tempo di volo.
18 Soluzione. considero l’equazione della traiettoria dei singoli proiettili di pietra: y = x tan ϑ0 − gx2 2 (v2 0 cos2 ϑ0) e per ottenere , dobbiamo risolvere rispetto alla velocità iniziale v2 0 (y + x tan ϑ0) = gx2 2 cos2 ϑ0 nel nostro caso si ha: y = −3300 m, x = 9400 m, ϑ0 = 35◦; sostituendo si ottiene v2 0 (3300 m + 9400 m tan 35◦) = 9.8 m s2 · 94002 m2 2 cos2 35◦ e risolvendo rispetto alla velocità, si ha v0 = 9400 m cos 35◦ s 9.8 m s2 2 (3300 m + 9400 m tan 35◦) = 256 m s il tempo di volo può essere ottenuto mediante la componente orizzontale della velocità iniziale, in quanto il proiettile, nella direzione orizzontale, si muove di moto rettilineo uniforme; calcoliamo vox = v0 cos 35◦= 210 m s .
Determiniamo ora il tempo impiegato per percorrere a tale velocità i 9400 m di lunghezza orizzontale t = 9400 m 210 m s = 45 s Esercizio 35. A quale velocità iniziale deve essere lanciata una palla, con un angolo di elevazione di 55◦, per centrare direttamente il canestro senza rimbalzo? (vedi figura) 19 Soluzione. utilizziamo anche qui la legge oraria del moto e risolviamola rispetto a v0. Da y −y0 = x tan ϑ0 − gx2 2v2 0 cos2 ϑ0 si ottiene v0 = s gx2 2 cos2 ϑ0 · 1 (y −y0) + x tan ϑ0 sostituendo i valori assegnati e facendo attenzione alla gestione dei segni, si ha v0 = s 9.8 m s2 · 5.52 m2 2x cos 55◦ · 1 5.5 m · x tan 55◦−0.9 m = 8 m s Esercizio 36. Un proiettile viene lanciato con velocità iniziale v0 = 30.0 m/s dal livello del suolo contro un bersaglio posto a una distanza orizzontale R = 20.0 m (vedi figura). Trovare i due possibili angoli, per la traiettoria alta e bassa.
Soluzione. Il valore assegnato R = 20.0 m è la gittata del proiettile; pertanto da R = v2 0 sin 2ϑ0 g si ottiene sin 2ϑ0 = Rg v2 0 sostituendo e risolvendo rispetto a ϑ0, si ha 2ϑ0 = arcsin 20.0 m · 9.8 m s2 30.02 m2 s2 !
= 12.6◦ Ne segue che gli angoli di elevazione saranno ϑ1 0 = 6.3◦ ϑ2 0 = 83.7◦ ricordando che si ottiene la stessa gittata per angoli tra loro complementari (cioè la cui somma è 90◦).
Esercizio 37. Qual è la massima altezza che può raggiungere una palla lanciata da un giocatore la cui massima gittata è di 60 m?
Soluzione. Ricordiamo il significato delle grandezze coinvolte: la gittata è la distanza massima orizzontale percorsa dal corpo per tornare alla stessa altezza del punto di partenza; essendo la traiettoria parabolica, la gittata è la distanza tra le due intersezioni della parabola con l’asse orizzontale passante per i due punti; l’altezza massima corrisponde all’ordinata del vertice della parabola, H = v2 0 sin2 ϑ0 2g tale altezza è massima per un angolo di 45◦; la gittata è espressa da R = v2 0 sin 2ϑ0 g 20 per un angolo di 45◦, diviene R = v2 0 g sostituendo in H, si ha H = R sin2 45◦ 2 = 60.0 m √ 2 2 2 2 = 15.0 m Esercizio 38. Un aeroplano, volando a 290 km/h con un angolo di 30◦verso il basso rispetto al piano orizzontale, sgancia un falso bersaglio radar, come in figura. La distanza orizzontale fra il punto di rilascio e quello in cui colpisce il suolo è di 690 m. Determinare l’altezza dell’aereo al momento dello sgancio e il tempo di volo del bersaglio.
Soluzione. Indichiamo con v la velocità dell’aereo e quindi anche del bersaglio ad esso solidale. Possiamo pertanto calcolare la componente orizzontale di tale velocità, conoscendo l’angolo.
vx = 290 3.6 m s · cos 30.0◦= 69.8 m s Il bersaglio percorre pertanto i 690 m in orizzontale con la velocità calcolata, per cui t = 690 m 69.8 m s = 9.9 s L’altezza alla quale l’aereo si trova, può essere ottenuta utilizzando la relazione che descrive la traiettoria del moto, considerando come negativo l’angolo rivolto verso il basso y0 −y = x tan ϑ −gx2 2v2 x = 690 tan (−30.0◦) −9.8 m s2 · 6902 m2 2 · (69.8)2 m2 s2 = 877 m Esercizio 39. Un pallone viene calciato in avanti con velocità iniziale di 20 m/s e un angolo di elevazione di 45◦. Contemporaneamente un attaccante, che si trova 54 m più avanti nella direzione del tiro, parte di scatto per raggiungere la palla. Quale deve essere la sua velocità media per raggiungere la palla subito prima che tocchi il terreno?
Soluzione. Calcoliamo innanzitutto la distanza che percorre la palla calciata; tale distanza corrisponde alla gittata (per l’angolo ϑ0 = 45◦, la gittata è massima, essendo sin 90◦= 1): R = v2 0 g = 202 m s 9.8 m s2 = 41 m 21 l’attaccante trovandosi oltre il punto di caduta della palla, dovrà raggiungerla percorrendo 54 −41 = 13 m.
Il tempo di volo della palla può essere ottenuto considerando il moto rettilineo uniforme della componente orizzontale v0x = v0 cos ϑ0 = 20 m s · √ 2 2 = 14.1 m s l’intervallo di tempo sarà quindi t = 41 m 14.1 m s = 2.9 s la velocità media del calciatore sarà pertanto vmedia = △s △t = 13 m 2.9 s = 4.5 m s Esercizio 40. Un aereo, picchiando a un angolo di 53◦rispetto alla verticale, sgancia un proiettile a una quota di 730 m dal suolo. Il proiettile colpisce il terreno dopo 5.00 s. Trovare la velocità dell’aereo, la distanza orizzontale percorsa dal proiettile durante la caduta e infine le componenti orizzontale e verticale della sua velocità all’istante in cui ha colpito il terreno.
Soluzione. nell’istante in cui il proiettile viene sganciato la velocità dell’aereo e del proiettile solidale è quella illustrata in figura. Pertanto il proiettile, mentre percorre in caduta i 730 m, verrà accelerato verso il basso dal suo peso incrementando così la sua velocità verticale iniziale, secondo la legge del moto uniformemente accelerato s = vyt + 1 2gt2 risolvendo rispetto a vy e sostituendo i valori assegnati, si ha vy = s −1 2gt2 t = 730 m −4.9 m s2 · 5.002 s2 5.00 s = 121.5 m s diretta verso il basso; dalla componente verticale è possibile ottenere la velocità, attraverso i teoremi della trigonometria, cioè l’ipotenusa di un triangolo rettangolo è uguale al prodotto di un cateto per il coseno dell’angolo adiacente v = 121.5 m s cos 53◦= 202 m s È ora possibile calcolare anche la componente orizzontale del moto, anche col th. di Pitagora, vx = p 2022 −121.52 = 161 m s Dalla componente orizzontale della velocità si ottiene la distanza orizzontale percorsa, in quanto tale moto può essere descritto dalle leggi del moto rettilineo uniforme x = vxt = 161 m s · 5.00 s = 806 m La componente orizzontale della velocità prima dell’impatto col terreno è sempre uguale a vx = 161 m s , mentre la componente verticale è quella ricavata pari a 121.5 m s si ricava dalle leggi del moto accelerato vf y = vy + gt = 121.5 m s + 9.81 m s2 · 5 s = 170.5 m s diretta verso il basso.
22 Esercizio 41. Una palla lanciata orizzontalmente dall’altezza di 20 m, tocca il suolo con una velocità tripla rispetto a quella iniziale. Trovare la velocità iniziale.
Soluzione. Soluzione se la palla è lanciata orizzontalmente, la sua velocità iniziale ha solo la componente orizzontale. La velocità iniziale, dovuta alla caduta libera della palla, può essere ottenuta dalla relazione v2 f = v2 0 + 2gy essendo vf = 3v0, sostituendo si ottiene 8v2 0 = 392 da cui v0 = s 392 m2 s2 8 = 7, 1 m s Esercizio 42. Un tennista serve la palla orizzontalmente da un’altezza sul terreno di 2.37 m a una velocità di 23.6 m/s. Con quale altezza la palla passa sopra la rete, alta 0.90 m, che si trova a una distanza di 12 m? Se il tennista servisse con un’inclinazione verso il basso di 5.00◦rispetto all’orizzontale, la palla passerà ancora sopra la rete ?
Soluzione. applichiamo la legge che descrive il moto parabolico che seguirà la pallina, determinando l’altezza raggiunta dopo aver percorso 12 m; l’altezza massima della pallina è quella al momento del lancio, per cui, essendo ϑ0 = 0 2.37 m −y = gx2 2v2 x = 9.8 m s2 · 122 m2 2 · 23.62 m2 s2 risolvendo rispetto a y, si ottiene y = 2.37 m −9.8 m s2 · 122 m2 2 · 23.62 m2 s2 = 1.10 m la pallina oltrepassa la rete di 20 cm. Se l’angolo iniziale è di 5.00◦verso il basso, bisogna calcolare la componente orizzontale della velocità, vx0 = 23.6 cos 5.00◦= 23.5 m s 2.37 m −y = 12 · tan 5.00◦+ 9.8 m s2 · 122 m2 2 · 23.52 m2 s2 risolvendo rispetto a y, si ha y = 0.43 cm quindi la palla non supera la rete, colpendo la rete a 4, 3 cm sopra il terreno.
Esercizio 43. Un treno si sta muovendo a 72 km/h quando una lanterna appesa all’estremità del treno, a 4, 9 m dal suolo, si stacca. Calcola la distanza percorsa dal treno nel tempo impiegato dalla lampada a cadere a terra.
Dove cade la lampada rispetto al treno e ai binari? Qual è il suo percorso rispetto al treno e ai binari?
Soluzione. Il treno si muove a velocità costante. La lampada, possiede una velocità orizzontale uguale a quella del treno al quale è appesa, e acquisterà una velocità verticale durante la caduta a causa della gravità g e percorrerà i 4, 9 m in h = 1 2gt2 da cui t = s 2h g = r2 × 4, 9 9, 8 = 1 s nello stesso tempo il treno percorre s = vt = 72 3, 6 m s × 1 s = 20 m La lampada, agganciata al treno, mantiene la sua velocità di 72 km/h e quindi cadrà perpendicolarmente al terreno rispetto al treno e 20 m più avanti rispetto ai binari, che sono ovviamente fermi. Il percorso rispetto al treno sarà rettilineo, mentre sarà parabolico rispetto ai binari.
23 Esercizio 44. Un cannone spara un proiettile con una velocità di 198 m/s. Determinare gli angoli per i quali il proiettile colpirà un bersaglio a 137 m di distanza e alto 5, 5 m.
Soluzione. Il proiettile deve colpire un bersaglio posto più in alto rispetto al livello da cui viene sparato (preso uguale a 0). Applichiamo la relazione che descrive la legge del moto del proiettile y −y0 = x tan α − g 2v2 0 cos2 αx2 sostituiamo i valori assegnati e risolviamo rispetto alle funzioni goniometriche 5, 5 = 137 tan α − 9, 8 2 × 1982 × cos2 α × 1372 5, 5 = 137 tan α − 2, 3 cos2 α moltiplichiamo per cos2 α 5, 5 cos2 α = 137 sin α cos α −2, 3 questa è un’equazione riconducibile ad una omogenea 5, 5 cos2 α = 137 sin α cos α −2, 3 sin2 α −2, 3 cos2 α 7, 8 cos2 α + 2, 3 sin2 α −137 sin α cos α = 0 da cui 2, 3 tan2 α −137 tan α + 7, 8 = 0 avremo tan α = 0, 06 59, 5 α = 3, 3◦ 89◦ Moto Circolare Uniforme Esercizio 45. In un modello di atomo di idrogeno, un elettrone orbita attorno al protone su un cerchio di raggio 5.28 · 10−11 m, alla velocità di 2.18 · 106 m/s. Calcolare l’accelerazione dell’elettrone.
Soluzione. l’esercizio presuppone che la velocità dell’elettrone rimanga sempre costante in modulo, in modo che il moto possa essere descritto dalle leggi del moto circolare uniforme. La velocità cambia ogni istante la propria direzione, ed essendo la velocità una grandezza vettoriale, tale variazione deve essere descritta dall’azione di una accelerazione, l’accelerazione centripeta, diretta cioè verso il centro. (tale accelerazione è dovuta alla forza elettrica tra la carica del nucleo e dell’elettrone).
ac = v2 r = 2.18 · 1062 m2 s2 5.28 · 10−11 = 9.00 · 1022 m s2 Esercizio 46. Calcolare la velocità angolare, la velocità lineare e l’accelerazione centripeta della luna, derivando la risposta dal fatto che la luna compie un giro completo in 28 giorni e che la distanza media dalla terra alla luna è di 3, 84 · 105 km.
Soluzione. Trasformiamo le unità di misura per renderle più facilmente applicabili.
28 giorni = 28 × 24 × 3600 s = 2, 42 · 106 s 3, 84 · 105 km = 3, 84 · 108 m La velocità angolare è ω = 2π rad T sec = 6, 28 2, 42 · 106 s = 2, 6 · 10−6 rad s la velocità lineare v = ωRT −L = 2, 6 · 10−6 rad s × 3, 84 · 108 m = 998 m s l’accelerazione centripeta ac = v2 RT −L = 9982 m2 s2 3, 84 · 108 m = 2, 6 · 10−3 m s2 24 Esercizio 47. Determinare modulo, direzione e verso dell’accelerazione di un velocista che corre a 10 m/s su una curva di raggio 25 m Soluzione. basta ricordare la relazione che lega l’accelerazione centripeta alle grandezze assegnate ac = v2 r = 10 m s 2 25 m = 4 m s2 l’accelerazione centripeta è sempre diretta verso il centro della curva.
Esercizio 48. La velocità angolare di un volano aumenta in modo uniforme da a 20 rad/s a 30 rad/s in 5 s.
Calcolare l’accelerazione e l’angolo totale di cui è ruotato in questo intervallo di tempo.
Soluzione. L’accelerazione sarà α = (30 −20) rad s 5 s = 2 rad s2 e l’angolo di cui è ruotato θ = θ0 + ω0 (t −t0) + 1 2α (t −t0)2 = 0 + 20 × 5 + 1 × 52 = 125 rad Esercizio 49. Un campo magnetico è in grado di costringere una particella carica a seguire un percorso circolare.
Supponiamo che un elettrone, per effetto di un certo campo magnetico, subisca un’accelerazione radiale di 3.0 · 1014 m/s2. Determinare la velocità dell’elettrone se il raggio del percorso circolare è di 15 cm e il periodo del moto circolare.
Soluzione. la relazione è la stessa del caso precedente; si tratta in questo caso di applicare la formula inversa v = √ar = r 3.0 · 1014 m s2 · 0.15 m = 6.7 · 106 m s il periodo, cioè il tempo di percorrenza di un intero giro, è dato da T = 2πr v = 2π · 0.15 m 6.7 · 106 m s = 1.4 · 10−7 s Esercizio 50. Un velocista corre alla velocità di 9.2 m/s su una pista circolare, con un’accelerazione centripeta di 3.8 m/s2. Determinare il raggio della pista e il periodo del moto.
Soluzione. esercizio di semplice applicazione delle relazioni che legano le grandezze cinematiche. Nel nostro caso, da a = v2 r si ricava r = v2 a = 9.2 m s 2 3.8 m s2 = 22.3 m il periodo è il tempo impiegato per percorrere un giro intero, cioè T = 2πr v = 2π · 22.3 m 9.2 m s = 15.2 s Esercizio 51. Un satellite terrestre viaggia su un’orbita circolare alla quota di 640 km sopra la superficie terrestre. Il periodo di rivoluzione è di 98.0 min. Calcolare la velocità del satellite e l’accelerazione di gravità a quella distanza.
25 Soluzione. In questo moto circolare, l’accelerazione di gravità è l’accelerazione centripeta che tiene vincolato il satellite nell’orbita. I dati sono forniti con diverse unità di misura e nel calcolo bisogna eseguire prima le opportune equivalenze. Nel calcolare la distanza bisogna tener conto anche del raggio terrestre, essendo la Terra un corpo solido esteso e l’ipotetico centro di rotazione si trova, appunto, nel centro della Terra v = 2πr T = 2π · 6.37 · 106 + 6.40 · 105 m 98.0 · 60 s = 7491 m s l’accelerazione è data da ac = v2 r = 74912 m2 s2 (6.37 · 106 + 6.40 · 105) m = 8.00 m s2 Esercizio 52. Se una sonda spaziale è in grado di sopportare un’accelerazione di 20 g, calcolare il minimo raggio di curvatura del percorso che può affrontare a una velocità pari a un decimo di quella della luce; determinare inoltre il tempo per compiere un quarto di giro.
Soluzione. la relazione che lega le tre grandezze date è ac = v2 r se la velocità è 3 · 107 m s , cioè un decimo della luce, si ha r = v2 ac = 9 · 1014 m2 s2 20 · 9.8 m s2 = 4.6 · 1012 m tale raggio equivale a circa 31 UA, cioè un’orbita simile a quella di Nettuno, supposta circolare. Il tempo è ottenibile da T = 2πr v = 2π · 4.6 · 1012 m 3.7 · 107 m s = 9.6 · 105 s cioè, dividendo per 4, T = 2.4 · 105 s ≈2.8 anni terrestri Esercizio 53. Un ventilatore compie 1200 giri al minuto. consideriamo un punto sul bordo esterno di una pala di raggio 0.15 m. Trovare la distanza che percorre questo punto ad ogni giro, la sua velocità e accelerazione.
Soluzione. in numero dei giri al minuto rappresenta la frequenza; calcoliamola in Hertz f = 1200 giri min 60 s min = 20 Hz la distanza che il punto percorre è rappresentata dalla lunghezza della circonferenza di raggio 0.15 m, cioè l = 2πr = 2π · 0.15 m = 0.94 m la velocità è data da v = 2πrf = 0.94 m · 20 Hz = 18.8 m s mentre l’accelerazione è a = v2 r = 18.82 m2 s2 0.15 m = 2369 m s2 Esercizio 54. Un treno viaggia alla velocità media di 216 km/h. Se abborda una curva a questa velocità e la massima accelerazione tollerabile dai passeggeri è 0.050g, determinare il minimo raggio ammissibile per le curve dei binari. Se una curva ha un raggio di 1.00 km a quale valore deve essere ridotta la velocità del treno per rispettare il limite dell’accelerazione?
26 Soluzione. un’accelerazione di 0.050g corrisponde a 0.050·9.8 m s2 = 0.49 m s2 . La velocità in m/s vale v = 216 3.6 m s = 60 m s . Poiché l’accelerazione vale a = v2 r , ricaviamo il raggio di curvatura, risolvendo la relazione rispetto a r r = v2 a = 602 m2 s2 0.49 m s2 = 7347 m se il raggio viene posto a 1000 m, allora, risolvendo rispetto a v, si ha v = √ar = r 0.49 m s2 · 1000 m = 22.1 m s = 80 km h Esercizio 55. Nell’esplosione di una supernova il suo nocciolo può contrarsi tanto da raggiungere un raggio di 20 km (stella di neutroni). Se una tale stella compie un giro al secondo, determinare la velocità di una particella posta sull’equatore e la sua accelerazione centripeta.
Soluzione. il fenomeno fisico viene affrontato in forma semplificata e può essere descritto quindi mediante le relazioni che trattano del moto circolare uniforme. Se la stella compie un giro al secondo, vuol dire che la frequenza del moto è di 1 Hz. Calcoliamo quindi la velocità v = 2πrf = 2π · 20000 m · 1 Hz = 125664 m s l’accelerazione sarà quindi a = v2 r = 1256642 m2 s2 20000 m = 789568 m s2 = 80568g Esercizio 56. Un astronauta sta girando in una centrifuga su un raggio di 5.0 m. Determinare la velocità se la sua accelerazione è di 7.0 g e la frequenza e il periodo corrispondenti.
Soluzione. il moto dell’astronauta può essere descritto secondo le leggi del moto circolare uniforme. Se sono noti il raggio e l’accelerazione, basta trasformare quest’ultima nell’unità del SI, cioè a = 7.0 · 9.8 m s2 ; sapendo che a = v2 r , si ottiene, risolvendo rispetto a v v = √ar = r 7.0 · 9.8 m s2 · 5.0 m = 18.5 m s essendo v = 2πrf, si può ottenere la frequenza f = v 2πr = 18.5 m s 2π · 5.0 m = 0.6 Hz il periodo è il reciproco della frequenza T = 1 0.6 Hz = 1.7 s Esercizio 57. Determinare l’accelerazione centripeta dovuta alla rotazione della Terra per un oggetto che si trova sull’equatore. Quale dovrebbe essere il periodo di rotazione della Terra affinché questa accelerazione sia uguale a 9.8 m/s2? Calcolare infine l’accelerazione per una persona posta ad una latitudine di 40◦N.
Soluzione. assumendo il raggio terrestre di 6.37 · 106 m e poiché l’equatore è un cerchio massimo, si ha a = v2 r = 2πr T 2 r = 4π2r T 2 ora sapendo che il periodo di rotazione corrisponde ad un giorno che contiene T = 24 · 3600 s, si ha a = 4π2 · 6.37 · 106 m 864002 s2 = 0.034 m s2 27 per avere una accelerazione pari a quella di gravità, il periodo dovrebbe essere T = r 4π2r a = s 4π2 · 6.37 · 106 m 9.8 m s2 = 5066 s = 84.4 minuti Se la persona è posta alla latitudine di 40◦N, descrive un cerchio (parallelo) più piccolo (si veda una rappresentazione schematica in figura).
Applicando il teorema dei triangoli rettangoli a PO′O, si ha PO′ = rparallelo = OP cos 40◦= 6.37 · 106 m · cos 40◦= 4.88 · 106 m Applicando la relazione scritta sopra, con periodo uguale (la velocità angolare della Terra è sempre la stessa), si ha a = 4π2 · 4.88 · 106 m 864002 s2 = 0.026 m s2 Esercizio 58. Una particella P viaggia a velocità costante su un cerchio di raggio 3.00 m, compiendo una rivoluzione ogni 20.0 s. Passa per il punto O all’istante t = 0. (a) Trovare modulo e direzione del vettore posizione rispetto a O per t = 5.00 s e t = 7.50 s. (b) Per l’intervallo di 5.00 s dalla fine del 5° alla fine del 10° secondo, trovare lo spostamento e velocità media. (c) Alla fine di questo intervallo, trovare la velocità e l’accelerazione istantanea.
Caso 1.
(a): Assumendo il punto O come riferimento e sapendo che il periodo è di 20.0 s, possiamo individuare la posizione del punto dopo 5.00 s. Esso si troverà a un quarto di giro dopo O (infatti 5.00 s sono la quarta parte del periodo). Il vettore spostamento sarà pertanto la corda OO′, cioè il lato del quadrato inscritto: OO′ = r √ 2 = 3.00 · √ 2 = 4.24 m 28 dopo 7.50 s il punto O si sarà spostato di modo che la corda OO′′ sottende un angolo al centro di 135◦. La sua lunghezza si può calcolare con il teorema della corda sapendo che l’angolo alla circonferenza sotteso dalla stessa corda è metà dell’angolo al centro, cioè 67.5◦; oppure attraverso il modello geometrico nella figura sotto: il triangolo OQP è isoscele, il triangolo CHP è rettangolo isoscele, per cui CH, lato del quadrato di diagonale CP, è CH = 3.00 m √ 2 = 2.12 m e l’altezza OH = 3.00 + 2.12 m = 5.12 m. Inoltre, osservando che CH = HP, si può ricavare OP con il th. di Pitagora OP = p (5.122 + 2.122) m2 = 5.54 m Caso 1.
(b): in questo intervallo, dalla fine del 5° alla fine del 10° secondo, la particella percorre un quarto di giro; lo spostamento è dato dal vettore che congiunge i punti O’y la cui intensità è pari al lato del quadrato inscritto, cioè s = 3.00 · √ 2 = 4.24 m l’angolo formato è di 135◦. La velocità vettoriale media è v = 4.24 m 5.00 s = 0.85 s nella stessa direzione dello spostamento Caso 2.
(c): la velocità istantanea alla fine di tale intervallo, cioè dopo 10 s, si ha v = 2πr T = 2π · 3.00 m 20.00 s = 0.94 m s l’accelerazione sarà a = v2 r = 0.942 m2 s2 3.00 m = 0.30 m s2 Esercizio 59. Un ragazzino fa ruotare un sasso legato a una corda lunga 1.5 m su un cerchio orizzontale a 2.0 m dal suolo. La corda si rompe, e il sasso si muove ora orizzontalmente andando a cadere a 10 m di distanza.
Calcolare l’accelerazione centripeta del sasso nel suo moto circolare.
Soluzione. L’esercizio raggruppa più aspetti sinora presentati. Il moto di rotazione legato alla corda è descritto dalle leggi del moto circolare uniforme; lo spezzarsi della corda serve a richiamare l’attenzione sul significato di velocità tangenziale, cioè la velocità che il sasso possiede se lasciato libero di muoversi di moto rettilineo uniforme. Infine la caduta indica che il sasso è soggetto alla forza di gravità che lo tira verso il basso facendogli descrivere un moto parabolico, cioè con componente della velocità orizzontale costante.
La corda rappresenta il raggio di curvatura costante del moto circolare. La velocità tangenziale sarà calcolabile dalle leggi del moto parabolico. Da y = gx2 2v2 0 , con ϑ0 = 0, si ha v = x r g 2y = 10 m · r 9.8 m s2 2 · 2.0 m = 15.65 m s 29 L’accelerazione sarà pertanto a = v2 r = 15.652 m2 s2 1.5 m = 163.3 m s2 Esercizio 60. Se il coefficiente di attrito statico tra i pneumatici e la strada è 0.25, determinare la velocità massima con la quale un’auto può affrontare una curva di raggio 47.5 m su un piano.
Soluzione. La forza centripeta che consente all’auto di fare la curva è rappresentata dalla forza di attrito statico tra strada e pneumatici (l’attrito dinamico non si deve considerare, perché l’auto non slitta). Le ulteriori forze in campo sono il peso dell’auto e la reazione vincolare, entrambe perpendicolari alla strada uguali in modulo e di verso opposto (il moto è infatti piano). Scriviamo la relazione della forza centripeta Fc = mv2 r fs = µsmg risolviamo rispetto a v, e sostituiamo i valori assegnati: v2 = µsmgr m v = √µsgr v = r 0.25 · 9.81 m s2 · 47.5 m = 11 m s ≃40 km h Esercizio 61. Durante una gara olimpica di bob, un equipaggio affronta una curva di raggio 7.5 m alla velocità di 100 km/h. Determinare l’accelerazione (in unità di g) alla quale sono sottoposti i componenti dell’equipaggio.
Soluzione. L’accelerazione alla quale sono sottoposti è quella radiale dovuta alla forza centripeta. L’accelerazione, detta centripeta, vale a = v2 r da cui, trasformando prima la velocità in unità SI, dà a = 100 3.6 2 m2 s2 7.5 m = 103 m s = 10.5 g Esercizio 62. Un’auto che pesa 10.7 kN e viaggia a 13.4 m/s tenta di affrontare una curva piana di raggio 61.0 m. Determinare la forza d’attrito minima necessaria per seguire un arco di cerchio di questo raggio e se con un coefficiente di attrito statico 0.35 fra pneumatici e strada, tale tentativo riesce.
Soluzione. l’auto ha una massa pari a m = P g = 10700 N 9.8 m s2 = 1092 kg anche in questo caso, la forza centripeta che consente all’auto di fare la curva è rappresentata dalla forza di attrito statico tra strada e pneumatici; per cui mv2 r = µsmg La forza centripeta è Fc = 1092 kg · (13.4)2 m2 s2 61.0 m = 3214 N Questa rappresenta la forza di attrito minima. Se µs = 0.35, la forza di attrito sarà Fs = 0.35 · 1092 · 9.8 = 3745 N La curva verrà quindi affrontata correttamente.
30 Esercizio 63. Un pendolo conico è costituito da un peso di massa 50 g attaccato a un filo di 1.2 m. La massa oscilla su un cerchio di raggio 25 cm. Trovare la velocità e accelerazione del peso e la tensione del filo.
Soluzione. Dalla figura si osserva che la somma dei due vettori è diretta lungo il raggio della circonferenza. Tale risultante rappresenta la forza centripeta che agisce sul corpo. Il periodo del pendolo (considerando la proiezione del moto del corpo sul diametro) è T = 2π s l g da cui, la velocità v = 2πr T = 2πr 2π q l g = r q l g = 0.25 m q 1.2 m 9.8 m s2 = 0.72 m s L’accelerazione centripeta, ac = v2 r , è ac = 0.72 m s 2 0.25 m = 2.1 m s2 La Tensione del filo è la differenza dei due vettori forza centripeta e peso. Pertanto P = 0.05 kg · 9.8 m s2 = 0.49 N la forza centripeta è FC = mac = 0.05 kg · 2.1 m s2 = 0.105 N Il modulo della tensione sarà T = p P 2 −F 2 c = p 0.492 −0.1052 = 0.48 N Esercizio 64. Nel modello dell’atomo di idrogeno di Bohr, l’elettrone descrive un’orbita circolare intorno al nucleo. Sapendo che il raggio dell’orbita è 5.3 ·−11 m e l’elettrone ruota con una frequenza di 6.6 · 1015 giri/s, trovare la sua velocità e la sua accelerazione; determinare poi la forza centripeta che agisce sull’elettrone. La massa dell’elettrone è 9.11 · 10−31 kg.
Soluzione. la velocità di rotazione sulla traiettoria circolare è data dal rapporto tra la distanza (lunghezza della circonferenza) e il tempo impiegato per percorrere un giro completo (periodo), cioè v = 2πr T ; sapendo che la frequenza f è definita come il reciproco del periodo, si può anche scrivere v = 2πrf, da cui v = 2πrf = 2π · 5.3 ·−11 m · 6.6 · 1015 n◦giri s = 2.2 · 106 m s 31 l’accelerazione è data da ac = v2 r = 2.2 · 106 m s 2 5.3 ·−11 m = 9.1 ·22 m s2 La forza centripeta è Fc = mac = 9.11 · 10−31 kg · 9.1 ·22 m s2 = 8.3 · 10−8 N Esercizio 65. Una massa m percorre una circonferenza sul piano privo di attrito di un tavolo e sostiene una massa M appesa ad un filo passante per un foro posto nel centro del del cerchio di rotazione. Trovare la velocità con cui deve muoversi m per poter sostenere la massa M.
Soluzione: Il peso della massa M rappresenta, attraverso la tensione del filo, la forza centripeta applicata sulla massa m. Cioè Fc = mv2 r = Mg da cui v = r Mgr m Esercizio 66. Uno stuntman guida un’auto su un dosso avente un raggio di curvatura di 250 m (vedi figura).
Trovare la massima velocità che può tenere senza che l’auto si stacchi dal fondo stradale nel punto più elevato.
Soluzione. la forza centripeta deve uguagliare il peso dell’auto: mg = mv2 r da cui v = √gr = r 9.8 m s2 · 250 m = 49.5 m s = 178 km h Esercizio 67. Una moneta è posta su un piatto orizzontale rotante intorno a un asse verticale, che compie tre giri in 3.14 s. Determinare: (a) la velocità della moneta quando gira senza slittare ad una distanza di 5.0 cm dal centro; (b) l’accelerazione; (c) l’intensità della forza di attrito rispetto al piatto se la moneta ha una massa di 2 g; (d) il coefficiente di attrito statico, se si osserva che la moneta si muove lungo la tangente quando è posta ad una distanza di oltre 10 cm dal centro.
Caso 1.
(a): se la moneta compie tre giri in 3.14 s, il suo periodo di rotazione è: T = 3.14 s 3 = 1.05 s e la sua frequenza f = 1 T = 1 1.05 = 0.96 Hz la velocità lineare di rotazione è: v = 2πr T = 2πrf = 2π · 0.05 m · 0.96 Hz = 0.3 m s 32 Caso 2.
(b): la sua accelerazione è ac = v2 r = 0.3 m s 2 0.05 m = 1.8 m s2 Caso 3.
(c): la forza di attrito è la forza centripeta che tiene la moneta sul piatto alla distanza di 5 cm è fs = mac da cui fs = 2.0 · 10−3 kg · 1.8 m s2 = 3.6 · 10−3 N Caso 4.
(d): quando r = 10 cm, la velocità diviene 0.6 m s e l’accelerazione ac = 0.6 m s 2 0.1 m = 3.6 m s2 pertanto µsmg = maC cioè µs = ac g = 3.6 m s2 9.8 m s2 = 0.37 Esercizio 68. Uno studente di peso di 66 kg ha un peso apparente di 55 kg quando si trova nel punto più alto di una ruota panoramica. Trovare il suo peso apparente nel punto più basso e il peso apparente nel punto più alto caso in cui la velocità della ruota raddoppia.
Soluzione. La figura mostra lo schema vettoriale delle due accelerazioni in gioco. quando il corpo si trova in alto, la sua accelerazione di gravità è ridotta dalla forza centripeta; viceversa, quando è nel punto più basso, il suo peso è aumentato dalla accelerazione centripeta. Sapendo che nel punto più alto, lo studente ha una riduzione del suo peso pari ai 5 6, possiamo scrivere g −ac = 5 6g da cui si ottiene ac = 1 6g Nel punto più basso questa accelerazione deve essere aggiunta a quella di gravità: g + 1 6g = 7 6g 33 Il peso dello studente di massa 66 kg sarà: P = mg = 66 kg · 7 6 · 9.8 m s2 = 755 N Se ora raddoppiamo la velocità della ruota, essendo aC = v2 r la sua accelerazione centripeta quadruplica, divenendo 1 6 · 2 3g = 2 3g; pertanto nel punto più alto avrà un peso apparente di m g −2 3g = 216 N Esercizio 69. Un’auto percorre una curva piana di raggio R = 220 m alla velocità di 94.0 km/h. Trovare la forza che il passeggero di massa m = 85.0 kg esercita sul sedile.
Soluzione. Il passeggero è sottoposto all’azione della gravità con una accelerazione diretta verso il basso e alla forza centripeta con una accelerazione diretta verso il centro della curva. Calcoliamo l’accelerazione centripeta, sapendo che v = 94.0 3.6 m s = 26.1 m s , ac = v2 R = 26.1 m s 2 220 m = 3.1 m s2 L’accelerazione totale sarà la somma vettoriale delle due atot = p 3.13 + 9.82 = 10.3 m s2 La forza esercitata contro il sedile sarà F = matot = 85.0 kg · 10.3 m s2 = 874 N Esercizio 70. Una pietra legata all’estremità di una corda è fatta girare in un piano verticale su una circonferenza di raggio R. Trovare la velocità critica al di sotto della quale la corda si allenterebbe raggiungendo il punto più elevato.
Soluzione. La forza centrifuga “avvertita” dalla pietra deve equilibrare il suo peso mg.
g = ac = v2 R da cui v = p gR Esercizio 71. Un filo può reggere senza rompersi una tensione massima di 4.0 kg. Se si attacca a un’estremità una pietra di 3.6 kg e, trattenendo il filo all’altro estremo, si fa girare la pietra descrivendo un cerchio verticale di 1.20 m di raggio, aumentando gradatamente la velocità di rotazione fino a che il filo si strappa. Determinare il punto dell’orbita in cui si trova la pietra al momento della rottura e la velocità della pietra in quel momento.
Soluzione. Nel momento in cui il filo si rompe la pietra si trova nel punto più basso della sua traiettoria, dove l’accelerazione di gravità si somma a quella centripeta. Per calcolare la velocità basta osservare che, quando la corda si rompe, vuol dire che si supera il limite massimo di trazione. Se la corda regge la tensione corrispondente a un corpo di massa 4.0 kg l’accelerazione centripeta fornirà l’incremento di peso corrispondente alla massa aggiuntiva di 4.0 −3.6 = 0.4 kg. La forza centripeta sarà pertanto mac = mv2 r = 0.4 · 9.8 N da cui v = s .4 · 9.8 N · 1.20 m 3.6 kg = 1.14 m s 34 Esercizio 72. Un campo magnetico esercita su una particella carica in moto una forza perpendicolare alla direzione del moto. Un elettrone in un tale campo è soggetto ad una accelerazione radiale di 3, 0 · 1014 m s2 . Trovare la sua velocità se il raggio della sua traiettoria curva è di 0, 15 m.
Soluzione. La relazione che lega queste grandezze è a = v2 R per cui v = √ aR = p 3, 0 · 1014 × 0, 15 = 6, 7 · 106 m s Esercizio 73. Trovare il valore dell’accelerazione centripeta di una particella che si trova sull’orlo di una affettatrice, ruotante con una frequenza di 1200 giri min e con un diametro di 0, 3 m.
Soluzione. La ruota dell’affettatrice ha una frequenza di 1200 60 = 20 Hz, cioè di 20 giri al secondo e il suo periodo (cioè il tempo per una rotazione) è T = 1 f = 0, 05 s. La circonferenza che caratterizza tale affettatrice è di crf = 0, 3π = 0, 9 m. La velocità, essendo costante in modulo, sarà v = ∆s ∆t = 0, 94 m 0, 05 s = 19 m s e l’accelerazione sarà a = v2 R = 360 0, 15 = 2400 m s2 Esercizio 74. Un satellite terrestre si muove su un’orbita circolare a 630 km sopra la superficie della terra. Il suo periodo è di 98 minuti. Determinare da questi dati l’accelerazione di gravità sull’orbita.
Soluzione. Troviamo la velocità con la quale viene percorsa l’orbita v = 2π (Rt + h) T = 2π 6, 36 · 106 + 6, 30 · 105 m 98 × 60 s = 7469 m s per cui a = g = v2 Rt + h = 74692 (6, 36 · 106 + 6, 30 · 105) = 7, 98 m s2 Esercizio 75. Le coordinate di un punto mobile sono x = 2 sin ωt, y = 2 cos ωt. Trovare l’equazione carte-siana della traiettoria. Calcolare il modulo della velocità. Calcolare le componenti tangenziale e centripeta dell’accelerazione. Identificare il tipo di moto descritto dall’equazione.
Soluzione. Eleviamo le relazioni della coordinate e sommiamo i due quadrati ottenuti x2 + y2 = 4 sin2 ωt + 4 cos2 ωt = 4 ricordando la relazione fondamentale della goniometria, sin2 ωt + cos2 ωt = 1. L’equazione descrive nel piano cartesiano una circonferenza di centro nell’origine e raggio uguale a 2.
Troviamo la velocità attraverso le sue componenti vx = 2ω cos ωt vy = −2ω sin ωt avremo v = q v2 x + v2 y = p 4ω2 cos2 ωt + 4ω2 sin2 ωt = √ 4ω2 = 2ω ciò indica che la velocità non dipende più dal tempo e l’accelerazione tangenziale sarà nulla. L’accelerazione centripeta è ac = v2 r = 4ω2 2 = 2ω2 35 Esercizio 76. Una particella si muove in un piano secondo le seguenti leggi orarie x = R sin ωt + ωRt y = R cos ωt + R dove ω e R sono costanti. La curva descritta, chiamata cicloide, è la traiettoria di un punto sul bordo di una ruota che rotola senza scivolare lungo l’asse x. Calcolare la velocità istantanea e l’accelerazione della particella in corrispondenza al massimo e al minimo valore di y.
Soluzione. La funzione y (t) è massima quando cos ωt = 1, cioè quando ωt = 0◦, per cui ymax = 2R; La funzione y (t) è minima quando cos ωt = 0, cioè quando ωt = 90◦, per cui ymax = R.
Calcoliamo il modulo della velocità x′ = Rω cos ωt + Rω y′ = −Rω sin ωt eleviamo al quadrato e sommiamo v p x ′2 + y ′2 = p R2ω2 cos2 ωt + R2ω2 + 2Rω cos ωt + R2ω2 sin2 ωt e ricordando la proprietà fondamentale della goniometria (sin2 α + cos2 α = 1), si ha v = √ 2Rω √ 1 + cos ωt per cui per ymax avremo vmax = 2Rω e per ymin, vmin = √ 2Rω Calcoliamo ora l’accelerazione x” = −Rω2 sin ωt y” = −Rω2 cos ωt elevando al quadrato e sommando avremo a = p x”2 + y”2 = √ R2ω4 = Rω2 che è indipendente da t.
Esercizio 77. Una persona si arrampica in 90 s su una scala mobile ferma. Mentre è fermo sulla stessa scala mobile, ora in movimento, è trasportato in alto in 50 s. Quanto tempo gli ci vuole per arrampicarsi sulla scala mobile in funzione?
Soluzione. La lunghezza della scala mobile non è determinante essendo in ogni caso sempre lo stesso; la indichiamo con l. Nel primo caso, scala mobile ferma e persona in movimento abbiamo l t = 90 s vsc = 0 vpe = l 90 nel secondo caso l t = 60 s vsc = l 60 vpe = 0 nel terzo caso sia la scala mobile che la persona sono in movimento nella stessa direzione e nello stesso verso, per cui la velocità relativa è data vsc + vpe = l 60 + l 90 = 5l 180 = l 36 Il tempo necessario per salire sarà quindi pari a 36 s. |
1349 | https://www.woodwellclimate.org/review-of-permafrost-science-in-ipccs-ar6-wg2/ | menu
Review of permafrost science in IPCC’s AR6 WG2
Policy Brief
by
Natalie Baillargeon
Susan M. Natali
Rachael Treharne
URL copied!
Exposed permafrost thaw in northeastern Siberia. / photo by Sue Natali
Download as PDF
Introduction
In February 2022, the second installment of the International Panel on Climate Change (IPCC)’s Sixth Assessment Working Group 2 Report (AR6 WG2) was released; it focuses on the latest findings on climate change impacts and associated vulnerabilities and adaptation needs.
The Arctic is experiencing the most dramatic warming on the globe and is already facing the impacts of climate change. Much of the Arctic is underlain by permafrost, which is ground that has been frozen for at least two consecutive years. Permafrost stores vast amounts of carbon, almost twice as much as in the atmosphere. As discussed in AR6 WG2, thawing permafrost is a cascading threat that significantly increases vulnerabilities and risks for Arctic nations and Indigenous Peoples. Below, we examine the permafrost science and the impacts of thawing permafrost in the AR6 WG2 report.
Research area
Arctic
Land subsidence caused by permafrost thaw in Nunapitchuk, Alaska.
Review
Impacts of permafrost thaw
Wildfires will further contribute to permafrost thaw and carbon emissions.
Thawing permafrost could increase vulnerabilities of Arctic communities and Indigenous Peoples by:
Changing hydrology (including both local wetting and drying) and increasing water insecurity;
Increasing the costs for maintenance and reconstruction of infrastructure;
Damaging or causing loss of cultural heritage sites and livelihoods;
Threatening economic activities, including food supplies and supply chains; and
Causing migration and/or forced relocation due to decreased standard of living or deterioration of land.
Other risks from thawing permafrost include gastrointestinal illnesses due to the reduced quality of surface waters, previously frozen disease (such as anthrax) outbreaks, and mercury entering food chains and water supplies.
Already, some areas of Arctic permafrost have shifted from carbon sinks to carbon sources due to thaw.
Future risks of permafrost thaw
Arctic ecosystems are among the “unique and threatened systems” that face high levels of risk without adaptation, with risk levels increasing further if temperatures reach 1.5°C.
Thawing permafrost is a self-reinforcing feedback as the resulting emissions will drive further warming. Overshooting warming targets substantially increases carbon emissions from permafrost thaw. While uncertainties remain on the timing and magnitude, thawing permafrost has the potential to become a tipping point. As a tipping point, thawing permafrost could transform socio-ecological systems on a permanent and irreversible timescale.
If warming reaches 2°C, the AR6 WG2 reports that it could reduce permafrost extent by 5-20%. Limiting warming to 1.5°C would prevent some losses. As stated, these projections may be underestimated as they do not account for deeper permafrost layers and abrupt permafrost collapse.
There may be adaptation limits to permafrost thaw, especially for coastal communities, resulting in permanent loss and damage of Arctic lands.
Conclusion
As the IPCC stated in the Summary for Policymakers, “the cumulative scientific evidence is unequivocal: climate change is a threat to human well-being and planetary health. Any further delay in concerted anticipatory global action on adaptation and mitigation will miss a brief and rapidly closing window of opportunity to secure a liveable and sustainable future for all.” This is especially true for Arctic nations and Indigenous Peoples, who are facing growing threats from climate change. Support for adaptation efforts in the Arctic should be prioritized alongside mitigation efforts. For more information on permafrost science in AR6 WG1, please read our fact sheet.
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Elementary Number Theory, 7th Edition
Textbook Solutions
Elementary Number Theory, 7th Edition Solutions
David Burton
2007
ISBN: 9780073383149
Elementary Number Theory Answers
Elementary Number Theory answers based on a simplistic approach get students equipped with all the information needed to solve the concepts covered in the textbook. It consists of multiple topics, including preliminaries, divisibility theory in the integers, Fermat's theorem, Euler's generalization of Fermat's theorem, quadratic reciprocity law, Fibonacci numbers, etc.
Chapter 1: Preliminaries
Problems 1.1 6
Problem 1 6
Problem 2 7
Problem 3 7
Problem 4 7
Problem 5 7
Problem 6 7
Problem 7 7
Problem 8 7
Problem 9 7
Problem 10 7
Problem 11 7
Problem 12 7
Problem 13 8
Problem 14 8
Problems 1.2 10
Problem 1 10
Problem 2 11
Problem 3 11
Problem 4 11
Problem 5 12
Problem 6 12
Problem 7 12
Problem 8 12
Problem 9 12
Problem 10 12
Chapter 2: Divisibility Theory in the Integers
Problems 2.1 15
Problem 1 15
Problem 2 15
Problem 3 15
Problem 4 16
Problem 5 16
Problem 6 16
Problem 7 16
Problem 8 16
Problem 9 16
Problem 10 16
Problem 11 16
Problems 2.2 19
Problem 1 19
Problem 2 19
Problem 3 19
Problem 4 19
Problem 5 19
Problem 6 19
Problem 7 19
Problem 8 19
Problem 9 19
Problem 10 19
Problem 11 19
Problems 2.3 24
Problem 1 24
Problem 2 24
Problem 3 24
Problem 4 24
Problem 5 24
Problem 6 25
Problem 7 25
Problem 8 25
Problem 9 25
Problem 10 25
Problem 11 25
Problem 12 25
Problem 13 25
Problem 14 25
Problem 15 25
Problem 16 25
Problem 17 25
Problem 18 25
Problem 19 25
Problem 20 25
Problem 21 26
Problem 22 26
Problem 23 26
Problems 2.4 31
Problem 1 31
Problem 2 31
Problem 3 31
Problem 4 31
Problem 5 31
Problem 6 31
Problem 7 31
Problem 8 31
Problem 9 31
Problem 10 31
Problem 11 31
Problem 12 32
Problems 2.5 37
Problem 1 37
Problem 2 37
Problem 3 37
Problem 4 38
Problem 5 38
Problem 6 38
Problem 7 38
Problem 8 38
Chapter 3: Primes and Their Distribution
Problems 3.1 43
Problem 1 43
Problem 2 43
Problem 3 43
Problem 4 43
Problem 5 43
Problem 6 43
Problem 7 43
Problem 8 43
Problem 9 43
Problem 10 43
Problem 11 43
Problem 12 43
Problem 13 43
Problem 14 44
Problem 15 44
Problem 16 44
Problem 17 44
Problem 18 44
Problem 19 44
Problems 3.2 49
Problem 1 49
Problem 2 49
Problem 3 49
Problem 4 49
Problem 5 49
Problem 6 49
Problem 7 49
Problem 8 49
Problem 9 50
Problem 10 50
Problem 11 50
Problem 12 50
Problem 13 50
Problem 14 50
Problems 3.3 58
Problem 1 58
Problem 2 58
Problem 3 58
Problem 4 58
Problem 5 58
Problem 6 58
Problem 7 58
Problem 8 58
Problem 9 58
Problem 10 58
Problem 11 58
Problem 12 58
Problem 13 58
Problem 14 58
Problem 15 58
Problem 16 58
Problem 17 59
Problem 18 59
Problem 19 59
Problem 20 59
Problem 21 59
Problem 22 59
Problem 23 59
Problem 24 59
Problem 25 59
Problem 26 59
Problem 27 60
Problem 28 60
Chapter 4: The Theory of Congruences
Problems 4.2 67
Problem 1 67
Problem 2 67
Problem 3 67
Problem 4 67
Problem 5 67
Problem 6 68
Problem 7 68
Problem 8 68
Problem 9 68
Problem 10 68
Problem 11 68
Problem 12 68
Problem 13 68
Problem 14 68
Problem 15 68
Problem 16 68
Problem 17 68
Problem 18 68
Problems 4.3 73
Problem 1 73
Problem 2 73
Problem 3 73
Problem 4 73
Problem 5 73
Problem 6 74
Problem 7 74
Problem 8 74
Problem 9 74
Problem 10 74
Problem 11 74
Problem 12 74
Problem 13 74
Problem 14 74
Problem 15 74
Problem 16 74
Problem 17 74
Problem 18 74
Problem 19 74
Problem 20 75
Problem 21 75
Problem 22 75
Problem 23 75
Problem 24 75
Problem 25 75
Problem 26 75
Problem 27 75
Problem 28 75
Problems 4.4 82
Problem 1 82
Problem 2 82
Problem 3 82
Problem 4 82
Problem 5 82
Problem 6 82
Problem 7 83
Problem 8 83
Problem 9 83
Problem 10 83
Problem 11 83
Problem 12 83
Problem 13 83
Problem 14 83
Problem 15 83
Problem 16 83
Problem 17 83
Problem 18 83
Problem 19 84
Problem 20 84
Chapter 5: Fermat's Theorem
Problems 5.2 92
Problem 1 92
Problem 2 92
Problem 3 92
Problem 4 92
Problem 5 92
Problem 6 92
Problem 7 92
Problem 8 92
Problem 9 92
Problem 10 92
Problem 11 92
Problem 12 93
Problem 13 93
Problem 14 93
Problem 15 93
Problem 16 93
Problem 17 93
Problem 18 93
Problem 19 93
Problem 20 93
Problem 21 93
Problems 5.3 96
Problem 1 96
Problem 2 96
Problem 3 96
Problem 4 96
Problem 5 96
Problem 6 96
Problem 7 97
Problem 8 97
Problem 9 97
Problem 10 97
Problem 11 97
Problem 12 97
Problem 13 97
Problem 14 97
Problem 15 97
Problem 16 97
Problem 17 97
Problem 18 97
Problems 5.4 102
Problem 1 102
Problem 2 102
Problem 3 102
Problem 4 102
Problem 5 102
Problem 6 102
Problem 7 102
Problem 8 102
Chapter 6: Number-Theoretic Functions
Problems 6.1 110
Problem 1 110
Problem 2 110
Problem 3 110
Problem 4 110
Problem 5 110
Problem 6 110
Problem 7 110
Problem 8 110
Problem 9 110
Problem 10 110
Problem 11 111
Problem 12 111
Problem 13 111
Problem 14 111
Problem 15 111
Problem 16 111
Problem 17 111
Problem 18 111
Problem 19 111
Problem 20 111
Problem 21 111
Problem 22 111
Problem 23 112
Problems 6.2 116
Problem 1 116
Problem 2 116
Problem 3 116
Problem 4 116
Problem 5 116
Problem 6 116
Problem 7 116
Problem 8 117
Problems 6.3 121
Problem 1 121
Problem 2 121
Problem 3 121
Problem 4 121
Problem 5 121
Problem 6 121
Problem 7 122
Problem 8 122
Problem 9 122
Problem 10 122
Problem 11 122
Problem 12 122
Problem 13 122
Problems 6.4 126
Problem 1 126
Problem 2 126
Problem 3 126
Problem 4 127
Problem 5 127
Problem 6 127
Chapter 7: Euler's Generalization of Fermat's Theorem
Problems 7.2 135
Problem 1 135
Problem 2 135
Problem 3 135
Problem 4 135
Problem 5 135
Problem 6 135
Problem 7 135
Problem 8 135
Problem 9 135
Problem 10 135
Problem 11 135
Problem 12 135
Problem 13 136
Problem 14 136
Problem 15 136
Problem 16 136
Problem 17 136
Problem 18 136
Problem 19 136
Problem 20 136
Problem 21 136
Problems 7.3 140
Problem 1 140
Problem 2 140
Problem 3 140
Problem 4 140
Problem 5 141
Problem 6 141
Problem 7 141
Problem 8 141
Problem 9 141
Problem 10 141
Problem 11 141
Problem 12 141
Problem 13 141
Problems 7.4 145
Problem 1 145
Problem 2 145
Problem 3 145
Problem 4 145
Problem 5 145
Problem 6 145
Problem 7 145
Problem 8 145
Problem 9 145
Problem 10 146
Problem 11 146
Problem 12 146
Problem 13 146
Problem 14 146
Problem 15 146
Problem 16 146
Chapter 8: Primitive Roots and Indices
Problems 8.1 151
Problem 1 151
Problem 2 151
Problem 3 151
Problem 4 151
Problem 5 151
Problem 6 151
Problem 7 152
Problem 8 152
Problem 9 152
Problem 10 152
Problem 11 152
Problem 12 152
Problem 13 152
Problems 8.2 157
Problem 1 157
Problem 2 157
Problem 3 158
Problem 4 158
Problem 5 158
Problem 6 158
Problem 7 158
Problem 8 158
Problem 9 158
Problem 10 158
Problem 11 158
Problem 12 158
Problems 8.3 162
Problem 1 162
Problem 2 162
Problem 3 162
Problem 4 162
Problem 5 162
Problem 6 162
Problem 7 162
Problem 8 162
Problem 9 163
Problem 10 163
Problem 11 163
Problems 8.4 167
Problem 1 167
Problem 2 167
Problem 3 167
Problem 4 167
Problem 5 167
Problem 6 167
Problem 7 167
Problem 8 167
Problem 9 167
Problem 10 167
Problem 11 167
Problem 12 168
Problem 13 168
Problem 14 168
Problem 15 168
Problem 16 168
Problem 17 168
Chapter 9: The Quadratic Reciprocity Law
Problems 9.1 173
Problem 1 173
Problem 2 174
Problem 3 174
Problem 4 174
Problem 5 174
Problem 6 174
Problem 7 174
Problem 8 174
Problem 9 174
Problem 10 174
Problem 11 174
Problem 12 174
Problem 13 174
Problems 9.2 184
Problem 1 184
Problem 2 184
Problem 3 184
Problem 4 184
Problem 5 184
Problem 6 184
Problem 7 184
Problem 8 185
Problem 9 185
Problem 10 185
Problem 11 185
Problem 12 185
Problem 13 185
Problem 14 185
Problem 15 185
Problem 16 185
Problem 17 185
Problem 18 185
Problems 9.3 190
Problem 1 190
Problem 2 190
Problem 3 190
Problem 4 190
Problem 5 190
Problem 6 191
Problem 7 191
Problem 8 191
Problem 9 191
Problem 10 191
Problem 11 191
Problem 12 191
Problem 13 191
Problem 14 191
Problem 15 191
Problem 16 191
Problem 17 192
Problem 18 192
Problem 19 192
Problem 20 192
Problems 9.4 195
Problem 1 195
Problem 2 195
Problem 3 195
Problem 4 195
Problem 5 195
Problem 6 195
Problem 7 195
Problem 8 195
Problem 9 196
Problem 10 196
Chapter 10: Introduction to Cryptography
Problems 10.1 207
Problem 1 207
Problem 2 207
Problem 3 207
Problem 4 208
Problem 5 208
Problem 6 208
Problem 7 208
Problem 8 208
Problem 9 208
Problem 10 208
Problem 11 209
Problem 12 209
Problem 13 209
Problem 14 209
Problem 15 209
Problems 10.2 214
Problem 1 214
Problem 2 214
Problem 3 214
Problem 4 214
Problem 5 214
Problem 6 214
Problem 7 214
Problems 10.3 218
Problem 1 218
Problem 2 218
Problem 3 218
Problem 4 218
Problem 5 218
Chapter 11: Numbers of Special Form
Problems 11.2 226
Problem 1 226
Problem 2 226
Problem 3 226
Problem 4 226
Problem 5 226
Problem 6 226
Problem 7 226
Problem 8 226
Problem 9 226
Problem 10 226
Problem 11 226
Problem 12 226
Problem 13 226
Problem 14 227
Problem 15 227
Problem 16 227
Problem 17 227
Problem 18 227
Problem 19 227
Problem 20 227
Problems 11.3 236
Problem 1 236
Problem 2 236
Problem 3 236
Problem 4 236
Problem 5 236
Problem 6 236
Problem 7 236
Problem 8 236
Problem 9 236
Problem 10 236
Problem 11 237
Problem 12 237
Problem 13 237
Problem 14 237
Problem 15 237
Problem 16 237
Problems 11.4 243
Problem 1 243
Problem 2 243
Problem 3 243
Problem 4 243
Problem 5 243
Problem 6 243
Problem 7 243
Problem 8 243
Problem 9 243
Problem 10 244
Problem 11 244
Problem 12 244
Problem 13 244
Problem 14 244
Problem 15 244
Problem 16 244
Problem 17 244
Problem 18 244
Problem 19 244
Problem 20 244
Chapter 12: Certain Nonlinear Diophantine Equations
Problems 12.1 251
Problem 1 251
Problem 2 251
Problem 3 251
Problem 4 251
Problem 5 251
Problem 6 251
Problem 7 251
Problem 8 251
Problem 9 251
Problem 10 251
Problem 11 251
Problem 12 251
Problem 13 252
Problems 12.2 259
Problem 1 259
Problem 2 259
Problem 3 259
Problem 4 259
Problem 5 259
Problem 6 259
Problem 7 259
Problem 8 259
Problem 9 260
Problem 10 260
Problem 11 260
Problem 12 260
Chapter 13: Representation of Integers as Sums of Squares
Problems 13.2 270
Problem 1 270
Problem 2 270
Problem 3 270
Problem 4 271
Problem 5 271
Problem 6 271
Problem 7 271
Problem 8 271
Problem 9 271
Problem 10 271
Problem 11 271
Problem 12 271
Problem 13 271
Problem 14 271
Problem 15 272
Problem 16 272
Problem 17 272
Problem 18 272
Problems 13.3 280
Problem 1 280
Problem 2 280
Problem 3 280
Problem 4 280
Problem 5 280
Problem 6 280
Problem 7 280
Problem 8 280
Problem 9 280
Problem 10 280
Problem 11 281
Problem 12 281
Problem 13 281
Problem 14 281
Problem 15 281
Problem 16 281
Problem 17 281
Problem 18 281
Chapter 14: Fibonacci Numbers
Problems 14.2 291
Problem 1 291
Problem 2 291
Problem 3 291
Problem 4 291
Problem 5 291
Problem 6 291
Problem 7 291
Problem 8 291
Problem 9 291
Problem 10 291
Problem 11 291
Problem 12 291
Problem 13 291
Problem 14 291
Problem 15 291
Problem 16 291
Problem 17 291
Problem 18 292
Problem 19 292
Problem 20 292
Problems 14.3 299
Problem 1 299
Problem 2 299
Problem 3 300
Problem 4 300
Problem 5 300
Problem 6 300
Problem 7 300
Problem 8 300
Problem 9 300
Problem 10 300
Problem 11 300
Problem 12 300
Problem 13 300
Problem 14 301
Problem 15 301
Problem 16 301
Problem 17 301
Problem 18 301
Problem 19 301
Problem 20 301
Problem 21 301
Problem 22 301
Problem 23 302
Problem 24 302
Problem 25 302
Problem 26 302
Problem 27 302
Problem 28 302
Chapter 15: Continued Fractions
Problems 15.2 318
Problem 1 318
Problem 2 318
Problem 3 318
Problem 4 318
Problem 5 318
Problem 6 319
Problem 7 319
Problem 8 319
Problem 9 319
Problem 10 319
Problem 11 319
Problem 12 319
Problems 15.3 332
Problem 1 332
Problem 2 333
Problem 3 333
Problem 4 333
Problem 5 333
Problem 6 333
Problem 7 333
Problem 8 333
Problem 9 333
Problem 10 333
Problem 11 333
Problem 12 334
Problem 13 334
Problems 15.4 337
Problem 1 337
Problem 2 337
Problem 3 337
Problem 4 337
Problem 5 337
Problem 6 337
Problem 7 337
Problems 15.5 350
Problem 1 350
Problem 2 350
Problem 3 351
Problem 4 351
Problem 5 351
Problem 6 351
Problem 7 351
Problem 8 351
Problem 9 351
Problem 10 351
Problem 11 351
Problem 12 351
Problem 13 351
Problem 14 352
Problem 15 352
Problem 16 352
Chapter 16: Some Modern Developments
Problems 16.1 356
Problem 1 356
Problem 2 356
Problem 3 356
Problem 4 356
Problem 5 356
Problems 16.2 370
Problem 1 370
Problem 2 370
Problem 3 370
Problem 4 370
Problem 5 370
Problem 6 370
Problem 7 371
Problem 8 371
Problem 9 371
Problem 10 371
Problem 11 371
Problem 12 371
Problem 13 371
Problems 16.3 374
Problem 1 374
Problem 2 374
Problem 3 375
Problem 4 375
Problem 5 375
Miscellaneous Problems
Miscellaneous Problems 384
Problem 1 384
Problem 2 384
Problem 3 384
Problem 4 384
Problem 5 384
Problem 6 384
Problem 7 384
Problem 8 384
Problem 9 384
Problem 10 384
Problem 11 384
Problem 12 385
Problem 13 385
Problem 14 385
Problem 15 385
Problem 16 385
Problem 17 385
Problem 18 385
Problem 19 385
Problem 20 385
Problem 21 385
Problem 22 385
Problem 23 385
Problem 24 385
Problem 25 385
Problem 26 385
Problem 27 385
Problem 28 385
Problem 29 385
Problem 30 385
Problem 31 385
Problem 32 385
Problem 33 385
Problem 34 385
Problem 35 385
Problem 36 386
Problem 37 386
Problem 38 386
Problem 39 386
Problem 40 386
Problem 41 386
Problem 42 386
Problem 43 386
Problem 44 386
Problem 45 386
Problem 46 386
Problem 47 386
Problem 48 386
Problem 49 386
Problem 50 386
Elementary Number Theory Chapters list with step by step answer key
Chapter 1: Preliminaries Answers
The first chapter of Elementary Number Theory textbook solutions covers several exercises related to the initial level topics, including mathematical induction that is a mathematical proof technique, and the binomial theorem that describes the algebraic expansion of powers of a binomial.
Chapter 2: Divisibility Theory in the Integers Answers
In this chapter, students learn early number theory, the division algorithm, the greatest common divisor, the Euclidean Algorithm, and the diophantine equation ax + by = c, along with questions based on it to evaluate conceptual clarity.
Chapter 3: Primes and Their Distribution Answers
Students will find important topics like the fundamental theorem of arithmetic, the sieve of Eratosthenes, and the Goldbach conjecture in this chapter, along with their properties and examples.
Chapter 4: The Theory of Congruences Answers
In this chapter of Elementary Number Theory exercise solutions, students learn the basic properties of congruence, binary and decimal representations of integers, and the relation between the linear congruence and the Chinese remainder theorem.
Chapter 5: Fermat's Theorem Answers
This chapter elaborates on multiple theorems with the help of examples and solutions to prove in mathematical terms, including Fermat's little theorem and pseudoprimes, Wilson's theorem, and the Fermat-Kraitchik factorization method based on the representation of an odd integer as the difference of two squares.
Chapter 6: Number-Theoretic Functions Answers
In this chapter, students get conceptual clarity on topics like the sum and number of divisors, the Mobius inversion formula, the greatest integer function, and an application to the calendar.
Chapter 7: Euler's Generalization of Fermat's Theorem Answers
This chapter of Elementary Number Theory exercise solutions includes multiple concepts like Euler's phi-function, Euler's theorem, and some properties of the phi-function in the context of Fermat's theorem that deals with the powers of integers modulo positive integers.
Chapter 8: Primitive Roots and Indices Answers
This chapter delves into concepts like the order of an integer modulo n, primitive roots for primes, composite numbers having primitive roots, the theory of indices, and multiple questions based on the concepts to bring better understanding and clarity.
Chapter 9: The Quadratic Reciprocity Law Answers
In this chapter, students deal with the quadratic reciprocity laws, including Euler's criterion, the Legendre symbol and its properties, quadratic reciprocity, and the quadratic congruences with composite moduli.
Chapter 10: Introduction to Cryptography Answers
This chapter introduces students to the topic of cryptography as the science of using mathematics to hide data behind encryption and covers Caesar cipher to public-key cryptography, the knapsack cryptosystem, and an application of primitive roots to cryptography.
Chapter 11: Numbers of Special Form Answers
This chapter of Elementary Number Theory solutions of numbers of special form delves into concepts like perfect numbers, Mersenne prime and amicable numbers, and Fermat numbers, including definitions, proofs, remarks on theorems, and examples.
Chapter 12: Certain Nonlinear Diophantine Equations Answers
In this chapter of Elementary Number Theory textbook solutions, students get insight on nonlinear diophantine equations like x² + y² = z² and Fermat's last theorem, with the help of examples and exercises to solve for practice, understanding, and further conceptual clarity.
Chapter 13: Representation of Integers as Sums of Squares Answers
In this solution of Elementary Number Theory by Burton, students focus on representing integers as sums of squares, including sums of two squares and more than two squares, with the help of definitions, propositions, theorems, and examples.
Chapter 14: Fibonacci Numbers Answers
This chapter gets students accustomed to the Fibonacci sequence and certain identities involving Fibonacci numbers with the help of examples and practice exercises.
Chapter 15: Continued Fractions Answers
This chapter elaborates on the concepts like finite continued fractions, infinite continued fractions, Farey fractions, Pell's equation to enhance the understanding of classic number theory.
Chapter 16: Some Modern Developments Answers
This chapter covers the latest developments in number theory, including primality testing and factorization, and an application to factoring: remote coin flipping, along with a focus on examples.
Miscellaneous Problems Answers
This chapter provides students with practice problems related to all the chapters covered in the textbook, intending to simulate the challenges they might face in exams.
Free Textbook Solutions
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Elementary Number Theory answers provide students with resources to understand mathematical concepts and their applications efficiently. Subject experts verify the solutions to develop problem-solving skills in students. The book also gives access to plenty of questions and well-structured exercises that build skills and mathematical techniques to add value to their course.
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Are the Elementary Number Theory textbook solutions updated as per the latest developments?
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Solution of Elementary Number Theory by Burton consists of all the newly developed concepts which provide students the essential tools to solve questions put forward to them with ease. The version is in line with the solutions provided for the edition published in 2021. Brainly sees that all the textbooks covered are timely updated to provide the best online guidance possible.
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The book has thorough concepts and exercises to prepare students for college-level exams, quizzes, and similar exams. The online resources consist of stepwise solutions of all the chapters that the book features to provide an enhanced learning experience. Even students searching for help concerning their homework can rest assured that they need to look nowhere else.
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1351 | https://unina2.on-line.it/sebina/repository/catalogazione/documenti/Carleson,%20Gamelin%20-%20Complex%20Dynamics.pdf | Lennart Carleson Theodore W. Gamelin Complex Dynamics With 28 Figures , Springer Lennart Carleson Department of Mathematics Royal Institute of Technology S-loo 44 Stockholm Theodore W. Gamelin Department of Mathematics University of California Sweden Los Angeles, CA 90024-1555 USA and Department of Mathematics University of California Los Angeles, CA 90024-1555 USA Editorial Board (North America): S. Axler Department of Mathematics Michigan State University East Lansing, MI 48824 USA F. W. Gehring Department of Mathematics Universtiy of Michigan Ann Arbor, MI 48109 USA P.R. Halmos Department of Mathematics Santa Clara University Santa Clara, CA 95053 USA On the cover: A filled-in Julia set with parabolic fixed point, attracting petals, and repelling arms. Mathematics Subject Classification (1991): 3OCxx, S8Fxx Library of Congress Cataloging-in-Publication Data Carleson, Lennart. Complex dynamics/by L. Carleson and T. Gamelin. p. cm. - (Universitext) Includes bibliographical references and index. ISBN-13: 978-0-387-97942-7 001: 10.1007/978-1-4612-4364-9 e-ISBN-13: 978-1-4612-4364-9 I. Functions of complex variables. 2. Mappings (Mathematics) 3. Fixed point theory. I. Title. QA33 1.7 . C37 1993 SIS'.9-dc20 92-32457 Printed on acid-free paper. © 1993 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereaf-ter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone. Production managed by Francine McNeill; manufacturing supervised by Vincent Scelta. Photocomposed copy prepared from the author's .A,A.1S-1EX file. 9 8 7 6 5 4 3 2 (Corrected second printing, 1995) Contents Preface v I. Conformal and Quasiconformal Mappings 1 1. Some Estimates on Conformal Mappings . 1 2. The Riemann Mapping. 5 3. Montel's Theorem .... 9 4. The Hyperbolic Metric . . 11 5. Quasiconformal Mappings 15 6. Singular Integral Operators 17 7. The Beltrami Equation. . . 19 II. Fixed Points and Conjugations 27 1. Classification of Fixed Points 27 2. Attracting Fixed Points ... 31 3. Repelling Fixed Points .... 32 4. Superattracting Fixed Points 33 5. Rationally Neutral Fixed Points. 35 6. lrrat ionally Neutral Fixed Points 41 7. Homeomorphisms of the Circle 47 viii Contents III. IV. V. VI. VII. Basic Rational Iteration 1. The Julia Set . . . . . . . 2. Counting Cycles .......... . 3. Density of Repelling Periodic Points 4. Polynomials . . . . . . . . . . . . . . Classification of Periodic Components 1. Sullivan's Theorem ..... 2. The Classification Theorem 3. The Wolff-Denjoy Theorem Critical Points and Expanding Maps 1. Siegel Disks . . . 2. Hyperbolicity . . . . . . . . . 3. Subhyperbolicity . . . . . . . 4. Locally Connected Julia Sets Applications of Quasiconformal Mappings 1. Polynomial-like Mappings 2. Quasicircles ...... . 3. Herman Rings. . . . . . . 4. Counting Herman Rings. 5. A Quasiconformal Surgical Procedure Local Geometry of the Fatou Set 1. Invariant Spirals 2. Repelling Arms 3. John Domains. . VIII. Quadratic Polynomials 1. The Mandelbrot Set . . 2. The Hyperbolic Components of M 3. Green's Function of Jc . . . . . . . 4. Green's Function of M . . . . . . . 5. External Rays with Rational Angles 6. Misiurewicz Points 7. Parabolic Points .......... . 53 53 58 63 65 69 69 74 79 81 81 89 91 93 99 99 .101 .103 .105 .106 109 .109 .113 .117 123 . 123 .133 .136 .139 .142 .148 . 153 Contents ix Epilogue 161 References 163 Index 171 Symbol Index 175 |
1352 | https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_7?srsltid=AfmBOoqn9oBPlQThoDtp-l91qClu-CZfORL8Dh0QRafwhj8noN3bNjEh | Art of Problem Solving
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2000 AIME I Problems/Problem 7
Contents
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5
7 Solution 6
8 Solution 7
9 See also
Problem
Suppose that and are three positive numbers that satisfy the equations and Then where and are relatively prime positive integers. Find .
Solution 1
We can rewrite as .
Substituting into one of the given equations, we have
We can substitute back into to obtain
We can then substitute once again to get Thus, , so .
Solution 2
Let .
Thus . So .
Solution 3
Since , so . Also, by the second equation. Substitution gives , , and , so the answer is 4+1 which is equal to .
Solution 4
(Hybrid between 1/2)
Because and . Substituting and factoring, we get , , and . Multiplying them all together, we get, , but is , and by the Identity property of multiplication, we can take it out. So, in the end, we get . And, we can expand this to get , and if we make a substitution for , and rearrange the terms, we get This will be important.
Now, lets add the 3 equations , and . We use the expand the Left hand sides, then, we add the equations to get Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus We move all constant terms to the right, and all linear terms to the left, to get , so which gives an answer of
-AlexLikeMath
Solution 5
Get rid of the denominators in the second and third equations to get and . Then, since , we have and . Then, since we know that , we can subtract these two equations to get that . The result follows that and , so , and the requested answer is
Solution 6
Rewrite the equations in terms of x.
becomes .
becomes
Now express in terms of x.
.
This evaluates to , giving us . We can now plug x into the other equations to get and .
Therefore, .
, and we are done. ~MC413551
Solution 7
First, we have been given the value of , so we should probably figure out a way to use that. Before that, we replace , because why deal with variables when you don't have to. We multiply these equations to get: . Plugging the value in for , we get: . Now, we need a way to get these other random terms out of there. We add the three equations to get . Plugging this back into the equation we found we get:
-jb2015007
See also
Erm, this is very similar to 2000 AMC 12 Q20 ackthually
2000 AIME I (Problems • Answer Key • Resources)
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1353 | https://www.reddit.com/r/learnmath/comments/1i17n1g/high_school_combinatorics_how_are_the_binomial/ | [high school(?) combinatorics] How are the binomial coefficient and exponents over 2 connected? : r/learnmath
Skip to main content[high school(?) combinatorics] How are the binomial coefficient and exponents over 2 connected? : r/learnmath
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Fancy_Card_3632
[high school(?) combinatorics] How are the binomial coefficient and exponents over 2 connected?
My question in short: can anyone explain why this here is true?
[;\sum_{k=0}^{n}\binom{n}{k}=2^{n};]
One question in our last math exam was "The school has a new lighting system consisting of 8 lamps that can all be turned on and off independently. How many different lighting combinations are possible?" (the exam wasn't in English, I might have mistranslated it a bit). While I wrote down [;2^{8}=256;], a classmate used (essentially) [;\sum_{k=0}^{8}\binom{8}{k}=256;].
After the exam we talked about it and agreed that both ways made sense in the given context and that the binomial coefficient and the exponent of 2 are in some way related. We then wrote down the generalization from above, which worked for all numbers we tried. We then wanted to find out why but weren't able to. To me it felt like I was missing something really obvious.
funnily enough, when I was reading a book yesterday I found the same expression as above with a comment saying that it "follows from the definition of the binomial coefficient" (again, the quote wasn't in English, I might have mistranslated it), confirming what we were thinking but sadly not explaining it.
I hope this is the right sub to ask this and someone can help me out, I've spent way too much time on this the last few days lol.
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1354 | https://www.youtube.com/watch?v=A-LjVDC4Mzc | Transforming Polynomial Functions
Ryne Roper
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Use this information to help you in your Algebra 2 class!
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transformation #polynomials #functions #RyneRoper
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Transcript:
[Music] hello everyone in this video we're going to talk about how to transform polom functions so the good news about transforming polinomial functions is that as you are watching this video I'm sure at some point either in earlier in Algebra 2 or earlier in Algebra 1 you have gone over these Transformations so hopefully in algebra one you started off by transforming linear functions and you learned what a translation reflection stretch and Shrink and then you transformed quadratic functions or exponentials or radicals or cubic functions right and they're all the same Transformations so that's the good news um the I guess there is no bad news but the news for today in this video is that we're just going to use these Transformations on polinomial functions so the Transformations that we see on the screen right now I have gone over in a previous video multiple videos actually uh but one video that has been um a popular video on my channel is how to transform linear functions and I'll link that the cards right now um it's a lengthy video because I go a little bit more in depth than I'm going to go in this video onto each of these Transformations because like I said that's kind of the first taste of Transformations is when you do it with linear functions and so that video go and check it out if you need some more help so let's start with the top just says a horizontal translation is going to shift the graph left or right so we're going to move our graph either to the left or to the right and the notation that we see right here is what it's going to look like so there's going to be um some addition or subtraction um happening directly with our X variable because X represents going left or right and we use the variable H for a horizontal translation and with horizontal we got to be careful it's kind of opposite of what it looks like so if we see x - 5 we are actually going to the right if we see x + 2 we are actually going to the left and the reason for that is our initial um notation is x - H right so if H is a positive number let's just say h is three well let's use the same number we did up here five so if H is five we just plug five in right there and it looks like x - 5 right well if H is -2 and we plug -2 in for H now we have x - -2 well we simplify that as x + 2 and that's why it kind of looks opposite of what we think it should look like um so usually we say oh minus we're going to go left plus uh addition we're going to go right but it's really the opposite with horizontal so that's really the big thing we need to remember with horizontal translations vertical translation we're going to shift the graph up or down we use the variable K and it is what it looks like so plus one we go up minus 4 we go down Reflections we could flip the graph over the x or the y- axis so if we are making our X opposite we will reflect over the opposite axis so that's the the Y AIS and remember F ofx here is like the same thing as y so if we make y opposite we will reflect over the x axis then we get into stretches and shrinks so horizontal stretch or Shrink is going to um stretch our graph away from or Shrink towards the Y AIS so that could be like a horizontal stretch or a horizontal shrink and in that case we there's going to be multiplication involved and we use the variable a and we're going to be multiplying directly with our X and just like horizontal translation was a little bit opposite of what we would think horizontal stretch and Shrink is opposite as well if we see a two right here um really what it is is a shrink by a factor of 1/2 if we see a 1/2 right here really it's a stretch by a factor of two and the reason why that's the case once again um is because we are stretching or shrinking by a factor of 1 over a okay so on our first example 2x to the 4th power then we would be um that two would become 1 over two so that's where our 1/2 comes in and so then 1/2 tells us it's a shrink okay if we have one divid 1/2 that becomes 1 2 which would be two right and that's why it's a stretch okay vertical stretch and strink much like vertical translation is what it looks like so we still have a here but now we are multiplying our outputs so if I have 8 x 4th it is a stretch by a factor of eight 1/4 shrink by a factor of 1/4 okay all right so let's do a couple examples example number one says we're going to describe the transformation of f ofx = x Cub represented by G ofx = x - 3 cubed - 1 and then graph each function so let's start off and let's identify our two Transformations that have happened so this x minus 3 in here we are adding or subtracting so in this case obviously we're subtracting but we are subtracting directly with the X so we know it's going to be a horizontal translation so now we just need to know and also we know it's three units so now we just need to think okay is it right or left well it says x minus 3 so remember maybe maybe we're thinking left but it's opposite of what it looks like because it's horizontal so we're going to translate three units to the right okay and then our ne1 right here tells us now okay that's subtraction again but it's not happening directly to X so it's a vertical translation and vertical does look like what it's supposed to or what we think it would so we're going to go down one unit okay and so now if we graph our parent function not our parent function but our original function we'll do that in our navy color so that's FX = X Cub so we know we'd have 0 Z 1 one negative 1 negative 1 and let's do we would we know 2 cubed would be eight so we could add that up here 2 CU would be eight uh we don't have uh room on our the bottom of our graph to put negative to negative 8 so we'll just do that for right now okay so now we know if we draw our graph kind of what the the shape of it should look like okay we're going down like that and so now in red we will graph our G of X function so let's take these four points that we had and we're going to translate them three units right and then one unit down so one two 3 and down one one 2 3 down one 1 2 3 down one and then 1 two 3 and down one so now we have our new function over here so our G of X function and that's what it look like so we had G of X there and we had F ofx right here there we go all right so that was a horizontal translation three units right and then a vertical translation one unit down so let's look at our last example um one example here but kind of two separate problems that we're going to do so it says describe the transformation of f represented by G and then graph each function so once again let's describe it so we have f ofx = x 4th and then it becomes G of x = 3x to 4th so in this case we see that we have our multiplication happening um notice the difference here on number two we have that multiplication 0.5 happening directly with X so that's going to be horizontal which tells us that on number one here this is going to be a vertical stretch by a factor of three okay so now we can go back and graph our f ofx = x to the 4th function so if we had 0 0 4th is 0 1 to 4th is 1 - 1 to 4th is negative 1 and so we're thinking oh it looks like a parabola well it kind of does but the bottom of our Parabola is a little more rounded out it's a little more flat at the bottom I guess you could say and so let's make it look a little bit like this okay so I'm going to kind of round out the bottom there a little bit all right so hopefully that does it justice okay so now in green we can graph a vertical stretch by a factor of three so our point on the origin is not going to change but our two points at 1 one and negative 1 one should move up by a factor of three so instead of being 1 one it's going to be 1 3 because 1 3 is 3 so we can move that point right there and also 13 okay so once again we can draw our graph here round it out a little bit at the bottom and bring it back up okay so that's what number one would look like with a vertical stretch of a factor of three okay all right so now let's look at number two we have FX = X 5th and we're going to G of X = 0.5x 5th + 2 so right here we see a multiplication um happening to X so we know it's going to be horizontal and so now we need to think okay is it a stretch or is it a shrink well it's a fraction so we might be thinking shrink but remember it's horizontal it's opposite of what it looks like so this is actually going to be a horizontal stretch and it's going to be by a factor of 1 over 1 12 which would be by a factor of two and then this plus two means we are going to translate up to units so this would be a vertical translation two units up okay so let's graph our um x to the 5ifth graph so we know we would have 0 0 and if we had one it would be one and- 1 to the 5th would be negative 1 okay so we might be thinking all right looks like our X cubed graph but once again we're going to be a little bit more um flat here at where our three points are are together so I'm going to bring bring this part up right here let it go a little bit closer a little bit flatter there and then go up there okay all right so now in red let's horizontally stretch this graph by a factor of two and then let's translate it up to so horizontal stretch we're going to stretch it out like this so we know that the Point that's on the origin is just going to stay there but we have our point at 1 one is going to move out by a factor of two so that's going to move out to one two and same thing for1 uh -2 is going to or NE 1 Nega 1 is going to move out to -21 so that's just us doing the stretch and so now we need to translate up to so we can move our origin point up to and then our other points up to okay so now I'm going to take away these two points over over here cuz that's not part of my final function that was just helping me get there and now we can draw our new function here so I'm going to draw my line up once again let it come be a little flat there and then go up like that okay so not a perfect drawing there by any means but hopefully it gives us an a little bit of an idea of what it should look like and we can see the stretch there because that middle portion is is more elongated it's stretched out and then we can see the vertical translation because we move those points up too and that's how you can transform some polinomial functions [Music] |
1355 | https://www.liveflow.com/product-guides/declining-balance-depreciation-in-excel-explained | Back to guides
Excel Tips
Declining Balance Depreciation in Excel: Explained
In this article, you will learn how to perform Declining balance depreciation in Excel.
What does the Declining balance depreciation method mean?
Declining balance depreciation, also known as the reducing balance method, is a commonly used depreciation method in accounting. In declining balance depreciation, a fixed depreciation rate is applied to the asset's net book value (cost of the asset minus accumulated depreciation) each year. The net book value is multiplied by the depreciation rate to calculate the depreciation expense. As a result, the depreciation expense decreases over time.
The formula to calculate declining balance depreciation for a specific period is:
Depreciation Expense = Net Book Value Depreciation Rate
The declining balance method allows for faster depreciation of assets in the early years, reflecting the assumption that assets typically lose their value more rapidly in the initial stages of their useful life. This method can be beneficial for assets that have higher maintenance and repair costs as they age or for assets that quickly become technologically obsolete.
How do you perform Declining balance depreciation in Excel?
To perform declining balance depreciation in Excel, you can use the built-in function called "DB.". The syntax of the DB function in Excel is as follows:
=DB(cost, salvage, life, period, [month])
The function takes the following arguments:
cost: This is the initial cost or value of the asset.
salvage: This is the estimated salvage value of the asset at the end of its useful life.
life: This represents the useful life of the asset in terms of periods (e.g., years).
period: This specifies the period for which you want to calculate the depreciation expense.
[month]: This is an optional argument and is used when depreciation is calculated on a monthly basis. If omitted, the default assumption is annual depreciation.
Additionally, the following formula is used to calculate the depreciation rate.
Rate = 1 – ((salvage / cost) ^ (1 / life))
Case Study: Declining Balance Depreciation
Suppose your company purchases a machine for $100,000 with an estimated salvage value of $15,000. The machine is expected to have a useful life of 7 years, and we want to calculate the declining balance depreciation expense for each year.
Here's how you can use the DB function in Excel to perform the depreciation calculation:
Step 1: In a spreadsheet, enter the following necessary information:
"Cost of Asset.", "Salvage Value.", "Useful Life." & "Period for calculating depreciation expense"
Step 2: In a cell where you want to display the depreciation expense, enter the following formula:
=DB(B1,B2,B3,A6)
Step 3: Press Enter to calculate the depreciation expense for year 1. Drag the formula down to calculate the depreciation expense for each year.
You can now observe the declining balance depreciation expenses for each year, reflecting the decreasing book value of the asset over its useful life.
Note: Make sure to format the cells appropriately (e.g., currency format for cost, salvage value, and depreciation expense) for better readability.
Learn how to do this step-by-step in the video below 👇
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1356 | https://courses.lumenlearning.com/suny-albany-chemistry/chapter/colligative-properties/ | 11.4 Colligative Properties
Learning Objectives
By the end of this section, you will be able to:
Express concentrations of solution components using mole fraction and molality
Describe the effect of solute concentration on various solution properties (vapor pressure, boiling point, freezing point, and osmotic pressure)
Perform calculations using the mathematical equations that describe these various colligative effects
Describe the process of distillation and its practical applications
Explain the process of osmosis and describe how it is applied industrially and in nature
The properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend only upon the total concentration of solute species, regardless of their identities. These colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module.
Mole Fraction and Molality
Several units commonly used to express the concentrations of solution components were introduced in an earlier chapter of this text, each providing certain benefits for use in different applications. For example, molarity (M) is a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species:
M=mol soluteL solutionM=mol soluteL solution
Because solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different at different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are mole fraction (introduced in the previous chapter on gases) and molality.
The mole fraction, X, of a component is the ratio of its molar amount to the total number of moles of all solution components:
XA=molAtotal mol of all componentsXA=molAtotal mol of all components
Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms:
m=mol solutekg solventm=mol solutekg solvent
Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module.
Example 1: Calculating Mole Fraction and Molality
The antifreeze in most automobile radiators is a mixture of equal volumes of ethylene glycol and water, with minor amounts of other additives that prevent corrosion. What are the (a) mole fraction and (b) molality of ethylene glycol, C2H4(OH)2, in a solution prepared from 2.22 × 103 g of ethylene glycol and 2.00 × 103 g of water (approximately 2 L of glycol and 2 L of water)?
Show Answer
(a) The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the unit definition.
molC2H4(OH)2=2220g×1molC2H4(OH)262.07gC2H4(OH)2=35.8molC2H4(OH)2molH2O=2000g×1molH2O18.02gH2O=11.1molH2OXethylene glycol=35.8molC2H4(OH)2(35.8+11.1)mol total=0.763molC2H4(OH)2=2220g×1molC2H4(OH)262.07gC2H4(OH)2=35.8molC2H4(OH)2molH2O=2000g×1molH2O18.02gH2O=11.1molH2OXethylene glycol=35.8molC2H4(OH)2(35.8+11.1)mol total=0.763
Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).
(b) To find molality, we need to know the moles of the solute and the mass of the solvent (in kg).
First, use the given mass of ethylene glycol and its molar mass to find the moles of solute:
2220gC2H4(OH)2(molC2H2(OH)262.07g)=35.8molC2H4(OH)22220gC2H4(OH)2(molC2H2(OH)262.07g)=35.8molC2H4(OH)2
Then, convert the mass of the water from grams to kilograms:
2000 gH2O(1kg1000g)=2 kgH2O2000 gH2O(1kg1000g)=2 kgH2O
Finally, calculate molarity per its definition:
molality=mol solutekg solventmolality=35.8molC2H4(OH)22kgH2Omolality=17.9mmolality=mol solutekg solventmolality=35.8molC2H4(OH)22kgH2Omolality=17.9m
Check Your Learning
What are the mole fraction and molality of a solution that contains 0.850 g of ammonia, NH3, dissolved in 125 g of water?
Show Answer
7.14 × 10-3; 0.399 m
Example 2: Converting Mole Fraction and Molal Concentrations
Calculate the mole fraction of solute and solvent in a 3.0 m solution of sodium chloride.
Show Answer
Converting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as:
3.0mol NaCl1.0kgH2O3.0mol NaCl1.0kgH2O
The numerator for this solution’s mole fraction is, therefore, 3.0 mol NaCl. The denominator may be computed by deriving the molar amount of water corresponding to 1.0 kg
1.0kgH2O(1000g1kg)(molH2O18.02g)=55molH2O1.0kgH2O(1000g1kg)(molH2O18.02g)=55molH2O
and then substituting these molar amounts into the definition for mole fraction.
XH2O=molH2Omol NaCl+molH2OXH2O=55molH2O3.0mol NaCl+55molH2OXH2O=0.95XNaCl=mol NaClmol NaCl+molH2OXNaCl=3.0molH2O3.0mol NaCl+55molH2OXNaCl=0.052XH2O=molH2Omol NaCl+molH2OXH2O=55molH2O3.0mol NaCl+55molH2OXH2O=0.95XNaCl=mol NaClmol NaCl+molH2OXNaCl=3.0molH2O3.0mol NaCl+55molH2OXNaCl=0.052
Check Your Learning
The mole fraction of iodine, I2, dissolved in dichloromethane, CH2Cl2, is 0.115. What is the molal concentration, m, of iodine in this solution?
Show Answer
1.50 m
Exercises
What are the mole fractions of H3PO4 and water in a solution of 14.5 g of H3PO4 in 125 g of water?
Outline the steps necessary to answer the question.
Answer the question.
What are the mole fractions of HNO3 and water in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
Outline the steps necessary to answer the question.
Answer the question.
Calculate the mole fraction of each solute and solvent:
583 g of H2SO4 in 1.50 kg of water—the acid solution used in an automobile battery
0.86 g of NaCl in 1.00 × 102 g of water—a solution of sodium chloride for intravenous injection
46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH
25 g of I2 in 125 g of ethanol, C2H5OH
Calculate the mole fraction of each solute and solvent:
0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0 °C
125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack
25 g of Cl2 in 125 g of dichloromethane, CH2Cl2
0.372 g of histamine, C5H9N, in 125 g of chloroform, CHCl3
Calculate the mole fractions of methanol, CH3OH; ethanol, C2H5OH; and water in a solution that is 40% methanol, 40% ethanol, and 20% water by mass. (Assume the data are good to two significant figures.)
What is the difference between a 1 M solution and a 1 m solution?
What is the molality of phosphoric acid, H3PO4, in a solution of 14.5 g of H3PO4 in 125 g of water?
Outline the steps necessary to answer the question.
Answer the question.
What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
Outline the steps necessary to answer the question.
Answer the question.
Calculate the molality of each of the following solutions:
583 g of H2SO4 in 1.50 kg of water—the acid solution used in an automobile battery
0.86 g of NaCl in 1.00 × 102 g of water—a solution of sodium chloride for intravenous injection
46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH
25 g of I2 in 125 g of ethanol, C2H5OH
Calculate the molality of each of the following solutions:
0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0°C
125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack
25 g of Cl2 in 125 g of dichloromethane, CH2Cl2
0.372 g of histamine, C5H9N, in 125 g of chloroform, CHCl3
The concentration of glucose, C6H12O6, in normal spinal fluid is 75mg100g75mg100g. What is the molality of the solution?
A 13.0% solution of K2CO3 by mass has a density of 1.09 g/cm3. Calculate the molality of the solution.
Show Selected Answers
The mole fractions are as follows:
Find number of moles of HNO3 and H2O in 100 g of the solution. Find the mole fractions for the components.
The number of moles of HNO3 is 68g63.01g/mol=1.079mol68g63.01g/mol=1.079mol. The number of moles of water is 32 g18.015 g/mol=1.776 mol32 g18.015 g/mol=1.776 mol. The mole fraction of HNO3 is 1.079(1.079+1.776)=0.3781.079(1.079+1.776)=0.378. The mole fraction of H2O is 1 − 0.378 = 0.622.
The mole fractions are as follows:
molNa2CO3=710gNa2CO3×1mol105.9886gNa2CO3=6.70molmolH2O=10,000g18.0153g/mol=555.08molmolNa2CO3=710gNa2CO3×1mol105.9886gNa2CO3=6.70molmolH2O=10,000g18.0153g/mol=555.08mol
Total number of moles = 555.08 mol + 6.70 mol = 561.78 mol
XNa2CO3=6.70mol561.78mol=0.0119XH2O=555.08mol561.78mol=0.988XNa2CO3=6.70mol561.78mol=0.0119XH2O=555.08mol561.78mol=0.988
molNH4NO3=125gNH4NO3×1mol80.0434gNH4NO3=1.56molmolH2O=275g18.0153g/mol=15.26molmolNH4NO3=125gNH4NO3×1mol80.0434gNH4NO3=1.56molmolH2O=275g18.0153g/mol=15.26mol
Total number of moles = 15.26 mol + 1.56 mol = 16.82 mol
XNH4NO3=1.56mol16.82mol=0.9927XH2O=15.26mol16.82mol=0.907XNH4NO3=1.56mol16.82mol=0.9927XH2O=15.26mol16.82mol=0.907
molCl2=25gCl2×1mol70.9054gCl2=0.35molmolCH2Cl2=125g84.93g/mol=1.47molmolCl2=25gCl2×1mol70.9054gCl2=0.35molmolCH2Cl2=125g84.93g/mol=1.47mol
Total number of moles = 1.47 mol + 0.35 mol = 1.82 mol
XCl2=0.35mol1.82mol=0.192XCH2Cl2=1.47mol1.82mol=0.808XCl2=0.35mol1.82mol=0.192XCH2Cl2=1.47mol1.82mol=0.808
molC5H9N=0.372gC5H9N×1mol83.1332gC5H9N=4.47×10−3molmolCHCl3=125g119.38g/mol=1.047molmolC5H9N=0.372gC5H9N×1mol83.1332gC5H9N=4.47×10−3molmolCHCl3=125g119.38g/mol=1.047mol
Total number of moles = 1.047 mol + 0.00447 mol = 1.05 mol
XC5H9N=0.00447mol1.05mol=0.00426XCHCl3=1.047mol1.05mol=0.997XC5H9N=0.00447mol1.05mol=0.00426XCHCl3=1.047mol1.05mol=0.997
6. In a 1 M solution, the mole is contained in exactly 1 L of solution. In a 1 m solution, the mole is contained in exactly 1 kg of solvent.
The molality is as follows:
Determine the molar mass of HNO3. Determine the number of moles of acid in the solution. From the number of moles and the mass of solvent, determine the molality.
Molar mass HNO3 = 63.01288 g mol–1If we assume 100 g of solution, then 68.0 g is HNO3 and 32.0 g is water.
molHNO3=68.0gHNO3×1mol63.02188gHNO3=1.08molmHNO3=1.08mol0.0320g=33.7mmolHNO3=68.0gHNO3×1mol63.02188gHNO3=1.08molmHNO3=1.08mol0.0320g=33.7m
The molality of each is as follows:
molNa2CO3=710gNa2CO3×1mol105.9886gNa2CO3molality ofNa2CO3=6.70mol10.0kg=6.70×10−1mmolNa2CO3=710gNa2CO3×1mol105.9886gNa2CO3molality ofNa2CO3=6.70mol10.0kg=6.70×10−1m
molNH4NO3=125gNH4NO3×1mol80.0434gNH4NO3=1.56molmolality ofNH4NO3=1.56mol0.275kg=5.67mmolNH4NO3=125gNH4NO3×1mol80.0434gNH4NO3=1.56molmolality ofNH4NO3=1.56mol0.275kg=5.67m
molCl2=25gCl2×1mol70.9054gCl2=0.35molmolCl2=25gCl2×1mol70.9054gCl2=0.35mol
molC5H9N=0.372gC5H9N×1mol83.1332gC5H9N=4.47×10−3molmolality ofC5H9N=4.47×10−3mol0.125kg=0.0358mmolC5H9N=0.372gC5H9N×1mol83.1332gC5H9N=4.47×10−3molmolality ofC5H9N=4.47×10−3mol0.125kg=0.0358m
Find the mass of K2CO3 and the mass of water in solution. Assume 100.0 mL of solution and that the density of water is 1.00 g cm–3. Then find the moles of K2CO3 and the molality.
Mass (solution)=100.0mL×1cm31mL×1.09gcm3=109.0gMass(K2CO3)=13.0%100%×109g=14.2gMass (solution)=100.0mL×1cm31mL×1.09gcm3=109.0gMass(K2CO3)=13.0%100%×109g=14.2g
Mass (H2O) = 109.0 g – 14.2 g = 94.8 g
mol(K2CO3)=14.2gK2CO3×1mol138.206gK2CO3=0.1027molmol(K2CO3)=14.2gK2CO3×1mol138.206gK2CO3=0.1027mol
Vapor Pressure Lowering
As described in the chapter on liquids and solids, the equilibrium vapor pressure of a liquid is the pressure exerted by its gaseous phase when vaporization and condensation are occurring at equal rates:
liquid⇌gas
Dissolving a nonvolatile substance in a volatile liquid results in a lowering of the liquid’s vapor pressure. This phenomenon can be rationalized by considering the effect of added solute molecules on the liquid’s vaporization and condensation processes. To vaporize, solvent molecules must be present at the surface of the solution. The presence of solute decreases the surface area available to solvent molecules and thereby reduces the rate of solvent vaporization. Since the rate of condensation is unaffected by the presence of solute, the net result is that the vaporization-condensation equilibrium is achieved with fewer solvent molecules in the vapor phase (i.e., at a lower vapor pressure) (Figure 1). While this kinetic interpretation is useful, it does not account for several important aspects of the colligative nature of vapor pressure lowering. A more rigorous explanation involves the property of entropy, a topic of discussion in a later text chapter on thermodynamics. For purposes of understanding the lowering of a liquid’s vapor pressure, it is adequate to note that the greater entropy of a solution in comparison to its separate solvent and solute serves to effectively stabilize the solvent molecules and hinder their vaporization. A lower vapor pressure results, and a correspondingly higher boiling point as described in the next section of this module.
Figure 1. The presence of nonvolatile solutes lowers the vapor pressure of a solution by impeding the evaporation of solvent molecules.
The relationship between the vapor pressures of solution components and the concentrations of those components is described by Raoult’s law: The partial pressure exerted by any component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
PA=XAP∘A
where PA is the partial pressure exerted by component A in the solution, P∘A is the vapor pressure of pure A, and XA is the mole fraction of A in the solution. (Mole fraction is a concentration unit introduced in the chapter on gases.)
Recalling that the total pressure of a gaseous mixture is equal to the sum of partial pressures for all its components (Dalton’s law of partial pressures), the total vapor pressure exerted by a solution containing i components is
Psolution=∑iPi=∑iXiP∘i
A nonvolatile substance is one whose vapor pressure is negligible (P° ≈ 0), and so the vapor pressure above a solution containing only nonvolatile solutes is due only to the solvent:
Psolution=XsolventP∘solvent
Example 3: Calculation of a Vapor Pressure
Compute the vapor pressure of an ideal solution containing 92.1 g of glycerin, C3H5(OH)3, and 184.4 g of ethanol, C2H5OH, at 40 °C. The vapor pressure of pure ethanol is 0.178 atm at 40 °C. Glycerin is essentially nonvolatile at this temperature.
Show Answer
Since the solvent is the only volatile component of this solution, its vapor pressure may be computed per Raoult’s law as:
Psolution=XsolventP∘solvent
First, calculate the molar amounts of each solution component using the provided mass data.
92.1gC3H5(OH)3×1molC3H5(OH)392.094gC3H5(OH)3=1.00molC3H5(OH)3184.4gC2H5OH×1molC2H5OH46.069gC2H5OH=4.000molC2H5OH
Next, calculate the mole fraction of the solvent (ethanol) and use Raoult’s law to compute the solution’s vapor pressure.
XC2H5OH=4.000mol(1.00mol+4.000mol)=0.800Psolv=XsolvP∘solv=0.800×0.178atm=0.142atm
Check Your Learning
A solution contains 5.00 g of urea, CO(NH2)2 (a nonvolatile solute) and 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution?
Show Answer
23.4 torr
Elevation of the Boiling Point of a Solvent
As described in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution’s boiling point will subsequently be increased. Compared to pure solvent, a solution, therefore, will require a higher temperature to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, ΔTb, is called boiling point elevation and is directly proportional to the molal concentration of solute species:
ΔTb=Kbm
where Kb is the boiling point elevation constant, or the ebullioscopic constant and m is the molal concentration (molality) of all solute species.
Boiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of Kb for several solvents are listed in Table 1.
| Table 1. Boiling Point Elevation and Freezing Point Depression Constants for Several Solvents | | | | |
--- ---
| Solvent | Boiling Point (°C at 1 atm) | Kb (Cm−1) | Freezing Point (°C at 1 atm) | Kf (Cm−1) |
| water | 100.0 | 0.512 | 0.0 | 1.86 |
| hydrogen acetate | 118.1 | 3.07 | 16.6 | 3.9 |
| benzene | 80.1 | 2.53 | 5.5 | 5.12 |
| chloroform | 61.26 | 3.63 | −63.5 | 4.68 |
| nitrobenzene | 210.9 | 5.24 | 5.67 | 8.1 |
The extent to which the vapor pressure of a solvent is lowered and the boiling point is elevated depends on the total number of solute particles present in a given amount of solvent, not on the mass or size or chemical identities of the particles. A 1 m aqueous solution of sucrose (342 g/mol) and a 1 m aqueous solution of ethylene glycol (62 g/mol) will exhibit the same boiling point because each solution has one mole of solute particles (molecules) per kilogram of solvent.
Example 4: Calculating the Boiling Point of a Solution
What is the boiling point of a 0.33 m solution of a nonvolatile solute in benzene?
Show Answer
Use the equation relating boiling point elevation to solute molality to solve this problem in two steps.
Calculate the change in boiling point: ΔTb=Kbm=2.53∘Cm−1×0.33m=0.83∘C
Add the boiling point elevation to the pure solvent’s boiling point: Boiling temperature=80.1∘C+0.83∘C=80.9∘C
Check Your Learning
What is the boiling point of the antifreeze described in Example 1?
Show Answer
109.2 °C
Example 5: The Boiling Point of an Iodine Solution
Find the boiling point of a solution of 92.1 g of iodine, I2, in 800.0 g of chloroform, CHCl3, assuming that the iodine is nonvolatile and that the solution is ideal.
Show Answer
We can solve this problem using four steps.
Convert from grams to moles of I2using the molar mass of I2in the unit conversion factor: 0.363 mol
Determine the molality of the solution from the number of moles of solute and the mass of solvent, in kilograms: 0.454 m
Use the direct proportionality between the change in boiling point and molal concentration to determine how much the boiling point changes: 1.65 °C
Determine the new boiling point from the boiling point of the pure solvent and the change: 62.91 °C. Check each result as a self-assessment.
Check Your Learning
What is the boiling point of a solution of 1.0 g of glycerin, C3H5(OH)3, in 47.8 g of water? Assume an ideal solution.
Show Answer
100.12 °C
Exercises
Why does 1 mol of sodium chloride depress the freezing point of 1 kg of water almost twice as much as 1 mol of glycerin?
What is the boiling point of a solution of 115.0 g of sucrose, C12H22O11, in 350.0 g of water?
Outline the steps necessary to answer the question
Answer the question
What is the boiling point of a solution of 9.04 g of I2 in 75.5 g of benzene, assuming the I2 is nonvolatile?
Outline the steps necessary to answer the question.
Answer the question.
What is the freezing temperature of a solution of 115.0 g of sucrose, C12H22O11, in 350.0 g of water, which freezes at 0.0 °C when pure?
Outline the steps necessary to answer the question.
Answer the question.
What is the freezing point of a solution of 9.04 g of I2 in 75.5 g of benzene?
Outline the steps necessary to answer the following question.
Answer the question.
Show Answer
The boiling point can be found as follows:
Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the boiling point of water and the boiling point of the solution; determine the new boiling point.
mol sucrose=115.0g342.300gmol−1=0.3360molmolality=0.3360molC12H22O110.3500kgH2O=0.9599m
ΔTb = Kbm = (0.512 °C m–1)(0.9599 m) = 0.491 °C
The boiling point of pure water at 100.0 °C increases 0.491 °C to 100.491 °C, or 100.5 °C.
4. The freezing temperature can be found as follows:
Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature.
mol sucrose=115.0g342.300gmol−1=0.336molm sucrose=0.336mol0.350kg=0.960m
ΔTb = Kbm = (1.86 °C m–1)(0.960 m) = 1.78 °C
The freezing temperature is 0.0 °C – 1.78 °C = –1.8 °C.
Distillation of Solutions
Distillation is a technique for separating the components of mixtures that is widely applied in both in the laboratory and in industrial settings. It is used to refine petroleum, to isolate fermentation products, and to purify water. This separation technique involves the controlled heating of a sample mixture to selectively vaporize, condense, and collect one or more components of interest. A typical apparatus for laboratory-scale distillations is shown in Figure 2.
Figure 2. A typical laboratory distillation unit is shown in (a) a photograph and (b) a schematic diagram of the components. (credit a: modification of work by “Rifleman82”/Wikimedia commons; credit b: modification of work by “Slashme”/Wikipedia)
Oil refineries use large-scale fractional distillation to separate the components of crude oil. The crude oil is heated to high temperatures at the base of a tall fractionating column, vaporizing many of the components that rise within the column. As vaporized components reach adequately cool zones during their ascent, they condense and are collected. The collected liquids are simpler mixtures of hydrocarbons and other petroleum compounds that are of appropriate composition for various applications (e.g., diesel fuel, kerosene, gasoline), as depicted in Figure 3.
Figure 3. Crude oil is a complex mixture that is separated by large-scale fractional distillation to isolate various simpler mixtures.
Depression of the Freezing Point of a Solvent
Figure 4. Rock salt (NaCl), calcium chloride (CaCl2), or a mixture of the two are used to melt ice. (credit: modification of work by Eddie Welker)
Solutions freeze at lower temperatures than pure liquids. This phenomenon is exploited in “de-icing” schemes that use salt (Figure 4), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an “antifreeze” in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below 0 °C (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans).
The decrease in freezing point of a dilute solution compared to that of the pure solvent, ΔTf, is called the freezing point depression and is directly proportional to the molal concentration of the solute
ΔTf=Kfm
where m is the molal concentration of the solute in the solvent and Kf is called the freezing point depression constant (or cryoscopic constant). Just as for boiling point elevation constants, these are characteristic properties whose values depend on the chemical identity of the solvent. Values of Kf for several solvents are listed in Table 1.
Example 6: Calculation of the Freezing Point of a Solution
What is the freezing point of the 0.33 m solution of a nonvolatile nonelectrolyte solute in benzene described in Example 2?
Show Answer
Use the equation relating freezing point depression to solute molality to solve this problem in two steps.
Calculate the change in freezing point: ΔTf=Kfm=5.12∘Cm−1×0.33m=1.7∘C
Subtract the freezing point change observed from the pure solvent’s freezing point: Freezing Temperature=5.5∘C−1.7∘C=3.8∘C
Check Your Learning
What is the freezing point of a 1.85 m solution of a nonvolatile nonelectrolyte solute in nitrobenzene?
Show Answer
−9.3 °C
Colligative Properties and De-Icing
Sodium chloride and its group 2 analogs calcium and magnesium chloride are often used to de-ice roadways and sidewalks, due to the fact that a solution of any one of these salts will have a freezing point lower than 0 °C, the freezing point of pure water. The group 2 metal salts are frequently mixed with the cheaper and more readily available sodium chloride (“rock salt”) for use on roads, since they tend to be somewhat less corrosive than the NaCl, and they provide a larger depression of the freezing point, since they dissociate to yield three particles per formula unit, rather than two particles like the sodium chloride.
Because these ionic compounds tend to hasten the corrosion of metal, they would not be a wise choice to use in antifreeze for the radiator in your car or to de-ice a plane prior to takeoff. For these applications, covalent compounds, such as ethylene or propylene glycol, are often used. The glycols used in radiator fluid not only lower the freezing point of the liquid, but they elevate the boiling point, making the fluid useful in both winter and summer. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and prevent the formation of more ice, which would be particularly dangerous if formed on the control surfaces of the aircraft (Figure 5).
Figure 5. Freezing point depression is exploited to remove ice from (a) roadways and (b) the control surfaces of aircraft.
Exercises
A 1.0 m solution of HCl in benzene has a freezing point of 0.4 °C. Is HCl an electrolyte in benzene? Explain.
A 12.0-g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at -1.94 °C. Calculate the molar mass of the substance.
Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of Na2SO4, and 0.030 mol of MgCl2, assuming complete dissociation of these electrolytes.
Show Answer
No. Pure benzene freezes at 5.5 °C, and so the observed freezing point of this solution is depressed by ΔTf = 5.5 − 0.4 = 5.1 °C. The value computed, assuming no ionization of HCl, is ΔTf = (1.0 m)(5.14 °C/m) = 5.1 °C. Agreement of these values supports the assumption that HCl is not ionized.
ΔTf = 1.94 °C
m=ΔTfKf=1.94∘C1.86∘C/m=1.04m
Moles of solute=1.04m×0.0800kg=0.0834mol
Molar mass=12.0g0.0834mol=144gmol-1
Molecular mass = 144 amu
0.010 mol NaCl contains 0.010 mol Na+ + 0.010 mol Cl–0.020 mol Na2SO4 contains 0.040 mol Na+ + 0.020 mol SO42−
0.030 mol MgCl2 contains 0.030 mol Mg2+ + 0.060 mol Cl–Total numbers of moles = 0.020 mol + 0.060 mol + 0.090 mol = 0.170 mol
ΔTb=Kbm=0.512∘C/m×0.170mol0.100kg= 0.870º C.
Phase Diagram for an Aqueous Solution of a Nonelectrolyte
The colligative effects on vapor pressure, boiling point, and freezing point described in the previous section are conveniently summarized by comparing the phase diagrams for a pure liquid and a solution derived from that liquid. Phase diagrams for water and an aqueous solution are shown in Figure 6.
Figure 6. These phase diagrams show water (solid curves) and an aqueous solution of nonelectrolyte (dashed curves).
The liquid-vapor curve for the solution is located beneath the corresponding curve for the solvent, depicting the vapor pressure lowering, ΔP, that results from the dissolution of nonvolatile solute. Consequently, at any given pressure, the solution’s boiling point is observed at a higher temperature than that for the pure solvent, reflecting the boiling point elevation, ΔTb, associated with the presence of nonvolatile solute. The solid-liquid curve for the solution is displaced left of that for the pure solvent, representing the freezing point depression, ΔTb, that accompanies solution formation. Finally, notice that the solid-gas curves for the solvent and its solution are identical. This is the case for many solutions comprising liquid solvents and nonvolatile solutes. Just as for vaporization, when a solution of this sort is frozen, it is actually just the solvent molecules that undergo the liquid-to-solid transition, forming pure solid solvent that excludes solute species. The solid and gaseous phases, therefore, are composed solvent only, and so transitions between these phases are not subject to colligative effects.
Osmosis and Osmotic Pressure of Solutions
A number of natural and synthetic materials exhibit selective permeation, meaning that only molecules or ions of a certain size, shape, polarity, charge, and so forth, are capable of passing through (permeating) the material. Biological cell membranes provide elegant examples of selective permeation in nature, while dialysis tubing used to remove metabolic wastes from blood is a more simplistic technological example. Regardless of how they may be fabricated, these materials are generally referred to as semipermeable membranes.
Consider the apparatus illustrated in Figure 7, in which samples of pure solvent and a solution are separated by a membrane that only solvent molecules may permeate. Solvent molecules will diffuse across the membrane in both directions. Since the concentration of solvent is greater in the pure solvent than the solution, these molecules will diffuse from the solvent side of the membrane to the solution side at a faster rate than they will in the reverse direction. The result is a net transfer of solvent molecules from the pure solvent to the solution. Diffusion-driven transfer of solvent molecules through a semipermeable membrane is a process known as osmosis.
Figure 7. Osmosis results in the transfer of solvent molecules from a sample of low (or zero) solute concentration to a sample of higher solute concentration.
When osmosis is carried out in an apparatus like that shown in Figure 7, the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the osmotic pressure (Π) of the solution. The osmotic pressure of a dilute solution is related to its solute molarity, M, and absolute temperature, T, according to the equation
Π=MRT
where R is the universal gas constant.
Example 7: Calculation of Osmotic Pressure
What is the osmotic pressure (atm) of a 0.30 M solution of glucose in water that is used for intravenous infusion at body temperature, 37 °C?
Show Answer
We can find the osmotic pressure, Π, using the formula Π = MRT, where T is on the Kelvin scale (310 K) and the value of R is expressed in appropriate units (0.08206 L atm/mol K).
Π=MRT=0.03mol/L×0.08206 L atm/mol K×310 K=7.6atm
Check Your Learning
What is the osmotic pressure (atm) a solution with a volume of 0.750 L that contains 5.0 g of methanol, CH3OH, in water at 37 °C?
Show Answer
5.3 atm
If a solution is placed in an apparatus like the one shown in Figure 8, applying pressure greater than the osmotic pressure of the solution reverses the osmosis and pushes solvent molecules from the solution into the pure solvent. This technique of reverse osmosis is used for large-scale desalination of seawater and on smaller scales to produce high-purity tap water for drinking.
Figure 8. Applying a pressure greater than the osmotic pressure of a solution will reverse osmosis. Solvent molecules from the solution are pushed into the pure solvent.
Reverse Osmosis Water Purification
In the process of osmosis, diffusion serves to move water through a semipermeable membrane from a less concentrated solution to a more concentrated solution. Osmotic pressure is the amount of pressure that must be applied to the more concentrated solution to cause osmosis to stop. If greater pressure is applied, the water will go from the more concentrated solution to a less concentrated (more pure) solution. This is called reverse osmosis. Reverse osmosis (RO) is used to purify water in many applications, from desalination plants in coastal cities, to water-purifying machines in grocery stores (Figure 9), and smaller reverse-osmosis household units. With a hand-operated pump, small RO units can be used in third-world countries, disaster areas, and in lifeboats. Our military forces have a variety of generator-operated RO units that can be transported in vehicles to remote locations.
Figure 9. Reverse osmosis systems for purifying drinking water are shown here on (a) small and (b) large scales. (credit a: modification of work by Jerry Kirkhart; credit b: modification of work by Willard J. Lathrop)
Examples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm. Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be isotonic with blood serum. If a less concentrated solution, a hypotonic solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture. This process is called hemolysis. When a more concentrated solution, a hypertonic solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called crenation. These effects are illustrated in Figure 10.
Figure 10. Red blood cell membranes are water permeable and will (a) swell and possibly rupture in a hypotonic solution; (b) maintain normal volume and shape in an isotonic solution; and (c) shrivel and possibly die in a hypertonic solution. (credit a/b/c: modifications of work by “LadyofHats”/Wikimedia commons)
Exercises
What is the osmotic pressure of an aqueous solution of 1.64 g of Ca(NO3)2 in water at 25 °C? The volume of the solution is 275 mL.
Outline the steps necessary to answer the question.
Answer the question.
What is osmotic pressure of a solution of bovine insulin (molar mass, 5700 g mol−1) at 18 °C if 100.0 mL of the solution contains 0.103 g of the insulin?
Outline the steps necessary to answer the question.
Answer the question.
Show Answer to Question 1
The osmotic pressure of an aqueous solution of 1.64 g of Ca(NO3)2 in water at 25 °C can be found as follows:
Determine the molar mass of Ca(NO3)2; determine the number of moles of Ca(NO3)2 in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure.
MCa(NO3)2=1.64g Ca(NO3)2×1mol/164.088g Ca(NO3)20.275L=0.363M
The molarity of the ions is three times the molarity of Ca(NO3)2. Therefore, multiply the molarity of Ca(NO3)2 by 3: Π = MRT = 3 × 0.0363 mol L–1 × 0.08206 L atm mol–1 K–1 × 298.15 K = 2.67 atm.
Determination of Molar Masses
Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements.
Example 8: Determination of a Molar Mass from a Freezing Point Depression
A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound?
Show Answer
We can solve this problem using the following steps.
Determine the change in freezing point from the observed freezing point and the freezing point of pure benzene (Table 1): ΔTf=5.5∘C−2.32∘C=3.2∘C
Determine the molal concentration from Kf, the freezing point depression constant for benzene (Table 1), and ΔTf: ΔTf=Kfmm=ΔTfKf=3.2∘C5.12∘Cm−1=0.63m
Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution: Moles of solute=0.62mol solute1.00kg solvent×0.0550kg solvent=0.035mol
Determine the molar mass from the mass of the solute and the number of moles in that mass: Molar mass=4.00g0.034mol=1.2×102g/mol
Check Your Learning
A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this compound?
Show Answer
1.8 × 102 g/mol
Example 9: Determination of a Molar Mass from Osmotic Pressure
A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin?
Show Answer
Here is one set of steps that can be used to solve the problem:
Convert the osmotic pressure to atmospheres, then determine the molar concentration from the osmotic pressure: Π=5.9torr×1atm760torr=7.8×10−3atmΠ=MRTM=ΠRT=7.8×10−3atm(0.08206L atm/mol K)(295K)=3.2×10−4M
Determine the number of moles of hemoglobin in the solution from the concentration and the volume of the solution:moles of hemoglobin=3.2×10−4mol1L solution×0.500L solution=1.6×10−4mol
Determine the molar mass from the mass of hemoglobin and the number of moles in that mass: molar mass=10.0g1.6×10−4mol=6.2×104g/mol
Check Your Learning
What is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C?
Show Answer
2.7 × 104 g/mol
Exercises
What is the molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp 76.8 °C; Kb = 5.02 °C/m) that boils at 81.5 °C at 1 atm?
Outline the steps necessary to answer the question.
Solve the problem.
A sample of an organic compound (a nonelectrolyte) weighing 1.35 g lowered the freezing point of 10.0 g of benzene by 3.66 °C. Calculate the molar mass of the compound.
Show Answer to Question 1
The molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp 76.8 °C; Kb = 5.02 °C/m) can be found as follows:
Determine the molal concentration from the change in boiling point and Kb; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute.
ΔTb = 81.5 − 76.8 = 4.7 °C, ΔTb = Kbm, so m=ΔTbKb=4.7∘C5.02∘C/m=0.94m. Moles of solute = molality × kg of solvent = 0.94 m 0.02500 kg = 0.024 mol;
molar mass=massmoles=5.00g0.024mol=2.1×102gmol−1
Molecular mass = 2.1 × 102 g mol−1
Colligative Properties of Electrolytes
As noted previously in this module, the colligative properties of a solution depend only on the number, not on the kind, of solute species dissolved. For example, 1 mole of any nonelectrolyte dissolved in 1 kilogram of solvent produces the same lowering of the freezing point as does 1 mole of any other nonelectrolyte. However, 1 mole of sodium chloride (an electrolyte) forms 2 moles of ions when dissolved in solution. Each individual ion produces the same effect on the freezing point as a single molecule does.
Example 10: The Freezing Point of a Solution of an Electrolyte
The concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Assume that each of the ions in the NaCl solution has the same effect on the freezing point of water as a nonelectrolyte molecule, and determine the freezing temperature the solution (which is approximately equal to the freezing temperature of seawater).
Show Answer
We can solve this problem using the following series of steps.
Convert from grams to moles of NaCl using the molar mass of NaCl in the unit conversion factor: 0.072 mol NaCl
Determine the number of moles of ions present in the solution using the number of moles of ions in 1 mole of NaCl as the conversion factor (2 mol ions/1 mol NaCl): 0.14 mol ions
Determine the molality of the ions in the solution from the number of moles of ions and the mass of solvent, in kilograms: 1.1 m
Use the direct proportionality between the change in freezing point and molal concentration to determine how much the freezing point changes: 2.0 °C
Determine the new freezing point from the freezing point of the pure solvent and the change: −2.0 °C. Check each result as a self-assessment.
Check Your Learning
Assume that each of the ions in calcium chloride, CaCl2, has the same effect on the freezing point of water as a nonelectrolyte molecule. Calculate the freezing point of a solution of 0.724 g of CaCl2 in 175 g of water.
Show Answer
−0.208 °C
Assuming complete dissociation, a 1.0 m aqueous solution of NaCl contains 1.0 mole of ions (1.0 mol Na+ and 1.0 mol Cl−) per each kilogram of water, and its freezing point depression is expected to be
ΔTf=2.0mol ions/kg water×1.86∘C kg water/mol ion=3.7∘C
When this solution is actually prepared and its freezing point depression measured, however, a value of 3.4 °C is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution.
To account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van’t Hoff is used. The van’t Hoff factor (i) is defined as the ratio of solute particles in solution to the number of formula units dissolved:
i=moles of particles in solutionmoles of formula units dissolved
Values for measured van’t Hoff factors for several solutes, along with predicted values assuming complete dissociation, are shown in Table 2.
| Table 2. Expected and Observed van’t Hoff Factors for Several 0.050 m Aqueous Electrolyte Solutions | | | |
--- --- |
| Electrolyte | Particles in Solution | i (Predicted) | i (Measured) |
| HCl | H+, Cl− | 2 | 1.9 |
| NaCl | Na+, Cl− | 2 | 1.9 |
| MgSO4 | Mg2+, SO42− | 2 | 1.3 |
| MgCl2 | Mg2+, 2Cl− | 3 | 2.7 |
| FeCl3 | Fe3+, 3Cl− | 4 | 3.4 |
| glucose | C12H22O11 | 1 | 1.0 |
Figure 11. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less.
In 1923, the chemists Peter Debye and Erich Hückel proposed a theory to explain the apparent incomplete ionization of strong electrolytes. They suggested that although interionic attraction in an aqueous solution is very greatly reduced by solvation of the ions and the insulating action of the polar solvent, it is not completely nullified. The residual attractions prevent the ions from behaving as totally independent particles (Figure 11). In some cases, a positive and negative ion may actually touch, giving a solvated unit called an ion pair. Thus, the activity, or the effective concentration, of any particular kind of ion is less than that indicated by the actual concentration. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less and less. Thus, in extremely dilute solutions, the effective concentrations of the ions (their activities) are essentially equal to the actual concentrations. Note that the van’t Hoff factors for the electrolytes in Table 2 are for 0.05 m solutions, at which concentration the value of i for NaCl is 1.9, as opposed to an ideal value of 2.
Exercises
Meat can be classified as fresh (not frozen) even though it is stored at −1 °C. Why wouldn’t meat freeze at this temperature?
An organic compound has a composition of 93.46% C and 6.54% H by mass. A solution of 0.090 g of this compound in 1.10 g of camphor melts at 158.4 °C. The melting point of pure camphor is 178.4 °C. Kf for camphor is 37.7 °C/m. What is the molecular formula of the solute? Show your calculations.
A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the salt dissolved in 100 g of water produces a solution that freezes at about −1.4 °C. What is the formula of the salt? Show your calculations.
Show Answer to Question 1
The ions and compounds present in the water in the beef lower the freezing point of the beef below −1 °C.
Key Concepts and Summary
Properties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. The total concentration of solute particles in a solution also determines its osmotic pressure. This is the pressure that must be applied to the solution to prevent diffusion of molecules of pure solvent through a semipermeable membrane into the solution. Ionic compounds may not completely dissociate in solution due to activity effects, in which case observed colligative effects may be less than predicted.
Key Equations
(PA=XAP∘A)
Psolution=∑iPi=∑iXiP∘i
Psolution=XsolventP∘solvent
ΔTb = Kbm
ΔTf = Kfm
Π = MRT
Exercises
Which is/are part of the macroscopic domain of solutions and which is/are part of the microscopic domain: boiling point elevation, Henry’s law, hydrogen bond, ion-dipole attraction, molarity, nonelectrolyte, nonstoichiometric compound, osmosis, solvated ion?
What is the microscopic explanation for the macroscopic behavior illustrated in Figure 7 in Solubility?
Sketch a qualitative graph of the pressure versus time for water vapor above a sample of pure water and a sugar solution, as the liquids evaporate to half their original volume.
A solution of potassium nitrate, an electrolyte, and a solution of glycerin (C3H5(OH)3), a nonelectrolyte, both boil at 100.3 °C. What other physical properties of the two solutions are identical?
A solution contains 5.00 g of urea, CO(NH2)2, a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution?
Arrange the following solutions in order by their decreasing freezing points: 0.1 m Na3PO4, 0.1 m C2H5OH, 0.01 m CO2, 0.15 m NaCl, and 0.2 m CaCl2.
How could you prepare a 3.08 m aqueous solution of glycerin, C3H8O3? What is the freezing point of this solution?
A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, CS2 (Kb = 2.43 °C/m). If the boiling point elevation was 0.107 °C, what is the formula of a sulfur molecule in carbon disulfide?
In a significant experiment performed many years ago, 5.6977 g of cadmium iodide in 44.69 g of water raised the boiling point 0.181 °C. What does this suggest about the nature of a solution of CdI2?
Lysozyme is an enzyme that cleaves cell walls. A 0.100-L sample of a solution of lysozyme that contains 0.0750 g of the enzyme exhibits an osmotic pressure of 1.32 × 10−3 atm at 25 °C. What is the molar mass of lysozyme?
The osmotic pressure of a solution containing 7.0 g of insulin per liter is 23 torr at 25 °C. What is the molar mass of insulin?
The osmotic pressure of human blood is 7.6 atm at 37 °C. What mass of glucose, C6H12O6, is required to make 1.00 L of aqueous solution for intravenous feeding if the solution must have the same osmotic pressure as blood at body temperature, 37 °C?
What is the freezing point of a solution of dibromobenzene, C6H4Br2, in 0.250 kg of benzene, if the solution boils at 83.5 °C?
What is the boiling point of a solution of NaCl in water if the solution freezes at −0.93 °C?
The sugar fructose contains 40.0% C, 6.7% H, and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 °C. The boiling point of ethanol is 78.35 °C, and Kb for ethanol is 1.20 °C/m. What is the molecular formula of fructose?
The vapor pressure of methanol, CH3OH, is 94 torr at 20 °C. The vapor pressure of ethanol, C2H5OH, is 44 torr at the same temperature.
Calculate the mole fraction of methanol and of ethanol in a solution of 50.0 g of methanol and 50.0 g of ethanol.
Ethanol and methanol form a solution that behaves like an ideal solution. Calculate the vapor pressure of methanol and of ethanol above the solution at 20 °C.
Calculate the mole fraction of methanol and of ethanol in the vapor above the solution.
The triple point of air-free water is defined as 273.15 K. Why is it important that the water be free of air?
A sample of HgCl2 weighing 9.41 g is dissolved in 32.75 g of ethanol, C2H5OH (Kb = 1.20 °C/m). The boiling point elevation of the solution is 1.27 °C. Is HgCl2 an electrolyte in ethanol? Show your calculations.
Show Selected Answers
The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Therefore, some regions will exist in which the water molecules will exclude oil molecules and other regions will exist in which oil molecules will exclude water molecules, forming a heterogeneous region.
Both form homogeneous solutions; their boiling point elevations are the same, as are their lowering of vapor pressures. Osmotic pressure and the lowering of the freezing point are also the same for both solutions.
The molality is m=0.107∘C2.34∘C/m=0.00457m
mol S = 4.57 m × 0.0178 kg = 8.13 × 10−4 mol
Molecular mass=0.210g8.13×10−4mol=285gmol−1
The atomic mass of sulfur is 32.066.
25832.066=8.05
The formula for the sulfur molecule is S8.
The molarity of the solution is:
M=ΠRT=1.32×10−3atm(0.08206L atmmol−1K−1)(298K)=5.40×10−5molL-1
Number of moles = 5.40 × 10-5 mol L−1 × 0.100 L = 5.40 × 10−6 mol
molar mass=0.0750g5.40×10−6mol=1.39×104gmol-1
Molecular mass = 1.39 × 104 amu.
The molarity of the solution is M=ΠRT=7.6atm(0.08206L atmmol−1K−1)(310K)=0.30mol/L
Number of moles = 0.30 mol/L × 1.00 L = 0.30 mol
Mass (glucose) = 180.157 g mol−1 × 0.30 mol = 54 g
Find the molality of the solution from the freezing point depression. Using that value, determine the boiling point elevation and then the boiling point.
ΔTf=|0.0∘C−0.93∘C|=0.93∘C=kfm=1.86∘Cm-1×mmNaCl=0.93∘C1.86∘Cm−1=0.50m
ΔTb = Kbm = 0.512 °C m−1 × 0.50 m = 0.256 °C
The boiling point of pure water is 100.00 °C. Addition gives 100.00 °C + 0.26 °C = 100.26 °C.
XA=XAXA+XB
CH3OH = 32.04246 g mol−1C2H5OH = 46.063 g mol−1molCH3OH=50.0g32.04216gmol−1=1.5604molmolC2H5OH=50.0g46.069gmol−1=1.0853molXCH3OH=1.56041.5604+1.0853=0.590XC2H5OH=1.08531.5604+1.0853=0.410
Vapor pressures are:
CH3OH: 0.590 × 94 torr = 55 torr
C2H5OH: 0.410 × 44 torr = 18 torr
The number of moles of each substance is proportional to the pressure, so the mole fraction of each component in the vapor can be calculated as follows:
CH3OH:55(55+18)=0.75
C2H5OH:18(55+18)=0.25
Δbp=Kbm=(1.20∘C/m)(9.41g×1mol HgCl2271.496g0.03275kg)=1.27∘C
The observed change equals the theoretical change; therefore, no dissociation occurs.
Glossary
activity: effective concentration of ions in solution; it is lower than the actual concentration, due to ionic interactions.
boiling point elevation: elevation of the boiling point of a liquid by addition of a solute
boiling point elevation constant: the proportionality constant in the equation relating boiling point elevation to solute molality; also known as the ebullioscopic constant
colligative property: property of a solution that depends only on the concentration of a solute species
crenation: process whereby biological cells become shriveled due to loss of water by osmosis
freezing point depression: lowering of the freezing point of a liquid by addition of a solute
freezing point depression constant: (also, cryoscopic constant) proportionality constant in the equation relating freezing point depression to solute molality
hemolysis: rupture of red blood cells due to the accumulation of excess water by osmosis
hypertonic: of greater osmotic pressure
hypotonic: of less osmotic pressure
ion pair: solvated anion/cation pair held together by moderate electrostatic attraction
isotonic: of equal osmotic pressure
molality (m): a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms
mole fraction (X): the ratio of a solution component’s molar amount to the total number of moles of all solution components
osmosis: diffusion of solvent molecules through a semipermeable membrane
osmotic pressure (Π): opposing pressure required to prevent bulk transfer of solvent molecules through a semipermeable membrane
Raoult’s law: the partial pressure exerted by a solution component is equal to the product of the component’s mole fraction in the solution and its equilibrium vapor pressure in the pure state
semipermeable membrane: a membrane that selectively permits passage of certain ions or molecules
van’t Hoff factor (i): the ratio of the number of moles of particles in a solution to the number of moles of formula units dissolved in the solution
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1986 AIME Problems/Problem 5
Contents
1 Problem
2 Video Solution by OmegaLearn
3 Solution 1
4 Solution 2 (Simple)
5 Solution 3
6 Solution 4
7 Solution 5 (Easy Modular Arithmetic)
8 Solution 6 (Easiest Factor)
9 See also
Problem
What is the largest positive integer for which is divisible by ?
Video Solution by OmegaLearn
~ pi_is_3.14
~ momeme
Solution 1
If , . Using the Euclidean algorithm, we have , so must divide . The greatest integer for which divides is ; we can double-check manually and we find that indeed .
Solution 2 (Simple)
Let , then . Then Therefore, must be divisible by , which is largest when and
Solution 3
In a similar manner, we can apply synthetic division. We are looking for . Again, must be a factor of .
Solution 4
The key to this problem is to realize that for all . Since we are asked to find the maximum possible such that , we have: . This is because of the property that states that if and , then . Since, the largest factor of 900 is itself we have:
~qwertysri987
Solution 5 (Easy Modular Arithmetic)
Notice that . Therefore
~asops
Solution 6 (Easiest Factor)
In the problem, we can see that has a cube. Immediately think of sum of cubes, which leads us to add 900, resulting in -->. Since the contains , this part is divisible by --> therefore the 900 must be divisible by . The largest that satisfies this is .
See also
1986 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 4Followed by
Problem 6
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
All AIME Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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What does K means in The Inverse Square Law? [closed]
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Today in class i have learned the inverse square law and i was given the equation I=K/d 2 I=K/d 2 for some constant K. What does K really means?? I know that it is a constant but how can you get K or where?
intensity
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asked Oct 14, 2018 at 9:12
Probability_SarahProbability_Sarah
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1 Depends on the physical problem in context.Avantgarde –Avantgarde 2018-10-14 09:20:39 +00:00 Commented Oct 14, 2018 at 9:20
1 Your problem is unclear. There are many kinds of inverse square laws. en.wikipedia.org/wiki/Inverse-square_law#Occurrences What exactly were you studying?Harshit Joshi –Harshit Joshi 2018-10-14 09:34:33 +00:00 Commented Oct 14, 2018 at 9:34
hyperphysics.phy-astr.gsu.edu/hbase/vision/isql.htmlHarshit Joshi –Harshit Joshi 2018-10-14 09:41:34 +00:00 Commented Oct 14, 2018 at 9:41
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K is not a universal constant, in this context you can understand it better by changing the equation to:
K=I∗d 2 K=I∗d 2
So for any given point source of radiation, you have that I∗d 2 I∗d 2 is a constant, it won’t be the same for all sources, but for the same source it will be same the same for any d d. So the statement is that:
I 1∗d 2 1=I 2∗d 2 2 I 1∗d 1 2=I 2∗d 2 2
And that’s what is used to calculate the intensity at different distances
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answered Oct 14, 2018 at 9:30
Hugo VHugo V
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...how can you get K K or where?
Let me briefly answer that with another question −− where can you get (your units of) d d from? That is, suppose you've observed some I I. And your distance from its source is d d. Then you just infer a K=I×d 2 K=I×d 2. But if you've measured d d in meters, then you must have inferred a different K K than if you'd measured d d in feet. So, if you believe that inverse square law you've just learned, K K merely expresses the relationship between your units of intensity measurements and distance measurements, both of which are pretty arbitrary to begin with.
To somewhat remove this kind of arbitrariness, natural units (sometimes called theoretical units, like when I was taught it) can be introduced, where typically values c=ℏ=G=1 c=ℏ=G=1 are assigned to those physical constants, and units of measurement then follow from the assigned numerical values. Lots more discussion at
So, you could likewise just stipulate K=1 K=1 (for whatever physical field we're talking about here), and determine your units of measurement from that. It's six-of-one, half-a-dozen of the other, i.e., either specify your units beforehand and then determine K K from the ratio of your measurements, or else specify K K (usually =1=1) beforehand and then express your measurements in commensurate units.
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edited Oct 14, 2018 at 11:56
answered Oct 14, 2018 at 10:42
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It depends: your question was somewhat vague. As far as I'm concerned, inverse square laws are related to gravitational and electric forces. These can be expressed respectively as
F g=−G m 1 m 2 r 2 r^F g=−G m 1 m 2 r 2 r^
F e=q 1 q 2 4 π ε 0 1 r 2 r^F e=q 1 q 2 4 π ε 0 1 r 2 r^
You could also consider the magnetic force, which can be expressed as
F m=μ 0 q 1 q 2 4 π 1 r 2 R(v 1,v 2,r)F m=μ 0 q 1 q 2 4 π 1 r 2 R(v 1,v 2,r)
Which is kind of an inverse square law as well.
The K K you mentioned could be any of the constants multiplying 1 r 2 1 r 2 in those equations. In general, they relate the properties of the bodies involved in those interactions to the force each one of them feels as a consequence of the latter.
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answered Oct 14, 2018 at 9:35
Vinícius PeixotoVinícius Peixoto
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Table of Contents
Table of Contents
What Is the Least Squares Method?
Understanding the Method
Pros and Cons
Example
FAQs
The Bottom Line
Least Squares Method: What It Means and How to Use It, With Examples
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Definition
The least squares method determines the line of best fit for a dataset represented on a graph.
What Is the Least Squares Method?
The least squares method is a form of mathematical regression analysis used to select the trend line that best represents a set of data in a chart. That is, it is a way to determine the line of best fit for a set of data. Each point of data represents the relationship between a known independent variable and an unknown dependent variable.
This method is often used by stock analysts and traders who want to identify opportunities and trends.
Key Takeaways
Least squares regression is used to predict the behavior of dependent variables.
The least squares method is used to find the best fit for a set of data points.
The method minimizes the sum of the offsets or residuals of points from the plotted curve.
The least squares method provides the overall rationale for the placement of the line of best fit among the data points being studied.
Understanding the Least Squares Method
The least squares method is a form of regression analysis that provides the overall rationale for the placement of the line of best fit among the data points being studied. It begins with a set of data points using two variables, which are plotted on a graph along the x- and y-axis. Traders and analysts can use this as a tool to pinpoint bullish and bearish trends in the market along with potential trading opportunities.
The most common application of this method is sometimes referred to as linear or ordinary. It aims to create a straight line that minimizes the sum of squares of the errors generated by the results of the associated equations, such as the squared residuals resulting from differences in the observed value and the value anticipated based on that model.
For instance, an analyst may use the least squares method to generate a line of best fit that explains the potential relationship between independent and dependent variables. The line of best fit determined from the least squares method has an equation that highlights the relationship between the data points.
If the data shows a lean relationship between two variables, it results in a least-squares regression line. This minimizes the vertical distance from the data points to the regression line. The term least squares is used because it is the smallest sum of squares of errors, which is also called the variance. A non-linear least-squares problem, on the other hand, has no closed solution and is generally solved by iteration.
Important
Dependent variables are illustrated on the vertical y-axis, while independent variables are illustrated on the horizontal x-axis in regression analysis. These designations form the equation for the line of best fit, which is determined from the least squares method.
Advantages and Disadvantages of the Least Squares Method
The best way to find the line of best fit is by using the least squares method. However, traders and analysts may come across some issues, as this isn't always a foolproof way to do so. Some of the pros and cons of using this method are listed below.
Advantages
One of the main benefits of using this method is that it is easy to apply and understand. That's because it only uses two variables (one that is shown along the x-axis and the other on the y-axis) while highlighting the best relationship between them.
Investors and analysts can use the least square method by analyzing past performance and making predictions about future trends in the economy and stock markets. As such, it can be used as a decision-making tool.
Disadvantages
The primary disadvantage of the least square method lies in the data used. It can only highlight the relationship between two variables. As such, it doesn't take any others into account. And if there are any outliers, the results become skewed.
Pros
Easy to apply and understand
Highlights relationship between two variables
Can be used to make predictions about future performance
Cons
Only highlights relationship between two variables
Doesn't account for outliers
Fast Fact
Equations from the line of best fit may be determined by computer software models, which include a summary of outputs for analysis, where the coefficients and summary outputs explain the dependence of the variables being tested.
Example of the Least Squares Method
Here's a hypothetical example to show how the least square method works. Let's assume that an analyst wishes to test the relationship between a company’s stock returns and the returns of the index for which the stock is a component. In this example, the analyst seeks to test the dependence of the stock returns on the index returns.
To achieve this, all of the returns are plotted on a chart. The index returns are then designated as the independent variable, and the stock returns are the dependent variable. The line of best fit provides the analyst with a line showing the relationship between dependent and independent variables.
What Is the Least Squares Method?
The least squares method is a mathematical technique that allows the analyst to determine the best way of fitting a curve on top of a chart of data points. It is widely used to make scatter plots easier to interpret and is associated with regression analysis. These days, the least squares method can be used as part of most statistical software programs.
How Is the Least Squares Method Used in Finance?
The least squares method is used in a wide variety of fields, including finance and investing. For financial analysts, the method can help quantify the relationship between two or more variables, such as a stock’s share price and its earnings per share (EPS). By performing this type of analysis, investors often try to predict the future behavior of stock prices or other factors.
What Is an Example of the Least Squares Method?
Consider the case of an investor considering whether to invest in a gold mining company. The investor might wish to know how sensitive the company’s stock price is to changes in the market price of gold. To study this, the investor could use the least squares method to trace the relationship between those two variables over time onto a scatter plot. This analysis could help the investor predict the degree to which the stock’s price would likely rise or fall for any given increase or decrease in the price of gold.
Who First Discovered the Least Squares Method?
Although the inventor of the least squares method is up for debate, the German mathematician Carl Friedrich Gauss claims to have invented the theory in 1795.
The Bottom Line
Traders and analysts have a number of tools available to help make predictions about the future performance of the markets and economy. The least squares method is a form of regression analysis that is used by many technical analysts to identify trading opportunities and market trends. It uses two variables that are plotted on a graph to show how they're related.
Article Sources
Investopedia requires writers to use primary sources to support their work. These include white papers, government data, original reporting, and interviews with industry experts. We also reference original research from other reputable publishers where appropriate. You can learn more about the standards we follow in producing accurate, unbiased content in our editorial policy.
Stigler, Stephen M. "Gauss and The Invention of Least Squares." The Annals of Statistics, vol. 9, no, 3, May 1981, pp. 465. Download PDF.
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Suppose you have a cube. How many different triangles can be formed by connecting three vertices of the cube?
A) 4
B) 6
C) 8
D) 12
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There are 12 different triangles that can be formed by connecting three vertices of a cube.
Explanation
To find the number of different triangles that can be formed by connecting three vertices of a cube, we first need to identify the number of vertices of a cube. A cube has 8 vertices. To form a triangle, we need to select any 3 vertices from these 8 vertices. This can be done in C(8, 3) ways, which is equal to 56. However, we need to consider that some of these triangles will be congruent to each other, meaning they will be identical. Since the cube has symmetry, there are 3 types of triangles: those in the plane of each face, those with one vertex above a face, and those with one vertex below a face. Each type has 4 distinct triangles. Therefore, the total number of different triangles that can be formed is 4 + 4 + 4 = 12 (option d).
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The total number of different triangles that can be formed by connecting three vertices of a cube is 12. This is calculated by considering various combinations of vertices while accounting for collinearity. Thus, the answer is option D (12).
Explanation
To determine how many different triangles can be formed by connecting three vertices of a cube, we first identify the basic properties of the cube. A cube has 8 vertices. To form a triangle, we need to choose 3 vertices out of these 8.Vertices in a cube can be labeled as follows:
A(0,0,0)
B(1,0,0)
C(1,1,0)
D(0,1,0)
E(0,0,1)
F(1,0,1)
G(1,1,1)
H(0,1,1)
The total ways to choose 3 vertices from 8 is given by the combination formula:
C(8,3)=3!(8−3)!8!=3×2×1 8×7×6=56
However, not all selections of 3 vertices will yield unique triangles, as some sets of three vertices may be collinear (lying on the same line). In a cube, there are several scenarios to consider for selecting vertices that form distinct triangles.
Triangles on the same face: There are 6 faces on a cube, and each face can provide 2 distinct triangles from its corners, giving us 12 triangles.
One vertex above and two vertices on the same face: This arrangement can produce additional triangles, oriented differently but still yielding previously counted configurations.
All vertices from different levels (not all vertices can be taken this way): This leads to no formation of triangles due to collinearity.
After analyzing these distinct formations, we find that the unique formations amount to 12 distinct triangles total. Therefore, the correct answer is option D (12).
Examples & Evidence
For instance, if we consider the vertices of one face of the cube like ABCD, we can form triangles ABC and ABD among others. Each combination yields unique triangles based on their placement in the cube's geometry.
The calculations adhere to the combination formula and geometric properties governing cubes, which competitively enumerate the potential triangle formations based on vertex selection.
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### Free Mathematics solutions and answers
Community Answer suppose you have a cube. how many different triangles can be formed by connecting three verticesof the cube?
Community Answer 1 how many different triangles can be formed by connecting 3 of the vertices of the cube?
Community Answer 5.0 3 Suppose a cube is given. How many non-congruent triangles can be formed by connecting 3 of the vertices of the cube?
Community Answer 5.0 Let ABCDEFGH be a cube. a) How many different segments can you form by connecting the vertices of the cube? b) How many different triangles can you form by connecting 3 of the vertices of the cube?
Community Answer 5.0 5 How many non-congruent triangles can be formed by connecting 3 of the vertices of the cube?
Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer
Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four?
Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish?
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Consider the following incomplete deposit slip: | Description | Amount ($) | | ---: | Cash, including coins | 150 | 75 | | Check 1 | 564 | 81 | | Check 2 | 2192 | 43 | | Check 3 | 4864 | 01 | | Subtotal | ???? | ?? | | Less cash received | ???? | ?? | | Total | 7050 | 50 | | | | | How much cash did the person who filled out this deposit slip receive?
Which sequence is generated by the function f(n+1)=f(n)−2 for f(1)=10? A. −2,8,18,28,38,… B. 10,8,6,4,2,… C. 8,18,28,38,48,… D. −10,−12,−14,−16,−18,…
i) Write down the expansion of (1+x)3. ii) Find the first four terms in the expansion (1−x)−4 in ascending powers of x. For what values is the expansion valid? iii) When the expansion is valid, find the values of a and b in the equation below. (1−x)4(1+x)3=1+7 x+a x 2+b x 3+……
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Solve the simultaneous equations: [ \begin{array}{r} 5x+y=21 \ x-3y=9 \end{array} ]
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in this video we're going to talk about how to add vectors so a vector is a quantity that has magnitude and direction so let's say if we have a force vector that's 100 newtons directed towards the east the magnitude is the size of the force it's 100 newtons and the direction is east and let's say if we want to add it to another force that's directed east as well let's say it's 50 newtons the resultant sum of these two forces is going to give us a net force of 150. whenever you have two vectors that are parallel to each other you can simply add the numbers to get the resultant sum now let's say if we have a 200 newton vector directed east and 120 newton vector directed west what is the resultant vector and also specified the direction so this is positive 200 because it's directed towards the right this is negative 120 because it's directed towards the left if you add these two you should get a net force of positive 80. now it's going to be smaller than the original one but it's still directed east because this vector is greater than this one now as another example let's say if we have a force vector that's 60 newtons directed east and another one that's 90 newtons directed west what is the resultant force vector and what's the direction so this is positive 60 that's negative 90 so we're gonna have a net force of negative 30 newtons or you could just say 30 newtons directed west now what if we have a force that's 80 newtons directed north and another one that's 120 direct itself so 120 minus 8 is 40. so the net force is going to be 40 itself and if you recall this is north south east west now sometimes you may need to add two vectors that are not parallel or anti-parallel to each other so let's say if we have a third newton force vector directed east and a 40 newton force vector directed north what is the resultant force vector if you have two vectors that are perpendicular to each other you could find the resultant force vector by finding the left of the hypotenuse of the right triangle that is formed so therefore the resulting force vector is going to be the square root of f1 squared plus f2 squared let's call this f1 and this force f2 so it's going to be the square root of 30 squared plus 40 squared if you're familiar with the 345 triangle then this is going to be the 30 40 50 triangle so the resultant force is going to be 50 newtons now sometimes in addition to finding the magnitude of the resultant force vector which is 50 you may need to find the direction as well so you got to find the angle theta to find it you can use this formula it's the inverse tangent of the y component divided by the x component so the force in the y direction is 40 and the x direction is 30. so to find the angle it's going to be our tangent the opposite side which is the y value of 40 over the adjacent side which is the x value of 30. hopefully you're familiar with sohcahtoa sine is opposite over hypotenuse cosine is adjacent over hypotenuse and tangent is opposite relative to theta over at the adjacent side the arc tangent of 40 over 30 is 53.1 degrees so we can say that the resultant force vector has a magnitude of 50 newtons and a direction of 53.1 degrees relative to the x-axis let's try another example so let's say if we have a force vector that's 50 newtons directed west and another one that's 120 newtons directed south calculate the magnitude of the resulting force factor and find the direction so let's draw a triangle so this side is 50 and this side is going to be 120 so the resultant force vector is the hypotenuse of this triangle so notice that it's in quadrant three now if you're familiar with the 5 12 13 triangle then the hypotenuse has to be 130. so the resultant force vector is going to be the square root of 50 squared plus 120 squared and that's going to turn out to be 130 newtons so that's the magnitude of the resultant force vector now all we need to do is find the angle so first let's find a reference angle to find a reference angle use the arctangent formula take the force in the y direction divided by that in the x direction but make it positive initially this will give you an acute angle between 0 and 90 and then you could adjust it later so our tan 120 over 50 will give you a reference angle of 67.4 so here's the reference angle it's inside the triangle now what we need is the angle relative to the positive x-axis so if this is 180 then the resultant vector is 180 plus 67.4 relative to the x-axis so it's at an angle of 247.4 so therefore this is the resultant force vector which has a magnitude of 130 but an angle of 247.4 degrees so that's the answer now let's say if we have a force vector of 45 newtons directed east and we wish to add it to a force vector of 60 newtons directed south find the magnitude of the resulting force vector and also the angle so let's draw a triangle so this is 45 and this is 60. so let's find the magnitude of the resultant force vector so it's going to be the square root of 45 squared plus 60 squared which is 75 newtons so that's the magnitude now we got to find the angle but let's find a reference angle first so it's going to be arc tangent the force in the y direction divided by the force in the x direction so that's 60 over 45 which is 53.1 now that's the reference angle the first force vector is positive it's directed east the second one is directed south so therefore the resultant force vector is in quadrant four now if this angle is 53.1 to find the angle relative to the positive x-axis it's going to be 360 minus 53.1 keep in mind a full circle is 360 but you need to go back 53.1 degrees so 360 minus 53.1 will give us an angle of 306.9 so the resultant force vector has a magnitude of 75 newtons and it's directed at an angle of 306.9 degrees relative to the positive x-axis so this is the answer so if you know the resultant force vector is going to be in quadrant one then the angle is going to be the same as the reference angle and keep in mind the reference angle can be calculated by taking the inverse tangent of the force in the y direction divided by the force in the x direction and if you always make these positive you're always going to get an acute angle between 0 and 90 which is the reference angle now if you know the resultant force vector is in quadrant two then to find that angle it's going to be 180 minus the reference angle if it's in quadrant three we cover this one already it's 180 plus the reference angle and if you get an answer that's in quadrant four it's going to be 360 minus the reference angle which was the case in the last example so it's helpful to know these things or you could just see what to do visually once you graph everything now sometimes you may need to add vectors that are not parallel or perpendicular to each other so let's say if we want to add this vector which is a 100 inch directed east plus another vector let's say it has a magnitude of 150 but it's directed at 30 degrees above the x-axis how can we find the resultant vector first let's re-list what we have the first vector has a force of a hundred newtons and its angle it's east so it's on the x-axis so it has an angle of zero degrees the second force vector has a magnitude of 150 newtons and it has an angle of 30 degrees relative to the x-axis now what you want to do is you want to add these vectors using the component method so you want to break these forces into the components add their respective x and y components and then you could find the results in force vector so let's say if this is a force vector this is the x component and this is the y component and here is the angle f of x is equal to f cosine theta and f of y is f sine theta f you can find it by using pythagorean theorem it's f of x squared plus f of y squared and the angle theta is the arc tangent of f of y divided by f of x so those are some formulas that you're going to find useful in this portion of the video so what we need to do is find f one x and f one y so f one x is gonna be a hundred cosine of zero cosine of zero is one so a hundred times one is simply a hundred f one y is going to be a hundred sine of zero sine of zero is zero a hundred times zero is going to be zero now let's do the same for f2x and f2y f2x is 150 times cosine of 30 degrees now if you type that in your calculator make sure it's in degree mode so this will give you as a decimal 129.9 newtons f2y that's going to be 150 times sine 30 which is just 75. now what we need to do is add up the x components so if we add those two values the sum of the forces in the x direction is going to be a hundred plus 129.9 so that's going to be 229.9 and then we need to find the sum of the forces in the y direction which is just 0 plus 75 so that's simply 75. so f of x is 229.9 units and f of y is simply 75 newtons now let's go ahead and draw these values on an xy plane so f of x is 229 and f of y is 75. the resultant force vector is the hypotenuse and to find it it's going to be the square root of f of x squared plus f of y squared so you should get 241.8 so that's the resulting force vector that's the magnitude of it now that we have the magnitude we need to find the angle so it's going to be arc tangent f of x i mean f of y which is 75 divided by f of x which is 229.9 so the angle which is already in quadrant one is 18.1 degrees so now we have the magnitude and the direction of the resultant force vector |
1362 | https://www.ub.edu/gdne/amaydeusp_archivos/modeling%20pref%20data%20handbook09.pdf | [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 264 264–282 12 Modeling Preference Data Alberto Maydeu-Olivares and Ulf Böckenholt INTRODUCTION A great deal of data in psychological research can be considered the result of a choice process. Familiar instances are a citizen deciding whether to vote, and if so for whom; a shopper contemplating various brands of a product category; and a physician deciding various treatment options. Less obvious examples of discrete choices are responses to multiple-choice items of a proficiency test in mathematics or to rating items in a personality questionnaire. Here, an individual’s answers may be viewed as her top choices among the alternatives presented. Choices may also be expressed in an ordinal and continuous fashion. Instances include decisions on how much food to consume, how much to invest in the stock market, or how much to pay in an online auction. Finally, multivariate choices may be observed when considering, for example, consumers’ choices of brands within different product categories and their respective quantity purchases.
Choice outcomes may be gathered in either natural or experimental settings. Both types of outcomes are of interest, as they can often complement one another. They are referred to as revealed and stated prefer-ences respectively (Louviere et al., 2000).
For instance, in an election, stated preferences (rankings of the candidates shortly before the election) may prove more useful for predicting the election outcome and provide more information about the motives than would such revealed preferences as past-voting behavior. As revealed preferences are frequently collected in observational studies, they are more difficult to interpret in an unambiguous way, and they can also provide considerably less information than stated preferences. Typically, revealed preferences are first choices. Second-best option or least-liked option, etc. are less commonly observed, although they may be critical for accurately forecasting future behavior. Moreover, the collection of stated preference data is also useful in situations when it is difficult or impossible to collect revealed preference data because choices are made infrequently, or because new-choice options offered in the studies are yet unavailable on the market. For example, in taste-testing studies, consumers may be asked to choose among existing as well as newly-developed products.
Often, standard models can be used to model preference data. Thus, to model how much to invest in the stock market, for example, as a function of economic variables or psychological factors we may use the [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 265 264–282 MODELING PREFERENCE DATA 265 regression models discussed in Chapter 3.
Or to model the responses to a multiple-choice test we may use the item-response theory methods discussed in Chapter 7.
Finally, to analyze the ratings in a personality questionnaire we may use the factor-analysis methods discussed in Chapter 6. However, we also note that care needs to be taken in the selection of an appropriate model because one needs to take into account the response process that leads to the observed-choice data. For example, the choice of a response category in a mathematics test may be driven by both the abilities of the respondents and the difficulties of the items.
In contrast, the decision on how much to invest may depend on investment knowledge, the budget available and the respondents’ perception of risk. In ability testing, much work has focused on the separability of item and person characteristics leading to item-response models. However, in preference analysis, it is frequently a foregone conclusion that item and person characteristics are not separable. As a result, statistical tools are needed that can identify how respondents differ in their perception and preferences for a set of choice options (Böckenholt and Tsai, 2006).
The choice models that we consider in this chapter include the logistic-regression model (Bock, 1969; Luce, 1957; McFadden, 2001) and Thurstone’s (1927) class of models for comparative data in the form of rankings or paired comparisons. Both classes of models are probabilistic in nature and focus on deci-sion problems with a finite number of options.
They allow predicting how observed and unobserved attributes of both decision makers and choice options determine decisions. It is important to note that these models focus mainly on choice outcomes and to a lesser extent on underlying-decision processes. As a result, their main purpose is to summarize the data at hand and to facilitate the forecasting of choices made by decision makers facing possibly new or different variants of the choice options.
The objective of this chapter is to provide a gentle overview of modeling choice data, with an emphasis on statistical models that allow treating both observed and unobserved effects due to the decision makers and choice options. Our discussion of how to model individualdifferencesintheevaluationaswell as selection of choice options will consider first the situation when decision makers express their preferences in the form of liking judgments or purchase intentions.These types of data are commonly collected in conjoint studies (Marshall and Bradlow, 2002), which aim at measuring preferences for product attributes. We will then consider applications that involve partial and/or incomplete ranking data (Bock and Jones, 1968). Incomplete ranking data are obtained when a decision maker considers only a subset of the stimuli.
For example, in the method of paired comparison, two stimuli are presented at a time, and the decision maker is asked to select the preferred one. In contrast, in a partial ranking task, a decision maker is confronted with all stimuli and asked to provide a ranking for a subset of the available options.
For instance, in the best–worst method, a decision maker is instructed to select the best and worst options out of the set of choice options offered.
Both partial and incomplete approaches can be combined by offering multiple, distinct subsets of the choice options and obtaining partial or complete rankings for each of them.
For instance, a judge may be presented with all possible stimulus pairs sequentially and asked to select the preferred stimulus in each case.
Presenting choice options in multiple blocks has several advantages. First, the judgmental task is simplified since only a few options need to be considered at a time. Second, as we show later, it is possible to investigate whether judgesareconsistentintheirevaluationsofthe stimuli. Third, obtaining multiple judgments from each decision maker simplifies analyses of how individuals differ in their preferences for the stimuli, as we illustrate in one of the examples. These advantages need to be balanced with the possible boredom and learning effects that may affect a person’s evaluation of the stimuli when the number of blocks is large.
[16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 266 264–282 266 SCALING Analyses of partial and/or incomplete ranking data require the additional speci-fication that choice outcomes are a result of a maximization process. In other words, decision makers are assumed to select or choose options that have the highest utility among the considered options. These utilities are not observed but can be inferred, at least partially, from the choices observed under the maximization assumption. Because less information is available about the underlying utilities in a choice task than in a rating setting, we discuss interpretational issues in the application of choice models for partial and/or incomplete ranking data as well.
A BASIC MODEL FOR CONTINUOUS PREFERENCES Suppose continuous preferences (i.e., ratings on a 0 to 100 scale) have been obtained in a sample of N individuals from the population we wish to investigate on n stimuli. The goal of these analyses is to understand how individuals differ in the way they weight observed or unobserved attributes of the stimuli in their overall preference judgment.
Consider the following two-level model: yi = ν + νi (1) νi = 1ϕi + Wγi + Bxi + ηi + εi (2) ϕi = αϕ + ζϕi (3) γi = αγ + ζγ i (4) ηi = αη + ζηi (5) Equations (1) to (3) can be expressed in the combined equation: yi = ν + 1 αϕ + ζϕi + W αγ + ζγ i + Bxi + αη + ζηi + εi (6) Equation (1) states that respondent’s i pref-erences for the n stimuli,yi, equals the mean preference for each stimulus in the population of respondents, ν, plus the difference between the population average and the respondent’s preferences νi. Equation (2) assumes that this difference νi depends linearly on: (1) an intercept varying across respondents but common to all stimuli, νi,which captures the respondent’s average preference across stimuli; (2) r observed attributes of the stimuli w weighted idiosyncratically by each respondent with weights γ i; (3) p observed characteristics of the respondent (e.g., gen-der, education, etc.), xi; (4) m unobserved characteristics of the respondents, ηi; and (5) an error term, εij, which captures the respondent’s preference not accounted for by themodelaswellasrandomfluctuationsofthe respondent’s preferences. Finally, Equations (3) to (5) state that an individual’s inter-cept, weights, and unobserved characteristics, depend on the population means α, plus an error term, ζi, which captures the difference between the individual and the means across individuals.
The m unobserved characteristics of the respondents are common factors (see Chapter 6). In turn, the observed character-istics of the respondents, x, and the observed characteristics of the stimuli, w, may be metric variables or dummy variables which represent categorical factors (see Chapter 13). This is a basic setup for modeling preferences in the sense that it accounts for observed and unobserved attributes of the decision makers and allows relating attributes of the stimuli to the overall preference judgment with person-specific regression weights.
Although fairly general, this model can be extended in two important ways. First, there may be interactions between the respondents’ characteristics, x, between stimuli attributes, w, or between the respondents’ character-istics and the stimuli attributes. Especially, the latter effect can be of great interest in preference modeling when investigating how respondents with different background characteristics differ in terms of the per-ception and evaluation of the same choice option. For example, in survey studies on US politicians involving thermometer ratings (Regenwetter et al., 1999), it is well known that Republican and Democrat voters may disagree in systematic ways on their evalu-ation of political programs endorsed by the politicians.
[16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 267 264–282 MODELING PREFERENCE DATA 267 Second, the relationship between prefer-ences and the respondents’ characteristics and stimulus attributes may be non-linear.
Asimpleexampleisthelikingofthesweetness of drink as a function of the number of spoonfuls of sugar used. Too much or too little sugar may lead to lower likings suggesting a quadratic relationship between these two variables. In this case, an ideal-point model may prove superior to a linear representation when individuals choose the option that is closest to their ‘ideal’or most preferred option (Böckenholt, 1998; Coombs, 1964; MacKay et al., 1995).
Both extensions can be incorporated straightforwardly in the basic model for the observed characteristics of the respondents and stimuli. It is therefore instructive to consider also special cases of Equation (6). A special case is obtained when no information on the attributes of the stimuli or the observed characteristics of the respondents are available. In addition, the respondent specific intercept, ϕi, is not generally included in the model – but see Maydeu-Olivares and Coffman (2006). This leads to: yi = ν + ηi + εi, ηi = αη + ζ ηi (7) Thus, in this model, preferences among the n stimuliareexplainedsolelybyasetofm unob-servable characteristics of the respondents η, which leads to a model with m common factors. The common factors are treated as random effects, and they are assumed to be uncorrelated with the random errors. The variances of the random errors are assumed to be equal across respondents within a stimulus, but typically they are allowed to be different across stimuli. Also, random errors are assumed to be mutually uncorrelated across stimuli, so their covariance matrix is diagonal.
Finally, the means of the random errors and common factors are specified to be zero.
An interesting special case of the factor model is obtained when it is assumed that the mean preferences depend on the means of the unobserved factors. In the factor- analysis model, ν and αη are not jointly identified.
However, αη can be estimated if it is assumed that the intercepts are equal for all stimuli, υ = υ1. With this assumption, the population mean and covariance matrices of the observed preferences are: µ = ν1 + α, = ′ + (8) The latter expression is the standard formula for the covariance structure of the factor-analysis model where and denote the covariance matrices of ζη and ε, respectively.
Anotherspecialcaseofthegeneralmodelof Equation (6) is obtained when no information on the stimuli’s attributes is available and no unobserved characteristics of the respondents are specified. Also, the respondent specific intercept, ϕi, is not included in the model. In this case we have: yi = ν + Bxi + εi (9) This equation is a multivariate (fixed effects) regressionmodelwheretherespondents’char-acteristics are used to explain the preferences.
Typically, the x are assumed to be fixed and the errors are assumed to be independent with mean zero and common variance within a stimuli, but variances may be different across stimuli and errors across stimuli may be correlated.
Finally, when no information on the respon-dents’ characteristics is available and no unobserved characteristics of the respondents are specified, we have: yi = ν + 1ϕ + Wγi + εi (10) One approach to specify model (10) is to treat ϕi and γ i as random effects, where the random intercepts ϕi and random slopes γ i are assumed to be mutually uncorrelated and uncorrelated with the random errors εi. In this case, Equation (10) is a multivariate random-effects regression model. As in the common-factor model, the mean of the random errors is specified to be zero and their covariance matrix is assumed to be diagonal. In fact, the random-effects multivariate regression model is closely related to the common-factor model. The key differences between these [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 268 264–282 268 SCALING two models are that in the random-effects regression model: (1) the number of latent factors is fixed, r + 1 (the additional latent factor is the random intercept); and (2) the factor loadings are fixed constants (given by the n × r design matrix W). Also, as in the factor-analysismodel,itisinterestingtoletthe mean preferences depend on the means of the random intercepts and slopes, αϕ and αγ . To do so, ν must be set to zero for identification, leading to: µ = W∗α∗, = W∗∗W∗′ + (11) where W∗= 1 W , α∗= αϕ αγ , ∗ denotes the covariance matrix of ϕi, γ ′ i ′, and denotes the covariance matrices of the random errors.
An alternative approach to specify the model is to treat ϕi and γ i as fixed effects.
Again, in this case ν cannot be estimated, but the parameters of interest, ϕi and γ i, can be estimated for each person separately. This approachistakeninclassicalconjointanalysis (Louviere et al., 2000). In this popular technique, preferences are modeled using (10) with ν = 0 on a case-by-case basis where the r stimuli attributes are generally expressed as factors (in the analysis of variance sense) using effect coding.
Some remarks on estimation structural equation modeling (see Chapter 21) provides a convenient way of estimating the general model and its special cases presented in this section. Assuming multivariate nor-mality of the random variables y, estimation maybeperformedusingmaximumlikelihood.
However, it suffices to assume that the distribution of the observed preferences y conditional on x is multivariate normal. This assumption enables the inclusion of non-normal exogenous variables in the model, such as dummy variables (for further technical details, see Browne and Arminger, 1995).
When the observed preferences are non-normally distributed, asymptotically-robust standard errors and goodness of fit tests for maximum likelihood estimates can be obtained; see Satorra and Bentler (1994) for further details.
NUMERICAL EXAMPLE 1: MODELING PREFERENCES FOR A NEW DETERGENT Hair et al., (2006) provide ratings of 18 detergents on a 7–point scale ranging from ‘not at all likely to buy’ to ‘certain to buy’ by 86 customers. We note that although Hair et al. (2006) report the results obtained using 100 respondents, the dataset available for download contains only 86 respondents. The detergents were obtained using a fractional design (see Chapter 2) involving five factors: 1.
Form of the product (premixed liquid, concen-trated liquid, or powder).
2.
Number of applications per container (50, 100, or 200).
3.
Addition to disinfectant (yes, or no).
4.
Biodegradable (no, or yes).
5.
Price per application (35, 49, or 79).
The first, second and fifth attributes of this conjoint analysis consist of three levels, whereas the other attributes consist of two levels. With k levels per attribute, only k −1 are mathematically independent. Here, arbitrarily, we shall estimate the effects corre-sponding to the first k −1 levels. Also, notice that the attributes with three levels could be treated metrically, using a linear or quadratic function, etc. Here, we shall estimate them as analysis of variance (ANOVA) factors.
Fixed effects modeling: conjoint analysis If ϕi and γ i are treated as fixed effects, they can be estimated for each respondent separately. Thus, for each respondent, there are 18 observations and nine parameters: one intercept, two parameters each for factors 1, 2 and 5, and one parameter for each of the remaining two factors. In conjoint-analysis terminology, the predicted responses ˆ yi are called utilities and the estimated regression [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 269 264–282 MODELING PREFERENCE DATA 269 (actually ANOVA) parameters ˆ γ i are called part-worth utilities. In fact, in conjoint analysis part-worth utilities are estimated for all factor levels using the constraint that all parameters for an attribute within a respondent add up to zero. Typically, the part-worth utilities are only of secondary interest.
Of primary interest are the importance of each attribute in determining choice and the proportion of times that an option will be chosen in the population of consumers (see Louviere et al., 2000 for details on how to compute these statistics).
Here, we shall focus on the parameter estimates. Table 12.1 provides the means and variances of the parameter estimates averaged across the individual regressions. No standard errors are readily available when population meansandvariancesareestimatedinthisfash-ion (but see Bollen and Curran, 2006: 25–33).
Random effects approach Alternatively, ϕi and γ i can be treated as random effects. This multivariate regression random-effects regression model can be estimated as a confirmatory factor-analysis model where the factor matrix is given by the design matrix employed. In this example, the design matrix W∗for the 18 stimuli is: W∗= ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ 1 0 1 −1 −1 1 1 1 0 1 −1 −1 −1 −1 1 1 1 0 1 1 0 0 1 1 −1 0 1 1 −1 −1 −1 −1 1 −1 0 1 1 −1 −1 1 0 1 1 −1 −1 1 0 1 −1 −1 −1 −1 −1 −1 1 1 0 0 1 1 1 −1 −1 1 1 0 −1 −1 1 1 0 1 1 −1 −1 0 1 −1 1 0 1 1 0 1 1 0 1 1 0 1 1 −1 −1 0 1 −1 1 1 0 1 0 1 0 1 1 1 −1 −1 1 1 0 −1 −1 −1 1 −1 −1 1 1 0 1 0 1 1 1 0 1 0 1 0 1 1 −1 1 0 1 1 0 1 0 −1 −1 1 0 1 0 1 1 0 −1 1 0 1 1 −1 −1 1 0 1 −1 −1 −1 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ .
(12) The first column corresponds to the random intercept. Columns 2 and 3 correspond to the first two levels of the first factor, columns 4 and 5 correspond to the first two levels of the second factor, column 6 to the first level of the third factor, and so on. Notice how effect coding has been used, as is customary in conjoint analysis. The variances and covariances of the nine random effects can be estimated as well as their means if ν = 0 (for identification).Also, the covariance matrix of the random errors is assumed to be a diagonal matrix. This two-level regression model can be readily estimated with any software package for structural equation or multilevel modeling.
Assuming normality of the observations, we obtained by maximum likelihood that the model fits very poorly, X2 = 270.65 on 117 df, p < .01, the RMSEA (see Chapter 21, and also Browne and Cudeck, 1993) is 0.12.
This is a valuable piece of information, as with the fixed effects approach we could not obtain an overall assessment of the model’s fit.
Rather, we obtained an R2 for each individual separately (which in most cases ranged from 0.75 to 0.95).
Table 12.2 provides the estimated popu-lation means and variances of ϕi and γ i.
Notice in this table that the estimated variance for the preferences for concentrated liquid detergents is very small (0.001) suggesting that individuals vary little in their weight of this factor. Comparing the parameter estimates across methods (fixed effects versus random effects), we see that the estimated means are rather similar. The estimated variances, in contrast, appear generally larger when estimated on a case-by-case fashion.
The reasons for the discrepancy in the variance estimates are probably threefold.
First, in view of the large number of parameters that are estimated for each person in the fixed-effects approach, it is not surprising that the variance estimates are much larger in this case. Second, the assump-tion that the random effects are normally distributed constrains the estimates of the random effects’ variances and covariances.
Similar constraints are not in place when [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 270 264–282 270 SCALING Table 12.1 Estimated means and variances in the conjoint analysis example: fixed effects (case-by-case) results Premixed Concent.
50 100 Price Price Intercept liquid liquid applicat.
applicat.
Disinfectant Biodegrad.
35̸c 49̸c Mean 3.74 –0.22 0.17 –0.35 0.02 0.51 –0.15 1.13 0.08 Var.
0.63 0.23 0.15 0.32 0.19 0.38 0.17 0.63 0.25 Table 12.2 Estimated means and variances in the conjoint analysis example: random effects results Premixed Concent.
50 100 Price Price Intercept liquid liquid applicat.
applicat.
Disinfectant Biodegrad.
35̸c 49̸c Mean 3.74 −0.19 0.14 −0.34 0.02 0.50 −0.13 1.13 0.10 (0.09) (0.05) (0.04) (0.06) (0.05) (0.07) (0.05) (0.09) (0.05) Var.
0.55 0.10 0.001 0.21 0.07 0.30 0.11 0.52 0.10 (0.10) (0.04) (0.02) (0.05) (0.04) (0.06) (0.03) (0.10) (0.04) estimating the regression coefficients for each person separately. The multivariate-normality assumption of the random effects may only be partially appropriate for this data set: The distribution of the fixed-effect estimates of the coefficients for the ‘addition to disinfectant’ factor appears to be bimodal. However, the distributions of the other coefficients appear roughly normal, except for a few outliers and some excess kurtosis. Third, the two methods make different assumptions about the residual error variances. Whereas the individual-regression approach assumes that the εi are constant across stimuli but different across respondents, our random-effects model assumes that the εi are constant across respondents but different across stimuli. This latter specification can be relaxed provided covariates are available that allow model-ing heteroscedasticity effects on the person level.
In closing this example, we note that the structural equation modeling of the random-effects specification facilitates the testing of a number of interesting hypotheses.
For instance, one may test whether the residual errors are correlated for some stimuli.
This consideration of local dependencies may be particularly useful when similarities among stimuli (caused, for example, by the same presentation formats) cannot be accounted for by individual differences.
Also, replicated stimuli are accommodated easily.
Hair et al.
(2006) provide two replicates for each respondent. Modeling both replicates simultaneously using the random-effects model requires using a 36 × 9 design matrix obtained by duplicating the matrix in Equation (12).
NUMERICAL EXAMPLE 2: MODELING PREFERENCES FOR SPANISH POLITICIANS In our first example, we saw an instance of the basic model where preferences were modeled as a function of observed characteristics of the stimuli using (10). In this example, we shall model instead preferences as a function of unobserved characteristics of the respondents using (7). The Centro de Investigaciones Sociológicas (CIS) of the Spanish Government periodically obtains a representative national sample of approval ratings on a scale from 0 to 10 for the Ministers of the Spanish Government along with the leaders of the opposition parties. Here, we used the October 2004 data and selected the eight politicians with the lowest amount of missing responses. Using listwise deletion, we obtained a final sample size of 576. The purpose of this example is to show how unobserved characteristics of the respondents can be used to predict the average approval rating of each politician. Table 12.3 shows the average-approval ratings for the eight politicians analyzed. As we can see from this [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 271 264–282 MODELING PREFERENCE DATA 271 Table 12.3 Results for the political ratings example Factor loadings Politician Centralism–peripherialism Left–right Nationalism–non-nationalism ¯ y ˆ µ R2 Zapatero .95 1.00 1.60 5.55 5.55 72% Solbes .57 1.05 1.54 5.29 5.28 62% Bono .56 .87 1.90 5.01 5.02 69% Rajoy –2.06 1.81 –.25 4.43 4.43 98% Duran 1.14 1.21 –.15 3.65 3.66 54% Llamazares 1.59 .92 .56 3.64 3.64 57% Carod 2.24 .98 –.32 2.95 2.94 72% Imaz 1.79 1.11 –.55 2.86 2.86 76% table, the average ratings range from 5.55 to 2.86. The Spanish president at the time of the study (equivalent to Prime Minister in other political systems), Rodriguez Zapatero, obtained the highest rating, and the leader of the main opposition party, Rajoy, the fourth highest rating. The second and third positions are for two ministers of Zapatero’s cabinet, Solbes and Bono. A lower rating is obtained by the leader of a leftist party, Llamazares.
Low ratings are also obtained by the leaders of smaller, regional parties, Duran, Carod and Imaz. The regional parties these three politicians represent focus on the national identityoftheautonomousregionswheretheir parties operate. The aim of these parties is to increase the power of their regions with respect to the central Spanish government, in some cases with the declared objective of achieving independence.
Our model postulates that respondents use a number of unobserved preference dimensions to rate these politicians. We use a factor-analysis model to uncover these dimensions. Since the observed rat-ings are not normally distributed, we used maximum likelihood with robust standard errors and Satorra-Bentler mean adjusted goodness of fit statistics. A model with one common factor, which we interpreted as right-left political affiliation, fits very poorly, Satorra-Bentler (SB) mean adjusted X2 = 801.7 on 20 df, RMSEA = .263.
Amodel with two common factors, which can be interpreted as centralism–peripheralism and non-nationalism–nationalism, also fits rather poorly, SB X2 = 114.2 on 13 df, RMSEA = .118. However, a model with three dimensions cannot be rejected at the 5% significance level, SB X2 = 13.8 on 7 df, p = .05, RMSEA = .042. Next, we constrain the mean ratings to depend on the common factors, while estimating the factor means. That is, according to this model, the population mean and covariance matrices are given by Equation (8).
The model fits the data adequately: SB X2 = 24.2 on 11 df, p = .01, RMSEA= .046, and yields interesting insights into the indi-vidual differences underlying the ratings of the politicians. The factor loadings for this mean-structured factor model are provided in Table 12.3. In this model, the factor loadings represent the position of the politicians in the preference space of the respondents. A plot of the factor loadings (i.e., a preference map) facilitates the interpretation of the dimensions. The preference map is provided in Figure 12.1. One of the dimensions can be interpreted as centralism – peripheralism.
High scores on this dimension indicate that politicians are perceived as favoring a weak central government and more political power for Spain’s autonomous regions. Another dimension can be interpreted as Left–Right; higher scores indicate that politicians are perceived as endorsing conservative views on social issues and liberal views on economic issues. The third dimension is slightly more difficult to interpret. It may be interpreted as nationalism–non-nationalism, lower scores indicate that a politician’s discourse is per-ceived as focusing on national-identity issues, although the target nation differs, it may be Spain for Rajoy, the Basque Country for Imaz, or Catalonia for Duran and Carod.
[16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 272 264–282 272 SCALING centralism - peripherialism 2.5 1.5 .5 −.5 −1.5 −2.5 left - right 2.5 1.5 .5 −.5 −1.5 −2.5 solbes bono zapatero rajoy llamaza imaz duran carod centralism - peripherialism 2.5 1.5 .5 −.5 −1.5 −2.5 nationalism - nonnationalism 2.5 1.5 .5 −.5 −1.5 −2.5 solbes bono zapatero rajoy llamaza imaz duran carod left - right 2.5 1.5 .5 −.5 −1.5 −2.5 nationalism - nonnationalism 2.5 1.5 .5 −.5 −1.5 −2.5 solbes bono zapatero rajoy llamaza imaz duran carod Figure 12.1 Three-dimensional preference map of political preferences in Spain.
[16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 273 264–282 MODELING PREFERENCE DATA 273 Figure 12.1 and Table 12.3 show that there is not much perceived variability along the Left–Right dimension. Rajoy is perceived as slightly more to the Right than the remaining politicians, who are clustered together along this dimension. There is more perceived variability on the nationalism–non-nationalism dimension, with the leaders of the three regional-nationalist parties (Duran, Imaz and Carod) clustered together at one extreme and the two ministers of the Socialist Party and their president (Bono, Solbes and Zapatero), clustered at the other end. The leader of the main opposition party, Rajoy, is perceived as a (Spanish) nationalist, and Llamazares’ position is perceived to fall between both clusters. The dimension with the highest observed variability is centralism– peripheralism, with Rajoy on one extreme and the leaders of two of the regional-nationalist parties on the other extreme.
The factor model reproduces well the observed average ratings (see Table 12.3 for the politicians’ approval ratings expected under the model). Also, the R2 are quite high.
They range from 54% for Duran and 57% for Llamazares to 98% for Rajoy. Thus, these three dimensions explain almost exactly the average rating for Rajoy, but a substantial portion of the ratings’ variance for Duran and Llamazares depends on variables other than the dimensions considered here.
The model can be used to predict the average ratings when a politician’s position changes in this preference map (and when everything else remains the same). Under the model the average ratings depend linearly on the politicians’ position in the map and the population means on these dimen-sions. The estimated population means for centralism–peripheralism, Left–Right, and nationalism–non-nationalism are 0.61, 6.11 and 1.91 respectively. This means that if respondents perceived that a politician’s posi-tion had increased by one unit towards periph-eralism, Right, or non-nationalism extremes, the politician’s average rating would increase by 0.61, 6.11 and 1.91 points respectively.
These predictions have to take into account the range of values obtained. Extrapolating beyond the observed range may be misleading as we do not know if the model is appropriate beyond that range. This means, for instance, that we cannot predict what the average rating of Rajoy would be if his perceived position increased further along the Right dimension because he already has the highest position on this dimension. Also, we need to bear in mind that the model assumes that ratings increase linearly in capturing a politician’s position on the map, which of course is impossible. Like any other linear model, the model fitted here can only be regarded at best as an approximation within the range of the observations. For problems of this kind, a model that specifies that ratings increase non-linearly as a function of the politicians’ position may be more appropriate. One such non-linear model is an ideal-point model (see McKay et al., 1995) which states that the closer a politician is to the preferred position of a respondent, the higher his or her rating.
In closing this example, it is interesting to compare the R2 for the politicians’ ratings obtained when preferences are expressed solely as a function of unobserved char-acteristics of the stimuli (as we just did), to the R2 obtained when the ratings are expressed solely as a function of the observed characteristics of the respondents – using Equation (9) In so doing, using the region where the respondent resides, gender, age, and a self-score along the Left–Right dimension as predictors, we obtain R2’s ranging from 11% (for Duran) to 41% (for Rajoy). Thus, in this example, using unobserved characteristics of the respondents predicts a substantially larger amount of variance of the ratings than using the observed characteristics of the respondents. Both sources of information can be combined using the basic model of Equation (6), but we will not pursue this possibility here.
MODELING DISCRETE OUTCOMES: COMPARATIVE DATA Asking individuals to rate all stimuli under investigation on a sufficiently fine scale is [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 274 264–282 274 SCALING cognitively a complex task and may cast doubt on the reliability of such ratings. In particular, the positioning of a stimulus along a rating scale may give rise to contextual effects induced by the use of arbitrary labels of the scale. Respondents may also differ in their interpretation of the rating categories or in their response scale usage, which can add difficult-to-control-for method variance to the data. In contrast, comparing stimuli with each other is a less abstract task which produces data that are not contaminated by idiosyncratic uses of a response scale.
Because comparative judgments require less cognitive effort on the part of the respondents and avoid interpretational issues introduced by the number of categories and labels of a rating scale, we view them as often preferable to ratings for the measurement of preferences.
One of the simplest approaches to gather comparative information is to solicit rankings.
Inthiscase,stimuliarecompareddirectlywith each other with the aim of ranking them from most to least preferred. These types of data can be analyzed with model structures that are similar to the ones used for ratings but they also differ in one important aspect.As we show below, the comparison process between two stimuli is based on a difference operation between the separate evaluations of these stimuli. Because only the outcome of this difference operation is observed but not the separate evaluations, information about the origin of the stimulus scale can no longer be inferred from the data (Böckenholt, 2004).
Similarly, only interaction effects of variables describing the decision makers with stimulus characteristics can be identified – not their main effects. We discuss the implications of these limitations below.
Ranking data When coding rankings, it is useful to express ranking patterns using binary dummy vari-ables. For any two stimuli, we let ui,k be a dummy variable involving the comparison of two stimuli, i and k. We assume that a respondent prefers item iover item k if her utility for item i is larger than for item k, and consequently ranks item i before item k: ui,k = 1 if yi ≥yk 0 if yi < yk (13) where the ys are the preferences in Equa-tion (6), which now are not observed. Notice that when ranking n items, there are ˜ n = n(n−1) 2 indicator variables u.
Alternatively, the response process (13) can be described by computing differences between the latent utilities y. Let u∗ i,k = yi−yk be a variable that represents the difference between choice alternatives i and k. Then: ui,k = 1 if u∗ i,k ≥0 0 if u∗ i,k < 0 (14) is equivalent to Equation (13). Also, we can write the set of ñ equations as: u∗= A y (15) where A is an ñ × n design matrix. Each col-umn of A corresponds to one of the n choice alternatives, and each row of A corresponds to one of the ñ paired comparisons. For example, when n = 2, A = 1 −1 , whereas when n = 3 and n = 4, n=3 n=4 A= ⎡ ⎣ 1 −1 0 1 0 −1 0 1 −1 ⎤ ⎦, and A= ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 −1 0 0 1 0 −1 0 1 0 0 −1 0 1 −1 0 0 1 0 −1 0 0 1 −1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ (16) respectively.
Now, if we assume that the random vari-ables ε are multivariate normal conditional on any exogenous variables, we obtain the class of two-level models proposed by Bock (1958) that were based on Thurstone’s (1927, 1931) approach to analyzing comparative-judgment data. Thurstone did not take into account individual differences, but this important limitation was overcome by Bock (1958) and, [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 275 264–282 MODELING PREFERENCE DATA 275 subsequently, generalized by Takane (1987).
We obtain Bock’s (1958) extension of the classical models proposed by Thurstone by letting νi = εi in Equation (2).As a result, the parameterstobeestimatedarethemeanvector ν and the covariance matrix of ε, .Thurstone (1927) proposed constraining the covariance matrix of ε to be diagonal (leading to the so-called ‘Case III’).Aconstrained version of the Case III model is the Case V model, where, in addition, the ε are assumed to have a common variance.
Some remarks on the identification of model parameters Based on comparative judgments it is not possible to recover the origin of stimulus eval-uations. One stimulus may be judged more positively than another but this result does not allow any conclusions about whether either of the stimuli is attractive or unattractive. To estimate the model parameters, it is therefore necessary to introduce parameter constraints that specify the scale origin. Typically, this is done by setting one of the individual stimulus parameters to zero. Thus, an unrestricted model can be identified by fixing one of the ν, fixing the variances of to be equal to 1, and introducing an additional linear constraint among the off-diagonal elements of (Maydeu-Olivares and Hernández, 2007).
Alternative identifications constraints can be chosen (Dansie, 1986; Tsai, 2000; 2003) that may prove more convenient in an application of the ranking model. However, it is important to keep in mind that the original covariance matrix underlying the utilities cannot be recovered from the data, but only a reduced rank version of it. Thus, the interpretation of the results cannot be based on the estimated covariance matrix alone; we also need to take intoaccounttheclassofalternativecovariance structures that yield identical fits of the data.
For example, consider for three stimuli, the two mean and covariance structures: ν1 = ⎛ ⎝ 2 5 0 ⎞ ⎠, 1 = ⎛ ⎝ 1 0 0 0 2 0 0 0 3 ⎞ ⎠, and ν2 = ⎛ ⎝ √ .8 √ 5 0 ⎞ ⎠, 2 = ⎛ ⎝ 1 .7 .6 .7 1 .5 .6 .5 1 ⎞ ⎠ Although seemingly different, these two mean and covariance structures yield the same ranking probabilities of the three stimuli.
Model 1 suggests that the stimuli give rise to different variances in the population of judges and are assessed independently. In contrast, Model 2 suggests that the variances of the stimuli are the same and the assessments of the stimuli are correlated in the population of judges. This example demonstrates that care needs to be taken in the interpretation of the estimated parameters of a comparative-judgment model because only the differences between the evaluations of the stimuli are observed.
Paired comparisons data The use of rankings assumes that respondents can assess and order the stimuli under study in a consistent manner. This need not be the case.
Rather, respondents may consider different attributes in their comparison of stimuli or use non-compensatory decision rules which in both cases can lead to inconsistent judgments.
For example, in a classical study Tversky (1969) showed that judges who applied a lex-icographic decision rule systematically made intransitive choices. The method of paired comparisons facilitates the investigation of inconsistent judgments because here judges are asked to consider the same stimulus in multiple comparisons to other stimuli. The repeated evaluation of the same stimulus in different pairs can give useful insights on how judges arrive at their preference judgments.
Consider two pairwise comparisons in which stimulus j is preferred to stimulus k and stimulus k is preferred to stimulus l. If the judge is consistent, we expect that in a comparison of stimuli j and l, j is preferred to l. If a judge selects stimulus l in the last pairwise comparison then this indicates an intransitive cycle which may be useful in understanding the judgmental process. For instance, in a large-scale investigation (with [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 276 264–282 276 SCALING over 4000 respondents) of Zajonc’s (1980) proposition that esthetic and cognitive aspects of mentality are separate, Bradbury and Ross (1990) demonstrated that the incidence of intransitive choices for colors declines through childhood from about 50% to 5%.
For younger children, the novelty of a choice optionplaysadecisiverole,withtheresultthat they tend to prefer the stimulus they have not seenbefore.Thereductionofthiseffectduring childhood and adolescence is an important indicator of the developmental transition from a pre-logical to a logical reasoning stage. The diagnostic value of the observed number of intransitive cycles is highest when it is known in advance, which option triple will produce transitivity violations (Morrison, 1963). If this information is unavailable, probabilistic-choice models are needed to determine whether intransitivities are systematic or reflective of the stochastic nature of choice behavior. Here, Thurstone’s (1927) paired-comparison model can be a helpful diagnostic tool. As a side result, it also allows iden-tifying respondents who are systematically inconsistent and may have difficulties in their evaluations.
Inconsistent pairwise responses caused by random factors can be accounted for by adding an error term e to each difference judgment (15): u∗= A y + e (17) The random errors e are assumed to be normally distributed with mean zero, uncorrelated across pairs, and uncorrelated with y.Theerrortermaccountsforintransitive responses by reversing the sign of the difference between the preference responses yi and yk. Also, since y and e are assumed to be normally distributed, the latent difference responses u∗are normally distributed. Their mean vector and covariance matrix are: µu∗= Aν, and u∗= A A’+ 2 (18) where 2 denotes the covariance matrix of the random errors e, and is the covariance matrix of ε. Clearly, the smaller the elements of the error covariance matrix 2, the more consistent the respondents are in evaluating the choice alternatives.
In the extreme case, when all the elements of 2 are zero, the paired comparison data are effectively rankings and no intransitivities would be observed in the data. A more restricted model that is often found to be useful in applications involves setting the error variances to be equal for all pairs (i.e., 2 = ω2I). This restriction implies that the number of intransitivities is approximately equal for all pairs, provided the mean differences are small.
Some remarks on estimation Paired-comparison and ranking models can be estimated by maximum likelihood methods. This estimation approach requires multidimensional integration, which becomes increasingly difficult as the number of items to be compared increases (Böckenholt, 2001a).
However, the models can also be straightforwardly estimated using the following sequential procedure (see Muthén, 1993; Maydeu-Olivares and Böckenholt, 2005). Since Thurstone’s model assumes that multivariate-normaldatahasbeencategorized according to some thresholds, in a first stage the thresholds and tetrachoric correlations underlying the observed discrete choice data are obtained. In a second stage, the model parameters are estimated from the estimated thresholds and tetrachoric correlations using unweighted least squares (ULS), or diagonally weighted least squares (DWLS).
Asymptotically correct standard errors and a goodness of fit of the model to the estimated thresholds and tetrachoric correlations are available.
NUMERICAL EXAMPLE 3: MODELING VOCATIONAL INTERESTS The data for this example is taken from Elosua (2007). Data were collected from 1069 adolescents in the Spanish Basque Country using the 16PF Adolescent Personality [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 277 264–282 MODELING PREFERENCE DATA 277 Questionnaire (APQ; Schuerger, 2001). We note that although the overall sample size reported in Elosua (2007) is 1221, only 1069 students completed the paired compar-isons task. The Work Activity Preferences section of this questionnaire includes a paired comparisons task involving the 6 types of Holland’s, ‘Realistic, Investigative, Artistic, Social, Enterprise, and Conventional’ (RIASEC) model (see Holland, 1997). For each of the 15 pairs, the student chose their future preferred work activity. We shall fit the sequence of models suggested in Maydeu-Olivares and Böckenholt (2005; see their Figure 4 for a flow chart). All models were estimated using DWLS with mean corrected SB goodness-of-fit tests. This is denoted as WLSM estimation in Mplus (Muthén and Muthén, 2007).
First, we fit an unrestricted model. The model fits well: Satorra-Bentler’s mean adjusted X2 = 135.98, df = 86, p < .01, RMSEA = 0.023.
Next, we investigate whether error variances can be set equal for all pairs (i.e., 2 = ω2I). We obtain X2 = 200.16, df = 100, RMSEA = 0.031.
The fit worsens suggesting that the number of intransitivities may not be approximately equal across pairs. We conclude that the equal variance restriction may not be suitable and allow from here on the error variances across pairs to be unconstrained.
Now, we investigate whether a model that specify that preferences for the six Holland types are independent (i.e., a Case III model) is consistent with the data. We obtain X2 = 523.64, df = 65, RMSEA = 0.065 indicating that a model with unequal stimulus variances alone cannot account for the data. Another indication that the Case III model is mis-specified for these data is that the estimate for one of the paired specific variances becomes negative. It appears that Holland’s types were not evaluated independently of each other and that respondents may have used one or several attributes in arriving at their preference judgments. We use a factor-analysis model (7) to ‘uncover’ latent attributes that systematically influenced the respondents’ judgments. That is, we use: u∗= A y + e = A (ν + η + ε) + e (19) See Maydeu-Olivares and Böckenholt (2005) for details on how to identify this model.
A one factor model yields X2 = 150.87, df = 90, RMSEA = 0.025, whereas a two factor model yields almost the same fit as an unrestricted model, X2 = 135.98, df = 86, RMSEA = 0.023.
Next, we introduce parameter constraints among the loadings of the two factor model so that the stimuli lie on a circumplex, as stated in Holland’s theory. Specifically we let: µu∗=Aν, and u∗=A ′+ A′+2 (20) λ2 j1+λ2 j2 =ρ2, j=1,...,n (21) where λjk denotes the factor loading for stimuli j and factor k and ρ denotes the radius of the circumference. To estimate the model, we fix the loadings for one of the stimuli. The model yields X2 = 182.41, df = 90, p < .01, RMSEA = 0.031. The model still has a good fit according to the criterion of Browne and Cudeck (1993). However, notice that it has the same number of parameters as the one-factor model, yielding a somewhat worse fit.
In Table 12.4 we provide the parameter estimates for the circumplex model, whereas in Figure 12.2 we provide a plot of the factor loadings. We conclude that the specification Table 12.4 Parameter estimates and standard errors for a circumplex model fitted to the vocational interests data; paired comparisons Holland’s type ν diag() R –.23 (.14) .60 (.05) .05 (.06) .77 (.32) I –.16 (.09) –.62 .83 (.07) .84 (12) A .46 (.10) –.45 (.10) .52 (.06) .68 (.15) C –.61 (.03) –.18 (.10) .08 (.05) .74 (.12) S .29 (.12) –.57 .99 (.08) 1.52 (.22) E –.40 (fixed) –.50 (fixed) .00 (fixed) 1.00 (fixed) N = 1069; standard errors in parentheses. The elements of the diagonal matrix 2 range from .20 (.17) for the pair {R,C} to 3.42 (.79) for the pair {C,E}.
[16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 278 264–282 278 SCALING Factor 1 .8 .6 .4 .2 0.0 −.2 −.4 −.6 −.8 Factor 2 .8 .6 .4 .2 0.0 −.2 −.4 −.6 −.8 E S C A I R Figure 12.2 Circumplex model fitted to the vocational interest data.
that the loading patterns follow a circumplex structure is not in complete agreement with the data but that the stimuli can be arranged in a two-dimensional space.
MODELING DISCRETE OUTCOMES: FIRST-CHOICE DATA First-choice data are ubiquitous in natural settings. Whenever an individual faced with K alternatives is asked to report her preferred choice, we obtain first- choice data. The data obtained is usually coded using a single variable consisting of K unordered or nominal categories. Alternatively, we can code the data using K dummy variables, one for each alternative. This alternative coding of the data provides us with useful insights into the model. When we consider K such dummy variables and consider expressing them as a function of characteristics of the respondents or the stimuli using our Equation (6) we see that only K −1 such equations are estimable, as one of the dummies is redundant given the information in the remaining K −1 variables.
Thereisyetanotherwaytocodefirst-choice data that gives us additional insight into the model to be used. Ranking data can be viewed as a special case of paired comparison data where intransitive patterns have probability zero (Maydeu-Olivares, 2001). This can be accommodated within a Thurstonian model by letting the variances of all paired specific errors, e, to be zero. In turn, first-choice data can be viewed as a special case of ranking data where the information on second, third, etc. most preferred choices is missing by design. Thus, first choice data can be coded ñ indicator variables with missing data. How can first-choice data be modeled? Because there are only K −1 pieces of information, only a model with K −1 parameters can be estimated, that is, Thurstone’s Case V model. To put it differently, when the full ranking of alternatives is available, a variety of models can be estimated, including models [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 279 264–282 MODELING PREFERENCE DATA 279 that parameterize the association among the different alternatives in the choice set. But, as less information is available for modeling, some of these models can no longer be identified. In the limit, when only first choices are available, the utilities underlying the alter-natives must be assumed to be independently distributed with common variance, because it is the only model that can be identified.
As a result, when considering first-choice data, interest lies not in modeling relations between the stimuli, but in mod-eling relations between the first choices and respondent and/or stimuli characteris-tics. Now, Thurstonian models are obtained when the random errors ε are assumed to be normally distributed conditional on the exogenous variables. Unfortunately, this normality assumption leads to multivariate probit-regression models, which are noto-riously difficult to estimate. However, if the random errors are assumed to be inde-pendently Gumbel distributed, we obtain a multinomial-regression model (Bock, 1969; Böckenholt, 2001b).
NUMERICAL EXAMPLE 4: MODELING THE EFFECT OF GRADE AND GENDER ON VOCATIONAL INTERESTS For this example, we shall consider again the data from Elosua (2007) on preferences for the six Holland’s types (Realistic, Inves-tigative, Artistic, Social, Enterprise, and Conventional). In all, 558 respondents out of 1069 yielded transitive paired comparisons patterns, meaning that their paired compar-isons can be turned into rankings. We fitted a Case V Thurstonian model to these rank-ing data (Maydeu-Olivares, 1999; Maydeu-Olivares and Böckenholt, 2005) where the underlying utilities for Holland’s types are assumed to depend on the respondents’school grade (7th to 12th grade) and gender. That is, we used: u∗= A y = A (ν + Bx + ε) (22) where the covariance matrix of the random errors ε is assumed to be diagonal with common variance as stated by Thurstone’s Case V model. Parameter estimates are provided in Table 12.5. Next, we used only the respondents’ first-choice selections and estimated the effect of school grade and gender on preferences for vocational type usingmultinomial-logisticregression.Results are also provided in Table 12.5. Notice that estimates for both models cannot be directly compared as they are on different scales (logistic and normal). However, it is interesting to compare the substantive results.
We see in Table 12.5 that the effect of gender on vocational preferences is similar in both cases.Femaleadolescentsaremorelikelythan men to prefer a social vocation to a business one, and less likely to prefer a scientific vocation to a business one. Interestingly, there are substantive differences on the impact of school grade on career preferences. When ranking data are analyzed, older students are more likely to choose a business vocation than any other type. However, when only first choices are available a business vocation is Table 12.5 Parameter estimates and standard errors for the vocational interests data; rankings and first choices Multinomial logistic regression applied to first choices Thurstone’s Case V model applied to rankings Holland’s type ν Grade Gender ν Grade Gender R 1.97 (.60) –.19 (.13) –1.83 (.50) 1.05 (.18) –.14 (.04) –1.05(.14) I 2.18 (.56) –.16 (.12) –.20 (.38) 1.56 (.18) –.15 (.04) –.25 (.13) A 1.51 (.60) –.27 (.13) .67 (.42) 1.07(.19) –.20 (.04) .21 (.14) C 1.07 (.65) –.34 (.14) .77 (.48) .44 (.16) –.13 (.04) .10 (.12) S 2.02 (.56) –.20 (.12) .87 (.38) 1.05 (.20) –.14 (.04) .61 (.14) N = 558; standard errors in parentheses. Enterprise was used as reference. Estimates significant at the 5% level are marked in boldface. Gender is coded as 1 = females.
[16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 280 264–282 280 SCALING only preferred over a conventional and social type by older students. Also, in general, the estimates/SE ratios are larger for the ranking model than for the first choice model. We attribute this effect to the loss of information incurred when using first choices only.
CONCLUDING COMMENTS This chapter presents an introduction to random-effects models for the analyses of both continuous and discrete choice data.
Juxtaposing the two approaches has allowed us to show the similarities but also the differ-ences between the statistical frameworks. The models presented for continuous data are well suited to describing relationships between person- and attribute-specific characteristics and the overall liking of a stimulus. These relationships can be used in predicting preferences for new stimuli or preference changes when stimuli are modified. Repeated evaluations of the same stimulus facilitate reliability analyses but no strong benchmarks are available that allow the assessment of the stability of judgments or whether some judges are better qualified to assess the stimuli under consideration than others. Importantly, however, it is possible to categorize stimuli as attractive or unattractive on the basis of the evaluative scale used for assessing the stimuli.
Probabilistic approaches for the analysis of discrete choices facilitate similar statistical decomposition of person-and attribute-specific effects, but because choices are viewed as a result of a maximization process, information about the underlying origin of the utility scale is lost. Thus, overall assessments of whether a stimulus is attractive or unattractive are not possible.
Instead of reliability analyses, more rigorous tests of the consistency of the choices can be conducted under the assumption that the measured utilities are both stable across time and situations. Stochastic transitivity tests are available as well as tests of expansion and contraction consistency (Block and Marshak, 1960; Falmagne, 1985). Under contraction consistency, if a set of stimuli is narrowed to a smaller set such that stimuli from the smaller set are also in the larger set, then no unchosen stimulus should be chosen and no previously chosen stimulus should be unchosen from the smaller set. Similarly, under expansion consistency, if a smaller choice set is extended to a larger one, then the probability of choosing a stimulus from the larger set should not exceed the probability of choosing a stimulus from the smaller set. The choice literature is full of examples demonstrating violations of both stochastic transitivity as well as expansion-and contraction-consistency conditions (Shafir and LeBouef, 2006). Contextual effects (e.g., relational features such as dominance among choice options), choice processes (e.g., decision strategies), presentation formats, frames as well as characteristics of the decision maker have been shown to affect choice processes in systematic ways. In view of this long list, we conclude that the assumption of stable utilities should be viewed as a hypothesis that needs to be tested and validated in any given application.
Many extensions of these two modeling frameworks for continuous and discrete data have been proposed in the literature (Böckenholt, 2006). They include models for time-dependent data (Keane, 1997), models for multivariate choices where stimuli are compared with respect to different attributes (Bradley, 1984), models for dependent choices where the same stimuli are compared by clustered judges (e.g., family members evaluating the same movie), models that allow for social interactions on choice (Brock and Durlauf, 2001) and models that consider choices among risky choice options (Manski, 2004). In addition, a great deal of work has focused on combining revealed and stated preference data (Ben-Akiva et al., 1997) and on developing structural equation models that allow the integration of both choice and choice-related variables (e.g., attitudes, values) to enrich our understanding of possible determinants of choice (Kalidas et al., 2002). The toolbox for analyzing choice data is certainly large, demonstrating both the [16:53 20/2/2009 5283-Millsap-Ch12.tex] Job No: 5283 Millsap: The SAGE Handbook of Quantitative Methods in Psychology Page: 281 264–282 MODELING PREFERENCE DATA 281 importance of this topic in many different disciplines and the ubiquitousness of choice situations in our life.
ACKNOWLEDGEMENT Alberto Maydeu-Olivares was supported by grant SEJ2006–08204 from the Spanish Ministry of Education. Ulf Böckenholt grate-fully acknowledges the support of the Social Sciences and Humanities Research Council of Canada. Correspondence concerning this article should be addressed to Alberto Maydeu-Olivares, Faculty of Psychology, University of Barcelona, P. Valle de Hebrón, 171, 08035 Barcelona (Spain). E-mail: amay-deu@ub.edu.
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1363 | https://math.stackexchange.com/questions/1548985/if-a-rational-function-is-even-then-the-numerator-and-the-denominator-have-same | polynomials - If a rational function is even then the numerator and the denominator have same parity - Mathematics Stack Exchange
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If a rational function is even then the numerator and the denominator have same parity
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Let K K be a field and let F=P Q∈K(X)F=P Q∈K(X) be a rational fraction, for simplicity we denote also by F F the rational function associated to the rational fraction F F. It is clear that if P P and Q Q are both even or both odd polynomial functions, then F F is an even rational function and we have also that if one of the two polynomial functions is even and the other is odd, then F F must be odd. Now is the converse true, I mean do we have that if F F is an even rational function then necessarely both P P and Q Q are both odd or both even and that if F F is odd then necesserely one is odd and the other is even ?
I think it is true, but i don't know how to prove it, suppose that F F is even then
P(x)Q(x)=P(−x)Q(−x)P(x)Q(x)=P(−x)Q(−x)
hence
P(x)Q(−x)=P(−x)Q(x)P(x)Q(−x)=P(−x)Q(x)
But how to go from here ?
polynomials
rational-functions
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asked Nov 27, 2015 at 18:27
paliopalio
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1 what if P=Q P=Q and they're neither even nor odd ?mercio –mercio 2015-11-27 18:31:10 +00:00 Commented Nov 27, 2015 at 18:31
I'm sorry I don't understand your comment !palio –palio 2015-11-27 18:34:48 +00:00 Commented Nov 27, 2015 at 18:34
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You may suppose that P P and Q Q have no commun divisors in K[x]K[x]. Your equality P(x)Q(−x)=P(−x)Q(x)P(x)Q(−x)=P(−x)Q(x) show then that as P(x)P(x) divide P(−x)Q(x)P(−x)Q(x) and is prime to Q(x)Q(x), P(x)P(x) must divide P(−x)P(−x). As they have the same degree, there exists c c such that P(−x)=c P(x)P(−x)=c P(x). Replacing x x by −x−x, we get c 2=1 c 2=1, hence c=1 c=1 or c=−1 c=−1 and we are done.
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answered Nov 27, 2015 at 18:48
KelennerKelenner
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You are welcome.Kelenner –Kelenner 2015-11-27 18:51:57 +00:00 Commented Nov 27, 2015 at 18:51
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Take Q=x+x 2 Q=x+x 2, P=x Q P=x Q. Then F=x F=x, and F F is odd, but Q Q isn't odd or even.
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answered Nov 27, 2015 at 18:40
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I see your point, is my claim true when the rational fraction is irreducible ?palio –palio 2015-11-27 18:46:42 +00:00 Commented Nov 27, 2015 at 18:46
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3If Rational function R(u,v)=R(−u,v)R(u,v)=R(−u,v), then R(u,v)=R 1(u 2,v)R(u,v)=R 1(u 2,v)
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1365 | https://en.wikipedia.org/wiki/Minimum_spanning_tree | Jump to content
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Contents
1 Properties
1.1 Possible multiplicity
1.2 Uniqueness
1.3 Minimum-cost subgraph
1.4 Cycle property
1.5 Cut property
1.6 Minimum-cost edge
1.7 Contraction
2 Algorithms
2.1 Classic algorithms
2.2 Faster algorithms
2.3 Linear-time algorithms in special cases
2.3.1 Dense graphs
2.3.2 Integer weights
2.4 Decision trees
2.5 Optimal algorithm
2.6 Parallel and distributed algorithms
3 MST on complete graphs with random weights
4 Fractional variant
5 Other variants
6 Applications
7 References
8 Further reading
9 External links
Minimum spanning tree
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From Wikipedia, the free encyclopedia
Least-weight tree connecting graph vertices
A minimum spanning tree (MST) or minimum weight spanning tree is a subset of the edges of a connected, edge-weighted undirected graph that connects all the vertices together, without any cycles and with the minimum possible total edge weight. That is, it is a spanning tree whose sum of edge weights is as small as possible. More generally, any edge-weighted undirected graph (not necessarily connected) has a minimum spanning forest, which is a union of the minimum spanning trees for its connected components.
There are many use cases for minimum spanning trees. One example is a telecommunications company trying to lay cable in a new neighborhood. If it is constrained to bury the cable only along certain paths (e.g. roads), then there would be a graph containing the points (e.g. houses) connected by those paths. Some of the paths might be more expensive, because they are longer, or require the cable to be buried deeper; these paths would be represented by edges with larger weights. Currency is an acceptable unit for edge weight – there is no requirement for edge lengths to obey normal rules of geometry such as the triangle inequality. A spanning tree for that graph would be a subset of those paths that has no cycles but still connects every house; there might be several spanning trees possible. A minimum spanning tree would be one with the lowest total cost, representing the least expensive path for laying the cable.
Properties
[edit]
Possible multiplicity
[edit]
If there are n vertices in the graph, then each spanning tree has n − 1 edges.
There may be several minimum spanning trees of the same weight; in particular, if all the edge weights of a given graph are the same, then every spanning tree of that graph is minimum.
Uniqueness
[edit]
If each edge has a distinct weight then there will be only one, unique minimum spanning tree. This is true in many realistic situations, such as the telecommunications company example above, where it's unlikely any two paths have exactly the same cost. This generalizes to spanning forests as well.
Proof:
Assume the contrary, that there are two different MSTs A and B.
Since A and B differ despite containing the same nodes, there is at least one edge that belongs to one but not the other. Among such edges, let e1 be the one with least weight; this choice is unique because the edge weights are all distinct. Without loss of generality, assume e1 is in A.
As B is an MST, {e1} ∪ B must contain a cycle C with e1.
As a tree, A contains no cycles, therefore C must have an edge e2 that is not in A.
Since e1 was chosen as the unique lowest-weight edge among those belonging to exactly one of A and B, the weight of e2 must be greater than the weight of e1.
As e1 and e2 are part of the cycle C, replacing e2 with e1 in B therefore yields a spanning tree with a smaller weight.
This contradicts the assumption that B is an MST.
More generally, if the edge weights are not all distinct then only the (multi-)set of weights in minimum spanning trees is certain to be unique; it is the same for all minimum spanning trees.
Minimum-cost subgraph
[edit]
If the weights are positive, then a minimum spanning tree is, in fact, a minimum-cost subgraph connecting all vertices, since if a subgraph contains a cycle, removing any edge along that cycle will decrease its cost and preserve connectivity.
Cycle property
[edit]
For any cycle C in the graph, if the weight of an edge e of C is larger than any of the individual weights of all other edges of C, then this edge cannot belong to an MST.
Proof: Assume the contrary, i.e. that e belongs to an MST T1. Then deleting e will break T1 into two subtrees with the two ends of e in different subtrees. The remainder of C reconnects the subtrees, hence there is an edge f of C with ends in different subtrees, i.e., it reconnects the subtrees into a tree T2 with weight less than that of T1, because the weight of f is less than the weight of e.
Cut property
[edit]
For any cut C of the graph, if the weight of an edge e in the cut-set of C is strictly smaller than the weights of all other edges of the cut-set of C, then this edge belongs to all MSTs of the graph.
Proof: Assume that there is an MST T that does not contain e. Adding e to T will produce a cycle, that crosses the cut once at e and crosses back at another edge e'. Deleting e' we get a spanning tree T∖{e' } ∪ {e} of strictly smaller weight than T. This contradicts the assumption that T was a MST.
By a similar argument, if more than one edge is of minimum weight across a cut, then each such edge is contained in some minimum spanning tree.
Minimum-cost edge
[edit]
If the minimum cost edge e of a graph is unique, then this edge is included in any MST.
Proof: if e was not included in the MST, removing any of the (larger cost) edges in the cycle formed after adding e to the MST, would yield a spanning tree of smaller weight.
Contraction
[edit]
If T is a tree of MST edges, then we can contract T into a single vertex while maintaining the invariant that the MST of the contracted graph plus T gives the MST for the graph before contraction.
Algorithms
[edit]
In all of the algorithms below, m is the number of edges in the graph and n is the number of vertices.
Classic algorithms
[edit]
The first algorithm for finding a minimum spanning tree was developed by Czech scientist Otakar Borůvka in 1926 (see Borůvka's algorithm). Its purpose was an efficient electrical coverage of Moravia. The algorithm proceeds in a sequence of stages. In each stage, called Boruvka step, it identifies a forest F consisting of the minimum-weight edge incident to each vertex in the graph G, then forms the graph G1 = G \ F as the input to the next step. Here G \ F denotes the graph derived from G by contracting edges in F (by the Cut property, these edges belong to the MST). Each Boruvka step takes linear time. Since the number of vertices is reduced by at least half in each step, Boruvka's algorithm takes O(m log n) time.
A second algorithm is Prim's algorithm, which was invented by Vojtěch Jarník in 1930 and rediscovered by Prim in 1957 and Dijkstra in 1959. Basically, it grows the MST (T) one edge at a time. Initially, T contains an arbitrary vertex. In each step, T is augmented with a least-weight edge (x,y) such that x is in T and y is not yet in T. By the Cut property, all edges added to T are in the MST. Its run-time is either O(m log n) or O(m + n log n), depending on the data-structures used.
A third algorithm commonly in use is Kruskal's algorithm, which also takes O(m log n) time.
A fourth algorithm, not as commonly used, is the reverse-delete algorithm, which is the reverse of Kruskal's algorithm. Its runtime is O(m log n (log log n)3).
All four of these are greedy algorithms. Since they run in polynomial time, the problem of finding such trees is in FP, and related decision problems such as determining whether a particular edge is in the MST or determining if the minimum total weight exceeds a certain value are in P.
Faster algorithms
[edit]
Several researchers have tried to find more computationally-efficient algorithms.
In a comparison model, in which the only allowed operations on edge weights are pairwise comparisons, Karger, Klein & Tarjan (1995) found a linear time randomized algorithm based on a combination of Borůvka's algorithm and the reverse-delete algorithm.
The fastest non-randomized comparison-based algorithm with known complexity, by Bernard Chazelle, is based on the soft heap, an approximate priority queue. Its running time is O(m α(m,n)), where α is the classical functional inverse of the Ackermann function. The function α grows extremely slowly, so that for all practical purposes it may be considered a constant no greater than 4; thus Chazelle's algorithm takes very close to linear time.
Linear-time algorithms in special cases
[edit]
Dense graphs
[edit]
If the graph is dense (i.e. m/n ≥ log log log n), then a deterministic algorithm by Fredman and Tarjan finds the MST in time O(m). The algorithm executes a number of phases. Each phase executes Prim's algorithm many times, each for a limited number of steps. The run-time of each phase is O(m + n). If the number of vertices before a phase is n', the number of vertices remaining after a phase is at most . Hence, at most logn phases are needed, which gives a linear run-time for dense graphs.
There are other algorithms that work in linear time on dense graphs.
Integer weights
[edit]
If the edge weights are integers represented in binary, then deterministic algorithms are known that solve the problem in O(m + n) integer operations. Whether the problem can be solved deterministically for a general graph in linear time by a comparison-based algorithm remains an open question.
Decision trees
[edit]
Given graph G where the nodes and edges are fixed but the weights are unknown, it is possible to construct a binary decision tree (DT) for calculating the MST for any permutation of weights. Each internal node of the DT contains a comparison between two edges, e.g. "Is the weight of the edge between x and y larger than the weight of the edge between w and z?". The two children of the node correspond to the two possible answers "yes" or "no". In each leaf of the DT, there is a list of edges from G that correspond to an MST. The runtime complexity of a DT is the largest number of queries required to find the MST, which is just the depth of the DT. A DT for a graph G is called optimal if it has the smallest depth of all correct DTs for G.
For every integer r, it is possible to find optimal decision trees for all graphs on r vertices by brute-force search. This search proceeds in two steps.
A. Generating all potential DTs
There are different graphs on r vertices.
For each graph, an MST can always be found using r(r − 1) comparisons, e.g. by Prim's algorithm.
Hence, the depth of an optimal DT is less than r2.
Hence, the number of internal nodes in an optimal DT is less than .
Every internal node compares two edges. The number of edges is at most r2 so the different number of comparisons is at most r4.
Hence, the number of potential DTs is less than
B. Identifying the correct DTs To check if a DT is correct, it should be checked on all possible permutations of the edge weights.
The number of such permutations is at most (r2)!.
For each permutation, solve the MST problem on the given graph using any existing algorithm, and compare the result to the answer given by the DT.
The running time of any MST algorithm is at most r2, so the total time required to check all permutations is at most (r2 + 1)!.
Hence, the total time required for finding an optimal DT for all graphs with r vertices is:
which is less than
See also: Decision tree model
Optimal algorithm
[edit]
Seth Pettie and Vijaya Ramachandran have found a provably optimal deterministic comparison-based minimum spanning tree algorithm. The following is a simplified description of the algorithm.
Let r = log log log n, where n is the number of vertices. Find all optimal decision trees on r vertices. This can be done in time O(n) (see Decision trees above).
Partition the graph to components with at most r vertices in each component. This partition uses a soft heap, which "corrupts" a small number of the edges of the graph.
Use the optimal decision trees to find an MST for the uncorrupted subgraph within each component.
Contract each connected component spanned by the MSTs to a single vertex, and apply any algorithm which works on dense graphs in time O(m) to the contraction of the uncorrupted subgraph
Add back the corrupted edges to the resulting forest to form a subgraph guaranteed to contain the minimum spanning tree, and smaller by a constant factor than the starting graph. Apply the optimal algorithm recursively to this graph.
The runtime of all steps in the algorithm is O(m), except for the step of using the decision trees. The runtime of this step is unknown, but it has been proved that it is optimal - no algorithm can do better than the optimal decision tree. Thus, this algorithm has the peculiar property that it is provably optimal although its runtime complexity is unknown.
Parallel and distributed algorithms
[edit]
Further information: Parallel algorithms for minimum spanning trees
Research has also considered parallel algorithms for the minimum spanning tree problem. With a linear number of processors it is possible to solve the problem in O(log n) time.
The problem can also be approached in a distributed manner. If each node is considered a computer and no node knows anything except its own connected links, one can still calculate the distributed minimum spanning tree.
MST on complete graphs with random weights
[edit]
Alan M. Frieze showed that given a complete graph on n vertices, with edge weights that are independent identically distributed random variables with distribution function satisfying , then as n approaches +∞ the expected weight of the MST approaches , where is the Riemann zeta function (more specifically is Apéry's constant). Frieze and Steele also proved convergence in probability. Svante Janson proved a central limit theorem for weight of the MST.
For uniform random weights in , the exact expected size of the minimum spanning tree has been computed for small complete graphs.
| Vertices | Expected size | Approximate expected size |
---
| 2 | 1/2 | 0.5 |
| 3 | 3/4 | 0.75 |
| 4 | 31/35 | 0.8857143 |
| 5 | 893/924 | 0.9664502 |
| 6 | 278/273 | 1.0183151 |
| 7 | 30739/29172 | 1.053716 |
| 8 | 199462271/184848378 | 1.0790588 |
| 9 | 126510063932/115228853025 | 1.0979027 |
Fractional variant
[edit]
There is a fractional variant of the MST, in which each edge is allowed to appear "fractionally". Formally, a fractional spanning set of a graph (V,E) is a nonnegative function f on E such that, for every non-trivial subset W of V (i.e., W is neither empty nor equal to V), the sum of f(e) over all edges connecting a node of W with a node of VW is at least 1. Intuitively, f(e) represents the fraction of e that is contained in the spanning set. A minimum fractional spanning set is a fractional spanning set for which the sum is as small as possible.
If the fractions f(e) are forced to be in {0,1}, then the set T of edges with f(e)=1 are a spanning set, as every node or subset of nodes is connected to the rest of the graph by at least one edge of T. Moreover, if f minimizes, then the resulting spanning set is necessarily a tree, since if it contained a cycle, then an edge could be removed without affecting the spanning condition. So the minimum fractional spanning set problem is a relaxation of the MST problem, and can also be called the fractional MST problem.
The fractional MST problem can be solved in polynomial time using the ellipsoid method.: 248 However, if we add a requirement that f(e) must be half-integer (that is, f(e) must be in {0, 1/2, 1}), then the problem becomes NP-hard,: 248 since it includes as a special case the Hamiltonian cycle problem: in an -vertex unweighted graph, a half-integer MST of weight can only be obtained by assigning weight 1/2 to each edge of a Hamiltonian cycle.
Other variants
[edit]
The Steiner tree of a subset of the vertices is the minimum tree that spans the given subset. Finding the Steiner tree is NP-complete.
The k-minimum spanning tree (k-MST) is the tree that spans some subset of k vertices in the graph with minimum weight.
A set of k-smallest spanning trees is a subset of k spanning trees (out of all possible spanning trees) such that no spanning tree outside the subset has smaller weight. (Note that this problem is unrelated to the k-minimum spanning tree.)
The Euclidean minimum spanning tree is a spanning tree of a graph with edge weights corresponding to the Euclidean distance between vertices which are points in the plane (or space).
The rectilinear minimum spanning tree is a spanning tree of a graph with edge weights corresponding to the rectilinear distance between vertices which are points in the plane (or space).
The distributed minimum spanning tree is an extension of MST to the distributed model, where each node is considered a computer and no node knows anything except its own connected links. The mathematical definition of the problem is the same but there are different approaches for a solution.
The capacitated minimum spanning tree is a tree that has a marked node (origin, or root) and each of the subtrees attached to the node contains no more than c nodes. c is called a tree capacity. Solving CMST optimally is NP-hard, but good heuristics such as Esau-Williams and Sharma produce solutions close to optimal in polynomial time.
The degree-constrained minimum spanning tree is a MST in which each vertex is connected to no more than d other vertices, for some given number d. The case d = 2 is a special case of the traveling salesman problem, so the degree constrained minimum spanning tree is NP-hard in general.
An arborescence is a variant of MST for directed graphs. It can be solved in time using the Chu–Liu/Edmonds algorithm.
A maximum spanning tree is a spanning tree with weight greater than or equal to the weight of every other spanning tree. Such a tree can be found with algorithms such as Prim's or Kruskal's after multiplying the edge weights by −1 and solving the MST problem on the new graph. A path in the maximum spanning tree is the widest path in the graph between its two endpoints: among all possible paths, it maximizes the weight of the minimum-weight edge. Maximum spanning trees find applications in parsing algorithms for natural languages and in training algorithms for conditional random fields.
The dynamic MST problem concerns the update of a previously computed MST after an edge weight change in the original graph or the insertion/deletion of a vertex.
The minimum labeling spanning tree problem is to find a spanning tree with least types of labels if each edge in a graph is associated with a label from a finite label set instead of a weight.
A bottleneck edge is the highest weighted edge in a spanning tree. A spanning tree is a minimum bottleneck spanning tree (or MBST) if the graph does not contain a spanning tree with a smaller bottleneck edge weight. A MST is necessarily a MBST (provable by the cut property), but a MBST is not necessarily a MST.
A minimum-cost spanning tree game is a cooperative game in which the players have to share among them the costs of constructing the optimal spanning tree.
The optimal network design problem is the problem of computing a set, subject to a budget constraint, which contains a spanning tree, such that the sum of shortest paths between every pair of nodes is as small as possible.
Applications
[edit]
Minimum spanning trees have direct applications in the design of networks, including computer networks, telecommunications networks, transportation networks, water supply networks, and electrical grids (which they were first invented for, as mentioned above). They are invoked as subroutines in algorithms for other problems, including the Christofides algorithm for approximating the traveling salesman problem, approximating the multi-terminal minimum cut problem (which is equivalent in the single-terminal case to the maximum flow problem), and approximating the minimum-cost weighted perfect matching.
Other practical applications based on minimal spanning trees include:
Taxonomy.
Cluster analysis: clustering points in the plane, single-linkage clustering (a method of hierarchical clustering), graph-theoretic clustering, and clustering gene expression data.
Constructing trees for broadcasting in computer networks.
Image registration and segmentation – see minimum spanning tree-based segmentation.
Curvilinear feature extraction in computer vision.
Handwriting recognition of mathematical expressions.
Circuit design: implementing efficient multiple constant multiplications, as used in finite impulse response filters.
Regionalisation of socio-geographic areas, the grouping of areas into homogeneous, contiguous regions.
Comparing ecotoxicology data.
Topological observability in power systems.
Measuring homogeneity of two-dimensional materials.
Minimax process control.
Minimum spanning trees can also be used to describe financial markets. A correlation matrix can be created by calculating a coefficient of correlation between any two stocks. This matrix can be represented topologically as a complex network and a minimum spanning tree can be constructed to visualize relationships.
References
[edit]
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Further reading
[edit]
Otakar Boruvka on Minimum Spanning Tree Problem (translation of both 1926 papers, comments, history) (2000) Jaroslav Nešetřil, Eva Milková, Helena Nesetrilová. (Section 7 gives his algorithm, which looks like a cross between Prim's and Kruskal's.)
Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. Introduction to Algorithms, Second Edition. MIT Press and McGraw-Hill, 2001. ISBN 0-262-03293-7. Chapter 23: Minimum Spanning Trees, pp. 561–579.
Eisner, Jason (1997). State-of-the-art algorithms for minimum spanning trees: A tutorial discussion. Manuscript, University of Pennsylvania, April. 78 pp.
Kromkowski, John David. "Still Unmelted after All These Years", in Annual Editions, Race and Ethnic Relations, 17/e (2009 McGraw Hill) (Using minimum spanning tree as method of demographic analysis of ethnic diversity across the United States).
External links
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Wikimedia Commons has media related to Minimum spanning trees.
Implemented in BGL, the Boost Graph Library
The Stony Brook Algorithm Repository - Minimum Spanning Tree codes
Implemented in QuickGraph for .Net
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1366 | https://byjus.com/english/types-of-adverbs/ | Having learned about adverbs and their functions in a sentence, you should definitely learn the different types of adverbs. This article will provide you with all that you need to know about the various types of adverbs in brief. Check out the examples given below as well to have a complete understanding.
Table of Contents
How Are Adverbs Classified?
Different Types of Adverbs with Examples
Test Your Understanding of Types of Adverbs
Frequently Asked Questions on Types of Adverbs in English
How Are Adverbs Classified?
Adverbs can be classified into different types in accordance with what kind of information they are providing or what they are describing.
The different types of adverbs are:
Adverbs of manner
Adverbs of time
Adverbs of place
Adverbs of frequency
Adverbs of degree
Conjunctive adverbs
Different Types of Adverbs with Examples
Let us now take a look at each of these adverb types in detail.
Adverbs of Manner: These adverbs are those that describe the manner in which an action is done. Basically, it can be said that the adverbs of manner answer the question ‘how’.
Examples of adverbs of manner:
Quickly, promptly, clearly, slowly, gradually, eventually, rapidly, seriously, instantly, keenly, etc.
Adverbs of Time: As the name suggests, the adverbs of time are used to tell the reader when some action is occurring. Adverbs of time include general time periods and specific times. We can identify an adverb of time by asking the question ‘when’.
Examples of adverbs of time:
Now, soon, today, tomorrow, the day after tomorrow, next month, recently, forever, etc.
Adverbs of Place: These adverbs are used to indicate where the action mentioned in the sentence is taking place. Adverbs of place can be identified by asking the question ‘where’.
Examples of adverbs of place:
Somewhere, anywhere, nowhere, here, outside, inside, wherever, elsewhere, left, right, north, east, south, west, etc.
Adverbs of Frequency: These adverbs are used to denote how often an action or event is happening. The adverbs of frequency can be recognised by asking the question ‘how often’.
Examples of adverbs of frequency:
Seldom, rarely, never, often, weekly, monthly, yearly, annually, usually, sometimes, occasionally, constantly, frequently, etc.
Adverbs of Degree: These adverbs are used to indicate how intense an action of quality is. It is used to describe adjectives and adverbs. For instance, an adverb of manner expresses how fast or how slow a vehicle is moving, how hot or cold the weather is, how interesting or boring a movie is and so on.
Examples of adverbs of degree:
Very, too, extremely, much, more, most, little, less, incredibly, totally, greatly, hardly, deeply, barely, etc.
Conjunctive Adverbs: Conjunctive adverbs perform a little differently from the other types of adverbs. These adverbs are seen to act like a conjunction to link two sentences or clauses together and hence the name, ‘conjunctive adverbs’.
Examples of conjunctive adverbs:
However, nevertheless, meanwhile, therefore, instead, likewise, notably, subsequently, rather, namely, on the other hand, incidentally, in addition to, etc.
Test Your Understanding of Types of Adverbs
Identify the adverbs in the following sentences and specify which type of adverbs they are:
She often visits the orphanages and old age homes in different parts of the country.
The tea was very hot, I almost burned my tongue.
It is most likely expected to rain in the month of September in India.
I find it difficult to keep going to the post office every week to check if my parcel has arrived.
I had not completed my assignment; therefore I decided to stay back home and complete it.
Can we go somewhere so that I can relax my mind?
The children became too restless because of the rumour.
Geetha was putting the children to sleep; meanwhile her brother cleaned the house.
My father went outside.
Next month, we will be in Chennai.
You should have guessed it all right. Let us check.
She often visits the orphanages and old age homes in different parts of the country. – Adverb of Frequency
The tea was very hot, I almost burned my tongue. – Adverb of Degree
It is most likely expected to rain in the month of September in India. – Adverb of Degree, Adverb of Manner
I find it difficult to keep going to the post office every week to check if my parcel has arrived. – Adverb of Time
I had not completed my assignment; therefore I decided to stay back home and complete it. – Conjunctive Adverb
Can we go somewhere so that I can relax my mind? – Adverb of Place
The children became too restless because of the rumour. – Adverb of Degree
Geetha was putting the children to sleep; meanwhile her brother cleaned the house. – Conjunctive Adverb
My father went outside. – Adverb of Place
Next month, we will be in Chennai. – Adverb of TIme
Frequently Asked Questions on Types of Adverbs in English
Q1
What are the different types of adverbs in the English language?
The six different types of adverbs are:
Adverbs of Manner
Adverbs of Time
Adverbs of Place
Adverbs of Frequency
Adverbs of Degree
Conjunctive Adverbs
Q2
How many types of adverbs are there in English?
There are six main types of adverbs namely adverbs of manner, adverbs of time, adverbs of place, adverbs of frequency, adverbs of degree and conjunctive adverbs.
Q3
Give some examples of the different types of adverbs.
Likely, however, yesterday, daily, seldom, last year, quarterly, quickly, eventually, inside, keenly, rarely, never, too, much, very, so, little, etc. are some examples of the different types of adverbs.
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1367 | https://en.wiktionary.org/wiki/celerity | Jump to content
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celerity
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From Wiktionary, the free dictionary
English
[edit]
WOTD – 22 July 2006
Etymology
[edit]
From Old French celeritee (compare French célérité), from Latin celeritas, from celer (“fast, swift”).
Pronunciation
[edit]
IPA(key): /sɪˈlɛɹɪti/
| | |
--- |
| Audio (US): | (file) |
Rhymes: -ɛɹɪti
Noun
[edit]
celerity (usually uncountable, plural celerities)
Speed, swiftness.
c. 1603–1604 (date written), William Shakespeare, “Measure for Measure”, in Mr. William Shakespeares Comedies, Histories, & Tragedies […] (First Folio), London: […] Isaac Iaggard, and Ed[ward] Blount, published 1623, →OCLC, [Act V, (please specify the scene number in lowercase Roman numerals)]:
: O most kind maid, / It was the swift celerity of his death, / Which I did think with slower foot came on, / That brain'd my purpose.
1842, [anonymous collaborator of Letitia Elizabeth Landon], chapter XXXV, in Lady Anne Granard; or, Keeping up Appearances. […], volume II, London: Henry Colburn, […], →OCLC, page 153:
: ...when a new medium for attraction was started in the bazaar to which we have alluded, and her letter was dispatched with all possible celerity, insisting that her daughters "should work day and night"—so ran the document—for three weeks,...
1851, Herman Melville, chapter 48, in Moby-Dick:
: The phantoms, for so they then seemed, were flitting on the other side of the deck, and, with a noiseless celerity, were casting loose the tackles and bands of the boat which swung there.
1937, Dorothy L. Sayers, chapter 11, in Busman’s Honeymoon:
: “My parsnip wine is really extra good this year. Dr Jellyfield always takes a glass when he comes—which isn’t very often, I’m pleased to say, because my health is always remarkably good.”
“That will not prevent me from drinking to it,” said Peter, disposing of the parsnip wine with a celerity which might have been due to eagerness but, to Harriet, rather suggested a reluctance to let the draught linger on the palate.
1960 March, “Talking of Trains: The Slough Derailment”, in Trains Illustrated, page 132:
: Warning messages were relayed to the signalmen at Slough East and Dolphin Junction boxes with such celerity that the up main signals were replaced to danger in front of the express before it finally stopped. The signalman at Dolphin Junction just had time to replace the down main signals and pull the emergency lever to lay detonators in front of the fast approaching 7 p.m. Paddington-Reading train.
1998, Hugo Adam Bedau, The Death Penalty in America: Current Controversies:
: The celerity of executions is a generally neglected issue in the empirical literature on deterrence and capital punishment.
2007, Rudolph Joseph Gerber, John M. Johnson, The Top Ten Death Penalty Myths: The Politics of Crime Control:
: As of 2006, celerity has disappeared entirely from the deterrence equation among both high- and low-volume executing jurisdictions.
2018 January 2, Adam Gopnik, “Never Mind Churchill, Clement Attlee Is a Model for These Times”, in The New Yorker:
: When Churchill and Roosevelt were considering their declaration of the Atlantic Charter, it was Attlee, acting with a celerity and a clarity of purpose that belied his reputation for caution, who insisted on including “freedom from want” as one of its aims, making economic rights and, with them, a decent life for all, one of the official aims of the war.
2. (oceanography, meteorology) The speed of an individual wave (as opposed to the speed of groups of waves); often denoted c.
1. (hydrology) The speed with which a perturbation to the flow propagates through the flow domain.
3. (telecommunications, dated) The speed of symbol transmission, now called baud rate.
1867, Taliaferro Preston Shaffner, The Telegraph Manual:
: Celerity of dispatching the Chappe telegraph [section title]
1885, The Wonders of the Universe, Cassell Publishing Company:
: ...and many endeavours have been made, not only to transmit signals with celerity, but also to transmit more than one communication at the same time along the same wire.
Related terms
[edit]
accelerate
celeripede
decelerate
Translations
[edit]
| |
| Bulgarian: бързина (bg) f (bǎrzina), скорост (bg) f (skorost) Esperanto: rapideco French: célérité (fr) f Ido: rapideso (io) Korean: (행동의) 민첩함, 기민함 ((haengdong'ui) mincheopham, giminham) Norwegian: seleritet Portuguese: celeridade (pt) f Sanskrit: जव (sa) m (java), जुवस् (sa) n (juvas) Spanish: celeridad (es) f Turkish: hız (tr), çabukluk (tr) Ukrainian: швидкість f (švydkistʹ) |
speed of individual waves
| |
| French: célérité (fr) f Malay: laju rambat Portuguese: celeridade (pt) f |
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1368 | https://en.wikipedia.org/wiki/Compton_edge | Compton edge - Wikipedia
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Greatest energy a photon scattered on an electron can transfer to it
This article may be too technical for most readers to understand. Please help improve it to make it understandable to non-experts, without removing the technical details.(September 2024) (Learn how and when to remove this message)
In gamma-ray spectrometry, the Compton edge is a feature of the measured gamma-ray energy spectrum that results from Compton scattering in the detector material. It corresponds to the highest energy that can be transferred to a weakly bound electron of a detector's atom by an incident photon in a single scattering process, and manifests itself as a ridge in the measured gamma-ray energy spectrum. It is a measurement phenomenon (meaning that the incident radiation does not possess this feature), which is particularly evident in gamma-ray energy spectra of monoenergetic photons.
When a gamma ray scatters within a scintillator or a semiconductor detector and the scattered photon escapes from the detector's volume, only a fraction of the incident energy is deposited in the detector. This fraction depends on the scattering angle of the photon, leading to a spectrum of energies corresponding to the entire range of possible scattering angles. The highest energy that can be deposited, corresponding to full backscatter, is called the Compton edge. In mathematical terms, the Compton edge is the inflection point of the high-energy side of the Compton region.
Background
[edit]
Gamma-ray spectrum of radioactive Am-Be source. The Compton continuum is due to scattering effects within the detector material. The highest energy that can be transferred by an incident photon in a single scattering process is referred to as the Compton edge. The photopeak after the Compton edge corresponds to the full deposition of the incident gamma-ray's energy (through a single process such as the photoelectric effect, or a sequence of various processes). Single- and double-escape peaks correspond to interactions involving pair production where the annihilation photons escape from the detector volume
In a Compton scattering process, an incident photon collides with a weakly bound electron, leading to its release from the atomic shell. The energy of the outgoing photon, E', is given by the formula:
E′=E 1+E m e c 2(1−cosθ){\displaystyle E^{\prime }={\frac {E}{1+{\frac {E}{m_{\text{e}}c^{2}}}(1-\cos \theta )}}}
E is the energy of the incident photon.
m e{\displaystyle m_{\text{e}}} is the mass of the electron.
c is the speed of light.
θ{\displaystyle \theta } is the angle of deflection of the photon.
(note that the above formula does not account for the electron binding energy, which can play a non-negligible role for low-energy gamma rays).
The energy transferred to the electron, E T{\displaystyle E_{T}}, varies with the photon's scattering angle. For θ{\displaystyle \theta } equal to zero there is no energy transfer, while the maximum energy transfer occurs for θ{\displaystyle \theta } equal to 180 degrees (backscattering).
E T=E−E′{\displaystyle E_{T}=E-E^{\prime }}E ComptonEdge=E T(max)=E(1−1 1+2 E m e c 2){\displaystyle E_{\text{ComptonEdge}}=E_{T}({\text{max}})=E\left(1-{\frac {1}{1+{\frac {2E}{m_{\text{e}}c^{2}}}}}\right)}
In a single scattering act, is impossible for the photon to transfer any more energy via this process; thus, there is a sharp cutoff at this energy, leading to the name Compton edge. If multiple photopeaks are present in the spectrum, each of them will have its own Compton edge. The part of the spectrum between the Compton edge and the photopeak is due to multiple subsequent Compton-scattering processes.
The continuum of energies corresponding to Compton scattered electrons is known as the Compton continuum.
References
[edit]
^ Jump up to: abcKnoll, Glenn F. Radiation Detection and Measurement 2000. John Wiley & Sons, Inc.
^ Jump up to: abcPrekeges, Jennifer (2010). Nuclear medicine instrumentation. Sudbury, Massachusetts: Jones and Bartlett Publishers. p.42. ISBN9781449611125.
See also
[edit]
Gamma spectroscopy
Compton suppression
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1369 | https://www.dictionary.com/browse/syntactics | Advertisement
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syntactics
[sin-tak-tiks]
noun
the branch of semiotics dealing with the formal properties of languages and systems of symbols.
syntactics
/ sɪnˈtæktɪks /
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(functioning as singular) the branch of semiotics that deals with the formal properties of symbol systems; proof theory
Word History and Origins
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1370 | https://www.mathworksheets4kids.com/multiplying-monomials.php | Multiplying Monomials Worksheets
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Multiplying Monomials Worksheets
Delve into our printable multiplying monomials worksheets for a wealth of practice in finding the product of any two monomials, a monomial by a binomial, and a monomial by a polynomial. Monomials, by the way, are algebraic expressions consisting of a single term. They can be univariate or multivariate expressions. Abounding in exercises on single-variable and multivariable expressions, these pdfs help high school students perform monomial multiplication with ease. Grab some worksheets for free!
Multiplying Monomials - Single Variable
In these multiplication of monomial by monomial pdf worksheets with 2 or 3 monomials of single variables, multiply the coefficients of one term with the other and add the exponents with the same base.
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Multiplying Monomials - Multivariable
Multiplying multivariable monomials is best perfected with scads of practice! Regroup the factors - the coefficients, up to three variables and constants, and find the product of the monomials here.
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Multiply Monomials by Binomials - Single Variable
Let your monomial multiplication skills be second to none! Multiply monomials with binomials - expressions with two terms. Multiply the monomial with each term inside the parentheses to obtain the product.
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Multiply Monomials by Binomials - Multivariable
This resource thrives on multiplication of multivariate monomials and binomial expressions. Bring the product rule of exponents into play, perform the operation, and determine the product.
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Multiplying Monomials by Polynomials - Single Variable
Celebrate the diversely gifted monomials with these pdf worksheets presenting univariate monomials and polynomials! Distribute the monomial to every term of the polynomial and multiply them.
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Multiplying Monomials by Polynomials - Multivariable
High school students apply the distributive property and determine the product of the multivariable monomial and polynomial expressions that our printable multiplying monomials worksheets are chock-full of.
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Missing Monomials
Let your skills shine with a challenge packed in this set of multiplication equations consisting of missing monomials. Complete the multiplication statement by plugging in the missing factor or product.
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1371 | https://www.jmap.org/IJMAP/LivingEnvironment/0816ExamLE.pdf | LIVING ENVIRONMENT LIVING ENVIRONMENT The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION LIVING ENVIRONMENT Thursday, August 18, 2016 — 12:30 to 3:30 p.m., only Student Name _________ School Name _________ Print your name and the name of your school on the lines above. A separate answer sheet for multiple-choice questions in Parts A, B–1, B–2, and D has been provided to you. Follow the instructions from the proctor for completing the student information on your answer sheet.
You are to answer all questions in all parts of this examination. Record your answers for all multiple-choice questions, including those in Parts B–2 and D, on the separate answer sheet. Record your answers for all open-ended questions directly in this examination booklet. All answers in this examination booklet should be written in pen, except for graphs and drawings, which should be done in pencil. You may use scrap paper to work out the answers to the questions, but be sure to record all your answers on the answer sheet or in this examination booklet as directed.
When you have completed the examination, you must sign the declaration printed on your separate answer sheet, indicating that you had no unlawful knowledge of the questions or answers prior to the examination and that you have neither given nor received assistance in answering any of the questions during the examination. Your answer sheet cannot be accepted if you fail to sign this declaration.
DO NOT OPEN THIS EXAMINATION BOOKLET UNTIL THE SIGNAL IS GIVEN.
The possession or use of any communications device is strictly prohibited when taking this examination. If you have or use any communications device, no matter how briefly, your examination will be invalidated and no score will be calculated for you.
Notice...
A four-function or scientific calculator must be made available for you to use while taking this examination.
1 Which sequence represents structures organized from most complex to least complex?
(1) chloroplast →guard cell →leaf →oak tree (2) guard cell →chloroplast →leaf →oak tree (3) oak tree →guard cell →leaf →chloroplast (4) oak tree →leaf →guard cell →chloroplast 2 Autotrophs differ from heterotrophs in that only autotrophs (1) require carbon dioxide for cellular respiration (2) release oxygen as a product of cellular respiration (3) synthesize nutrients using carbon dioxide and water (4) break down sugars to assemble other molecules 3 Burmese pythons are large snakes that have been introduced into the Florida Everglades ecosystem. Burmese pythons and alligators hunt the same prey. One likely effect of the introduction of the pythons is that (1) alligators will have more prey available (2) pythons will become native to the Everglades (3) alligator populations will decline (4) pythons will become an endangered species 4 Which activity enables humans to produce new genetic combinations in other organisms?
(1) selecting and breeding the organisms for specific traits (2) increasing the number of enzymes available to the organisms (3) growing organisms that reproduce asexually (4) decreasing the amount of DNA in the diet of the organisms 5 Before a cell divides, an exact copy of each chromosome is made by the process of (1) genetic engineering (3) mutation (2) replication (4) recombination 6 Some time ago, there were thousands of California condors in North America. Large numbers were poisoned from lead in bullets that were used to kill the animals the condors fed on.
An effort was made to help save this large scavenger. There are now more than 350 California condors in North America. The condors most likely increased in number because humans decided to (1) produce lead-resistant condors through asexual reproduction (2) pass laws against using lead bullets to kill animals used by condors for food (3) introduce plants that didn’t absorb the lead from discharged bullets (4) produce lead-resistant prey for the condors through genetic engineering 7 In humans, the maintenance of a stable internal temperature is a direct result of (1) detection of and reaction to stimuli in the environment (2) digestion of starches and absorption of protein from the internal environment (3) diffusion of water and excretion of glucose to the external environment (4) transport of ATP and locomotion through the environment 8 Which molecules are needed to cut and copy segments of DNA?
(1) reproductive hormones (2) carbohydrates (3) antibodies (4) biological catalysts 9 Evolution can occur at different rates; however, for evolution to occur, there must be (1) variations within a species (2) extinction of the species (3) asexual reproduction (4) no change in the genes of an organism Living Environment–Aug. ’16 Part A Answer all questions in this part. Directions (1–30): For each statement or question, record on the separate answer sheet the number of the word or expression that, of those given, best completes the statement or answers the question.
10 The two reactions illustrated in the diagrams below often occur when a foreign substance enters the body.
The cells labeled A and B are examples of cells known as (1) guard cells (3) white blood cells (2) reproductive cells (4) specialized skin cells 11 The diagram below represents a protein molecule present in some living things.
This type of molecule is composed of a sequence of (1) amino acids arranged in a specific order (2) simple sugars alternating with starches arranged in a folded pattern (3) large inorganic subunits that form chains that interlock with each other (4) four bases that make up the folded structure 12 Three human actions that have been made possible in recent times are: • Doctors are able to diagnose and treat some fetal problems prior to the birth of a child.
• Cloning can produce large numbers of plants that are resistant to drought.
• Male insects can be sterilized with radiation to prevent them from mating successfully.
Which statement summarizes these three actions?
(1) Reproductive technology has medical, agricultural, and ecological applications.
(2) Development is a highly regulated process involving mitosis and differentiation.
(3) Reproduction and development are subject to environmental effects.
(4) Human development, birth, and aging should be viewed as a predictable pattern of events.
Foreign substance Specific proteins Cell A Cell B produces Living Environment–Aug. ’16 [OVER] 13 Natural selection is best described as (1) a change in an organism in response to a need of that organism (2) a process of nearly constant improvement that leads to an organism that is nearly perfect (3) differences in survival rates as a result of different inherited characteristics (4) inheritance of characteristics acquired during the life of an organism 14 Which statement best describes a situation where competition occurs in an ecosystem?
(1) A deer outruns an attacking wolf.
(2) A deer, during the winter, consumes tree bark.
(3) A deer and a rabbit consume grass in a field.
(4) A deer and a rabbit are both startled by a hawk flying overhead.
15 A woman changes her hair color to red; however, her children will not inherit this red hair color because the woman does not have (1) genes for red hair in her skin (2) genes for red hair in her sex cells (3) proteins for red hair in the placenta (4) proteins for red hair in her egg cells 16 Fossils provide evidence that (1) life on Earth millions of years ago was more complex than life is today (2) the changes that will occur in species in the future are easy to predict (3) many species of organisms that lived long ago are now extinct (4) most species of organisms that lived long ago are exactly the same today 17 A male frigatebird displays to the female by inflating its large red throat sac, throwing its head back, vibrating its wings, and producing a “drumming” sound with its throat sac. For the frigatebird, this behavior has most likely resulted in (1) hiding from predators (2) greater reproductive success (3) locating new sources of food (4) reduced population growth 18 Plant species X lives in a hot, dry environment.
Slowly, over hundreds of years, the climate becomes wetter. Fungi attack species X and cause the population of species X to decrease. However, plant species X could survive if the plants (1) try to mutate quickly and synthesize new proteins (2) are watered often and fertilized with extra nutrients (3) can adapt to the new conditions by mating with the fungus (4) have a few members of the population that are fungus-resistant 19 The diagram below represents reproduction in a yeast cell. The genes in the bud are identical to the genes in the parent.
This type of production of offspring is a form of (1) sexual reproduction (2) asexual reproduction (3) gene manipulation (4) genetic engineering 20 A human cell that contains all of the information necessary for the growth and development of a complete organism is (1) a sperm cell (3) a zygote (2) a gamete (4) an egg cell 21 When would exposure to a potentially harmful substance be most likely to damage many organs in a developing embryo?
(1) during the last three months of pregnancy (2) during the early stages of pregnancy (3) during the formation of the zygote (4) during meiosis in both males and females Nucleus Bud Living Environment–Aug. ’16 22 The human female reproductive system is represented below.
Within which structure does the placenta normally develop?
(1) A (3) C (2) B (4) D 23 An energy-rich organic compound needed by organisms is (1) water (3) oxygen (2) salt (4) glucose 24 SCIDS (Severe Combined Immunodeficiency Syndrome) is a disorder where a genetic mutation inhibits the production and functioning of T-cells. T-cells are special types of white blood cells that play a role in the body’s immune response. A possible symptom of SCIDS would be an increase in the (1) number of antigens produced (2) red blood cell count (3) number of infections by pathogens (4) ability to maintain homeostasis 25 An organ, such as a kidney, used for transplant needs to be tested for compatibility with the person who is to receive the organ. If this is not done, the (1) donated organ might attack the body (2) donated organ might attack the immune system (3) immune system might attack its own body cells (4) immune system might attack the donated organ 26 A researcher concludes from a 10-year study that the biodiversity of an ecosystem had increased. Which set of observations represents evidence for this claim?
(1) There were more niches and greater stability in the ecosystem.
(2) There were more niches and less energy lost as heat in the ecosystem.
(3) There were fewer niches for decomposers and less stability.
(4) There were fewer niches for consumers and greater cycling of materials.
27 The final consumers in many food webs are (1) autotrophs (3) herbivores (2) hosts (4) carnivores 28 Abandoned farmland that once grew corn is now covered with bushes and small trees. These observed changes resulted directly from (1) evolutionary change (2) ecological succession (3) loss of biodiversity (4) selective breeding 29 The overuse of chemical fertilizers has resulted in the growth of some lawns in which decomposers cannot live. This would interfere most directly with the ability of the lawn ecosystem to (1) recycle energy (2) recycle nutrients (3) maintain atmospheric pH (4) reduce biodiversity 30 On Long Island, several businesses use geothermal technology. This involves taking heat from within Earth and using it to heat buildings.
One benefit of this technology is that it (1) contributes to global warming (2) reduces the ozone shield (3) reduces dependence on fossil fuels (4) decreases resources for many species B A C D Living Environment–Aug. ’16 [OVER] Living Environment–Aug. ’16 Part B–1 Answer all questions in this part. Directions (31–43): For each statement or question, record on the separate answer sheet the number of the word or expression that, of those given, best completes the statement or answers the question.
31 Recently, researchers from Stanford University have changed mouse skin cells into mouse nerve cells. This was accomplished by inserting genes that control the synthesis of certain proteins into the skin cells. This type of research is often successful in advancing knowledge regarding the functioning of human cells because (1) cells present in humans often function in similar ways to cells present in other organisms (2) cells from different types of organisms function differently when transplanted into humans (3) the cells in all complex organisms contain the same genes and function in similar ways (4) cellular research using mice can always be applied to human cells since all complex organisms produce the same proteins 32 In the experimental setup below, which substance would be used to prove that the gas produced by the yeast in the vacuum bottle could change the pH of the liquid in the flask?
(1) an indicator (2) a chemical messenger (3) an enzyme (4) a salt solution 33 Which statement best expresses a basic scientific assumption?
(1) Interpretation of experimental results has provided explanations for all natural phenomena.
(2) If a conclusion is valid, similar investigations by other scientists should result in the same conclusion.
(3) For any conclusion to be valid, the design of the experiment requires that only two groups be compared.
(4) After a scientist formulates a conclusion based on an experiment, no further investigation is necessary.
34 The diagram below represents a cycle that occurs in nature.
Which phrase describes a human activity that could have a negative effect on this cycle?
(1) a decrease in the amount of sulfates given off by motor vehicles (2) an increase in recycling programs for plastics and metals (3) the continued deforestation and removal of forest resources (4) development of programs to conserve wildlife Carbon dioxide in the atmosphere Organic compounds in autotrophs Bacterial activity Autotrophic activity Organic compounds in heterotrophs and their wastes Metabolic activity Rubber tube Flask Thermometer Yeast-glucose solution Vacuum bottle Living Environment–Aug. ’16 [OVER] Base your answers to questions 35 through 37 on the information and graph below and on your knowledge of biology.
A farmer growing potatoes notices aphids, a type of insect, feeding on the plants. An insecticide was sprayed on the plants several times over a two-year period. The graph represents samples of three different generations of insecticide-resistant and nonresistant aphids over this time period.
35 The resistance gene was present in the aphid population as a result of (1) the need of the potatoes to become resistant to the insecticide (2) changes in the aphids’ local habitat by the insecticide (3) a recombination of the proteins in the potato cells (4) a random change in the aphids’ DNA sequence 36 In year three, the farmer discontinued the use of the insecticide. Which statement would best predict the population in generation 4?
(1) The nonresistant aphid would become extinct.
(2) The nonresistant aphid population would likely increase.
(3) The resistant aphid would mutate to a nonresistant aphid.
(4) The plants would be free of insect populations.
37 One negative consequence of using an insecticide is that it (1) selects for insecticide-resistant organisms (2) keeps a balance of organic compounds (3) encourages biodiversity in plants (4) gives the nonresistant aphids a survival advantage Generations of Aphids Generation 1 Generation 2 Generation 3 Non Resistant Resistant Non Resistant Resistant Population of Aphids Non Resistant Resistant 38 The activity of a single-celled organism is represented in the diagram below.
Which concept is best illustrated by this diagram?
(1) The life functions performed by single-celled organisms are different from the life functions performed by complex organisms.
(2) Single-celled organisms carry out life functions that are essential for survival.
(3) Since single-celled organisms lack organs, they can survive only in moist environments.
(4) Single-celled organisms contain one organelle that performs all the life functions.
39 The chart below provides information about two scientific discoveries in the field of biology.
Which statement is the best interpretation of the material presented in the chart?
(1) Scientific explanations are built by combining evidence that can be observed with what people already know.
(2) Inquiry involves making judgments about the reliability of the source and relevance of the information.
(3) Science provides information, but values are also essential to making ethical decisions.
(4) Hypotheses are valuable even if they turn out not to be true, because they may lead to further investigation.
Early Discovery Later Discovery People living near swamps are more likely to get malaria than people who do not live near swamps. Burning swamps early in the summer reduces the amount of malaria.
Mosquitoes breed and lay their eggs in swamps and other pools of still water.
Mosquitoes are the carriers of the organisms that cause malaria.
Dark-staining bodies called chromosomes can be seen only in dividing cells. The number of chromosomes doubles during cell division.
Chromosomes contain DNA, which is able to copy itself. DNA carries the genetic code, which is passed from a parent cell to two or more daughter cells.
Molecules of A Food molecules Waste molecules B Energy for metabolism Compound C Living Environment–Aug. ’16 Base your answers to questions 40 through 43 on the diagram below and on your knowledge of biology. This diagram represents the roles of different parts of the human body in keeping blood sugar at a balanced, normal level over time.
40 The diagram shows human body structures that are coordinated to maintain homeostasis. Which row correctly identifies the functions of these structures?
41 When body system X releases too much sugar into the blood, the body can maintain homeostasis by making (1) more hormone A, only (3) more hormone A and more hormone B (2) more hormone B, only (4) no hormone A and no hormone B 42 If organ Y becomes unable to produce enough hormone B, then homeostasis would be disrupted. To restore homeostasis and compensate for the lack of hormone B, one useful action would be to (1) increase the production of hormone A (3) reduce the carbohydrates in body system X (2) remove organ Y from the body surgically (4) reduce the synthesis of enzymes in organ Y 43 If body system X temporarily stops releasing sugar into the blood, a likely response of the body would be to (1) stop using enzymes in body system X (3) start to increase synthesis of hormone B (2) stop organ Y from producing hormone A (4) start to increase synthesis of hormone A Row Body System X Organ Y (1) Digestion Regulation (2) Circulation Synthesis (3) Excretion Transport (4) Locomotion Nutrition Homeostasis of Blood Sugar Level Decreases sugar level in the blood Increases sugar level in the blood Body System X Organ Y Breaks down carbohydrates and releases sugar into the blood Makes hormone A, which releases sugar stored in the liver so it enters the blood Makes hormone B, which moves sugar out of the blood into cells Living Environment–Aug. ’16 [OVER] Living Environment–Aug. ’16 Base your answers to questions 44 through 47 on the information, diagram, and table below and on your knowledge of biology.
A concentrated starch solution was placed in a thistle tube with a semi-permeable membrane covering the wide opening. It was then placed in a beaker of water. The height of the solution in the tube was measured every 5 minutes for 25 minutes. The setup and the data collected are shown below.
Height of Liquid in Thistle Tube Time (min) Height (cm) 0 2 5 3 10 6 15 8 20 10 25 11 Before Membrane After 25 minutes Thistle tube Water Starch solution Part B–2 Answer all questions in this part. Directions (44–55): For those questions that are multiple choice, record on the separate answer sheet the number of the choice that, of those given, best completes each statement or answers each question. For all other questions in this part, follow the directions given and record your answers in the spaces provided in this examination booklet.
Directions (44–45): Using the information in the data table, construct a line graph on the grid below, following the directions below.
44 Mark an appropriate scale, without any breaks in the data, on each labeled axis. 45 Plot the data for height on the grid. Connect the points and surround each point with a small circle. 46 Explain why the height of the solution in the thistle tube increased during the 25-minute period. Note: The answer to question 47 should be recorded on your separate answer sheet.
47 The experiment was repeated, and an amber-colored solution was added to the water in the beaker. After 10 minutes, the water in the beaker remained amber-colored and the starch solution had turned blue-black.
The most likely reason for this observation is that (1) starch molecules moved out of the thistle tube (2) water molecules moved into the thistle tube (3) amber-colored solution moved into the thistle tube (4) water molecules moved out of the thistle tube Time (min) Height (cm) Height of Liquid in Thistle Tube Example: Living Environment–Aug. ’16 [OVER] Base your answer to question 48 on the information and diagrams below and on your knowledge of biology.
When fish of certain species are injured, a chemical substance stored in skin cells of the fish is released into the water. This chemical causes an alarm response among other fish of the same species in the area. Nearby fish of this species become more alert and group together near the bottom.
48 Explain why the chemical released from the injured fish may not cause an alarm response in other fish species. No Alarm Response Alarm Response Living Environment–Aug. ’16 Living Environment–Aug. ’16 [OVER] Base your answers to questions 49 through 51 on the diagram below and on your knowledge of biology.
Note: The answer to question 49 should be recorded on your separate answer sheet.
49 The process represented in the diagram is (1) DNA replication (3) gel electrophoresis (2) natural selection (4) genetic engineering Note: The answer to question 50 should be recorded on your separate answer sheet.
50 The original gene for the production of a human hormone was most likely removed from a (1) chromosome (3) mitochondrion (2) ribosome (4) cell membrane 51 State one possible reason why a gene for the production of a human hormone would be placed in bacterial DNA. Cut bacterial DNA Gene for a human hormone Inserted into bacterial cell Cell division Base your answers to questions 52 through 55 on the passage below and on your knowledge of biology.
The lake sturgeon is a fish that often grows over six feet long and can weigh close to two hundred pounds. It is currently an endangered species in the Great Lakes area, although the species has lived in those lakes and rivers for millions of years. Now, there is a program to increase the sturgeon population by reintroducing lake sturgeon to areas where they have disappeared.
Like the lake sturgeon, bloater fish are also found in the Great Lakes. Both find their food on or near the bottoms of lakes. They eat a variety of small organisms, including insect larvae, worms, and clams. These small organisms feed on algae.
52 Identify one population that will decrease in size after the lake sturgeon are added to the new ecosystems.
Support your answer. 53 Part of the food web of a lake ecosystem is represented in the diagram below. Indicate the position in the food web where each organism listed below would be placed, by writing the name of each in the appropriate box. algae bloater fish clams 54 Identify which population, other than lake sturgeon, will increase in size after the lake sturgeon are added to the new ecosystems. Support your answer. 55 State what the arrows in the food web represent. Worms Insect larvae Lake sturgeon Living Environment–Aug. ’16 Base your answers to questions 56 and 57 on the information and drawing below and on your knowledge of biology. The drawing represents a salamander. Salamanders are small amphibians that live in a variety of environments.
Two species of salamander inhabit an island. The habitat on each side of the island is different. One side tends to be wet; the other side tends to be dry. Researchers want to know if the salamanders will survive equally well on either side of the island. Species A lives on the wet side of the island, while Species B lives on the dry side of the island. Researchers develop two artificial habitats, one that simulates conditions on the wet side and one that simulates conditions on the dry side.
56 Explain why researchers would put the salamanders in an artificial environment, as opposed to conducting the experiment in their natural habitat. 57 Researchers put three salamanders of each species in each of the two different artificial environments.
Why would other scientists question the validity of the conclusions based on this setup? Living Environment–Aug. ’16 [OVER] Part C Answer all questions in this part.
Directions (56–72): Record your answers in the spaces provided in this examination booklet.
Base your answers to questions 58 through 61 on the information and diagram below and on your knowledge of biology.
Green Roofs People in Albany and New York City are using “green roofs” to improve the environment. A green roof can be added to many buildings that have large, flat roofs. Green roofs have three parts: a protective layer to separate plant roots from the roof of the building, an absorptive layer to catch and hold rainwater, and a layer of plants. Often, green roofs use Sedum, a short, desert plant, because it is efficient at storing water in its leaves and can withstand the colder climate.
A green roof saves energy, reduces carbon dioxide in the atmosphere, and prevents rainwater and melting snow from overloading sewer systems. It can also protect the roof of a building from damage. However, green roofs can be expensive to install, and require care and maintenance.
58 Most varieties of Sedum are not native plants in Albany or New York City. State one reason why it may be dangerous to introduce a new species to an established ecosystem. 59 State one reason why a green roof reduces the amount of carbon dioxide in the atmosphere. 60 State one reason why it is important to reduce the amount of carbon dioxide in the atmosphere. 61 State one disadvantage of a green roof. Roof of building The “Green Roof” Plants Absorptive layer Protective layer Living Environment–Aug. ’16 Living Environment–Aug. ’16 [OVER] 62 A scientist took samples from a culture of E. coli bacteria and placed them in each of 100 petri dishes.
Once the bacteria began to grow in the dishes, she exposed 50 of the dishes to x-ray radiation and 50 to natural light. After five days, she examined samples of DNA from the bacteria and recorded any differences she found between the DNA of the two groups. State one hypothesis the experiment would test. Base your answers to questions 63 and 64 on the passage below and on your knowledge of biology.
Super Vaccine Could Eliminate Flu Every flu season, vaccine makers must bet on which strain of influenza A will pose the greatest threat to the public, and millions of Americans must decide whether to get a shot.
In August, virologist Gary Nabel at the National Institutes of Health (NIH) announced progress toward a universal flu vaccine: two shots of it could provide years of protection from every known influenza A virus.
“We use a prime-boost strategy, meaning that we immunize with two vehicles that deliv-er the vaccine in different ways,” Nabel says. In their experimental treatment, he and his colleagues injected mice, ferrets, and monkeys with viral DNA, causing their muscle cells to produce hemagglutinin, a protein found on the surface of all flu viruses. The animals’ immune systems then began making antibodies that latch onto the protein and disable the virus. The researchers followed the DNA injection with a traditional seasonal flu shot, which contains dead viruses. This one-two punch protected the test subjects against influenza A viruses that had emerged in 1934 and 2007, and other experiments showed that the antibodies it generated successfully neutralized a wide variety of flu strains. Nabel’s col-leagues at the NIH are already testing similar approaches in humans.
Source: Rowe, A. “Super Vaccine Could Eliminate Flu.” Discover, Jan./Feb. 2011, p. 37.
63 Identify one specific difference, other than it is a two-step vaccination, between Nabel’s vaccination and a traditional flu vaccine. 64 Explain how injecting dead or weakened viruses into a person can help to fight against future infections from that virus. Living Environment–Aug. ’16 Base your answers to questions 65 through 68 on the information below and on your knowledge of biology.
Bald Eagle Facts • Bald eagles eat primarily fish, carrion (dead animals), smaller birds, and rodents. Their most important non-carrion food is fish, which they catch by swooping down and grabbing fish that are near the surface of the water.
• The number of nesting pairs in the lower 48 United States increased from fewer than 450 in the early 1960s to more than 4,500 adult bald eagle nesting pairs in the 1990s. Today, there are an estimated 9,789 nesting pairs of bald eagles.
• Bald eagles are found in large numbers in certain areas during the winter (known as roosts). These winter roosts are located in areas where prey are plentiful. Winter roosts are protected under federal law, and managed with a buffer zone to reduce human disturbance. As winter ends, the eagles return to their summer nesting/hunting areas.
Bald Eagle Research In the winter of 2009, volunteers from an Audubon group conducted a survey of roosting bald eagles at four locations in an area in the lower Hudson River Valley. The data below show the average number of eagles sighted and the number of visits made by the volunteers each month. Among the other data collected were percent ice cover and percent cloud cover on the surface of the water. The eagles fly freely between these four sites, depending on a variety of conditions.
Some of the data are shown in the table below.
Bald Eagles Sighted at Four Hudson Valley Locations in 2009 Location January February March Average Number of Eagles Number of Visits Average Number of Eagles Number of Visits Average Number of Eagles Number of Visits Croton Reservoir 22.86 7 47.88 8 9.17 6 George’s Island Park 27.00 7 18.38 8 5.00 4 George’s Island North 12.29 7 4.43 7 2.20 5 Stony Point 3.57 7 3.63 8 0.00 5 Living Environment–Aug. ’16 [OVER] 65 State one reason why the percent ice cover is important to the ability of eagles to obtain food. 66 What inference can be made about the percent ice cover at Croton Reservoir between January and February 2009? Support your answer. 67 State one reason why the number of eagles sighted showed a change at all four sites between February and March. 68 State one possible reason why a popular hiking trail in this area is closed during eagle roosting seasons. Base your answers to questions 69 through 72 on the information below and on your knowledge of biology.
Coral Reef Ecosystems There are many ecological interactions that maintain the biodiversity present in coral reefs. In addition to coral, microscopic algae, seaweed, sea grasses, sponges and worms, and a variety of fish are among the organisms that live in reef ecosystems. Ocean currents often link different reef systems and move organisms from one reef area to another. This movement is a factor in repopulating a reef that has been damaged by environmental changes.
One environmental change involves an increased growth of seaweed. When the population of seaweed increases, the reef shifts from a coral-dominated ecosystem to a seaweed-dominated ecosystem. This change disrupts the relationships between the organisms that live there.
Studies have shown that, as the density of seaweed in a reef area increases, the number of fish that eat the seaweed in that area decreases. This may be due to the presence of more predators, or the taste of the more mature plants. The fish move to areas where there is less seaweed growth. As this trend continues, the reef areas are taken over by the seaweed.
Once this happens, it is very hard to remove the seaweed and restore the reef to a healthy ecosystem.
In addition to this problem, temperature changes are threatening the ocean currents that connect the reef systems. A change in the currents would reduce the movement of fish larvae from one area to another. This contributes to the seaweed problem.
69 State the role of the sea grasses in the reef ecosystem. 70 Identify one abiotic factor that is affecting the stability of the coral reef ecosystems and state how the factor identified is important to the coral reef ecosystems. Abiotic factor: Effect: 71 State one reason why it is important to maintain the stability of the coral reefs. 72 State one advantage of the fish larvae moving by ocean currents into a damaged reef ecosystem. Living Environment–Aug. ’16 Base your answers to questions 73 and 74 on the diagram below, which represents a process that occurs in living cells, and on your knowledge of biology.
Note: The answer to question 73 should be recorded on your separate answer sheet.
73 The process shown in the diagram is (1) cellular respiration (3) gene recombination (2) cellular reorganization (4) protein synthesis Note: The answer to question 74 should be recorded on your separate answer sheet.
74 Structure X is a (1) mitochondrion (3) nucleus (2) vacuole (4) ribosome CUA A C C A C C Newly formed molecule C C G mRNA Glycine Glycine Arginine Proline Alanine G A A G G U U G G U G G G G C Structure X Living Environment–Aug. ’16 [OVER] Part D Answer all questions in this part.
Directions (73–85): For those questions that are multiple choice, record on the separate answer sheet the number of the choice that, of those given, best completes the statement or answers the question. For all other questions in this part, follow the directions given and record your answers in the spaces provided in this examination booklet.
Base your answers to questions 75 through 77 on the diagram below and on your knowledge of biology.
Note: The answer to question 75 should be recorded on your separate answer sheet.
75 A finch that picks small insects out from cracks in the bark of trees would most likely have a beak that is (1) sharp and thin (3) rounded and thin (2) sharp and thick (4) rounded and thick Note: The answer to question 76 should be recorded on your separate answer sheet.
76 Which statement is a basic assumption from The Beaks of Finches lab?
(1) The type of beak indicates the type of food the finch eats.
(2) Different birds have different songs.
(3) Birds with larger beaks can find mates more easily.
(4) Nesting behavior of finches is an inherited trait.
77 State two reasons why the large ground finch and sharp-billed ground finch could live on the same island but not compete for food, even though they both eat mainly plant food. Reason 1: Reason 2: Large ground finch Vegetarian finch Large tree finch Small tree finch Woodpecker finch Warbler finch Cactus finch Sharp-billed ground finch Small ground finch Medium ground finch E d g e c r u s h i n g B i t i n g t i p s C r u s h i n g b i l l s Mainly plant food Mainly animal food G r a s p i n g b i l l s All animal food P r o b i n g P r o b i n g b i l l s from: Galapagos: A Natural History Guide Variations in Beaks of Galapagos Islands Finches Living Environment–Aug. ’16 Base your answer to question 78 on the table below and on your knowledge of biology. The table shows which of four enzymes are present in three related plant species.
Comparison of Four Enzymes The tree diagrams below show two possible evolutionary relationships between the three species.
78 In the space below, write the number of the tree diagram that shows the most probable evolutionary relationship between the three species. Support your answer. Tree diagram: _ Plant Species Enzyme W Enzyme X Enzyme Y Enzyme Z Species A present present absent present Species B absent absent present absent Species C present present absent present Tree 1 Tree 2 A B C A C B Living Environment–Aug. ’16 [OVER] Base your answers to questions 79 and 80 on the diagram below, which represents the shrinking of a cell in response to an increase in the concentration of a substance outside of the cell.
79 Identify substance A. 80 Identify one likely substance in the environment of the cell that caused this response. Note: The answer to question 81 should be recorded on your separate answer sheet.
81 A student lifted weights after school and found that his muscles started to burn. He couldn’t continue to lift the weights after prolonged exercising. This muscle fatigue is most likely due to (1) the heart beating too fast and tiring out (2) the lungs accumulating oxygen (3) lack of oxygen and build up of waste in the muscles (4) lack of carbon dioxide in the muscles A A A A Living Environment–Aug. ’16 Base your answers to questions 82 and 83 on the diagram below and on your knowledge of biology. The diagram represents three sections of a cell membrane showing three different methods involved in the transport of various molecules across the membrane.
Note: The answer to question 82 should be recorded on your separate answer sheet.
82 Methods A and B are classified as methods of passive transport because they do not require (1) ATP (3) light (2) carbon dioxide (4) DNA 83 Using information from the diagram, state one reason why the movement of molecules in method C represents active transport. Method A Method B Method C Living Environment–Aug. ’16 [OVER] 84 State one reason why some species might have similar body structures even if they are not closely related. 85 A student went out to the school track and walked two laps, ran two laps, and then walked two more laps.
On the grid below, draw a line that shows what most likely happened to the pulse rate of the student during these activities. Pulse Rate Time Effect of Activity on Pulse Rate Living Environment–Aug. ’16 LIVING ENVIRONMENT LIVING ENVIRONMENT Printed on Recycled Paper FOR TEACHERS ONLY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION LIVING ENVIRONMENT Thursday, August 18, 2016 — 12:30 to 3:30 p.m., only SCORING KEY AND RATING GUIDE Directions to the Teacher: Refer to the directions on page 2 before rating student papers.
Updated information regarding the rating of this examination may be posted on the New York State Education Department’s web site during the rating period. Check this web site at: and select the link “Scoring Information” for any recently posted information regarding this examination. This site should be checked before the rating process for this examination begins and several times throughout the Regents Examination period.
Multiple Choice for Parts A, B–1, B–2, and D Allow 1 credit for each correct response.
1 . . . . . . 4 . . . . . . 9 . . . . . . 1 . . . . . . 17 . . . . . . 2 . . . . . . 25 . . . . . . 4 . . . . . .
2 . . . . . . 3 . . . . . . 10 . . . . . . 3 . . . . . . 18 . . . . . . 4 . . . . . . 26 . . . . . . 1 . . . . . .
3 . . . . . . 3 . . . . . . 11 . . . . . . 1 . . . . . . 19 . . . . . . 2 . . . . . . 27 . . . . . . 4 . . . . . .
4 . . . . . . 1 . . . . . . 12 . . . . . . 1 . . . . . . 20 . . . . . . 3 . . . . . . 28 . . . . . . 2 . . . . . .
5 . . . . . . 2 . . . . . . 13 . . . . . . 3 . . . . . . 21 . . . . . . 2 . . . . . . 29 . . . . . . 2 . . . . . .
6 . . . . . . 2 . . . . . . 14 . . . . . . 3 . . . . . . 22 . . . . . . 4 . . . . . . 30 . . . . . . 3 . . . . . .
7 . . . . . . 1 . . . . . . 15 . . . . . . 2 . . . . . . 23 . . . . . . 4 . . . . . .
8 . . . . . . 4 . . . . . . 16 . . . . . . 3 . . . . . . 24 . . . . . . 3 . . . . . .
Part A Part B–1 Part B–2 Part D 31 . . . . . . 1 . . . . . . 35 . . . . . . 4 . . . . . . 39 . . . . . . 1 . . . . . . 43 . . . . . . 4 . . . . . .
32 . . . . . . 1 . . . . . . 36 . . . . . . 2 . . . . . . 40 . . . . . . 1 . . . . . .
33 . . . . . . 2 . . . . . . 37 . . . . . . 1 . . . . . . 41 . . . . . . 2 . . . . . .
34 . . . . . . 3 . . . . . . 38 . . . . . . 2 . . . . . . 42 . . . . . . 3 . . . . . .
47 . . . . . . 3 . . . . . . 49 . . . . . . 4 . . . . . . 50 . . . . . . 1 . . . . . .
73 . . . . . . 4 . . . . . . 75 . . . . . . 1 . . . . . . 81 . . . . . . 3 . . . . . .
74 . . . . . . 4 . . . . . . 76 . . . . . . 1 . . . . . . 82 . . . . . . 1 . . . . . .
Directions to the Teacher Follow the procedures below for scoring student answer papers for the Regents Examination in Living Environment. Additional information about scoring is provided in the publication Information Booklet for Scoring Regents Examinations in the Sciences.
Do not attempt to correct the student’s work by making insertions or changes of any kind. If the student’s responses for the multiple-choice questions are being hand scored prior to being scanned, the scorer must be careful not to make any marks on the answer sheet except to record the scores in the designated score boxes. Marks elsewhere on the answer sheet will interfere with the accuracy of the scanning.
Allow 1 credit for each correct response. At least two science teachers must participate in the scoring of the Part B–2, Part C, and Part D open-ended questions on a student’s paper. Each of these teachers should be responsible for scoring a selected number of the open-ended questions on each answer paper. No one teacher is to score more than approximately one-half of the open-ended questions on a student’s answer paper. Teachers may not score their own students’ answer papers.
Students’ responses must be scored strictly according to the Scoring Key and Rating Guide. For open-ended questions, credit may be allowed for responses other than those given in the rating guide if the response is a scientifically accurate answer to the question and demonstrates adequate knowledge as indicated by the examples in the rating guide. On the student’s separate answer sheet, for each question, record the number of credits earned and the teacher’s assigned rater/scorer letter. Fractional credit is not allowed. Only whole-number credit may be given for a response. If the student gives more than one answer to a question, only the first answer should be rated. Units need not be given when the wording of the questions allows such omissions.
For hand scoring, raters should enter the scores earned in the appropriate boxes printed on the separate answer sheet. Next, the rater should add these scores and enter the total in the box labeled “Total Raw Score.” Then the student’s raw score should be converted to a scale score by using the conversion chart that will be posted on the Department’s web site at: on Thursday, August 18, 2016. The student’s scale score should be entered in the box labeled “Scale Score” on the student’s answer sheet.
The scale score is the student’s final examination score.
Schools are not permitted to rescore any of the open-ended questions on this exam after each question has been rated once, regardless of the final exam score. Schools are required to ensure that the raw scores have been added correctly and that the resulting scale score has been determined accurately.
Because scale scores corresponding to raw scores in the conversion chart may change from one administration to another, it is crucial that, for each administration, the conversion chart provided for that administration be used to determine the student’s final score.
Living Environment Rating Guide – Aug. ’16 Part B–2 44 Allow 1 credit for marking an appropriate scale, without any breaks in the data, on each labeled axis.
45 Allow 1 credit for correctly plotting the data, connecting the points, and surounding each point with a small circle.
Example of a 2-credit graph for questions 44–45: Note: Allow credit if points are correctly plotted, but not circled.
Do not assume that the intersection of the x- and y-axes is the origin (0,0), unless it is labeled. An appropriate scale only needs to include the data range in the data table.
Do not allow credit if points are plotted that are not in the data table, e.g., (0,0), or for extending lines beyond the data points.
46 Allow 1 credit. Acceptable responses include, but are not limited to: — Water diffused (moved) into the thistle tube.
— Water went from an area of high concentration to an area of low concentration.
— Osmosis took place.
— Water diffused through the membrane.
47 MC on scoring key Time (min) Height (cm) 12 11 10 9 8 7 6 5 4 3 2 1 0 0 5 10 15 20 25 Height of Liquid in Thistle Tube Living Environment Rating Guide – Aug. ’16 48 Allow 1 credit. Acceptable responses include, but are not limited to: — Other species lack the receptors for the chemical.
— The alarm chemicals are specific to the species.
— Other species lack the ability to sense/recognize the chemical.
49 MC on scoring key 50 MC on scoring key 51 Allow 1 credit. Acceptable responses include, but are not limited to: — The bacteria will produce the human hormone quickly.
— to increase the production of insulin or some other hormone — The human hormone produced can be given to people who need it, and the number of allergic reactions will be reduced.
— to get the bacteria to produce a human hormone 52 Allow 1 credit for identifying one correct population and supporting the answer. Acceptable responses include, but are not limited to: — Populations of insect larvae [or clams or worms] will decrease in size because they will be eaten by both lake sturgeon and bloater fish.
— The bloater fish population will decrease in size because it will have to compete with the lake sturgeon for food resources.
— clams because they are eaten by the sturgeon 53 Allow 1 credit for: Worms Insect larvae Lake sturgeon Bloater fish Algae Clams Living Environment Rating Guide – Aug. ’16 54 Allow 1 credit for algae and supporting the answer. Acceptable responses include, but are not limited to: — Algae, because there will be fewer organisms that feed on them.
— More algae will survive and reproduce because populations of insect larvae, clams, and worms that eat the algae will decrease in size because they will be eaten by both lake stur-geon and bloater fish.
Note: Allow credit for an answer that is consistent with the student’s response to question 53.
55 Allow 1 credit. Acceptable responses include, but are not limited to: — The arrows show the direction of energy movement from one organism to another.
— The arrows show the flow of energy.
— movement of nutrients through the food web — what each organism feeds on Living Environment Rating Guide – Aug. ’16 Part C 56 Allow 1 credit. Acceptable responses include, but are not limited to: — too hard to track in natural habitat — The salamanders are easier to watch.
— The conditions the salamanders are kept in can be better controlled.
— to eliminate the possibility of them becoming an invasive species — to avoid predators 57 Allow 1 credit. Acceptable responses include, but are not limited to: — The sample size is small.
— The environment is artificial.
— The salamanders are not in their natural habitat.
— The experiment was not repeated.
58 Allow 1 credit. Acceptable responses include, but are not limited to: — New species might compete with native species for limited resources.
— Native species can have their space, light, and water taken away by invasive plants.
— New plant species could also have nonnative insects/pathogens on them.
59 Allow 1 credit. Acceptable responses include, but are not limited to: — Plants take in carbon dioxide to perform photosynthesis.
— Plants use CO2 to produce food.
— Plants take in carbon dioxide.
60 Allow 1 credit. Acceptable responses include, but are not limited to: — Carbon dioxide adds to the “Greenhouse Effect.” — Excessive carbon dioxide increases the temperature.
— CO2 has been associated with global warming/global climate change.
61 Allow 1 credit. Acceptable responses include, but are not limited to: — Green roofs are expensive to install.
— Green roofs need more maintenance.
— Green roofs might be too heavy for the building structure.
Living Environment Rating Guide – Aug. ’16 62 Allow 1 credit. Acceptable responses include, but are not limited to: — Bacteria exposed to x-ray radiation will show a greater number of mutations than bacteria exposed to natural light.
— Bacteria exposed to x rays will show less DNA damage than bacteria exposed to natural light.
— If bacteria are exposed to x rays, they will show more mutations than bacteria not exposed to x rays.
— The DNA of bacteria exposed to x rays will change more than that of bacteria not exposed to x rays.
— X rays affect bacterial DNA.
Note: Do not allow credit for a hypothesis written in the form of a question.
63 Allow 1 credit. Acceptable responses include, but are not limited to: — protects from every known influenza A virus — directly injects viral DNA — adds years of protection — It makes muscle cells produce hemagglutinin.
— It causes muscle cells to make a protein found on the surface of all flu viruses.
64 Allow 1 credit. Acceptable responses include, but are not limited to: — Your body responds by making antibodies that protect you from future infections.
— There are antigens on the surface of the dead virus, which then trigger the body to produce antibodies specific to that antigen.
— The body makes memory cells that can fight the virus in the future.
65 Allow 1 credit. Acceptable responses include, but are not limited to: — The access to fish is limited by the amount of ice cover.
— They have to search for foods other than fish.
— The ice cover might limit the number of organisms that are available for the eagles to eat.
66 Allow 1 credit for stating what inference can be made about the percent ice cover at Croton Reservoir between January and February 2009 and supporting the answer. Acceptable responses include, but are not limited to: — The ice cover was less in February because there were more eagles.
— Since there were fewer eagles in January, there must have been more ice.
Living Environment Rating Guide – Aug. ’16 67 Allow 1 credit. Acceptable responses include, but are not limited to: — Winter is ending, so the eagles returned to their summer nesting areas.
— As winter ended, the eagles migrated away.
— They fly freely between the sites, so their numbers would vary.
— The number of visits by volunteers varied.
— There were fewer visits in March than in January or February.
— There were more late-season storms/ice in March than February.
68 Allow 1 credit. Acceptable responses include, but are not limited to: — Winter roosts are protected under federal law and managed with a buffer zone to reduce human disturbance.
— Humans might disturb the eagles.
69 Allow 1 credit. Acceptable responses include, but are not limited to: — Their role is to convert inorganic (raw) materials into organic matter/food.
— They capture radiant energy.
— They serve as the producer in this food web/ecosystem.
— They serve as food for fish.
— produce oxygen 70 Allow 1 credit. Acceptable responses include, but are not limited to: — temperature: causes a change in currents — light: used by plants for photosynthesis — ocean currents: carry organisms from one reef to another — carbon dioxide: used by plants for photosynthesis 71 Allow 1 credit. Acceptable responses include, but are not limited to: — in order to maintain the food webs that exist there — because many organisms depend on them for food or shelter — to maintain biodiversity — keep oceans healthy — to prevent extinction of reef species 72 Allow 1 credit. Acceptable responses include, but are not limited to: — The fish larvae could repair/repopulate the damaged reefs.
— The reef might become more stable.
— They might keep the seaweed under control/eat the seaweed.
— It would provide the fish larvae with food/shelter.
— The fish larvae would have less competition.
Living Environment Rating Guide – Aug. ’16 Part D 73 MC on scoring key 74 MC on scoring key 75 MC on scoring key 76 MC on scoring key 77 Allow 1 credit for stating two reasons. Acceptable responses include, but are not limited to: — They have different beak structures.
— They might eat different types of plants.
— They might eat at different times of day or night.
— They might live in different areas of the island.
78 Allow 1 credit for tree 2 and supporting the answer. Acceptable responses include, but are not lim-ited to: — Species B has enzyme Y, which is not present in either species A or C.
— Enzyme W (and/or X) is present in species A and C, but not in B.
— A and C have enzymes W, X, and Z, but B doesn’t.
— A and C have 3 enzymes in common: W, X, and Z.
— A and C have more in common.
79 Allow 1 credit for water.
80 Allow 1 credit. Acceptable responses include, but are not limited to: — salt — sugar — seawater 81 MC on scoring key Living Environment Rating Guide – Aug. ’16 82 MC on scoring key 83 Allow 1 credit. Acceptable responses include, but are not limited to: — C shows the transport of molecules from an area of low concentration to an area of higher concentration.
— The molecules are moving against the concentration gradient.
84 Allow 1 credit. Acceptable responses include, but are not limited to: — They may have evolved in the same or similar environments, and these structures and traits were advantageous adaptations.
— They have two separate mutations that produce the same appearance.
— These structures might help them use similar food sources/adapt to their environment.
— The species inhabit similar environments.
— They occupy similar niches.
— natural selection — convergent evolution 85 Allow 1 credit.
Example of a 1-credit response: Note: Allow credit for a graph that shows a gradual increase followed by a gradual decrease.
Pulse Rate Time Effect of Activity on Pulse Rate Living Environment Rating Guide – Aug. ’16 Living Environment Rating Guide – Aug. ’16 The Chart for Determining the Final Examination Score for the August 2016 Regents Examination in Living Environment will be posted on the Department’s web site at: on Thursday, August 18, 2016. Conversion charts provided for previous administrations of the Regents Examination in Living Environment must NOT be used to determine students’ final scores for this administration.
Online Submission of Teacher Evaluations of the Test to the Department Suggestions and feedback from teachers provide an important contribution to the test development process. The Department provides an online evaluation form for State assessments. It contains spaces for teachers to respond to several specific questions and to make suggestions. Instructions for completing the evaluation form are as follows: 1. Go to
2. Select the test title.
3. Complete the required demographic fields.
4. Complete each evaluation question and provide comments in the space provided.
5. Click the SUBMIT button at the bottom of the page to submit the completed form.
Living Environment Rating Guide – Aug. ’16 Map to Core Curriculum August 2016 Living Environment Standards Question Numbers Part A 1–30 Part B–1 31–43 Part B–2 44–55 Part C 56–72 Standard 1 — Analysis, Inquiry and Design Key Idea 1 39 47 Key Idea 2 56, 57, 62 Key Idea 3 33, 36, 38 44, 45 Appendix A (Laboratory Checklist) 32 Standard 4 Key Idea 1 1, 14 31, 40 46, 48, 52, 53, 54, 55 69 Key Idea 2 4, 5, 8, 11 35 49, 50, 51 Key Idea 3 9, 13, 15, 16, 17, 18 37 Key Idea 4 12, 19, 20, 21, 22 Key Idea 5 2, 7, 10, 23, 24, 25 41, 42, 43 63, 64 Key Idea 6 26, 27, 28 65, 66, 67, 70, 71, 72 Key Idea 7 3, 6, 29, 30 34 58, 59, 60, 61, 68 Part D 73–85 Lab 1 73, 74, 78, 84 Lab 2 81, 8 Lab 3 75, 76, 77 Lab 5 79, 80, 82, 83 5 Living Environment Conversion Chart - Aug. '16 1 of 1 Raw Scale Raw Scale Raw Scale Score Score Score Score Score Score 85 100 56 78 27 50 84 98 55 77 26 48 83 97 54 76 25 47 82 96 53 75 24 46 81 95 52 75 23 44 80 95 51 74 22 43 79 94 50 73 21 41 78 93 49 73 20 40 77 92 48 72 19 38 76 91 47 71 18 37 75 91 46 70 17 35 74 90 45 69 16 33 73 89 44 68 15 32 72 88 43 68 14 30 71 88 42 67 13 28 70 87 41 66 12 26 69 86 40 65 11 24 68 86 39 64 10 22 67 85 38 63 9 20 66 84 37 62 8 18 65 84 36 61 7 16 64 83 35 59 6 14 63 82 34 58 5 12 62 82 33 57 4 10 61 81 32 56 3 7 60 80 31 55 2 5 59 80 30 54 1 3 58 79 29 52 0 0 57 78 28 51 The State Education Department / The University of the State of New York To determine the student’s final examination score, find the student’s total test raw score in the column labeled “Raw Score” and then locate the scale score that corresponds to that raw score. The scale score is the student’s final examination score. Enter this score in the space labeled “Scale Score” on the student’s answer sheet.
Schools are not permitted to rescore any of the open-ended questions on this exam after each question has been rated once, regardless of the final exam score. Schools are required to ensure that the raw scores have been added correctly and that the resulting scale score has been determined accurately.
Because scale scores corresponding to raw scores in the conversion chart change from one administration to another, it is crucial that for each administration the conversion chart provided for that administration be used to determine the student’s final score. The chart above is usable only for this administration of the Regents Examination in Living Environment.
Chart for Converting Total Test Raw Scores to Final Examination Scores (Scale Scores) Regents Examination in Living Environment – August 2016 |
1372 | https://www.youtube.com/watch?v=jYbt9fT9a00 | Trigonometric Identity with Double Angle Formula sin2x cos 2x
Anil Kumar
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Posted: 14 Oct 2016
SumToProductTrigIdentity:
Double Angle Identities:
Anil Kumar Math Classes: anil.anilkhandelwal@gmail.com
Challenge Examples on Compound Angles:
#SumToProductTrigIdentity #CompoundAngleIdentities #EquivalentTrigonometricExpressions #mhf4utrigonometry #CofunctionIdentity #relatedAcuteAngle #mhf4u_trigonometry #trigonometric_identities_Test #CompoundAngle #LinearTrigonometricEquations #QuadraticTrigonometricEquations
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Transcript:
I'm Adele Kumar we are working with trigonometric identities using double angle let me give you the formulas first sine 2x that could be written as 2 sine x cos x and cos 2 X can be written in many different ways you could write this as cos square X minus sine square X you would also write this as 2 cos square X minus 1 or 1 minus 2 sine square X appropriate use of the formula helps you to solve the question in minimal number of steps that is kind of critical so let's begin from the right side the reason right side is more complicated so we could simplify it towards the left that's the whole idea right the right side for us is sine 2x over sine X minus cos 2x over cos x now at this stage you could take common denominator of sine X cosine X right so let me write down sine X cosine X as common denominator cross multiply so we get cos x times sine 2x minus sine x times cos 2x cream now sine 2x can be written as 2 times sine x cos x so we could write this as cos x times 2 times sine x cos x right and here cos 2 X could be written as we could use one of the formulas which we are using since we have cos square X here I'll prefer to use 2 cos square X minus 1 so instead of cos 2 X I am writing 2 cos square X minus 1 right denominator for us is sine x cos x there are many different ways of doing this question how I've adopted this technique now the numerator we can factor sine X K so let us factor sine X from the numerator so we have factored the terms sine X that is this one what are we left with we're left with 2 cos square X let's put them in brackets 2 cos square X from here minus 2 cos square X minus 1 that becomes plus right because this is minus all signs will change to cos square X plus 1 crate divided by sine X cos x now here you can cancel sine X with sine X what is left in the brackets 2 cos square X minus 2 cos square X is 0 so left with 1 so you get 1 over cos x + 1 / cos X is secant X and that is the right side which we need to prove right so like this you can very easily prove this identity right some of you can actually simplify the stage a bit right so dividing this and this you can simplify a bit and then continue but I hope this is also a good way of proving I'm Anil Kumar and I hope that helps thank you and all the best |
1373 | https://lscanlonscience.weebly.com/uploads/1/0/9/2/109247633/half_reaction_practice_key.pdf | Half Reaction Practice - KEY Equation #1: Zn + Fe3+ à Zn2+ + Fe 1. Zinc goes from 0 to +2 and iron goes from +3 to 0. The changes in oxidation state indicate a redox reaction occurred. 2. Zinc is losing electrons, while iron is gaining electrons. 3. Zinc is oxidized because the oxidation number increases, meaning electrons are lost. Iron is reduced because the oxidation number decreases, meaning electrons are gained. 4. Zn0 is the reducing agent (because it is oxidized) and Fe3+ is the oxidizing agent (because it is reduced). 5. Half reactions: a. Oxidation: Zn0 à Zn2+ + 2 e- BALANCE 3(Zn0 à Zn2+ + 2 e-) = 3 Zn0 à 3 Zn2+ +6 e- b. Reduction: Fe3+ + 3 e- à Fe0 BALANCE 2(Fe3+ + 3e- à Fe0) = 2 Fe3+ + 6 e- à 2 Fe0 6. 3 Zn + 2 Fe3+ à 3 Zn2+ + 2 Fe 7. The above balanced equation shows conservation of mass because there are 2 moles of iron on each sides of the equation, and 3 moles of zinc on each side of the equation. It shows conservation of charge because the sum of the charges on the left side is (+6), and the sum of the charges on the right side is (+6). Both sides of the equation have the same net charge. In addition, from the balanced half-reactions, we see the electrons lost by Zn0 is equal to the electrons gained by Fe3+. Equation #2: Al + Ni2+ à Al3+ Ni 1. Aluminum goes from 0 to +3 and nickel goes from +2 to 0. The changes in oxidation state indicate a redox reaction occurred. 2. Aluminum is losing electrons, while nickel is gaining electrons. 3. Aluminum is oxidized because the oxidation number increases, meaning electrons are lost. Nickel is reduced because the oxidation number decreases, meaning electrons are gained. 4. Al0 is the reducing agent (because it is oxidized) and Ni2+ is the oxidizing agent (because it is reduced). 5. Half reactions: a. Oxidation: Al0 à Al3+ + 3 e- BALANCE 2(Al0 à Al3+ + 3 e-) = 2 Al0 à 2 Al3+ + 6 e- b. Reduction: Ni2+ + 2 e- à Ni0 BALANCE 3(Ni2+ + 2 e- à Ni0) = 3 Ni2+ + 6 e- à 3 Ni0 6. 2 Al + 3 Ni2+ à 2 Al3+ + 3 Ni 7. The above balanced equation shows conservation of mass because there are 2 moles of aluminum on each sides of the equation, and 3 moles of nickel on each side of the equation. It shows conservation of charge because the sum of the charges on the left side is (+6), and the sum of the charges on the right side is (+6). Both sides of the equation have the same net charge. In addition, from the balanced half-reactions, we see the electrons lost by Al0 is equal to the electrons gained by Ni2+. Equation #3: Cu + Ag+ à Cu2+ + Ag 1. Copper goes from 0 to +2 and silver goes from +1 to 0. The changes in oxidation state indicate a redox reaction occurred. 2. Copper is losing electrons, while silver is gaining electrons. 3. Coper is oxidized because the oxidation number increases, meaning electrons are lost. Silver is reduced because the oxidation number decreases, meaning electrons are gained. 4. Cu0 is the reducing agent (because it is oxidized) and Ag+ is the oxidizing agent (because it is reduced). 5. Half reactions: a. Oxidation: Cu0 à Cu2+ + 2 e- BALANCE no balancing needed = Cu0 à Cu2+ + 2 e- b. Reduction: Ag+ + e- à Ag0 BALANCE 2(Ag+ + e- à Ag0) = 2 Ag+ + 2 e- à 2 Ag0 6. Cu + 2 Ag+ à Cu2+ + 2 Ag 7. The above balanced equation shows conservation of mass because there are 2 moles of silver on each sides of the equation, and 1 mole of copper on each side of the equation. It shows conservation of charge because the sum of the charges on the left side is (+2), and the sum of the charges on the right side is (+2). Both sides of the equation have the same net charge. In addition, from the balanced half-reactions, we see the electrons lost by Cu0 is equal to the electrons gained by Ag+. Equation #4: Ag+ + Pb à Pb2+ + Ag 1. Lead goes from 0 to +2 and silver goes from +1 to 0. The changes in oxidation state indicate a redox reaction occurred. 2. Lead is losing electrons, while silver is gaining electrons. 3. Lead is oxidized because the oxidation number increases, meaning electrons are lost. Silver is reduced because the oxidation number decreases, meaning electrons are gained. 4. Pb0 is the reducing agent (because it is oxidized) and Ag+ is the oxidizing agent (because it is reduced). 5. Half reactions: a. Oxidation: Pb0 à Pb2+ + 2 e- BALANCE no balancing needed = Pb0 à Pb2+ + 2 e- b. Reduction: Ag+ + e- à Ag0 BALANCE 2(Ag+ + e- à Ag0) = 2 Ag+ + 2 e- à 2 Ag0 6. Pb + 2 Ag+ à Pb2+ + 2 Ag 7. The above balanced equation shows conservation of mass because there are 2 moles of silver on each sides of the equation, and 1 mole of lead on each side of the equation. It shows conservation of charge because the sum of the charges on the left side is (+2), and the sum of the charges on the right side is (+2). Both sides of the equation have the same net charge. In addition, from the balanced half-reactions, we see the electrons lost by Pb0 is equal to the electrons gained by Ag+. Equation #5: Zn + Cr3+ à Zn2+ + Cr 1. Zinc goes from 0 to +2 and chromium goes from +3 to 0. The changes in oxidation state indicate a redox reaction occurred. 2. Zinc is losing electrons, while chromium is gaining electrons. 3. Zinc is oxidized because the oxidation number increases, meaning electrons are lost. Chromium is reduced because the oxidation number decreases, meaning electrons are gained. 4. Zn0 is the reducing agent (because it is oxidized) and Cr3+ is the oxidizing agent (because it is reduced). 5. Half reactions: a. Oxidation: Zn0 à Zn2+ + 2 e- BALANCE 3(Zn0 à Zn2+ + 2 e-) = 3 Zn0 à 3 Zn2+ +6 e- b. Reduction: Cr3+ + 3 e- à Cr0 BALANCE 2(Cr3+ + 3e- à Cr0) = 2 Cr3+ + 6 e- à 2 Cr0 6. 3 Zn + 2 Cr3+ à 3 Zn2+ + 2 Cr 7. The above balanced equation shows conservation of mass because there are 2 moles of chromium on each sides of the equation, and 3 moles of zinc on each side of the equation. It shows conservation of charge because the sum of the charges on the left side is (+6), and the sum of the charges on the right side is (+6). Both sides of the equation have the same net charge. In addition, from the balanced half-reactions, we see the electrons lost by Zn0 is equal to the electrons gained by Cr3+. Equation #6: Ag+ + Ni à Ag + Ni2+ 1. Nickel goes from 0 to +1 and silver goes from +1 to 0. The changes in oxidation state indicate a redox reaction occurred. 2. Nickel is losing electrons, while silver is gaining electrons. 3. Nickel is oxidized because the oxidation number increases, meaning electrons are lost. Silver is reduced because the oxidation number decreases, meaning electrons are gained. 4. Ni0 is the reducing agent (because it is oxidized) and Fe3+ is the oxidizing agent (because it is reduced). 5. Half reactions: a. Oxidation: Ni0 à Ni2+ + 2 e- BALANCE no balancing needed = Ni0 à Ni2+ + 2 e- b. Reduction: Ag+ + e- à Ag0 BALANCE 2(Ag+ + e- à Ag0) = 2 Ag+ + 2 e- à 2 Ag0 6. Ni + 2 Ag+ à Ni2+ + 2 Ag 7. The above balanced equation shows conservation of mass because there are 2 silver of iron on each sides of the equation, and 1 mole of nickel on each side of the equation. It shows conservation of charge because the sum of the charges on the left side is (+2), and the sum of the charges on the right side is (+2). Both sides of the equation have the same net charge. In addition, from the balanced half-reactions, we see the electrons lost by Ni0 is equal to the electrons gained by Ag+. Equation #7: Cu2+ + Fe à Cu + Fe3+ 1. Iron goes from 0 to +3 and copper goes from +2 to 0. The changes in oxidation state indicate a redox reaction occurred. 2. Iron is losing electrons, while copper is gaining electrons. 3. Iron is oxidized because the oxidation number increases, meaning electrons are lost. Copper is reduced because the oxidation number decreases, meaning electrons are gained. 4. Fe0 is the reducing agent (because it is oxidized) and Cu2+ is the oxidizing agent (because it is reduced). 5. Half reactions: a. Oxidation: Fe0 à Fe3+ + 3 e- BALANCE 2(Fe0 à Fe3+ + 3 e-) = 2 Fe0 à 2 Fe3+ +6 e- b. Reduction: Cu2+ + 2 e- àCu0 BALANCE 3(Cu2+ + 2 e- à Cu0) = 3 Cu2+ + 6 e- à 3 Cu0 6. 2 Fe + 3Cu2+ à 2 Fe3+ + 3 Cu 7. The above balanced equation shows conservation of mass because there are 2 moles of iron on each sides of the equation, and 3 moles of copper on each side of the equation. It shows conservation of charge because the sum of the charges on the left side is (+6), and the sum of the charges on the right side is (+6). Both sides of the equation have the same net charge. In addition, from the balanced half-reactions, we see the electrons lost by Fe0 is equal to the electrons gained by Cu2+. Equation #8: Cu + Al3+ à Cu2+ + Al 1. Copper goes from 0 to +2 and aluminum goes from +3 to 0. The changes in oxidation state indicate a redox reaction occurred. 2. Copper is losing electrons, while aluminum is gaining electrons. 3. Copper is oxidized because the oxidation number increases, meaning electrons are lost. Aluminum is reduced because the oxidation number decreases, meaning electrons are gained. 4. Cu0 is the reducing agent (because it is oxidized) and Al3+ is the oxidizing agent (because it is reduced). 5. Half reactions: a. Oxidation: Cu0 à Cu2+ + 2 e- BALANCE 3(Cu0 à Cu2+ + 2 e-) = 3 Cu0 à 3 Cu2+ +6 e- b. Reduction: Al3+ + 3 e- à Al0 BALANCE 2(Al3+ + 3e- à Al0) = 2 Al3+ + 6 e- à 2 Al0 6. 3 Cu + 2 Al3+ à 3 Cu2+ + 2 Al 7. The above balanced equation shows conservation of mass because there are 2 moles of aluminum on each sides of the equation, and 3 moles of copper on each side of the equation. It shows conservation of charge because the sum of the charges on the left side is (+6), and the sum of the charges on the right side is (+6). Both sides of the equation have the same net charge. In addition, from the balanced half-reactions, we see the electrons lost by Cu0 is equal to the electrons gained by Al3+. |
1374 | https://www.mathway.com/popular-problems/Calculus/593636 | Find the Local Maxima and Minima y=x^2-x-1 | Mathway
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Calculus Examples
Popular Problems
Calculus
Find the Local Maxima and Minima y=x^2-x-1
y=x 2−x−1 y=x 2-x-1
Step 1
Write y=x 2−x−1 y=x 2-x-1 as a function.
f(x)=x 2−x−1 f(x)=x 2-x-1
Step 2
Find the first derivative of the function.
Tap for more steps...
Step 2.1
Differentiate.
Tap for more steps...
Step 2.1.1
By the Sum Rule, the derivative of x 2−x−1 x 2-x-1 with respect to x x is d d x[x 2]+d d x[−x]+d d x[−1]d d x[x 2]+d d x[-x]+d d x[-1].
d d x[x 2]+d d x[−x]+d d x[−1]d d x[x 2]+d d x[-x]+d d x[-1]
Step 2.1.2
Differentiate using the Power Rule which states that d d x[x n]d d x[x n] is n x n−1 n x n-1 where n=2 n=2.
2 x+d d x[−x]+d d x[−1]2 x+d d x[-x]+d d x[-1]
2 x+d d x[−x]+d d x[−1]2 x+d d x[-x]+d d x[-1]
Step 2.2
Evaluate d d x[−x]d d x[-x].
Tap for more steps...
Step 2.2.1
Since −1-1 is constant with respect to x x, the derivative of −x-x with respect to x x is −d d x[x]-d d x[x].
2 x−d d x[x]+d d x[−1]2 x-d d x[x]+d d x[-1]
Step 2.2.2
Differentiate using the Power Rule which states that d d x[x n]d d x[x n] is n x n−1 n x n-1 where n=1 n=1.
2 x−1⋅1+d d x[−1]2 x-1⋅1+d d x[-1]
Step 2.2.3
Multiply−1-1 by 1 1.
2 x−1+d d x[−1]2 x-1+d d x[-1]
2 x−1+d d x[−1]2 x-1+d d x[-1]
Step 2.3
Differentiate using the Constant Rule.
Tap for more steps...
Step 2.3.1
Since −1-1 is constant with respect to x x, the derivative of −1-1 with respect to x x is 0 0.
2 x−1+0 2 x-1+0
Step 2.3.2
Add 2 x−1 2 x-1 and 0 0.
2 x−1 2 x-1
2 x−1 2 x-1
2 x−1 2 x-1
Step 3
Find the second derivative of the function.
Tap for more steps...
Step 3.1
By the Sum Rule, the derivative of 2 x−1 2 x-1 with respect to x x is d d x[2 x]+d d x[−1]d d x[2 x]+d d x[-1].
f''(x)=d d x(2 x)+d d x(−1)f′′(x)=d d x(2 x)+d d x(-1)
Step 3.2
Evaluate d d x[2 x]d d x[2 x].
Tap for more steps...
Step 3.2.1
Since 2 2 is constant with respect to x x, the derivative of 2 x 2 x with respect to x x is 2 d d x[x]2 d d x[x].
f''(x)=2 d d x(x)+d d x(−1)f′′(x)=2 d d x(x)+d d x(-1)
Step 3.2.2
Differentiate using the Power Rule which states that d d x[x n]d d x[x n] is n x n−1 n x n-1 where n=1 n=1.
f''(x)=2⋅1+d d x(−1)f′′(x)=2⋅1+d d x(-1)
Step 3.2.3
Multiply 2 2 by 1 1.
f''(x)=2+d d x(−1)f′′(x)=2+d d x(-1)
f''(x)=2+d d x(−1)f′′(x)=2+d d x(-1)
Step 3.3
Differentiate using the Constant Rule.
Tap for more steps...
Step 3.3.1
Since −1-1 is constant with respect to x x, the derivative of −1-1 with respect to x x is 0 0.
f''(x)=2+0 f′′(x)=2+0
Step 3.3.2
Add 2 2 and 0 0.
f''(x)=2 f′′(x)=2
f''(x)=2 f′′(x)=2
f''(x)=2 f′′(x)=2
Step 4
To find the local maximum and minimum values of the function, set the derivative equal to 0 0 and solve.
2 x−1=0 2 x-1=0
Step 5
Find the first derivative.
Tap for more steps...
Step 5.1
Find the first derivative.
Tap for more steps...
Step 5.1.1
Differentiate.
Tap for more steps...
Step 5.1.1.1
By the Sum Rule, the derivative of x 2−x−1 x 2-x-1 with respect to x x is d d x[x 2]+d d x[−x]+d d x[−1]d d x[x 2]+d d x[-x]+d d x[-1].
d d x[x 2]+d d x[−x]+d d x[−1]d d x[x 2]+d d x[-x]+d d x[-1]
Step 5.1.1.2
Differentiate using the Power Rule which states that d d x[x n]d d x[x n] is n x n−1 n x n-1 where n=2 n=2.
2 x+d d x[−x]+d d x[−1]2 x+d d x[-x]+d d x[-1]
2 x+d d x[−x]+d d x[−1]2 x+d d x[-x]+d d x[-1]
Step 5.1.2
Evaluate d d x[−x]d d x[-x].
Tap for more steps...
Step 5.1.2.1
Since −1-1 is constant with respect to x x, the derivative of −x-x with respect to x x is −d d x[x]-d d x[x].
2 x−d d x[x]+d d x[−1]2 x-d d x[x]+d d x[-1]
Step 5.1.2.2
Differentiate using the Power Rule which states that d d x[x n]d d x[x n] is n x n−1 n x n-1 where n=1 n=1.
2 x−1⋅1+d d x[−1]2 x-1⋅1+d d x[-1]
Step 5.1.2.3
Multiply−1-1 by 1 1.
2 x−1+d d x[−1]2 x-1+d d x[-1]
2 x−1+d d x[−1]2 x-1+d d x[-1]
Step 5.1.3
Differentiate using the Constant Rule.
Tap for more steps...
Step 5.1.3.1
Since −1-1 is constant with respect to x x, the derivative of −1-1 with respect to x x is 0 0.
2 x−1+0 2 x-1+0
Step 5.1.3.2
Add 2 x−1 2 x-1 and 0 0.
f'(x)=2 x−1 f′(x)=2 x-1
f'(x)=2 x−1 f′(x)=2 x-1
f'(x)=2 x−1 f′(x)=2 x-1
Step 5.2
The first derivative of f(x)f(x) with respect to x x is 2 x−1 2 x-1.
2 x−1 2 x-1
2 x−1 2 x-1
Step 6
Set the first derivative equal to 0 0 then solve the equation 2 x−1=0 2 x-1=0.
Tap for more steps...
Step 6.1
Set the first derivative equal to 0 0.
2 x−1=0 2 x-1=0
Step 6.2
Add 1 1 to both sides of the equation.
2 x=1 2 x=1
Step 6.3
Divide each term in 2 x=1 2 x=1 by 2 2 and simplify.
Tap for more steps...
Step 6.3.1
Divide each term in 2 x=1 2 x=1 by 2 2.
2 x 2=1 2 2 x 2=1 2
Step 6.3.2
Simplify the left side.
Tap for more steps...
Step 6.3.2.1
Cancel the common factor of 2 2.
Tap for more steps...
Step 6.3.2.1.1
Cancel the common factor.
2 x 2=1 2 2 x 2=1 2
Step 6.3.2.1.2
Divide x x by 1 1.
x=1 2 x=1 2
x=1 2 x=1 2
x=1 2 x=1 2
x=1 2 x=1 2
x=1 2 x=1 2
Step 7
Find the values where the derivative is undefined.
Tap for more steps...
Step 7.1
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
Step 8
Critical points to evaluate.
x=1 2 x=1 2
Step 9
Evaluate the second derivative at x=1 2 x=1 2. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
2 2
Step 10
x=1 2 x=1 2 is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.
x=1 2 x=1 2 is a local minimum
Step 11
Find the y-value when x=1 2 x=1 2.
Tap for more steps...
Step 11.1
Replace the variable x x with 1 2 1 2 in the expression.
f(1 2)=(1 2)2−(1 2)−1 f(1 2)=(1 2)2-(1 2)-1
Step 11.2
Simplify the result.
Tap for more steps...
Step 11.2.1
Simplify each term.
Tap for more steps...
Step 11.2.1.1
Apply the product rule to 1 2 1 2.
f(1 2)=1 2 2 2−(1 2)−1 f(1 2)=1 2 2 2-(1 2)-1
Step 11.2.1.2
One to any power is one.
f(1 2)=1 2 2−(1 2)−1 f(1 2)=1 2 2-(1 2)-1
Step 11.2.1.3
Raise 2 2 to the power of 2 2.
f(1 2)=1 4−1 2−1 f(1 2)=1 4-1 2-1
f(1 2)=1 4−1 2−1 f(1 2)=1 4-1 2-1
Step 11.2.2
Find the common denominator.
Tap for more steps...
Step 11.2.2.1
Multiply 1 2 1 2 by 2 2 2 2.
f(1 2)=1 4−(1 2⋅2 2)−1 f(1 2)=1 4-(1 2⋅2 2)-1
Step 11.2.2.2
Multiply 1 2 1 2 by 2 2 2 2.
f(1 2)=1 4−2 2⋅2−1 f(1 2)=1 4-2 2⋅2-1
Step 11.2.2.3
Write −1-1 as a fraction with denominator 1 1.
f(1 2)=1 4−2 2⋅2+−1 1 f(1 2)=1 4-2 2⋅2+-1 1
Step 11.2.2.4
Multiply−1 1-1 1 by 4 4 4 4.
f(1 2)=1 4−2 2⋅2+−1 1⋅4 4 f(1 2)=1 4-2 2⋅2+-1 1⋅4 4
Step 11.2.2.5
Multiply−1 1-1 1 by 4 4 4 4.
f(1 2)=1 4−2 2⋅2+−1⋅4 4 f(1 2)=1 4-2 2⋅2+-1⋅4 4
Step 11.2.2.6
Multiply 2 2 by 2 2.
f(1 2)=1 4−2 4+−1⋅4 4 f(1 2)=1 4-2 4+-1⋅4 4
f(1 2)=1 4−2 4+−1⋅4 4 f(1 2)=1 4-2 4+-1⋅4 4
Step 11.2.3
Combine the numerators over the common denominator.
f(1 2)=1−2−1⋅4 4 f(1 2)=1-2-1⋅4 4
Step 11.2.4
Simplify the expression.
Tap for more steps...
Step 11.2.4.1
Multiply−1-1 by 4 4.
f(1 2)=1−2−4 4 f(1 2)=1-2-4 4
Step 11.2.4.2
Subtract 2 2 from 1 1.
f(1 2)=−1−4 4 f(1 2)=-1-4 4
Step 11.2.4.3
Subtract 4 4 from −1-1.
f(1 2)=−5 4 f(1 2)=-5 4
Step 11.2.4.4
Move the negative in front of the fraction.
f(1 2)=−5 4 f(1 2)=-5 4
f(1 2)=−5 4 f(1 2)=-5 4
Step 11.2.5
The final answer is −5 4-5 4.
y=−5 4 y=-5 4
y=−5 4 y=-5 4
y=−5 4 y=-5 4
Step 12
These are the local extrema for f(x)=x 2−x−1 f(x)=x 2-x-1.
(1 2,−5 4)(1 2,-5 4) is a local minima
Step 13
y=x 2−x−1 y=x 2-x-1
y y 12 12 1 x 2 1 x 2
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⎡⎢⎣x 2 1 2√π∫x d x⎤⎥⎦[x 2 1 2 π∫x d x]
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1375 | https://physics.stackexchange.com/questions/73028/the-relation-between-gausss-law-and-coulomb-law-and-why-is-it-important-that-th | electromagnetism - The relation between Gauss's law and Coulomb law and why is it important that the electric field decrease proportionally to $\frac{1}{r^{2}}$? - Physics Stack Exchange
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The relation between Gauss's law and Coulomb law and why is it important that the electric field decrease proportionally to 1 r 2 1 r 2?
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My question relates to the third MIT's video lecture about Electricity and Magnetism, specifically from 21:18−22:00 21:18−22:00 :
I have watched the development of Gauss's law, but I still don't quite understand the link between Gauss's law and Coulomb law: How does Gauss's law change if Coulomb law would of been a different one.
I also don't understand why is it so important for Gauss's law that the electric field decrease proportionally to 1 r 2 1 r 2 ?
For example, what would of happened if the electric field decrease proportionally to 1 r 1 r , or 1 r 3 1 r 3 ?
electromagnetism
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asked Aug 2, 2013 at 21:47
BelgiBelgi
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Essentially a duplicate of physics.stackexchange.com/q/47084/2451 , physics.stackexchange.com/q/47193/2451 and links therein.Qmechanic –Qmechanic♦ 2013-08-02 22:08:04 +00:00 Commented Aug 2, 2013 at 22:08
Possible duplicate of How is Gauss' Law (integral form) arrived at from Coulomb's Law, and how is the differential form arrived at from that?garyp –garyp 2016-01-06 14:19:45 +00:00 Commented Jan 6, 2016 at 14:19
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Gauss' law and Coulomb's law are equivalent - meaning that they are one and the same thing. Either one of them can be derived from the other. The rigorous derivations can be found in any of the electrodynamics textbooks, for eg., Jackson. For eg., consider a point charge q. As per Coulomb's law, the electric field produced by it is given by
E⃗=k q r 2 r^E→=k q r 2 r^
, where k=1 4 π ϵ 0 k=1 4 π ϵ 0. Now, consider a sphere of radius r r centred on charge q. So, for the surface S S of this sphere you have:
∫S E⃗.d s→=∫S k q r 2 d s=k q r 2∫S d s=k q r 2(4 π r 2)=4 π k q=q ϵ 0∫S E→.d s→=∫S k q r 2 d s=k q r 2∫S d s=k q r 2(4 π r 2)=4 π k q=q ϵ 0
, which is Gauss' law. Note that if the r 2 r 2 in the expression for the surface area of the sphere in the numerator did not exactly cancel out the r 2 r 2 in the denominator of Coulomb's law, the surface intergral would actually depend on r r. Hence you would not have the result that the surface integral is independent of the area of the surface, which is what is implied by Gauss' law. Though this result has been derived for a sphere, it can be derived for any arbitrary shape and size of the surface, you can refer to Jackson for eg., for the rigorous derivation. Note that by performing these steps in reverse, you can also derive Coulomb's law from Gauss' law, thus demonstrating that they are equivalent.
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edited Aug 2, 2013 at 23:15
answered Aug 2, 2013 at 23:08
guruguru
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The Gauss' law/inverse square connection is explained by Frederic above. But allow me to expand on the fundamental issue.
In a nutshell, if the inverse square law doesn't hold, then the photon must have mass, and hence a finite lifetime. This is explained well in Jackson's Classical Electrodynamics (look in the index for Proca Lagrangian, in the second edition). The use of a massive photon model has been pushed through theoretically and the resulting observations allow one to put an upper bound on the photon mass. If the photon mass were measureable, we would have to rethink all sorts of fundamental issues, especially in cosmology.
This is dry, but it's readily available.
Interestingly there is a deep connection with pure mathematics too; the residue of a function is the 1/z term of the Laurent expansion. Why? That's because this is the only term that survives to give a finite contribution to an integral at infinity. 1/r would be the force associated with a 1/r^2 potential.
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answered Aug 2, 2013 at 22:49
user27777 user27777
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The 1/|r|2 1/|r|2 dependence is fundamentally geometrical in nature, stemming from our 3d world.
Differential equations, even PDEs, are sometimes said to have Green's functions. Consider the differential equation
∇⋅K=α∇⋅K=α
for some vector field K K and some scalar field α α. This is structurally identical to Gauss's law. This equation has a Green's function G G that satisfies
∇⋅G=δ∇⋅G=δ
where δ δ is the Dirac delta function. This describes, in essence, a point source δ δ generating a field G G.
The Green's function is given by
G(r)=r^4 π|r|2 G(r)=r^4 π|r|2
Thus, any differential equation of the form ∇⋅K=α∇⋅K=α has the same Green's function in 3d--a point source always generates the same basic field.
In short, it is a physical statement to say ∇⋅E=ρ/ϵ 0∇⋅E=ρ/ϵ 0, but once that has been said, Coulomb's law, which describes a point charge, must inevitably follow due to the mathematical structure of the equations and the geometrical nature of 3d space.
(Of course, you're free to work the other way around.)
Edit: crucially, in 2d and 1d, the Green's functions are different. You probably already know these solutions. The 2d Green's function is something familiar from the field of a uniform line charge, and has 1/|r|1/|r| dependence. The 1d Green's function has constant magnitude but changes direction on opposite sides of the "point" charge--this is the basic feature of a uniformly charged sheet.
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edited Aug 3, 2013 at 4:53
answered Aug 3, 2013 at 1:45
MuphridMuphrid
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Gauss's law states that the ratio of charge and the dielectric constant is given by a (two-dimensional) surface integral over the electric field:
∫E⋅d A=Q ϵ 0,∫E⋅d A=Q ϵ 0,
where I have omitted vector notation for simplicity. It can be linked to Coulomb's law by assuming spherical symmetry of the electric field and performing the integration.
The other way around, you can start by assuming that Coulomb's law,
E=1 4 π Q r 2,E=1 4 π Q r 2,
holds, and take the divergence on both sides of the equation. This leads to the differential form of Gauss's law:
∇⋅E=ρ ϵ 0,∇⋅E=ρ ϵ 0,
where ρ ρ is the charge density. In order to reproduce its integral form, just integrate both sides of the equation. Detailed calculations can be found on the wikipedia page on Gauss's law. If you really want to understand how it all comes together (which I assume to be true, otherwise you wouldn't ask here), I would recommend reproducing the calculation on your own. Then you can also try to see what happens if you assume other forms of Coulomb's law.
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answered Aug 2, 2013 at 22:22
Frederic BrünnerFrederic Brünner
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1376 | https://www.youtube.com/watch?v=crxFsKRGEqk | Volume and Surface Area of Cones - GCSE Maths
1st Class Maths
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Posted: 16 Mar 2025
This video is for students aged 14+ studying GCSE Maths.
A video explaining how to find the volume and surface area of a cone.
Exam Question Booklets:📝
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0:00 Intro
0:10 Volume of a Cone Formula
0:59 Volume Example
1:36 Practice Questions (Volume)
2:42 Surface Area of a Cone Formula
3:45 Surface area Example
4:34 Practice Question (Surface Area)
5:48 Working backwards Example 1
7:32 Working backwards Example 2
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Intro [Music] Volume of a Cone Formula in the previous video we learned how to find the volume and surface area of a pyramid in this video we're going to do the same ideas but for a cone now a cone isn't technically a pyramid but it's a bit like a pyramid where rather than having a square base we have a circular base the volume of a pyramid was 1/3 multiplied by its base area multiplied by its perpendicular height and this formula still holds for the volume of a cone but since the base area is a circle we know its area will be > r s the area of a circle and for the perpendicular height which is the height from the top of the cone vertically down to the base we're going to call it h so we end up with the formula 1/3 > r^ 2 H or 1/3 PK r^ 2 H good news this formula will be given to you in the exam let's use this Volume Example formula to find the volume of this cone here so we would do volume equals 1/3 multiplied pi multiplied by R 2 and we can see for this cone the radius is four so 4^ s and then multiplied by H the perpendicular height which we can see from this cone is N9 so multiply by 9 now assuming this question appears on a calculator paper we can just type this into the calculator and you'll get this number here and it might say for example round your answer to one decimal place in which case we'd get 150. and since it's a volume the units would be cm cubed here are two more Practice Questions (Volume) cones for you to try and find the volume of be careful on the second cone I have given you the diameter and not the radius and make sure you give both answers to one decimal place so for the first one we do volume equal 1/3 pi R 2 and we can see the radius is 3 so 3^ squ and then multiplied by H and the perpendicular height is 8 so multi by 8 if you type this into the calculator you get this number here which would round to one decimal place as 75.4 cm cubed the second cone we do volume equals 1/3 multipli pi and then we need to multiply by R squ but for this cone you can see I've given you the diameter Instead at 20 the radius is half of the diameter so if we half 20 we find the radius is actually 10 so multi by 10^ S and then multiplied by H which we can see is 17 so multiply by 17 if you type this into the calculator you'll get this number here and if you round that to one decimal place like the question asks you get 17802 cm cubed so that's the volume Surface Area of a Cone Formula covered now let's take a look at surface area before we look at surface area we need to understand one more length on a cone so far we've worked with this length here which is the radius and this one here which is the perpendicular height which we called H we also need to know about this height here which goes on a slope we call it the slant height and we give it the letter L now that we know this we're ready to have a look at the formula for the surface area the surface area of a cone is made up of two faces we have this circular face on the bottom and we know the area of a circle is p piun r squ so the surface area formula starts with PK r s but there's another face on this cone which rather than being flat is curved we call it the curved surface area of the cone and its formula is pi r l where L was the slant height that we talked about earlier so if the curved surface area is p r l then the total surface area must be PK R S Plus P RL so it's the area of the circular base and then we add to this the curved surface area let's use this formula to find the Surface area Example surface area of a cone and we'll do this one here and we're going to do it to two decimal places so we start with surface area equals Pi R 2 and we can see from this cone the radius is 10 10 so 10 2 and then we add to this Pi R which is 10 and multiplied by L which is the slant height which is 26 so multiplied by 26 notice how we don't need the perpendicular height 24 or toall to work out the surface area so all we would do now is type this into the calculator and we'd end up with this number here and this question says to two decimal places so 130. N7 since we're doing a surface area we'll use the unit cm squar and now here Practice Question (Surface Area) are two more cones for you to find the surface area of be careful to give your answers to one decimal place and also notice I've given you the diameter on the second cone again and not the radius so for the first one using the formula we do surface area equals pi multiplied by the radius squared and the radius of this cone is three so 3^ s and then we add to this pi multiplied R which is 3 multiplied by L which is the slant height which for this cone is five so multiplied by five typing this into your calculator will give this number here and to one decimal place that's 75.4 CM squ for the second cone we do surface area equals pi multiplied R 2 and this time the diameter has been given as 3.6 if you half 3.6 you find the r is 1.8 so multiplied by 1.8 s and then we add to this the curved surface area so pi multiplied by R which is 1.8 again multiplied by L which for this cone we can see is 8.2 so multiplied by 8.2 typing all of this into your calculator gives you this number here and to one decimal place that's 56.5 CM squ now let's have a look at Working backwards Example 1 some trickier questions involving cones so for this question here we have a cone and we're told the volume of the cone is 700 cm cubed but notice we've not been given a H the question wants us to work out H the perpendicular height of the cone to one decimal place so if we were working out the volume of the cone we would use the volume formula 1/3 r^ 2 H so if we were to work out the volume we do 1/3 piun multiplied by R 2 and we can see the radius is 7 so 7^ squ then we would normally multiply by H but we don't know H so we'll just multiply by H but we do know what this calculation needs to give it's the volume of the cone which we're told in the question is 700 cm cubed so we can say that this is equal to 700 now we just have an equation to solve to do this I would type all of this part into the calculator which is everything apart from the h on the left hand side your calculator should give you back 49 over 3 piun so we have 49 over 3 piun H which we could write as 49 Pi H over 3 so we're going to solve this equation now we'll multiply both sides by three on the left hand side the 3es will then cancel so we just have 49 Pi H and on the right hand side 700 3 is 2,100 then we can divide both sides by 49 Pi remembering that 49 Pi is just a number on the left hand side this will cancel the 49 Pi so we just have H and on the right hand side you'll need to use your calculator to do 2,100 divide 49 Pi but that will give you this number here the question says to give it to one decimal place so we'd say h is 13.6 cm M for this question we have a different Working backwards Example 2 cone and notice we're missing R this time we're told that the volume of this cone is 1,300 cm cubed and we're going to work out R the radius of the cone to one decimal place so we'll do this in a very similar way we'll start with the formula for the volume of a cone 1/3 multiplied Pi then multiplied by r s but we don't know R so we'll just write multip R 2 and then multiplied by H but we do know h H is 18 and this must equal the volume we given in the question 1,300 so it equals 1,300 for this one I would change around the order of the multiplications so it's 1/3 piun 18 R 2 then we'll work out this part here you can type this into your calculator and you'll see it gives you 6 Pi so this is just the same as 6 Pi so we have 6 piun r^ 2 which is 6k r^ 2 so we have this equation to solve this time we'll start by dividing both sides by 6 Pi on the left hand side the 6 Pi will cancel so we have R squ and on the right hand side 1,300 / 6 Pi gives you this number here now this is not the value of R but the value of R squar so we need to square root both sides of the equation if we square root the left hand side the square root of R squ is r and if we square root the right hand side we square root this number on your calculator and you learn that with this number here the question says to give it to one decimal place so R is 8.3 thank you for watching this video I hope you found it useful check out the Y and I think you should watch next subscribe so you don't miss out on future videos and why not try the exam questions in this video's description |
1377 | http://www.xiongan.gov.cn/2018-08/13/c_129931972.htm | 实用贴！天气预报里的专业名词了解一下！-中国雄安官网
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实用贴！天气预报里的专业名词了解一下！
2018-08-13 15:06:14 来源：新华网
晴天要防晒，雨天要带伞……天气多变，总要看看天气预报才好放心出门。天气预报里的那些专业名词你了解吗？ 下面这些常见的天气术语了解一下。
天空状况有几种？
天空状况是以天空云量多少和阳光强弱来决定的，分晴天、少云、多云、阴天四种情况。
晴天：天空无云，或有零星云块，但中、低云量不到1成，高云量在4成以下，通常用语“晴”。
少云：天空中有1～3成的中、低云层，或有4～5成的高云，即以晴天为主，用语“晴间多云”。
多云：天空中有4～7成的中、低云层，或有6～10成的高云，用语“多云”。
阴天：中低云云量占天空8成及以上，天空阴云密布，或稍有云隙,而仍感到阴暗，用语“阴”或“阴天间多云”。
“间”“转”是什么意思？
“间”是间或，有时的意思。如晴间多云，指天空晴天为主，间或出现少量云朵；阴间多云，指天空以阴为主，间或天空云层有缝隙，透出部分蓝天。
“转”的意思为转变,当出现“转”，表明大气环流有转变，有天气系统影响，如晴转多云是指天空云量增多。
“阴转多云”说明未来一定时间内，天气形势逐渐朝变好的方向发展，且趋势比较明显，表现为天空由阴暗、满天云层变化为云层逐渐抬高，云量减少。
早晨、傍晚等具体是什么时间段？
小雨有多小、大雨有多大？
在气象上通常用某一段时间内降水量的多少来划分降水强度。根据国家气象部门规定的降水量标准，降雨可分为小雨、中雨、大雨、暴雨、大暴雨和特大暴雨六种。
（综合/中国气象局网站、中国天气网）
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1378 | http://home.ustc.edu.cn/~zyx240014/USTCProbability/files/Recitation1%20of%20Probability.pdf | 概率论第1 次习题课讲义 宗语轩 2022 秋, USTC 1 集合运算, 事件与概率 方法论: 事件运算← →集合运算 对于事件A, B, 有 事件交, 并, 余, 差← →A ∩B, A ∪B, Ac(对立事件), A\B ω ∈A ∩B ⇐ ⇒ω ∈A且ω ∈B ⇐ ⇒A, B同时发生 ω ∈A ∪B ⇐ ⇒ω ∈A或ω ∈B ⇐ ⇒A发生或B发生 ω ∈Ac ⇐ ⇒ω / ∈A ⇐ ⇒Ac发生⇐ ⇒A不发生 ω ∈A\B ⇐ ⇒ω ∈A且ω / ∈B ⇐ ⇒A发生但同时B不发生 1.1 随机事件(集合) 的运算 设{Ai, i ∈I} 为一列事件族, 类比数列上下确界的定义, 我们定义记号 sup i∈I Ai := ∪ i∈I Ai = {x : ∃i ∈I, x ∈Ai} , inf i∈I Ai := ∩ i∈I Ai = {x : ∀i ∈I, x ∈Ai} .
集合运算的三个基本性质: 交换律, 结合律, 分配律. 分配律可表示为: B ∩ (∪ i∈I Ai ) = ∪ i∈I (B ∩Ai), B ∪ (∩ i∈I Ai ) = ∩ i∈I (B ∪Ai) 定理1.1 (De.Morgan 法则). 设{Ai, i ∈I} 为一列事件族, 则有 (1) (∪ i∈I Ai )c = ∩ i∈I Ac i; 0个人主页:
发现错误欢迎联系: zyx240014@mail.ustc.edu.cn.
1 概率论第1 次习题课讲义 2 (2) (∩ i∈I Ai )c = ∪ i∈I Ac i.
关于 n ∪ i=1 Ai, 我们有如下分划: 命题1.1. 设{Ai, i = 1, 2, · · · , n} 为一列事件族(n ⩽+∞), 则有 n ∪ i=1 Ai = n ⊔ i=1 ( Ai\ i−1 ∪ k=1 Ak ) .
其中 ⊔ 表示无交并. 通过上述方式, 我们可以将 ∪ i∈I Ai 转化成事件的无交并(互不相容), 并利 用概率测度的可列可加性进行计算.
我们考虑事件序列{An} 的极限: (1) 上限事件: lim sup n→+∞An := ∞ ∩ n=1 ∞ ∪ m=n Am = { ω : 有无穷多个Ak包含ω } , 即{An} 发生无穷多次; (2) 下限事件: lim inf n→+∞An := ∞ ∪ n=1 ∞ ∩ m=n Am = { ω : 有限个Ak不包含ω } , 即{An} 不发生有限次.
注. 利用 { ∞ ∪ m=n Am }∞ n=1 的递减性和 { ∞ ∩ m=n Am }∞ n=1 的递增性, 我们亦有: lim sup n→+∞An = lim n→∞ ∞ ∪ m=n Am, lim inf n→+∞An = lim n→∞ ∞ ∩ m=n Am.
定义1.1. 我们称事件序列{An} 收敛, 若lim sup n→+∞An = lim inf n→+∞An. 记收敛的极限为A := lim n→+∞An.
接上, 不难验证A ∈F, 且满足概率连续性: P(A) = lim n→+∞P(An) (注意到 ∞ ∩ m=n Am ⊆An ⊆ ∞ ∪ m=n Am).
例1.1. 设{fn(x)} 及f(x) 是定义在R 上的实值函数, 则使fn(x) 不收敛于f(x) 的一切点x 所形 成的集合D 可表示为 D = ∞ ∪ k=1 ∞ ∩ N=1 ∞ ∪ n=N { x : |fn(x) −f(x)| ⩾1 k } .
引理1.1. 概率测度的一些性质: (1) P(Ac) = 1 −P(A); (2) 若A ⊂B, 则P(B) = P(A) + P(B\A) ≥P(A); (3) P(A1 ∪A2 ∪· · · ∪An) = n ∑ i=1 P ( Ai\ i−1 ∪ k=1 Ak ) , n ⩽+∞; 概率论第1 次习题课讲义 3 (4) P(A ∪B) = P(A) + P(B) −P(A ∩B); (5) 次(σ) 可加性: P(A1 ∪A2 ∪· · · ∪An) ⩽ n ∑ i=1 P(Ai), n ⩽+∞; (6) Jordan 公式: P ( n ∪ i=1 Ai ) = n ∑ k=1 (−1)k−1 ∑ i1<··· 0, 则 P(A) = n ∑ i=1 P(Bi)P(A | Bi) 引理1.3 (贝叶斯公式). 设A1, A2, · · · , An 为Ω的一个划分, P(Ai) > 0, ∀i, 则当P(B) > 0 时有: P(Ai | B) = P(Ai)P(B | Ai) P(B) = P(Ai)P(B | Ai) ∑n j=1 P(Aj)P(B | Aj) 引理1.4. 设A1, A2, · · · , An 为Ω的一个划分, P(A1A2 · · · An−1) > 0, 则有 P(A1A2 · · · An) = P(A1)P(A2|A1)P(A3|A1A2) · · · P(An|A1A2 · · · An−1) 例1.4. A 与B 两个小孩从自己装有红与黄两种颜色积木的袋子里各摸出一块, A 摸出红色与黄色 的概率分别是0.8 与0.2, B 摸出红色与黄色的概率分别是0.9 与0.1. 现有一块红色积木, 求其为 A 取出的概率.
解. 记事件X 表示积木是由A 摸出的, 事件Y 表示摸出的积木是红色的. 则有 P(X|Y ) = P(Y |X)P(X) P(Y |X)P(X) + P(Y |Xc)P(Xc) = 0.8 · 0.5 0.8 · 0.5 + 0.9 · 0.5 = 8 17.
例1.5. 设有甲和乙两个罐子, 甲罐中有m 个红球和n 个黑球, 乙罐中有n 个红球和m 个黑球, 且 m > n. 随机选取一个罐子再从中随机抽取一球, 发现为红球, 将其放回后并摇匀. 若再次在该罐中 随机抽取一球, 问该球仍为红色的概率是否比1 2 大?
解. 记事件X 表示球是由甲取出的, 事件R1 表示第一次取出的球是红球, 事件R2 表示第二次取 出的球是红球. 则有 P(R2|R1) = P(R1R2) P(R1) = P(R1R2|X)P(X) + P(R1R2|Xc)P(Xc) P(R1|X)P(X) + P(R1|Xc)P(Xc) = 1 2(( m m+n)2 + ( n m+n)2) 1 2( m m+n + n m+n) = m2 + n2 (m + n)2 > 1 2.
这里不能取等是因为m > n.
例1.6. 平面上有n 个不同的点, 编号分别为1, 2, · · · , n. 现有一个质点在这些点上做随机游动, 假 设每次它在某点上停留片刻之后就会在其余所有点中等概率选择一个并移动到该点上. 设其初始位 置为点1 上, 求它在第一次返回此点之前访问过点2 的概率.
解. 设所求概率对应的事件C1, 事件Ai 表示在初始位置为点i 下, 第一次返回点1 之前访问过点2 (访问过的点包括初始位置), 事件Bij 表示初始位置为点i 下首次到j. 则有 P(C1) = P(B12)P(C1|B12) + n ∑ k=3 P(B1k)P(C1|B1k) = P(B12)P(A2) + n ∑ k=3 P(B1k)P(Ak) 概率论第1 次习题课讲义 5 = 1 n −1P(A2) + n −2 n −1P(Ak) = 1 n −1 + n −2 n −1P(A3).
由对称性知, P(A3) = P(Ac 3) = 1 2. 所以P(A1) = n 2(n −1).
例1.7. 甲乙两坛子中各装一只白球和一只黑球, 从两坛中各取出一球交换后放入另一坛中. 记事 件An, Bn, Cn 分别表示第n 次交换后甲坛的白球数是2, 1, 0, 记pn, qn, rn 表示其对应的概率, 求 pn, qn, rn 的关系式, 并讨论当n →+∞时的情形.
解. 我们有 pn+1 = P(An+1) = P(An+1|An)P(An) + P(An+1|Bn)P(Bn) + P(An+1|Cn)P(Cn) = 1 4P(Bn) = 1 4qn 类似地, 有 qn+1 = P(Bn+1) = P(Bn+1|An)P(An) + P(Bn+1|Bn)P(Bn) + P(Bn+1|Cn)P(Cn) = P(An) + 1 2P(Bn) + P(Cn) = pn + 1 2qn + rn.
rn+1 = P(An+1) = P(Cn+1|An)P(An) + P(Cn+1|Bn)P(Bn) + P(Cn+1|Cn)P(Cn) = 1 4P(Bn) = 1 4qn 将上述三式联立, 利用q0 = 1, q1 = 1 2, 消去pn, rn, 得 qn+1 −qn = −1 2(qn −qn−1) = ⇒qn = 2 3 ( 1 − ( −1 2 )n+1) 进而 pn = rn = 1 6 ( 1 − ( −1 2 )n) 令n →+∞, 可得p := lim n→+∞pn = 1 6, r := lim n→+∞rn = 1 6, q := lim n→+∞qn = 2 3.
例1.8 (配对问题). n 对夫妇面对面随机占座, 求恰好有k 对夫妇面对面坐的概率P (n) k .(0 ⩽k ⩽n).
解. 把n 位男士和n 位女士按序标号, 且同对夫妇标号相同. 记Ai 表示标号i 的夫妇面对面坐. 先 看k = 0 时的情形. 对∀i1 < i2 < · · · < ik, 有 P(Ai1Ai2 · · · Aik) = (n −k)!
n!
利用Jordan 公式, 有 P (n) 0 = P ( n ∩ k=1 Ac k ) = 1 − ( n ∪ k=1 Ak ) = 1 − n ∑ k=1 (−1)k−1 ∑ i1<···<ik P(Ai1 · · · Aik) 概率论第1 次习题课讲义 6 = 1 − n ∑ k=1 (−1)k−1 (n k )(n −k)!
n!
= n ∑ k=0 (−1)k k!
.
再看1 ⩽k ⩽n 时的情形. 考虑仅标号为第i1, i2, · · · ik 的夫妇面对面坐的概率: P(Ai1 · · · AikAc ik+1 · · · Ac in) = P(Ai1)P(Ai2|Ai1) · · · P(Aik|Ai1 · · · Aik−1)P(Aic k+1 · · · Ac in|Ai1 · · · Aik) = 1 n 1 n −1 · · · 1 n −(k −1)P (n−k) 0 = (n −k)!
n!
P (n−k) 0 故有 P (n) k = ∑ i1<···<ik P(Ai1 · · · AikAc ik+1 · · · Ac in) = (n k )(n −k)!
n!
P (n−k) 0 = 1 k!
n−k ∑ i=0 (−1)i i!
.
注. 我们发现, lim n→+∞P (n) k = e−1 k! , 该值即为参数为1 的泊松分布的随机变量取k 值的概率(后续章 节会讲到).
变式1.8. 考试时共有N 张考签, n 个学生参加考试(n ⩾N), 被抽过的考签立刻放回, 求在考试结 束之后, 恰好有k 张没有被抽到的概率.
答案: N−k ∑ i=0 (−1)i N!
k!i!(N −k −i)!
(N −k −i)n N n .
2 补充习题 注. 关于“摸球” 问题, 常见方法主体为如下几种: • 若为古典概型(样本空间有限+ 各事件发生等概率) 问题, 可以通过排列组合方法, 计算样本 空间和事件A 的样本点个数, 然后按定义计算: 对于事件A ⊆Ω, 有 P(A) = |A| |Ω|.
• 利用全概率公式, 贝叶斯公式, 教材习题1.4.2 的公式等条件概率方法, 以及利用基本的概率 测度运算性质.
• 若题干中涉及到含参整变量, 可结合上条方法得到递推关系, 转化成数列递推方法解决.
概率论第1 次习题课讲义 7 1 证明Bonferroni 不等式: P ( n ∪ r=1 Ar ) ⩾ n ∑ r=1 P(Ar) − ∑ 1⩽r<k⩽n P(Ar ∩Ak).
提示. 注意到 P (n+1 ∪ r=1 Ar ) = P ( n ∪ r=1 Ar ) + P(An+1) −P ( n ∪ r=1 (Ar ∩An+1) ) ⩾P ( n ∪ r=1 Ar ) + P(An+1) − n ∑ r=1 P(Ar ∩An+1), 再利用归纳法即可.
2 甲、乙两人轮流抛掷一枚均匀的骰子. 甲先掷, 一直到掷出了1 点, 交给乙掷, 而到乙掷出了 1 点, 再交给甲掷, 并如此一直下去. 求第n 次抛掷时由甲掷的概率.
解. 记事件An 表示第n 次抛掷时由甲掷, 记pn = P(An). 利用全概率公式, 有 pn = P(An) = P(An−1)P(An|An−1) + P(Ac n−1)P(An|Ac n−1) = 5 6pn−1 + 1 6(1 −pn−1) = 2 3pn−1 + 1 6.
利用p1 = 1, 整理并求解: pn −1 2 = 2 3(pn−1 −1 2) = ⇒pn = 1 2 (2 3 )n−1 + 1 2.
3 100 名乘客登上一架正好有100 个座位的飞机, 每名乘客对应一个座位. 第一位乘客先随机 选择一个座位坐. 第二位乘客如果自己的座位空着就坐自己的座位, 否则就在其他空余的座位中随 机选择一个座位坐. 第三位乘客如果自己的座位空着就坐自己的座位, 否则就在其他空余的座位中 随机选择一个座位坐. 这个过程一直持续到所有的100 名乘客都登机为止. 求最后一名乘客坐自己 的座位的概率.
提示. 不妨对任意i, 第i 名乘客对应第i 个座位. 把上述100 一般化为n(n ⩾2), 记最后一名乘客 坐自己的座位的概率是pn. 用归纳法证明pn = 1 2. 注意到: 若第1 位乘客选择第1 个座位, 则最后 一名乘客一定坐到自己的座位; 若第1 位乘客选择第i 个座位(2 ⩽i ⩽n −1), 则第2 位至第i −1 位乘客都坐到自己的座位, 之后的情形等同于原题中乘客座位数量为n −i + 1 时的情形; 若第1 位 乘客选择第n 个座位, 则最后一名乘客一定坐不到自己的座位. 因此, 通过对第一位乘客选择的座 位号取条件概率可以得到 pn+1 = 1 n + 1 + 1 n + 1 n ∑ k=2 pk.
利用归纳假设即得概率是1 2.
概率论第1 次习题课讲义 8 4 涂色问题 平面上的n 个点和连接各点之间的连线叫做一个完全图, 记作G. 点称作图的顶点, 顶点之间的连线叫做边, 共有 (n 2 ) 条. 给定一个整数k, G 中任意k 个顶点连同相应的边构成一个 有k 个顶点的完全子图, G 中共有 (n k ) 个这样的子图, 记作Gi, i = 1, · · · , (n k ) . 现将图G 的每 条边涂成红色或蓝色. 当 (n k ) < 2 k(k−1) 2 −1 时, 问: 是否有一种涂色方法, 使得没有一个子图Gi 的 (k 2 ) 条边是同一颜色的.
解. 上述组合问题我们通过概率方法解决. 我们对G 的边进行随机涂色, 每条边为红色和蓝色的概 率是1 2. 记事件Ai 表示子图Gi 各边的颜色相同, 则 n ∩ i=1 Ac i 表示没有一个子图Gi 的 (k 2 ) 条边是 同一颜色的. 利用 P(Ai) = P(Gi的各边均为红色) + P(Gi的各边均为蓝色) = 2 2(k 2) = 2 k(k−1) 2 −1 及 (n k ) < 2 k(k−1) 2 −1, 有 P ( n ∪ i=1 Ai ) ⩽ P ∑ i=1 (Ai) = (n k ) 2 k(k−1) 2 −1 < 1 故P ( n ∩ i=1 Ac i ) > 0. 从而存在一种涂色方法, 使得没有一个子图Gi 的 (k 2 ) 条边是同一颜色的.
3 乘积空间浅引∗ 问题: 同时抛掷2 枚硬币与抛掷1 枚硬币两次的概率空间的关系？ 事实: ∀i ∈I, Fi ⊂2Ω为σ 代数, 则 ∩ i∈I Fi 亦为σ 代数.
例3.1. Ω= {1, 2, · · · , n}, Ai = {i}(n > 2), Fi = {∅, Ω, Ai, Ac i}, 则F1 ∪F2 = {∅, Ω, A1, Ac 1, A2, Ac 2}, 明显A1 ∪A2 = {1, 2} / ∈F1 ∪F2, 非σ 代数.
由小σ 代数到大σ 代数, 方式之一: 对F, G ⊂2Ω, 称包含F, G 的最小σ 代数为F 与G 生成 的σ 代数, 记为σ(F ∪G) .
典型例子: 1 维Borel 域: R 上形如(a, b] 区间生成的σ 代数, 记为B(R), 其中每个元素称为 Borel 集. {b} = ∩ n ( b −1 n, b ] , (a, b) = (a, b]{b}, [a, b] = {a} ∪(a, b], [a, b) = {a} ∪(a, b). 即一切开 区间, 闭区间, 半开半闭区间均为Borel 集.
n 维Borel 域: Rn 上形如(a1, b1] × · · · × (an, bn] 生成的σ 域, 记为B(Rn) 0本节来自19 级数院何家志同学整理的往年笔记, 这部分大家有兴趣了解即可 概率论第1 次习题课讲义 9 乘积空间: (Ω1, F1, P1), (Ω2, F2, P2) 构造更大空间?
Ω1 × Ω2 = {(ω1, ω2) : ω1 ∈Ω1, ω2 ∈Ω2} F1 × F2 = {A1 × A2 : Ai ∈Fi, i = 1, 2} 未必为σ 代数.
例3.2. 接例子(3.1). F1 × F2 = {∅, Ω× Ω, A1 × Ac 2, Ac 1 × A2, Ac 1 × Ac 2, · · · }, 显见A1 × A2 ∈F1 × F2, 但 (A1 × A2)c = Ω× Ω{1, 2} = {2, 3, · · · , n} × Ω∪Ω× {1, 3, 4, · · · , n} / ∈F1 × F2 记F = σ(F1 × F2) 看作Ω= Ω1 × Ω2 的σ 代数, 概率测度? 引入F1 × F2 上函数 P12 : F1 × F2 →R, P12(A1 × A2) := P1(A1)P2(A2) 特别地, F1 × F2 中元素的不交并, 若Ω1, Ω2 有限, Fi = 2Ωi, 这时P12 良定.
对应: Ai ∈Fi, A1 ← →A1 × Ω2, A2 ← →Ω1 × A2, 在(Ω1 × Ω2, σ(F1 × F2), P12) 看A1 × A2 P12(A1 × A2) = P1(A1)P2(A2) = P12(A1 × Ω2) · P12(Ω1 × A2) 表明A1 × Ω2 与Ω1 × A2 独立.
例3.3. 掷硬币2 次. Ω= {HH, HT, TH, TT} = {H, T} × {H, T} P12(HH) = P1(H)P2(H) = 1 4, P12(HT) = 1 4. |
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1380 | https://www.quora.com/How-do-you-do-log-on-a-Casio-scientific-calculator | How to do log on a Casio scientific calculator - Quora
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How do you do log on a Casio scientific calculator?
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1y
To calculate logarithms on a Casio scientific calculator, follow these steps:
Turn on the calculator: Press the ON button.
Select the log function:
For base 10 logarithm (common log), press the log button.
For natural logarithm (base e), press the ln button.
Enter the number: Type the number you want to find the logarithm for.
Calculate: Press the = button to see the result.
Example:
To calculate the log base 10 of 100:
Press ON
Press log
Type 100
Press =
The display should show 2, since [math]\log_{10}(100) = 2[/math].
If you need to calculate logarithms with bases other than 10 or e, you can us
Continue Reading
To calculate logarithms on a Casio scientific calculator, follow these steps:
Turn on the calculator: Press the ON button.
Select the log function:
For base 10 logarithm (common log), press the log button.
For natural logarithm (base e), press the ln button.
Enter the number: Type the number you want to find the logarithm for.
Calculate: Press the = button to see the result.
Example:
To calculate the log base 10 of 100:
Press ON
Press log
Type 100
Press =
The display should show 2, since [math]\log_{10}(100) = 2[/math].
If you need to calculate logarithms with bases other than 10 or e, you can use the change of base formula:
[math]\log_b(a) = \frac{\log_k(a)}{\log_k(b)}[/math]
Where [math]k[/math] can be 10 or e. For example, to calculate [math]\log_2(8)[/math]:
Calculate [math]\log_{10}(8)[/math] using log.
Calculate [math]\log_{10}(2)[/math] using log.
Divide the results: [math]\frac{\log_{10}(8)}{\log_{10}(2)}[/math].
This method will give you the logarithm in any base.
Upvote ·
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How do you put formulas in a Casio calculator?
How do you do a log base on a calculator?
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Paul Guillory
High school and community college math instructor (2010–present) · Author has 247 answers and 393.5K answer views
·3y
I really love the Casio scientific calculators, and very much because of the log buttons. The Casio handles log calculations SO MUCH BETTER than almost any calculator on the market. Here’s why:
And it’s this button that really makes a difference:
This button lets you enter a log with the base! You can calculate logs directly without having to use the change of base formula. Here’s what that looks like:
The calculator will even return a fraction for an answer if you want it to:
There’s lots of other good qualities about the Casio fx115ES and fx300ES line of calculators, and I think they are the ide
Continue Reading
I really love the Casio scientific calculators, and very much because of the log buttons. The Casio handles log calculations SO MUCH BETTER than almost any calculator on the market. Here’s why:
And it’s this button that really makes a difference:
This button lets you enter a log with the base! You can calculate logs directly without having to use the change of base formula. Here’s what that looks like:
The calculator will even return a fraction for an answer if you want it to:
There’s lots of other good qualities about the Casio fx115ES and fx300ES line of calculators, and I think they are the ideal classroom calculators.
Upvote ·
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Eric Shockey
Studied at Montcalm Community College · Author has 137 answers and 203.8K answer views
·1y
There are two ways depending on what log you are talk about.
Base 10 or log - you press log you should get a “log(“ icon without the “ type in your number put the “)” with out the “ press enter or = key
It should split out the base 10 log of that number
To get the number back - you get the 10^x key it accessed by pressing the shift key and the log
And it will give you the anti-log of it
Natural log or ln key
Press the ln key it should display the ln( icon
Type in your number press the ) key press enter or = depending on entry mode
It will split out the natural log of that number
To get the anti log you
Continue Reading
There are two ways depending on what log you are talk about.
Base 10 or log - you press log you should get a “log(“ icon without the “ type in your number put the “)” with out the “ press enter or = key
It should split out the base 10 log of that number
To get the number back - you get the 10^x key it accessed by pressing the shift key and the log
And it will give you the anti-log of it
Natural log or ln key
Press the ln key it should display the ln( icon
Type in your number press the ) key press enter or = depending on entry mode
It will split out the natural log of that number
To get the anti log you press shift or the e^x key
It should get you the antilog of it.
Also a neat trick if you accidentally press ln instead of log - you can get the log value by dividing the ln value by ln( 10) to get the log value
Example ln(2)
.69314718
To get the base 10 log of 2- do ln(2)/ln(10) to get
.30102999 which is log(2)
This works due the change of base rule of logarithms.
Upvote ·
9 1
Anthony Hawken
Author has 10.8K answers and 3.8M answer views
·4y
Originally Answered: How do I use a log on a scientific calculator? ·
You press the log button followed by the number you want to find the logarithm of.
e.g. log 3 =
will give you 0.477121254
This is the logarithm of 3 using base 10 logarithms.
Upvote ·
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Related questions
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Is Casio a scientific calculator?
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1381 | https://www.excel-easy.com/examples/number-text-filters.html | Number and Text Filters in Excel - Step by Step Tutorial
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Arrows in the column headers appear.
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Enter 10,000 and click OK.
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1383 | https://moodle2.units.it/pluginfile.php/363370/mod_resource/content/1/cinematica_2020_parte1.pdf | Cinematica del punto materiale Studia il moto dei corpi senza riferimento alle sue cause e alla natura dei corpi stessi. Il corpo è descritto come un punto materiale senza dimensioni proprie La cinematica necessita di un SISTEMA DI RIFERIMENTO: ogni moto e’ relativo e va riferito ad un particolare sistema di riferimento con una origine e un sistema di assi cartesiani ❑ Le grandezze fondamentali della cinematica sono: ✓ spostamento ✓ velocità ✓ accelerazione ❑ Queste grandezze, come molte altre grandezze fisiche, in generale non sono numeri, ma vettori Traiettoria Traiettoria: linea continua luogo geometrico costituita da tutte le posizioni occupate nel tempo dal punto materiale in istanti successivi. NON da’ informazioni su come essa e’ percorsa (cioe’ sulla velocita’ e accelerazione) I moti possono essere rettilinei o curvilinei Direzione del moto: tangente alla traiettoria (con verso). Origine del moto: punto di partenza, posizione al tempo zero Legge del moto (diagramma orario) E’ il grafico spazio-tempo Relazione fra TEMPO e SPAZIO percorso. Il tempo e’ la variabile indipendente Con la legge oraria si puo’ determinare ad ogni istante la posizione del mobile sulla traiettoria Per esempio in un moto unidimensionale dobbiamo conoscere la funzione x = x(t) Ogni tipo di moto ha una particolare legge oraria. La legge oraria non da’ informazioni sulla traiettoria Grandezze scalari e vettoriali ✓Esempio: se vogliamo informare qualcuno su quanti siamo alti, basta dire: “sono alto 180 cm”, non serve aggiungere altro, poiché l’altezza è una grandezza scalare. Se invece vogliamo informarlo sulla nostra posizione nello spazio un numero potrebbe non essere sufficiente, poiché la posizione è un VETTORE ✓Di quanti numeri (o COMPONENTI) è composto un vettore ? Dipende dalla dimensione dello spazio in cui si muove il punto ✓Un vettore è sempre definito rispetto ad un sistema di riferimento specifico, ovvero un sistema di assi cartesiani x y z x x y una dimensione (1D) 3 dimensioni (3D) 2 dimensioni (2D) Grandezze scalari e vettoriali In Fisica tutte le grandezze si suddividono in quantità scalari e vettoriali (o ancora più complesse, ovvero matrici) A = B A e B debbono essere sempre grandezze omogenee: scalare = scalare oppure vettore = vettore ✓grandezze scalari: grandezze esprimibili mediante un singolo numero; temperatura, massa, lunghezza sono grandezze fisiche scalari ✓grandezze vettoriali: grandezze che richiedono più di un singolo numero per essere completamente definite; esempi sono la posizione nello spazio, lo spostamento, la velocità, l’accelerazione, la forza Come conseguenza di questa classificazione in ogni legge fisica del tipo: Il vettore Graficamente un vettore si rappresenta mediante una freccia. Il vettore è definito da 3 proprietà: ✓lunghezza o modulo ✓direzione ovvero la retta su cui il vettore giace ✓verso indicato dalla punta della freccia La freccetta sopra il simbolo v indica che si tratta di un vettore e non di uno scalare A volte nelle formule invece della freccetta si usa indicare il vettore scrivendo il simbolo in grassetto: in tal caso v indica il vettore, mentre v indica il suo modulo Per rappresentare concretamente un vettore in 2D o 3D è necessario specificare un sistema di riferimento cartesiano Il vettore Esempio #1: le forze (frecce rosse) che agiscono sulla sfera blu hanno stesso modulo (ovvero stessa lunghezza) ma diversa direzione, una verticale e l’altra orizzontale; dunque sono vettori diversi F F F F Esempio #2: adesso le due sono uguali in modulo e anche in direzione, ma hanno verso opposto; dunque sono ancora diverse tra loro Rappresentazione del vettore in componenti cartesiane Un vettore può essere decomposto in componenti usando un sistema di riferimento, detto anche riferimento Cartesiano (dal grande matematico francese René Descartés) ✓In 2D un sistema di riferimento è rappresentato da due assi x e y ortogonali (ovvero perpendicolari) che si incrociano in un punto detto origine ✓il vettore si indica mediante le sue componenti (o coordinate cartesiane) vx, vy, ovvero le proiezioni del vettore lungo gli assi, nel modo seguente: La Haye en Touraine 1569 ( ) v , x y v v = x y v v y vx Dalla punta del vettore, tracciamo due rette parallele agli assi: l’intersezione di queste rette con i due assi cartesiani ci dà il valore delle componenti vx, vy, ovvero le proiezioni del vettore lungo gli assi Rappresentazione del vettore in componenti cartesiane In 3 dimensioni (3D) un sistema di riferimento è rappresentato da tre assi ortogonali x, y, z La Haye en Touraine 1569 x y z v v y vx vz ( ) v , , x y z v v v = ✓dalla punta del vettore, tracciamo due rette (in rosso), una retta parallela al piano (x,y) ed una ortogonale al piano (x,y); la prima interseca l’asse z e dà la proiezione vz, la seconda interseca il piano x,y e dà la componente planare del vettore vp ✓la componente planare vp ovvero è semplicemente un vettore 2D nel piano (x,y), dunque possiamo procedere come in precedenza per scomporre vp nelle sue componenti vx, vy v p Indipendenza dei vettori dall’origine del riferimento cartesiano Un vettore NON dipende dall’origine del sistema di riferimento: può essere traslato nello spazio, rimanendo del tutto uguale a sé stesso esempio: consideriamo un vettore nello spazio 2D: v = (vx, vy) Supponiamo che inizialmente la coda del vettore sia sull’origine del riferimento (vettore blu); le coordinate del vettore sono: Trasliamo (ovvero spostiamo rigidamente) il vettore nello spazio; sia v’ (in rosso) il vettore traslato; calcolando le proiezioni di v’ lungo gli assi, si vede facilmente che: v 4 2 2 4 6 6 ' v 1 3 5 0 1 3 5 x y v = (4, 2) V’ = (4, 2) Il vettore è lo stesso, prima e dopo la traslazione; v e v’ sono del tutto identici, poiché uguali in modulo, direzione e verso Somma dei vettori x x x y y y S F G S F G = + = + Siano dati 2 vettori F = (Fx, Fy) e G = (Gx, Gy). Calcolariamo il vettore somma dei due vettori; chiamiamolo S = (Sx, Sy) ( ) G , x x y y S F F G F G = + = + + La somma di F e G è un vettore S le cui componenti sono la somma delle componenti corrispondenti di F e G F G 4 2 G G 2 4 6 6 S G ESERCIZIO in 2D: dati F e G, calcolare il vettore somma S = F + G F = (4, 2) G = (2, 3) S = (6, 5) 1 3 5 0 1 3 5 Differenza di vettori F G x F y F G G x G y G D G Notiamo che la componente Dy è negativa: infatti, il vettore D punta verso le coordinate negative dell’asse y ( ) ( ) ( ) ( ) G , G , , , x y x y x y x x y y D F F F F G G D S S F G F G = − = = = = − − F = (4, 2) G = (2, 3) D = (2, -1) La differenza di F e G è un vettore D le cui componenti sono la differenza delle componenti corrispondenti di F e G: Esempio numerico: Modulo del vettore 2 2 2 z y x F F F F + + = 38 .
5 29 9 16 4 3 4 2 2 2 2 = = + + = + + = F Ad esempio, nel caso in figura si ha Fx = 2, Fy = 4, Fz = 3 ; il modulo è dato da: F G 6 3 Nel caso bidimensionale a fianco F = (6,3) 7 .
6 45 9 36 = = + = F x y z F G 2 4 3 Il modulo di un vettore è uguale alla radice quadrata della somma dei quadrati delle componenti lungo gli assi.
Moto rettilineo • Si svolge lungo una retta su cui si definisce la coordinata x, la cui origine (x=0) e il cui verso sono arbitrari • Anche lorigine dei tempi (t=0) earbitraria • Il moto del corpo e descrivibile con una sola funzione x(t) • La funzione puo` essere rappresentata sul cosiddetto diagramma orario, sul cui asse delle ascisse poniamo t e su quello delle ordinate x 5 x t O O Moto rettilineo |
1384 | https://www.uvm.edu/~cvincen1/files/teaching/spring2017-math255/hw3sol.pdf | Math 255 - Spring 2017 Homework 3 Solutions 1. First note that for the equality to really be true, we would need to write |ab| = lcm(a, b) gcd(a, b).
This is because both lcm(a, b) and gcd(a, b) are defined to be positive, whereas a and b could be negative. (The problem arises when exactly one of a or b is negative.) To avoid this small sign problem, we assume that a and b are positive, which will prove even the correct assertion above because lcm(a, b) = lcm(|a|, |b|) and gcd(a, b) = gcd(|a|, |b|).
We consider the quantity ab gcd(a, b) and show that it is the least common multiple of a and b. This will prove the claim.
By definition of the greatest common divisor, there are r, s ∈Z such that a = gcd(a, b)r and b = gcd(a, b)s.
We note here that gcd(b, r) = 1 (if this were not the case gcd(a, b) gcd(b, r) would be a common divisor of a and b which is strictly greater than gcd(a, b), a contradiction).
Therefore, ab gcd(a, b) = gcd(a, b)rs.
From this we may conclude that ab gcd(a,b) is a multiple of both a and b, since ab gcd(a, b) = as and ab gcd(a, b) = br.
We now show that it is the smallest positive integer that is a multiple of a and b. Let m be another positive common multiple of a and b. Because m is a multiple of a, there is t ∈Z such that m = at, and therefore we may write m = gcd(a, b)rt.
Because m is also a multiple of b by assumption, we have that b divides gcd(a, b)rt.
Because gcd(b, r) = 1, as remarked above, we can conclude that b divides gcd(a, b)t.
1 Therefore, because all of the integers in this problem are positive, we have b ≤gcd(a, b)t br ≤gcd(a, b)rt ab gcd(a, b) ≤m.
Since ab gcd(a,b) is smaller than any other positive multiple of a and b, it must be lcm(a, b) and the claim is proved.
2. This follows immediately from the transitive property of “divides:” Since gcd(a, b)|a and a| lcm(a, b), it follows that gcd(a, b)| lcm(a, b).
2 3. We begin by proving that gcd(a + b, a −b) is a common divisor of 2a and 2b, from which we conclude that gcd(a + b, a −b) ≤gcd(2a, 2b).
We then show that gcd(2a, 2b) = 2 gcd(a, b).
After this we are done since gcd(a + b, a −b) ≤gcd(2a, 2b) = 2 gcd(a, b) = 2.
Since gcd(a+b, a−b) is a positive integer by definition, it follows that gcd(a+b, a−b) = 1 or 2.
We first prove our first claim. Recall that gcd(a + b, a −b) divides any integer linear combination (a + b)x + (a −b)y of a + b and a −b. If we choose x = y = 1, we get (a + b)x + (a −b)y = (a + b) + (a −b) = 2a.
Therefore gcd(a + b, a −b) divides 2a. We now choose x = 1, y = −1 and obtain (a + b)x + (a −b)y = (a + b) −(a −b) = 2b.
Therefore gcd(a + b, a −b) also divides 2b. Since gcd(a + b, a −b) is a common divisor of 2a and 2b, gcd(a + b, a −b) ≤gcd(2a, 2b).
We now prove the second claim. To show that 2 gcd(a, b) is gcd(2a, 2b), we must show that 2 gcd(a, b) divides both 2a and 2b and that if c is another common divisor of 2a and 2b, then c ≤2 gcd(a, b).
As usual, let r, s ∈Z be such that a = gcd(a, b)r and b = gcd(a, b)s. Then we have 2a = 2 gcd(a, b)r and 2b = 2 gcd(a, b)s, and 2 gcd(a, b) is shown to divide both 2a and 2b.
Now let c be a common divisor of 2a and 2b. We treat the case of c even and odd separately. Let’s start with c odd. In this case, gcd(c, 2) = 1, and therefore c|2a implies c|a. Similarly, c|2b implies c|b. Therefore when c is odd, c is a common divisor of a and b, from which it follows that c ≤gcd(a, b) < 2 gcd(a, b).
Now consider the case of c even. Since c divides 2a, we may choose u ∈Z such that 2a = cu. Dividing both sides by 2, we obtain a = c 2u, 3 where c 2 ∈Z since c is even. Therefore c 2 divides a. In the same way, writing 2b = cv for v ∈Z (recall that c divides 2b as well), we obtain b = c 2v and c 2 also divides b. Therefore c 2 is a common divisor of a and b, from which it follows that c 2 ≤gcd(a, b) or c ≤2 gcd(a, b).
Therefore whether c is even or odd, if it is a common divisor of 2a and 2b, we obtain that c ≤2 gcd(a, b). This completes our proof that gcd(2a, 2b) = 2 gcd(a, b).
As mentioned above this completes the proof since now we have established that gcd(a + b, a −b) ≤gcd(2a, 2b) = 2 gcd(a, b) = 2, and since gcd(a + b, a −b) is a positive integer by definition, gcd(a + b, a −b) = 1 or 2.
4 4. Let N be the number of coins, x be the number of coins in a full pile when attempting to divide into 77 piles, and y be the number of coins in each pile when the coins are divided into 78 piles. Then we have N = 77x −50 and N = 78y.
(Note that it is also acceptable to use the equation N = 77t+27 instead of N = 77x−50.
However, in that case t is not the number of coins in any pile; t = x −1 which is one less than the number of coins in each full pile.) This gives us the equation 77x −50 = 78y or 77x −78y = 50.
Since the greatest common divisor of 77 and 78 is 1 (consecutive integers always have a gcd of 1), this equation does have integer solutions.
We first solve the equation 77x −78y = gcd(77, 78) = 1.
By inspection, this has solution x0 = −1 and y0 = −1. Therefore the equation 77x −78y = 50 has particular solution xp = −50 and yp = −50.
We may now apply our theorem to obtain that all integer solutions of the equation 77x −78y = 50 are x = −50 −78t y = −50 −77t, for t ∈Z.
Since x and y are quantities of coins, they both must be nonnegative. We now solve for the values of t that give x ≥0 and y ≥0. We first look at x only: 0 ≤x = −50 −78t 78t ≤−50 t ≤−50 78 .
Since t is an integer, this implies t ≤−1. We now look at y only: 0 ≤y = −50 −77t 77t ≤−50 t ≤−50 77 .
5 Since t is an integer, this implies t ≤−1.
Thankfully it is easy to reconcile the conditions given by x and y: If t ≤−1 and t ≤−1, then t ≤−1!
Now this means that any value of t that is −1, −2, −3, −4 . . . will give a valid solution to the problem. Therefore, the best solution to the problem is “The number of coins I have is any of the following: N = 78(−50 −77t) = 6006t −3900, where t is any negative integer,” but N = 2106 (this is N when t = −1) or any other one of the valid answers will be accepted for full credit.
6 |
1385 | https://www.wyzant.com/resources/answers/704099/explain-the-relationship-between-the-terms-of-the-quadratic-formula-and-the | Log in Sign up
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Explain the relationship between the terms of the quadratic formula and the graph of a quadratic function.
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The quadratic formula is used to find the zeros or x-intercepts of a quadratic function. On the graph of the quadratic functions the x intercepts should match what the quadratic formula gives you.
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Isidro L. answered • 08/02/19
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Great job Denise !! it is fear to add : Giving y = ax^2 +bx +C, if the coefficient of x^2. , in this case (a )is negative the function or parabola opens down and if the coefficient is positive. or greater than zero the function or parabola opens up and for the (c) term, when x=0 that becomes the y-intercept of the function.
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The Quadratic Formula is used to find the roots of a quadratic equation (ax^2 + bx + c = 0).
Thus, x = ( -b ± √(b2 - 4ac) ) / 2a
can produce the two points where the parabola crosses the x-axis. Note that the discriminant (underneath the square root) helps to tell whether the parabola actually crosses the x-axix (and thus has real roots).
Using multiple items in the Quadratic Formula, you may find the vertex and determine the axis of symmetry. These can be labeled on the graph of a quadratic formula.
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4: Exponential and Logarithmic Functions
Precalculus 1e (OpenStax)
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4.4: Graphs of Logarithmic Functions
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4.4: Graphs of Logarithmic Functions
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4.3: Logarithmic Functions
4.5: Logarithmic Properties
Page ID 1355
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents
1. Learning Objectives
2. Finding the Domain of a Logarithmic Function
1. Given a logarithmic function, identify the domain
2. Example 4.4.1: Identifying the Domain of a Logarithmic Shift
1. Solution
3. Exercise 4.4.1/04:_Exponential_and_Logarithmic_Functions/4.04:_Graphs_of_Logarithmic_Functions#Exercise_.5C(.5CPageIndex.7B1.7D.5C))
4. Example 4.4.2: Identifying the Domain of a Logarithmic Shift and Reflection/04:_Exponential_and_Logarithmic_Functions/4.04:_Graphs_of_Logarithmic_Functions#Example_.5C(.5CPageIndex.7B2.7D.5C):_Identifying_the_Domain_of_a_Logarithmic_Shift_and_Reflection)
1. Solution/04:_Exponential_and_Logarithmic_Functions/4.04:_Graphs_of_Logarithmic_Functions#Solution_2)
5. Exercise 4.4.2/04:_Exponential_and_Logarithmic_Functions/4.04:_Graphs_of_Logarithmic_Functions#Exercise_.5C(.5CPageIndex.7B2.7D.5C))
Graphing Logarithmic Functions
CHARACTERISTICS OF THE GRAPH OF THE PARENT FUNCTION, F(X)=LOG B(X)
Given a logarithmic function with the form f(x)=log b(x), graph the function.
Example 4.4.3: Graphing a Logarithmic Function with the Form f(x)=log b(x)
Solution
Exercise 4.4.3
Graphing Transformations of Logarithmic Functions
Graphing a Horizontal Shift of f(x)=log b(x)
HORIZONTAL SHIFTS OF THE PARENT FUNCTION Y=LOG B(X)
Given a logarithmic function with the form f(x)=log b(x+c), graph the translation.
Example 4.4.4: Graphing a Horizontal Shift of the Parent Function y=log b(x)
Solution
Exercise 4.4.4
Graphing a Vertical Shift of y=log b(x)
VERTICAL SHIFTS OF THE PARENT FUNCTION Y=LOG B(X)
Given a logarithmic function with the form f(x)=log b(x)+d, graph the translation.
Example 4.4.5: Graphing a Vertical Shift of the Parent Function y=log b(x)
Solution
Exercise 4.4.5
Graphing Stretches and Compressions of y=log b(x)
VERTICAL STRETCHES AND COMPRESSIONS OF THE PARENT FUNCTION Y=LOG B(X)
Given a logarithmic function with the form f(x)=alog b(x), a>0,graph the translation.
Example 4.4.6: Graphing a Stretch or Compression of the Parent Function y=log b(x)
Solution
Exercise 4.4.6
Example 4.4.7: Combining a Shift and a Stretch
Solution
Exercise 4.4.7
Graphing Reflections of f(x)=log b(x)
REFLECTIONS OF THE PARENT FUNCTION Y=LOG B(X)
Given a logarithmic function with the parent function f(x)=log b(x), graph a translation.
Example 4.4.8: Graphing a Reflection of a Logarithmic Function
Solution
Exercise 4.4.8
Given a logarithmic equation, use a graphing calculator to approximate solutions.
Example 4.4.9: Approximating the Solution of a Logarithmic Equation
Solution
Exercise 4.4.9
Summarizing Translations of the Logarithmic Function
TRANSLATIONS OF LOGARITHMIC FUNCTIONS
Example 4.4.10: Finding the Vertical Asymptote of a Logarithm Graph
Solution
Exercise 4.4.10
Example 4.4.11: Finding the Equation from a Graph
Solution
Exercise 4.4.11
Media: Is it possible to tell the domain and range and describe the end behavior of a function just by looking at the graph?
Media
Key Equations
Key Concepts
Learning Objectives
Identify the domain of a logarithmic function.
Graph logarithmic functions.
In the Section on Graphs of Exponential Functions, we saw how creating a graphical representation of an exponential model gives us another layer of insight for predicting future events. How do logarithmic graphs give us insight into situations? Because every logarithmic function is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation. In other words, logarithms give the cause for an effect.
To illustrate, suppose we invest $2500 in an account that offers an annual interest rate of 5%, compounded continuously. We already know that the balance in our account for any year t can be found with the equation A=2500e 0.05t.
But what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation. By graphing the model, we can see the output (year) for any input (account balance). For instance, what if we wanted to know how many years it would take for our initial investment to double? Figure 4.4.1 shows this point on the logarithmic graph.
Figure 4.4.1
In this section we will discuss the values for which a logarithmic function is defined, and then turn our attention to graphing the family of logarithmic functions.
Finding the Domain of a Logarithmic Function
Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined.
Recall that the exponential function is defined as y=b x for any real number x and constant b>0, b≠1, where
The domain of y is (−∞,∞).
The range of y is (0,∞).
In the last section we learned that the logarithmic function y=log b(x) is the inverse of the exponential function y=b x. So, as inverse functions:
The domain of y=log b(x) is the range of y=b x: (0,∞).
The range of y=log b(x) is the domain of y=b x: (−∞,∞).
Transformations of the parent function y=log b(x) behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, stretches, compressions, and reflections—to the parent function without loss of shape.
In Graphs of Exponential Functions we saw that certain transformations can change the range of y=b x. Similarly, applying transformations to the parent function y=log b(x) can change the domain. When finding the domain of a logarithmic function, therefore, it is important to remember that the domain consists only of positive real numbers. That is, the argument of the logarithmic function must be greater than zero.
For example, consider f(x)=log 4(2x−3). This function is defined for any values of x such that the argument, in this case 2x−3,is greater than zero. To find the domain, we set up an inequality and solve for x:
2x−3>0 Show the argument greater than zero 2x>3 Add 3 x>1.5 Divide by 2
In interval notation, the domain of f(x)=log 4(2x−3) is (1.5,∞).
Given a logarithmic function, identify the domain
Set up an inequality showing the argument greater than zero.
Solve for x.
Write the domain in interval notation.
Example 4.4.1: Identifying the Domain of a Logarithmic Shift
What is the domain of f(x)=log 2(x+3)?
Solution
The logarithmic function is defined only when the input is positive, so this function is defined when x+3>0. Solving this inequality,
x+3>0 The input must be positive x>−3 Subtract 3
The domain of f(x)=log 2(x+3) is (−3,∞).
Exercise 4.4.1
What is the domain of f(x)=log 5(x−2)+1?
Answer
(2,∞)
Example 4.4.2: Identifying the Domain of a Logarithmic Shift and Reflection
What is the domain of f(x)=log(5−2x)?
Solution
The logarithmic function is defined only when the input is positive, so this function is defined when 5–2x>0. Solving this inequality,
5−2x>0 The input must be positive−2x>−5 Subtract 5 x<5 2 Divide by -2 and switch the inequality
The domain of f(x)=log(5−2x) is (–∞,5 2).
Exercise 4.4.2
What is the domain of f(x)=log(x−5)+2?
Answer
(5,∞)
Graphing Logarithmic Functions
Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function y=log b(x) along with all its transformations: shifts, stretches, compressions, and reflections.
We begin with the parent function y=log b(x). Because every logarithmic function of this form is the inverse of an exponential function with the form y=b x, their graphs will be reflections of each other across the line y=x. To illustrate this, we can observe the relationship between the input and output values of y=2 x and its equivalent x=log 2(y) in Table 4.4.1.
Table 4.4.1| x | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
| 2 x=y | 1 8 | 1 4 | 1 2 | 1 | 2 | 4 | 8 |
| log 2(y)=x | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
Using the inputs and outputs from Table 4.4.1, we can build another table to observe the relationship between points on the graphs of the inverse functions f(x)=2 x and g(x)=log 2(x). See Table 4.4.2.
Table 4.4.2| f(x)=2 x | (−3,1 8) | (−2,1 4) | (−1,1 2) | (0,1) | (1,2) | (2,4) | (3,8) |
| g(x)=log 2(x) | (1 8,−3) | (1 4,−2) | (1 2,−1) | (1,0) | (2,1) | (4,2) | (8,3) |
As we’d expect, the x- and y-coordinates are reversed for the inverse functions. Figure 4.4.2 shows the graph of f and g.
Figure 4.4.2:Notice that the graphs of f(x)=2 x and g(x)=log 2(x) are reflections about the line y=x.
Observe the following from the graph:
f(x)=2 x has a y-intercept at (0,1) and g(x)=log 2(x) has an x- intercept at (1,0).
The domain of f(x)=2 x, (−∞,∞), is the same as the range of g(x)=log 2(x).
The range of f(x)=2 x, (0,∞), is the same as the domain of g(x)=log 2(x).
CHARACTERISTICS OF THE GRAPH OF THE PARENT FUNCTION, F(X)=LOG B(X)
For any real number x and constant b>0, b≠1, we can see the following characteristics in the graph of f(x)=log b(x):
one-to-one function
vertical asymptote: x=0
domain: (0,∞)
range: (−∞,∞)
x - intercept: (1,0) and key point (b,1)
y-intercept: none
increasing if b>1
decreasing if 0<b<1
See Figure 4.4.3.
1, and the second graph shows the line when 0<1.">
Figure 4.4.3
Figure 4.4.4 shows how changing the base b in f(x)=log b(x) can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (Note: recall that the function ln(x) has base e≈2.718.)
Figure 4.4.4: The graphs of three logarithmic functions with different bases, all greater than 1.
Given a logarithmic function with the form f(x)=log b(x), graph the function.
Draw and label the vertical asymptote, x=0.
Plot the x- intercept, (1,0).
Plot the key point (b,1).
Draw a smooth curve through the points.
State the domain, (0,∞),the range, (−∞,∞), and the vertical asymptote, x=0.
Example 4.4.3: Graphing a Logarithmic Function with the Form f(x)=log b(x)
Graph f(x)=log 5(x). State the domain, range, and asymptote.
Solution
Before graphing, identify the behavior and key points for the graph.
Since b=5 is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical asymptote x=0, and the right tail will increase slowly without bound.
The x-intercept is (1,0).
The key point (5,1) is on the graph.
We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points (see Figure 4.4.5).
Figure 4.4.5
The domain is (0,∞), the range is (−∞,∞), and the vertical asymptote is x=0.
Exercise 4.4.3
Graph f(x)=log 1 5(x). State the domain, range, and asymptote.
Answer
Figure 4.4.6
The domain is (0,∞), the range is (−∞,∞), and the vertical asymptote is x=0.
Graphing Transformations of Logarithmic Functions
As we mentioned in the beginning of the section, transformations of logarithmic graphs behave similarly to those of other parent functions. We can shift, stretch, compress, and reflect the parent function y=log b(x) without loss of shape.
Graphing a Horizontal Shift of f(x)=log b(x)
When a constant c is added to the input of the parent function f(x)=log b(x), the result is a horizontal shift c units in the opposite direction of the sign on c. To visualize horizontal shifts, we can observe the general graph of the parent function f(x)=log b(x) and for c>0 alongside the shift left, g(x)=log b(x+c), and the shift right, h(x)=log b(x−c). See Figure 4.4.7.
Figure 4.4.7
HORIZONTAL SHIFTS OF THE PARENT FUNCTION Y=LOG B(X)
For any constant c,the function f(x)=log b(x+c)
shifts the parent function y=log b(x) left c units if c>0.
shifts the parent function y=log b(x) right c units if c<0.
has the vertical asymptote x=−c.
has domain (−c,∞).
has range (−∞,∞).
Given a logarithmic function with the form f(x)=log b(x+c), graph the translation.
Identify the horizontal shift:
If c>0,shift the graph of f(x)=log b(x) left c units.
If c<0,shift the graph of f(x)=log b(x) right c units.
Draw the vertical asymptote x=−c.
Identify three key points from the parent function. Find new coordinates for the shifted functions by subtracting c from the x coordinate.
Label the three points.
The Domain is (−c,∞),the range is (−∞,∞), and the vertical asymptote is x=−c.
Example 4.4.4: Graphing a Horizontal Shift of the Parent Function y=log b(x)
Sketch the horizontal shift f(x)=log 3(x−2) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.
Solution
Since the function is f(x)=log 3(x−2), we notice x+(−2)=x–2.
Thus c=−2, so c<0. This means we will shift the function f(x)=log 3(x) right 2 units.
The vertical asymptote is x=−(−2) or x=2.
Consider the three key points from the parent function, (1 3,−1), (1,0),and (3,1).
The new coordinates are found by adding 2 to the x coordinates.
Label the points (7 3,−1), (3,0),and (5,1).
The domain is (2,∞),the range is (−∞,∞),and the vertical asymptote is x=2.
Figure 4.4.8
Exercise 4.4.4
Sketch a graph of f(x)=log 3(x+4) alongside its parent function. Include the key points and asymptotes on the graph. State the domain, range, and asymptote.
Answer
Figure 4.4.9
The domain is (−4,∞),the range (−∞,∞),and the asymptote x=–4.
Graphing a Vertical Shift of y=log b(x)
When a constant d is added to the parent function f(x)=log b(x),the result is a vertical shift d units in the direction of the sign on d. To visualize vertical shifts, we can observe the general graph of the parent function f(x)=log b(x) alongside the shift up, g(x)=log b(x)+d and the shift down, h(x)=log b(x)−d.See Figure 4.4.10.
Figure 4.4.10
VERTICAL SHIFTS OF THE PARENT FUNCTION Y=LOG B(X)
For any constant d,the function f(x)=log b(x)+d
shifts the parent function y=log b(x) up d units if d>0.
shifts the parent function y=log b(x) down d units if d<0.
has the vertical asymptote x=0.
has domain (0,∞).
has range (−∞,∞).
Given a logarithmic function with the form f(x)=log b(x)+d, graph the translation.
Identify the vertical shift:
If d>0, shift the graph of f(x)=log b(x) up d units.
If d<0, shift the graph of f(x)=log b(x) down d units.
Draw the vertical asymptote x=0.
Identify three key points from the parent function. Find new coordinates for the shifted functions by adding d to the y coordinate.
Label the three points.
The domain is (0,∞), the range is (−∞,∞),and the vertical asymptote is x=0.
Example 4.4.5: Graphing a Vertical Shift of the Parent Function y=log b(x)
Sketch a graph of f(x)=log 3(x)−2 alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
Solution
Since the function is f(x)=log 3(x)−2,we will notice d=–2. Thus d<0.
This means we will shift the function f(x)=log 3(x) down 2 units.
The vertical asymptote is x=0.
Consider the three key points from the parent function, (1 3,−1), (1,0),and (3,1).
The new coordinates are found by subtracting 2 from the y coordinates.
Label the points (1 3,−3), (1,−2), and (3,−1).
The domain is (0,∞),the range is (−∞,∞), and the vertical asymptote is x=0.
Figure 4.4.11
The domain is (0,∞),the range is (−∞,∞),and the vertical asymptote is x=0.
Exercise 4.4.5
Sketch a graph of f(x)=log 2(x)+2 alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
Answer
Figure 4.4.12
The domain is (0,∞),the range is (−∞,∞), and the vertical asymptote is x=0.
Graphing Stretches and Compressions of y=log b(x)
When the parent function f(x)=log b(x) is multiplied by a constant a>0, the result is a vertical stretch or compression of the original graph. To visualize stretches and compressions, we set a>1 and observe the general graph of the parent function f(x)=log b(x) alongside the vertical stretch, g(x)=alog b(x) and the vertical compression, h(x)=1 alog b(x).See Figure 4.4.13.
1 is the translation function with an asymptote at x=0. The graph note the intersection of the two lines at (1, 0). This shows the translation of a vertical stretch." src="/@api/deki/files/12175/fig_6.5.13.jpg">
Figure 4.4.13
VERTICAL STRETCHES AND COMPRESSIONS OF THE PARENT FUNCTION Y=LOG B(X)
For any constant a>1,the function f(x)=alog b(x)
stretches the parent function y=log b(x) vertically by a factor of a if a>1.
compresses the parent function y=log b(x) vertically by a factor of a if 0<a<1.
has the vertical asymptote x=0.
has the x-intercept (1,0).
has domain (0,∞).
has range (−∞,∞).
Given a logarithmic function with the form f(x)=alog b(x), a>0,graph the translation.
Identify the vertical stretch or compressions:
If |a|>1, the graph of f(x)=log b(x) is stretched by a factor of a units.
If |a|<1, the graph of f(x)=log b(x) is compressed by a factor of a units.
Draw the vertical asymptote x=0.
Identify three key points from the parent function. Find new coordinates for the shifted functions by multiplying they y coordinates by a.
Label the three points.
The domain is (0,∞), the range is (−∞,∞), and the vertical asymptote is x=0.
Example 4.4.6: Graphing a Stretch or Compression of the Parent Function y=log b(x)
Sketch a graph of f(x)=2log 4(x) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
Solution
Since the function is f(x)=2log 4(x),we will notice a=2.
This means we will stretch the function f(x)=log 4(x) by a factor of 2.
The vertical asymptote is x=0.
Consider the three key points from the parent function, (1 4,−1), (1,0), and (4,1).
The new coordinates are found by multiplying the y coordinates by 2.
Label the points (1 4,−2), (1,0), and (4,2).
The domain is (0,∞), the range is (−∞,∞), and the vertical asymptote is x=0. See Figure 4.4.14.
Figure 4.4.14
The domain is (0,∞), the range is (−∞,∞), and the vertical asymptote is x=0.
Exercise 4.4.6
Sketch a graph of f(x)=1 2log 4(x) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
Answer
Figure 4.4.15
The domain is (0,∞), the range is (−∞,∞),and the vertical asymptote is x=0.
Example 4.4.7: Combining a Shift and a Stretch
Sketch a graph of f(x)=5log(x+2). State the domain, range, and asymptote.
Solution
Remember: what happens inside parentheses happens first. First, we move the graph left 2 units, then stretch the function vertically by a factor of 5, as in Figure 4.4.16. The vertical asymptote will be shifted to x=−2. The x-intercept will be (−1,0). The domain will be (−2,∞). Two points will help give the shape of the graph: (−1,0) and (8,5). We chose x=8 as the x-coordinate of one point to graph because when x=8, x+2=10, the base of the common logarithm.
Figure 4.4.16
The domain is (−2,∞), the range is (−∞,∞),and the vertical asymptote is x=−2.
Exercise 4.4.7
Sketch a graph of the function f(x)=3log(x−2)+1. State the domain, range, and asymptote.
Answer
Figure 4.4.17
The domain is (2,∞),the range is (−∞,∞), and the vertical asymptote is x=2.
Graphing Reflections of f(x)=log b(x)
When the parent function f(x)=log b(x) is multiplied by −1,the result is a reflection about the x-axis. When the input is multiplied by −1,the result is a reflection about the y-axis. To visualize reflections, we restrict b>1, and observe the general graph of the parent function f(x)=log b(x) alongside the reflection about the x-axis, g(x)=−log b(x) and the reflection about the y-axis, h(x)=log b(−x).
1 is the translation function with an asymptote at x=0. The graph note the intersection of the two lines at (1, 0). This shows the translation of a reflection about the x-axis." src="/@api/deki/files/12180/fig_6.5.18.jpg">
Figure 4.4.18
REFLECTIONS OF THE PARENT FUNCTION Y=LOG B(X)
The function f(x)=−log b(x)
reflects the parent function y=log b(x) about the x-axis.
has domain, (0,∞), range, (−∞,∞), and vertical asymptote, x=0, which are unchanged from the parent function.
The function f(x)=log b(−x)
reflects the parent function y=log b(x) about the y-axis.
has domain (−∞,0).
has range, (−∞,∞), and vertical asymptote, x=0, which are unchanged from the parent function.
Given a logarithmic function with the parent function f(x)=log b(x), graph a translation.
Table 4.4.3| If f(x)=−log b(x) | If f(x)=log b(−x) |
--- |
| 1. Draw the vertical asymptote, x=0. | 1. Draw the vertical asymptote, x=0. |
| 1. Plot the x- intercept, (1,0). | 1. Plot the x- intercept, (1,0). |
| 1. Reflect the graph of the parent function f(x)=log b(x) about the x-axis. | 1. Reflect the graph of the parent function f(x)=log b(x) about the y-axis. |
| 1. Draw a smooth curve through the points. | 1. Draw a smooth curve through the points. |
| 1. State the domain, (0,∞), the range, (−∞,∞), and the vertical asymptote x=0. | 1. State the domain, (−∞,0), the range, (−∞,∞), and the vertical asymptote x=0. |
Example 4.4.8: Graphing a Reflection of a Logarithmic Function
Sketch a graph of f(x)=log(−x) alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
Solution
Before graphing f(x)=log(−x), f(x)=log(−x),identify the behavior and key points for the graph.
Since b=10 is greater than one, we know that the parent function is increasing. Since the input value is multiplied by −1, f(x) is a reflection of the parent graph about the y - axis. Thus, f(x)=log(−x) will be decreasing as x moves from negative infinity to zero, and the right tail of the graph will approach the vertical asymptote x=0.
The x-intercept is (−1,0).
We draw and label the asymptote, plot and label the points, and draw a smooth curve through the points.
Figure 4.4.19
The domain is (−∞,0), the range is (−∞,∞), and the vertical asymptote is x=0.
Exercise 4.4.8
Graph f(x)=−log(−x). State the domain, range, and asymptote.
Answer
Figure 4.4.20
The domain is (−∞,0),the range is (−∞,∞),and the vertical asymptote is x=0.
Given a logarithmic equation, use a graphing calculator to approximate solutions.
Press [Y=]. Enter the given logarithm equation or equations as Y 1= and, if needed, Y 2=.
Press [GRAPH] to observe the graphs of the curves and use [WINDOW] to find an appropriate view of the graphs, including their point(s) of intersection.
To find the value of x, we compute the point of intersection. Press [2ND]then [CALC]. Select “intersect” and press [ENTER] three times. The point of intersection gives the value of x,for the point(s) of intersection.
Example 4.4.9: Approximating the Solution of a Logarithmic Equation
Solve 4ln(x)+1=−2ln(x−1) graphically. Round to the nearest thousandth.
Solution
Press [Y=] and enter 4ln(x)+1 next to Y 1=. Then enter −2ln(x−1) next to Y 2=. For a window, use the values 0 to 5 for x\0 and(–10 to 10 for y. Press [GRAPH]. The graphs should intersect somewhere a little to right of x=1.
For a better approximation, press [2ND]then [CALC]. Select [5: intersect] and press [ENTER] three times. The x-coordinate of the point of intersection is displayed as 1.3385297. (Your answer may be different if you use a different window or use a different value for Guess?) So, to the nearest thousandth, x≈1.339.
Exercise 4.4.9
Solve 5log(x+2)=4−log(x) graphically. Round to the nearest thousandth.
Answer
x≈3.049
Summarizing Translations of the Logarithmic Function
Now that we have worked with each type of translation for the logarithmic function, we can summarize each in Table 4.4.4 to arrive at the general equation for translating exponential functions.
1 and a compression is 0<|a|<1. Note its form is f(x)=alog_b(x). The third translation is a reflection about the x-axis with the form f(x) = -log_b(x). The fourth translation is a reflection about the y-axis with the form f(x)=log_b(-x). The general equation for all translations is f(x)=alog_b(x+c)+d.">
Table 4.4.4| Translations of the Parent Function y=log b(x) |
| Translation | Form |
| Shift Horizontally c units to the left Vertically d units up | y=log b(x+c)+d |
| Stretch and Compress Stretch if |a|>1 Compression if |a|<1 | y=alog b(x) |
| Reflect about the x-axis | y=−log b(x) |
| Reflect about the y-axis | y=log b(−x) |
| General equation for all translations | y=alog b(x+c)+d |
TRANSLATIONS OF LOGARITHMIC FUNCTIONS
All translations of the parent logarithmic function, y=log b(x), have the form
f(x)=alog b(x+c)+d
where the parent function, y=log b(x), b>1,is
shifted vertically up d units.
shifted horizontally to the left c units.
stretched vertically by a factor of |a| if |a|>0.
compressed vertically by a factor of |a| if 0<|a|<1.
reflected about the x- axis when a<0.
For f(x)=log(−x), the graph of the parent function is reflected about the y-axis.
Example 4.4.10: Finding the Vertical Asymptote of a Logarithm Graph
What is the vertical asymptote of f(x)=−2log 3(x+4)+5?
Solution
The vertical asymptote is at x=−4.
Analysis
The coefficient, the base, and the upward translation do not affect the asymptote. The shift of the curve 4 units to the left shifts the vertical asymptote to x=−4.
Exercise 4.4.10
What is the vertical asymptote of f(x)=3+ln(x−1)?
Answer
x=1
Example 4.4.11: Finding the Equation from a Graph
Find a possible equation for the common logarithmic function graphed in Figure 4.4.21.
Figure 4.4.21
Solution
This graph has a vertical asymptote at x=–2 and has been vertically reflected. We do not know yet the vertical shift or the vertical stretch. We know so far that the equation will have form:
f(x)=−alog(x+2)+k
It appears the graph passes through the points (–1,1) and (2,–1). Substituting (–1,1),
1=−alog(−1+2)+k Substitute(−1,1)1=−alog(1)+k Arithmetic 1=klog(1)=0
Next, substituting in (2,–1),
−1=−alog(2+2)+1 Substitute(2,−1)−2=−alog(4)Arithmetic a=2 log(4)Solve for a
This gives us the equation f(x)=–2 log(4)log(x+2)+1.
Analysis
We can verify this answer by comparing the function values in Table 4.4.5 with the points on the graph in Figure 4.4.21.
Table 4.4.5| x | −1 | 0 | 1 | 2 | 3 |
| f(x) | 1 | 0 | −0.58496 | −1 | −1.3219 |
| x | 4 | 5 | 6 | 7 | 8 |
| f(x) | −1.5850 | −1.8074 | −2 | −2.1699 | −2.3219 |
Exercise 4.4.11
Give the equation of the natural logarithm graphed in Figure 4.4.22.
Figure 4.4.22
Answer
f(x)=2ln(x+3)−1
Media: Is it possible to tell the domain and range and describe the end behavior of a function just by looking at the graph?
Yes, if we know the function is a general logarithmic function. For example, look at the graph in Figure 4.4.22. The graph approaches x=−3 (or thereabouts) more and more closely, so x=−3 is, or is very close to, the vertical asymptote. It approaches from the right, so the domain is all points to the right, x|x>−3. The range, as with all general logarithmic functions, is all real numbers. And we can see the end behavior because the graph goes down as it goes left and up as it goes right. The end behavior is that as x→−3+, f(x)→−∞ and as x→∞, f(x)→∞.
Media
Access these online resources for additional instruction and practice with graphing logarithms.
Graph an Exponential Function and Logarithmic Function
Match Graphs with Exponential and Logarithmic Functions
Find the Domain of Logarithmic Functions
Key Equations
General Form for the Translation of the Parent Logarithmic Function f(x)=log b(x)f(x)=alog b(x+c)+d
Key Concepts
To find the domain of a logarithmic function, set up an inequality showing the argument greater than zero, and solve for x. See Example 4.4.1 and Example 4.4.2
The graph of the parent function f(x)=log b(x) has an x- intercept at (1,0),domain (0,∞),range (−∞,∞),vertical asymptote x=0,and
if b>1,the function is increasing.
if 0<b<1, the function is decreasing.
See Example 4.4.3.
The equation f(x)=log b(x+c) shifts the parent function y=log b(x) horizontally
left c units if c>0.
right c units if c<0.
See Example 4.4.4.
The equation f(x)=log b(x)+d shifts the parent function y=log b(x) vertically
up d units if d>0.
down d units if d<0.
See Example 4.4.5.
For any constant a>0, the equation f(x)=alog b(x)
stretches the parent function y=log b(x) vertically by a factor of a if |a|>1.
compresses the parent function y=log b(x) vertically by a factor of a if |a|<1.
See Example 4.4.6 and Example 4.4.7.
When the parent function y=log b(x) is multiplied by −1, the result is a reflection about the x-axis. When the input is multiplied by −1, the result is a reflection about the y-axis.
The equation f(x)=−log b(x) represents a reflection of the parent function about the x- axis.
The equation f(x)=log b(−x) represents a reflection of the parent function about the y- axis.
See Example 4.4.8.
A graphing calculator may be used to approximate solutions to some logarithmic equations See Example 4.4.9.
All translations of the logarithmic function can be summarized by the general equation f(x)=alog b(x+c)+d. See Table 4.4.4.
Given an equation with the general form f(x)=alog b(x+c)+d,we can identify the vertical asymptote x=−c for the transformation. See Example 4.4.10.
Using the general equation f(x)=alog b(x+c)+d, we can write the equation of a logarithmic function given its graph. See Example 4.4.11.
This page titled 4.4: Graphs of Logarithmic Functions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.
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4.5: Logarithmic Properties
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1387 | https://math.stackexchange.com/questions/1971660/how-to-use-the-binomial-theorem-to-calculate-binomials-with-a-negative-exponent | Stack Exchange Network
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How to use the binomial theorem to calculate binomials with a negative exponent
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I'm having some trouble with expanding a binomial when it is a fraction. eg $(a+b)^{-n}$ where $n$ is a positive integer and $a$ and $b$ are real numbers.
I've looked at several other answers on this site and around the rest of the web, but I can't get it to make sense. From what I could figure out from the Wikipedia page on the subject, $(a+b)^n$ where $n>0$
Any help will be much appreciated.
binomial-theorem
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edited Oct 16, 2016 at 21:29
Michael Hardy
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asked Oct 16, 2016 at 21:22
Wilhelm ErasmusWilhelm Erasmus
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$\begingroup$ Use the generalized binomial theorem. This is equivalent to the Taylor series expansion, which might be a way forward that is more familiar to you. $\endgroup$
Mark Viola
– Mark Viola
2016-10-16 21:26:57 +00:00
Commented Oct 16, 2016 at 21:26
$\begingroup$ @Dr.MV Would it be something like ${\frac {1}{(1-x)^{s}}}=\sum _{k=0}^{\infty }{s+k-1 \choose k}x^{k}\equiv \sum _{k=0}^{\infty }{s+k-1 \choose s-1}x^{k}.$ then? In that case, how would one calculate the other term, e.g. the 1 in this case. Obviously that 1 won't make a difference, but say it was another variable. What would it be called? The series is infinite, so there would be no way to subtract it from the total as per usual. $\endgroup$
Wilhelm Erasmus
– Wilhelm Erasmus
2016-10-16 21:39:08 +00:00
Commented Oct 16, 2016 at 21:39
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For $\lvert x\rvert<1$ and a real number $\alpha$, you can write $(1+x)^{\alpha}$ as the convergent series $$(1+x)^{\alpha}=\sum_{k=0}^\infty \binom{\alpha}{k} x^k$$
Were $\binom\alpha k=\frac{\alpha(\alpha-1)(\alpha-2)\cdots (\alpha-k+1)}{k!}$. For instance \begin{align}\binom{1/2}{4}&=\frac{\frac12\cdot\left(-\frac12\right)\cdot\left(-\frac32\right)\cdot\left(-\frac52\right)}{24}=-\frac{15}{16\cdot24}\\binom{-n}{k}&=\frac{-n\cdot(-n-1)\cdots(-n-k+1)}{k!}=(-1)^k\frac{n(n+1)\cdots(n+k-1)}{k!}=\&=(-1)^k\frac{(n+k-1)!}{(n-1)!k!}=(-1)^k\binom{n+k-1}{k}\end{align}
Now, to use this effectively when $(a+b)^{-n}$, you need at least one between $a$ and $b$ not to be zero and $\lvert a\rvert\ne\lvert b\rvert$. Then, pick the one with the highest absolute value and factor it out. If it is $a$, this means writing all as
$$a^{-n}\left(1+\frac ba\right)^{-n}=a^{-n}\sum_{k=0}^\infty \binom{-n}{k}\frac{b^k}{a^k}=\sum_{k=0}^\infty \binom{n+k-1}{k}(-b)^ka^{-k-n}$$
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edited Oct 16, 2016 at 21:46
answered Oct 16, 2016 at 21:39
user228113user228113
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$\begingroup$ Thinking about it, $\lvert a\rvert \ne\lvert b\rvert$ already implies that at least one of them is non-zero, but better be safe than sorry... $\endgroup$
user228113
– user228113
2016-10-16 21:43:27 +00:00
Commented Oct 16, 2016 at 21:43
$\begingroup$ Thanks a lot :) It makes a lot more sense now. I didn't consider making the coefficient of "a" 1 $\endgroup$
Wilhelm Erasmus
– Wilhelm Erasmus
2016-10-16 21:44:55 +00:00
Commented Oct 16, 2016 at 21:44
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Note that $(a+b)^{-n} = \frac{1}{(a+b)^n}$.
Now, apply $(a+b)^n = \sum_{i=0}^n \binom{n}{i} a^i b^{n-i}$ to calculate the denominator.
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answered Oct 16, 2016 at 21:26
BatmanBatman
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$\begingroup$ is $\dfrac 1 {\text{something}}$ an "expansion"? $\qquad$ $\endgroup$
Michael Hardy
– Michael Hardy
2016-10-16 21:30:26 +00:00
Commented Oct 16, 2016 at 21:30
$\begingroup$ @MichaelHardy Depends how you look at it. :) $\endgroup$
Simply Beautiful Art
– Simply Beautiful Art
2016-10-17 00:45:59 +00:00
Commented Oct 17, 2016 at 0:45
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You will always get a series, and for my own part, I find the story easiest to understand when $a=1$. Then, the standard formula applies: $$ (1+b)^t=1+tb+\frac{t(t-1)}2b^2+\sum_{k=3}^\infty\binom tkb^k\,, $$ where $\binom tk=t(t-1)(t-2)\cdots (t-k+1)\big/k!$ . This is the Generalized Binomial Theorem mentioned in the comment of @Dr.MV, and its often proved in second-term Calculus. The formula is good and convergent whenever $|b|<1$, no matter whether $t$ is a negative integer or a rational number, or even a real number. When $a\ne1$, youll be mentioning $a^t$ everywhere. Note that this series expansion does not treat $a$ and $b$ symmetrically.
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answered Oct 16, 2016 at 21:41
LubinLubin
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When $n$ is a negative integer and so is $k$, then we can write \begin{align} \binom n k & = \frac{n!}{(n-k)! k!} \tag 1 \[10pt] & = \frac{n(n-1)(n-2)(n-3)\cdots(n-k+1)}{k!}. \tag 2 \end{align} Note that although line $(1)$ assumes $n$ is a nonnegative integer, line $(2)$ does not. Line $(2)$ makes sense if $n$ is negative or is a non-integer.
Now suppose among $a$ and $b$ the one with the smaller absolute value is $b$, i.e. we have $0 \le |b| < |a|.$ That implies $\left|\dfrac b a\right|<1,$ and we will need that to assure convergence of a series we will see below.
Now suppose we want to expand $(a+b)^m$ as a sum of powers of $a$ and powers of $b$, and $m$ is not necessarily positive and not necessarily an integer. Then $$ (a+b)^m = a^m \left( 1 + \frac b a \right)^m = a^m \sum_{k=0}^\infty \binom m k \left(\frac b a \right)^k = \sum_{k=0}^\infty \binom m k a^{m-k} b^k. $$ This is "Newton's binomial theorem."
Note that
All of the powers to which $b$ is raised are nonnegative integers, whereas those of $a$ need not be;
If $m$ happens to be a nonnegative integer then all of the terms in which $k>m$ are $0$, since $\dbinom m k=0$ in that case.
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answered Oct 16, 2016 at 21:42
Michael HardyMichael Hardy
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1388 | https://web.mit.edu/wwmath/calculus/differentiation/notation.html | The Notation of Differentiation
Suggested Prerequesites: The definition of the derivative
Often the most confusing thing for a student introduced to differentiation is the notation associated with it. Here an attempt will be made to introduce as many types of notation as possible.
A derivative is always the derivative of a function with respect to a variable. When we write the definition of the derivative as
we mean the derivative of the function f(x) with respect to the variable x.
One type of notation for derivatives is sometimes called prime notation. The function f´(x), which would be read f-prime of x'', means the derivative of f(x) with respect to x. If we say y = f(x), then y´ (ready-prime'') = f´(x). This is even sometimes taken as far as to write things such as, for y = x4 + 3x (for example), y´ = (x4 + 3x)´. Higher order derivatives in prime notation are represented by increasing the number of primes. For example, the second derivative of y with respect to x would be written as
Beyond the second or third derivative, all those primes get messy, so often the order of the derivative is instead writen as a roman superscript in parenthesis, so that the ninth derivative of f(x) with respect to x is written as f(9)(x) or f(ix)(x).
A second type of notation for derivatives is sometimes called operator notation. The operator Dx is applied to a function in order to perform differentiation. Then, the derivative of f(x) = y with respect to x can be written as Dxy (read D -- sub -- x of y'') or as Dxf(x (readD -- sub x -- of -- f(x)''). Higher order derivatives are written by adding a superscript to Dx, so that, for example the third derivative of y = (x2+sin(x)) with respect to x would be written as
Another commonly used notation was developed by Leibnitz and is accordingly called Leibnitz notation. With this notation, if y = f(x), then the derivative of y with respect to x can be written as
(his is read as dy -- dx'', but notdy minus dx'' or sometimes ``dy over dx''). Since y = f(x), we can also write
This notation suggests that perhaps derivatives can be treated like fractions, which is true in limited ways in some circumstances. (For example with the chain rule.) This is also called differential notation, where dy and dx are differentials. This notation becomes very useful when dealing with differential equations.
A variation of Leibnitz's differential notation is written instead as
which resembles the above operator notation, with (d/dx as the operator).
Higher order derivatives using leibnitz notation can be written as
The exponents may seem to be in strange places in the second form, but it makes sense if you look at the first form.
So, those are the most commonly used notations for differentiation. It's possible that there exist other, obscure notations used by a some, but these obscure forms won't be included here. It's helpful to be familiar with the different notations. When is a function differentiable?
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watko@mit.edu Last updated August 24, 1998 <\body> |
1389 | https://www.youtube.com/watch?v=QoUPXE6Ep9E | Group Factoring with Splitting and Product Sum Method Factor Trinomials IB SL Math
Anil Kumar
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How to Factor Monomial:
Factor Trinomials:
Learn From Anil Kumar: anil.anilkhandelwal@gmail.com
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Transcript:
[Music] [Music] I am Anil Kumar welcome to my series on factoring trinomials all those who want to take up IB or AP level of courses in high school for them this series on factoring trinomials will be very beneficial in this particular video we'll discuss how do we group Factor the trinomial of the form ax² + BX + C where a is greater than one we are going to find two magic numbers which can help us factor using product and sum method and then we'll split the center term and then finally Factor the trinomial let us Master this with nine examples just before you let me also thank all the viewers and subscribers who are watching my videos posting excellent comments and are also supporting us every dollar you support with helps us to reach Millions worldwide as you know for last 15 years we have been serving you absolutely free with 177,000 videos and these are watched over 80 countries your support can help now let's also look into some examples which we have done in past you'll find hundreds of solved examples on factoring with four different techniques so so these are some practice questions with their answers you can always copy the troms and practice after this video Let's now begin with the solution of the very first so we'll talk about concept before getting into details of solving I hope that will help our new viewers we need to factor 12 x² - 2 x - 4 as you can see all even numbers you can always factor out two which we call a group factoring so first thing you should notice can you group factor or not very few questions will be there where group factoring will be involved here you have it right so you just factor out two and then I prefer to use square brackets when you do group Factor dividing each number the co coefficients by 2 so you get 6 x² - x - 2 in the square bracket so the trinomial in the square bracket now is to be factored further and for that we're going to use our strategy of product and sum first we need to find two numbers M and N so that their product and sum can match some way or the other right so let's see what is the product product when we talk about is the product of the leading coefficient and the constant term which in this case is 12 and minus 4 but you know we factored and therefore we'll only consider between the product of 6 and minus 2 do you see that so this factoring helps bringing the numbers smaller 6 -2 is -12 correct so now we looking for two numbers M and N whose sum is the middle term which is minus1 you see that so we are looking for two numbers whose product M n is -12 and whose sum m + n is minus1 which is the coefficient of the center term which is x in this case coefficient of x is minus1 so as you can see what are the two numbers well minus number will be greater we'll always take M as a bigger number minus one has to be negative one has to be positive since the product is negative right and then smaller number will be positive 4 and 3 - 4 + three is the right combination okay so we can use minus 4 and 3 so what we now do is we split the center term and that is where the splitting comes right so first you group factor and then you split split the middle term right so we'll split the middle term and then rewrite the expression as 6 x² the first term but the middle term will be written as - 4x + 3x see - 4x + 3x is - x correct makes sense and then we have min-2 there correct at this stage we have split the center terms so that we have now four terms and now you can group two of each right first two and the last two and then again Factor so once you do this factoring what do you get we get this as two outside 6 X2 - 4x you see 2x is common and you're left with x - 2 plus 3x - 2 sorry 6 / 2 is 3 so you left with 3x - 2 in 3x - 2 1 is common there is no need to write one here we'll write 3x - 2 in the bracket and now you can see that 3x - 2 is common and we get 2x + 1 as the other Factor so we have factored the trinomial perfect makes sense right so that is how we are going to use this strategy so you can now clearly see how we have done and let's now move forward with the rest eight questions we have nine in all so this time we have 4x2 + 14x + 6 I like you to pause the video answer the question those of you who want to learn directly from me send an email on global maath Institute at gmail.com now as you can observe we have some common factors here also right so we have 4x² + 14x + 6 again two is common so we can take two outside we're left with 2X + 7 x + 3 now what is the product well the product is 2 3 which is six and the sum is seven and everything is positive in this case so how do you get sum of seven well 6 1 right 6 1 so m is 6 for you n is 1 you get the idea so we can split the center term which is 7 x into 2 which is 6X + x and that makes it 7x makes sense right and now we can common factor from the first two terms 2x is common so we left with x + 3 we already have x + 3 as the other two terms x + 3 right so now you can take x + 3 as a common factor and you're left with 2x + 1 makes sense right you can drop the square brackets last time I did not use the square brackets in the last statement right but you may or may not use it another important thing is keep on writing equal to sign right that also helps sometimes that the expression is equal to the previous expression so I hope now the steps are coming together right so you know factoring as such but we have added a component which is group factoring and then splitting well you can watch other videos in this playlist where we are discussing the Australian method and any other methods four in all to factor trinomials here is the next question for you so we have 3x Cub right so first copy the question and then Factor it 3x Cub - 3x2 - 60x clearly three is common right so you're left with even X is common so not only the numbers the variables could also be common so 3 x is common makes sense so we're left with X2 - x and here we are left with minus dividing by 3 gives me 20 now here we have leading coefficient of one and the product should be Min - 20 sum should be min-1 bigger number is negative smaller number is positive and the numbers are 5 4 so - 5 + 4 makes sense right so we left with x² we'll write this Asus 5x + 4x - 20 right so 3x in the first two terms X is common we left with x - 5 and now four is common we again get x - 5 it is a must that these things should be common otherwise you cannot Factor further so if you don't get this common at this stage it means some mistake has happened or the trinomial cannot be factored further doesn't make sense to you right okay great so equal to sign I'm just placing them all it's a good practice right so let me write this once again x - 5 is common factor we left with x + 4 correct now you can drop the square brackets square brackets was for better communication right now you may or may not use it perfect so we have factored it finally so I hope by now you have learned the method completely No Doubt right feel free to write a comment share your views and continue with this video solve these questions yourself 16 X2 + 17 x + 90 how will you do it well let's copy the question first 16 x² + 76 x + 90 okay now what is common uh can you divide it by two yes you can but not by four because 90 can be divided by four right uh how about there's no other number but so two you could divide with right so we write this as 8 x² + half of this so 2 3 38 right + 45 now in this case we land up with very big numbers we're looking for a product which is 8 45 positive and 38 at has the sum so what could be these numbers so when you land up in such situation and you have to work it out without calculator uh I think the best way is not really to multiply but to figure out what numbers to work with so split 8 into its factors which is 2 2 2 and 45 is 9 5 right so you can say 3 3 5 correct so what I've done here is it into prime factors and now we know when we multiply them we will get the required product Hub we are looking for two numbers if you add them then you get 38 so we want a sum of 38 so what is that combination which is going to give us sum of 38 right so we see that 3 3 is 9 9 2 is 18 do you see this and 5 4 is 20 that will give a sum of 38 you see that so now we can split it into those two magic numbers which are clearly for US 20 and 18 do you see that so we will split this as 20x + 18x do you see how we got this combination it was very easy since we split it into its prime factors and now the small numbers to work with is easier than a very big number to work with right so let me rewrite all of the terms right so 8 x² bracket 2 outside do you see how beautifully we did it it's a very good example and the tip for you is that prime factorization helps right now in the first two terms you see 4 x is common right so let's with 2x + 5 and here you see 9 is common and and so you're left with 2x + 5 so you can write down 2x + 5 as a common factor you're left with 4x + 9 and final answer so this is a very good test question I hope it makes sense feel free to write a comment share your views in case you want to learn from me you can always send an email on global math Institute gmail.com our students right there on the top we can be part of your success story very important great so we are always um you know Midway here already uh let's look forward for more difficult challenging questions 6 x² - 90x + 324 slightly bigger numbers for you to work at this stage once again I'll request you to pause the video and then look into uh do yourself the solution and then look into my solution is it clear okay so this question seems to be very difficult we are numbers with 6 x² - 90x right and then we have + 324 what could be a common factor well six can be a common factor right so let's take the common factor six and we left with x² dividing 90 by 6 what do you get you get 15 right 15x and this number 6 T 5 is 30 and 4 so we get x² - 15x + 54 now in this case we're looking for a product of 54 and sum of minus5 what could be these numbers right so uh 6 and 9 right so 54 you can split into two numbers both negative right so - 9 and - 6 correct when you add them you get -5 when you multiply them you get 54 clear so we're going to split this into - 9x and - 6X right and keep the other terms as such so we have now x² - 9x - 6X + 54 just to make you clear that we're looking into group with splitting strategies and then following the four steps to factor the trinomial correct so now we can Rewrite by group factoring between x² and - 9 x x is common we left with x - 9 and here 6 is common we're left with x - 9 let's put Min - 6 common right and minus and plus will make it minus and so we have x - 9 as a common factor with x - 6 as the other Factor so I dropped the square brackets and wrote the answer directly perfect so that is how you're going to do it I hope it is making sense to you so I hope the concept is absolutely clear let's move on and take up the next question it is again with a big numbers and we have X Cube this time correct can you please pause the video and answer this question and then look into my suggestions in case you want to learn from me feel free to send an email on global maath Institute at gmail.com now first thing is we see that 2x Cub 21x 2 - 36x what is common here well factor out x correct X is common so let's do that part it's a cubic equation so once we take out x what do we get we get a quadratic equation inside the square brackets correct okay so we get this quadratic equation now here what is the sum and what is the product well the product is -72 Right which is minus 36 2 and the sum is + 21 doesn't make sense to you that means we're looking for two numbers one of them is positive the other one is negative positive number is bigger how do we get these numbers that is the next challenge for us let's look into this so the prime factorization technique which we learned should help right so 36 could be written as 6 6 right but 6 can be further factored as 3 2 3 2 right so now we have smaller numbers to work with we need a combination in which we get 21 one of the number which is bigger is positive the smaller number is negative how can we get that but if you look carefully what we could do is we could take one number as -3 and and the other as the combination of these right so we get 2 3 6 12 24 24 - 3 is indeed 21 positive right so we'll split this into two which is + 24 x and - 3x doesn't make sense to you you see the technique right and now we copied the other terms as such very very important step now this is where you know this factoring prime factorization all comes together perfect now we'll group Factor next step great so 2x² 2x is common we left with x + 12 and here Min - 3 is common we again left with x + 12 x + 12 is common right so I'm dropping the square brackets writing x + 12 directly and we get 2x - 3 doesn't make sense to you that is how we're going to factor very good example I hope you have understood it right so another three questions to go 4X CU + 6 X2 + 2x is the next one copy it first 4X Cub + 6 X2 + 2x so what is common well 2x is common so we get 2x² + 3x uh plus one right because we took 2x common we're looking for a product which is number 2 1 which is two and the sum is + three right so the two numbers are + 2 and + 1 clear so we could now write within the square brackets 2x² + 2x + x for 3x correct now group Factor the first two terms where 2 x is common you get x + 1 the other two x + 1 put them in Brackets and now write down x + 1 as a common factor left with 2x + 1 to finish it off is that clear to you next example 8 x² let's copy the question as such which is 8 X2 - 34x + 36 all are even numbers so two is common right right not four because 34 cannot be divided by 4 so 2 is common we left with 4x² - 17 x + 18 right so we looking for a big product this time once again which is 4 18 both positive right and the number is 17 negative that means M and N both should be negative right and we need a sum of -7 how do you get this so once again we will fall back upon prime factorization technique uh 4 can be written as 2 2 18 is 9 2 so 2 3 3 correct makes sense you need the sum of two numbers to be Min -7 one is even the other one is odd correct that also helps to figure out perfect so what could be these two numbers both negative right so we could have eight and nine that can work right because we figured out that one has to be odd the other one has to be even right so so that works and we'll split this into - 8 and - 9 so this time I wrote Min - 8X first well 2x is common between 4x² and 8x we are left with Sorry 4 x is common okay 4X is common so we get x - 2 here - 9 is common we get x - 2 and clearly we have x - 2 as a common factor left with 4x - 9 Mak sense so that is how we have done this and here we have learned another beautiful technique and that is if you're looking for an odd number as an addition that means the numbers are even plus odd right and that helped us to quickly get the two magic numbers m&n right everything counts in a test paper in case you want to learn from me feel free to send an email on global maath Institute gmail.com we can be part of your success story here is the last question perfect take it as a test problem i' like you to pause the video answer this right so we given the question 8 x Cub + 2x² - 15x X is common correct so we left with 8 x² + 2x - 15 the product is minus because 15 8 let me write - 15 8 and the sum is just two so number which is right in between very close right so let's look into possible numbers - 15 8 8 could be written as 3 tws right and 15 is 5 3 correct now what two numbers one positive one negative positive is bigger negative is smaller cor but the difference is only of two okay so we could combine 3 4 is 12 and 5 2 is 10 makes sense right so we'll write this as 8 x² + 12 x - 10 x - 15 make sense right so now what is common between 8 x² and 12 x 4x is common right 4X x + 3 and here we could take five common so 2x + 3 right so because I could see 2x + 3 here that means I missed something right that can happen so you have to be very careful so you can take 2X + 3 as a common factor and you get 4x - 5 as the other one and now you get your answer x 2x + 3 4x - 5 makes sense so that is how we are going to do it so I hope the concept is absolutely clear in case you want to support your own channel we are actually solving your questions almost daily and I hope it is making sense feel free to send your donations every dollar count by email Global maath Institute at gmail.com so we are at the end of this particular video I hope you have understood all these questions my request is try some more on your own right that will definitely help thanks for your time and all the best for |
1390 | https://www.askiitians.com/forums/11-grade-biology-others/the-cloaca-in-a-frog-is-a-common-chamber-for-the-u_465139.htm | The cloaca in a frog is a common chamber for the urinary tract, repro - askIITians
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11 grade biology others>The cloaca in a frog is a common chamber ...
The cloaca in a frog is a common chamber for the urinary tract, reproductive tract and A. Alimentary canal B. Respiratory systemC. Hepatic portal vesselsD. Notochord E. Lymphatic system
Aniket Singh , 6 Months ago
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1 Answers
Askiitians Tutor Team
Last Activity: 6 Months ago
The correct answer is A. Alimentary canal.
The cloaca is a common chamber in many animals, including frogs, where the urinary tract, reproductive tract, and alimentary canal (digestive system) all open. It is a multipurpose chamber that serves different bodily functions.
Functions and structure of the cloaca:
Alimentary canal: The alimentary canal is the digestive tract, where food is processed, nutrients are absorbed, and waste is excreted. In frogs, the undigested food and waste materials from digestion are expelled through the cloaca.
Urinary tract: The cloaca also serves as an opening for the excretory system, where urine from the kidneys is excreted. The waste products from the kidneys pass into the cloaca before being excreted outside the body.
Reproductive tract: In frogs, the cloaca is also involved in reproduction. The male and female frogs release their reproductive cells (sperm and eggs) into the cloaca, which then passes them to the external environment for fertilization.
Thus, the cloaca in frogs serves as a common chamber for all three of these systems — the urinary tract, reproductive tract, and alimentary canal.
Why the other options are incorrect:
• B) Respiratory system: The respiratory system (gills in tadpoles and lungs in adult frogs) does not empty into the cloaca. Instead, gas exchange occurs through the skin and lungs.
• C) Hepatic portal vessels: The hepatic portal vessels are part of the circulatory system that transports blood from the digestive organs to the liver. They do not interact with the cloaca.
• D) Notochord: The notochord is a cartilaginous rod found in the embryonic stage of all vertebrates and does not pass through the cloaca.
• E) Lymphatic system: The lymphatic system, which involves the circulation of lymph, does not have any direct connection to the cloaca.
Therefore, the correct answer is A. Alimentary canal, as the cloaca is a common chamber for the urinary tract, reproductive tract, and alimentary canal in frogs.
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1391 | https://www.sciencedirect.com/science/article/abs/pii/S0960977607002792 | Skip to article
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Abstract
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References (19)
Cited by (23)
The Breast
Volume 17, Issue 3, June 2008, Pages 309-313
Original article Microdochectomy for single-duct pathologic nipple discharge and normal or benign imaging and cytology
Author links open overlay panel, , , , ,
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Abstract
This study evaluates microdochectomy as a means for diagnosis and treatment of patients with pathological nipple discharge (PND) but with benign or normal imaging and cytology.
From 1999 until 2006, in St. Mary's Hospital, 76 patients with the aforementioned condition underwent microdochectomy because of the presence of epithelial cells on nipple smear or for symptomatic relief. Most of the patients had intraductal papillomas (48.7%), duct ectasia (15.8%) or a combination of the two (13.2%). Other benign causes occurred in 11.8% of the patients. Eight patients, including one who was operated for symptomatic relief, had cancer. Of those patients with benign condition, 98% had symptomatic relief while PND recurred twice in one patient.
Pre-operative workup and imaging may not be suspicious in patients with single-duct PND and underlying malignancy, therefore, microdochectomy should be considered in such cases.
Introduction
Nipple discharge is the third most common breast complaint after breast lumps and pain1, 2, 3, 4 and is the second most common indication for a breast operation. The incidence is 512%2, 3, 4, 5, 6, 7 and usually the underlying cause is benign. Pathological nipple discharge (PND) is defined as spontaneous, persistent, unilateral and coming from a single duct during a non-lactational period.1, 2, 3, 4, 5, 6 The most common causes of PND are either the presence of a benign intraductal papilloma (IDP) or mammary duct ectasia (MDE).1 Nevertheless, in a small percentage of patients, the cause of nipple discharge is an underlying malignancy (1015%) usually DCIS.1, 8
Some patients with PND have either suspicious cytology or suspicious findings on mammography and therefore, the decision to proceed to microdochectomy or other diagnostic procedures is straight-forward. Precise preoperative diagnosis, however, is impossible in most cases. Microdochectomy is used for diagnosis and symptomatic relief of women with PND. The fact that the majority of patients have benign pathology, encourages a more conservative approach with annual mammography and repeated smear cytology in those patients who initially have normal or benign findings. In this study, we have sought to clarify the rationale of our approach of performing diagnostic microdochectomy on all patients with normal or benign epithelial cells in their nipple smears. We have also included in the study those patients who underwent microdochectomy for symptomatic relief despite having acellular cytological smears.
Section snippets
Patients and methods
We reviewed the clinical and operating case notes, the imaging, as well as cytology and histology results of the patients who underwent microdochectomy for diagnostic purposes or symptomatic relief at St. Mary's Hospital between the years 1999 and 2006. Of these patients 76 had normal or benign imaging and preoperative cytology, their presenting complaint was PND and these were enrolled in the study. Twelve of these patients underwent microdochectomy for symptomatic relief despite acellular
Results
The mean age of the patients was 48.86 years (range 2275 years). Thirty-seven patients had IDPs (48.7%), 12 had duct ectasia (15.8%) and 10 had both (13.2%). Other benign causes occurred in nine (11.8%) patients.
With a mean follow-up of 42.76 months (698) among the patients with benign condition, there was one recurrence of PND. This patient recurred twice and underwent further surgery. There were no major complications recorded. The mean operation time was 33 min (2050) and all patients were
Discussion
PND is usually caused by a benign IDP (48%)1, 4, 6, 7 while MDE is the second most common cause accounting for 1520% of the cases.1 Malignancy is associated with this condition in 1015%.1, 6, 8 The nature of the secretion can be bloody, serous, serosanguinous or watery and it is not significant for the differential diagnosis, i.e. cancer can present with any colour of discharge1, 2, 3, 4, 5, 9 but is more frequently associated with bloody, serosanguinous and serous secretions.2, 9, 10 Our
Conclusion
Pre-operative workup and imaging may not be suspicious in patients with PND and underlying malignancy. Microdochectomy is a safe and accurate approach for patients with PND and therefore, it should be offered as an option to all patients with this condition irrespective of age or colour of the discharge.
Conflict of interest
We declare there is not any financial or personal relationship with other people or organizations that could inappropriately influence this work.
Acknowledgements
We would like to thank the St. Mary's NHS IT department and theatre archives for providing us the data. We would also like to thank the pathology department and secretary for providing us the list of the patients underwent microdochectomy and the histology and cytology reports for these patients. We would also like to thank the radiology archive for providing us information and hard copies of the images for these patients. Finally, we would like to thank Mrs. C.L. Hadjiminas for providing us
References (19)
N. Cabioglu et al.
### Surgical decision making and factors determining a diagnosis of breast carcinoma in women presenting with nipple discharge
### J Am Coll Surg
(2003)
N. Sharma et al.
### Intraoperative intraductal injection of methylene blue dye to assist in major duct excision
### Am J Surg
(2006)
L.D. Louie et al.
### Identification of breast cancer in patients with pathologic nipple discharge: does ductoscopy predict malignancy?
### Am J Surg
(2006)
K.E. West et al.
### Correlation of nipple aspiration and ductal lavage cytology with histopathologic findings for patients before scheduled breast biopsy examination
### Am J Surg
(2006)
S. Masood et al.
### Nipple fluid cytology
### Clin Lab Med
(2005)
A.N. Hussain et al.
### Evaluating nipple discharge
### Obstet Gynecol Surv
(2006)
M.G. Florio et al.
### Surgical approach to nipple discharge: a ten-year experience
### J Surg Oncol
(1999)
S. Lau et al.
### Pathologic nipple discharge: surgery is imperative in postmenopausal women
### Ann Surg Oncol
(2005)
W. Al Sarakbi et al.
### Breast papillomas: current management with a focus on a new diagnostic and therapeutic modality
### Int Semin Surg Oncol
(2006)
There are more references available in the full text version of this article.
Cited by (23)
Nipple discharge and the efficacy of duct cytology in evaluating breast cancer risk
2010, Surgeon
Citation Excerpt :
Microdochectomy is the conventional surgical technique aiding definitive diagnosis and symptomatic relief of patients with pathological nipple discharge. It is less invasive than the radical subareolar (Hadfields) duct excision with lower morbidity rates.19,20 Ductography and fiberoptic ductoscopy allow pre-operative determination of the number, location and extent of underlying lesions.
Nipple discharge accounts for up to 5% of referrals to breast surgical services. With the vast majority of breast carcinomas originating in the ductal system, symptomatic dysfunction of this system often raises disproportionate clinical concern. The aim of this study is firstly, to evaluate the clinical importance of nipple discharge as an indicator of underlying malignancy and secondly, to assess the diagnostic application of duct cytology in patients presenting with nipple discharge.
We performed a retrospective analysis of all patients presenting with nipple discharge as their primary symptom to the symptomatic breast unit at a tertiary referral center over a 30-month period (n = 313). The Hospital Inpatient Enquiry (HIPE) System and BreastHealth database were used to identify our study cohort. Parameters evaluated included patient demographics, clinical presentation, clinical evaluation, radiological assessment and histological/cytological analysis.
Three-hundred and thirteen patients presented with nipple discharge as their primary complaint. Invasive breast carcinoma was diagnosed by Triple Assessment in 5% of patients. 24% of patients presenting with nipple discharge underwent nipple aspiration and cytological analysis. Duct cytology was diagnostic of the underlying breast carcinoma in 50% of triple assessment diagnosed carcinoma. Four risk factors were identified as having a significant association with breast carcinoma, these included (a) age >50 years (p < 0.0001), (b) bloody nipple discharge (p < 0.008), (c) presence of a breast lump (p < 0.0001) and (d) single duct discharge (p < 0.006).
Nipple discharge is a poor indicator of an underlying malignancy. Use of nipple aspiration and duct cytology for the assessment of nipple discharge is of limited diagnostic benefit. However, by utilizing the systematic, gold standard approach of Triple Assessment (clinical, radiological and cytological evaluation), the risk of underlying carcinoma can be accurately defined.
### Can we see what is invisible? The role of MRI in the evaluation and management of patients with pathological nipple discharge
2019, Breast Cancer Research and Treatment
### Ductoscopic detection of intraductal lesions in cases of pathologic nipple discharge in comparison with standard diagnostics: The german multicenter study
2014, Oncology Research and Treatment
### Bloody nipple discharge is a predictor of breast cancer risk: A meta-analysis
2012, Breast Cancer Research and Treatment
### Nipple Discharge Screening
2010, Women S Health
### Selective microdochectomy after ductoscopic wire marking in women with pathological nipple discharge
2009, BMC Cancer
View all citing articles on Scopus
View full text
Crown copyright © 2007 Published by Elsevier Ltd. All rights reserved. |
1392 | https://www.youtube.com/watch?v=5L2a6cuP_4s | Negative bases and exponents classification example | Algebra I | Khan Academy
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Transcript:
We're asked to
categorize the value of each exponential expression. And we categorize
each expression into either the bucket
less than negative 1, between negative 1 and
0, and between 0 and 1. And I encourage you
to pause this video and try this out on your own. So I assume you've
given a go at it. Now let's work
through it together. So I just copy and pasted
the exact same exercise right over here. And let's think about
each of these expressions. And actually, before
doing that, I actually want to draw a number line to
think about these categories. So it looks like
the important things to consider-- you have a 0. You have 1. And then you have negative 1. This first category right over
here, less than negative 1, that's all of these
values right over there. Then you have the category
between negative 1 and 0. That's these values
right over there. And then finally, you have
the category between 0 and 1. That's this category
right over here. So they didn't give us a
bucket for greater than 1, so I guess it's safe to
assume that none of these are going to be greater than 1. But let's go each expression
by each expression. So this first expression, 7
to the negative fourth power, we've already talked
about several times, that when you raise something
to a negative exponent, this isn't somehow negative 7
to the fourth power or anything like that. When you see a negative
number in an exponent, you should immediately
think reciprocal. So this is going to
be equal to 1 over 7 to the positive fourth power. And you don't even have to
know what 7 to the fourth actually is. It's 2,401, if
you're interested. But you don't even
have to know that. You just know that
this right over here is going to be a large
positive number, so 1 over that is going to be a very
small positive number. It's going to be
between 0 and 1. This is actually one 1/2,401,
but you don't even have to know that. This is a large
number, and so you say, well, this is going
to get pretty close. It's positive, so it's
going to be above 0. But it's not going to
be too much above 0. It's going to sit on the
number line-- actually, I'm not even doing
justice right over there. The way I drew it, at
scale, it's not even a pixel to the right
of 0 right over there. So this is going to
be between 0 and 1. And actually, I'll
draw lines like this, to show roughly on our number
line, where it would sit. Now let's think about
negative 7 to the third power. So negative 7 to the
third power, well, this is the same thing
as negative 7 times negative 7 times negative 7. And you could also think
about this as negative 1 times 7 times negative 1 times
7 times negative 1 times 7, which is the
same thing as-- you have three negative 1's
being multiplied together-- so negative 1 to the third power
times 7 to the third power. Well, what's negative
1 to the third power? Well, it's negative
1 times negative 1, which would be positive 1, but
then times another negative 1. So this is negative
1 right over here. So this is going to be equal
to-- let me write it this way. You could write it either
as-- I could write it like this-- negative
7 to the third power. So 7 to the third,
once again, we don't even have to
know what it is. This is going to be a
large positive value. But then we're negating it. If we had 7 to the third,
it would be greater than 1. But negative 7 to the
third is definitely less than negative 1. So this fits into that
bucket right over there. Now we have negative 7
to the negative fourth. So let me write it over
here-- so negative 7 to the negative fourth. So this could be
a little daunting because you have a negative base
raised to a negative exponent. But whenever I see
a negative exponent, the first thing I worry about or
the first thing my brain wants to do is take a reciprocal. So this is going to be 1 over
negative 7 to the fourth power. And once again, the same
idea as we just saw over here, we could view this
as 1 over negative 1 to the fourth power times 7
to the fourth power, which is going to be equal to--
negative 1 to the fourth power, this is just going
to be equal to 1. So this is going to be equal to
1 over 7 to the fourth power. Well, we've already
thought about what 1 over 7 to the fourth power is. It is between 0 and 1. So this right over here
is between 0 and 1. Another way you
can think about it, this is going to be 1
over negative 7 times negative 7 times negative
7 times negative 7. You have a negative times a
negative times a negative times a negative. Well, that's going
to be a positive. So it's positive 7
to the fourth power. Now let's think about this one
right over here-- negative 7 to the negative fifth power. So let me write this
out-- negative 7 to the negative fifth power. Well once again, as soon as
you see that negative exponent, I want to just take
the reciprocal, so I don't have a
negative exponent anymore. So it's 1 over negative
7 to the fifth power. And we could rewrite
this as 1 over negative 1 to the fifth times 7
to the fifth, which is equal to-- well, what's a
negative 1 to the fifth power? Well a negative 1
raised to an odd power is going to be a negative 1. So this is going to be negative
1 over 7 to the fifth power. So where is 1 over 7
to the fifth power? Well, that would be
really close to 0. This would be positive. 1 over 7 to the fifth would be
a very small positive number. But now we're taking
the negative of that. So we're going to jump on
to the other side of 0. So this is going to be
between negative 1 and 0. And then finally, negative
7 to the fifth power, well that's just going
to be-- this, once again, you could view that
as negative 1 times 7 to the fifth power, which is
the same thing as negative 1 to the fifth times 7
to the fifth, which is the same thing as
negative 7 to the fifth. 7 to the fifth is a very
large number, larger than 1. So the negative of that is going
to be a very negative number. It's going to be
less than negative 1. So this is going to be less
than negative 1 right over here. So now let's take our results. And since I have
a bad memory, I'm essentially probably going to
have to think through it again. And let's actually put
them in the buckets. 7 to the negative
fourth power-- this is not going to be
a negative value. This is 1 over 7 to the fourth. This is going to
be between 0 and 1. Negative 7 to the third
power-- that's essentially, if you take 7 to
the third power, you get a very large
positive number. But then, since that's
going to be times negative 1 to the third power,
it's going to be the negative version of that. So that's going to be
less than negative 1. Negative 7 to the
negative fourth-- well, we already determined,
especially because this ends up being 1 over negative
7 to the fourth power and that's an even exponent,
it ends up being the same as 7 to the negative fourth. And then you had negative 7
to the negative fifth power. So once again, that was
going to be 1 over negative 7 to the fifth power. And so we found that to be
between negative 1 and 0. And then finally,
this right over here, it's going to be negative. And it's going to be a
very negative number. It's less than negative 1. And so those are our choices. And we can check our
answer, and we got it right. |
1393 | https://m.fx361.cc/news/2021/0318/10151336.html | 例谈立体几何中直线与平面平行的证明策略_参考网
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例谈立体几何中直线与平面平行的证明策略
2021-03-18 林巧红
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摘 要：每年的数学高考中，6道解答题大题，立体几何必是其中一道，立体几何大题通常会设置2问，其中第一小问多为证明题，证明平行或垂直关系，而平行证明题型主要考点是线面证明，因为在证明线面平行的过程中经常会涉及两个考点，其一是线线平行的证明，其二是面面平行的证明。本文借助泉州质检题中一道线面平行解答题的探讨，进行一题多解，一题多变，从变中总结解题方法，从变中发现解题规律，从变中发现不变，引导学生多思多想，养成在学中求异，学中求变的习惯，使学生学一道题，会一类题，加深对问题实质性的掌握，增强应变能力。
关键词：线面平行;线线平行;面面平行;几何法;向量法
一、问题提出
证明线面平行，最容易想到的是使用它的判定定理：如果平面外一条直线和平面内的一条直线平行，那么这条直线和这个平面平行。关键是如何在平面内找到一条与已知直线平行的直线。既然要证明直线a//平面α，那么直线a肯定是平行平面α的，若a//α，那么经过直线a的平面β，要么与平面α相交，要么与平面α平行。当时，依据线面平行的性质定理知;当时，依据平行平面的性质知。因此过已知直线作平面，既可以构造交线使线线平行也可构造面面平行从而证明线面平行。
证明线面平行，除了可利用上述提到的这两种几何构造法外，还可以利用空间向量的线性关系或数量积关系，通过代数运算来解决。下面一起探索空间线面平行关系证明中的常用策略。
二、例题示范
例（2020年泉州5月质检）如：图四棱锥P-ABCD的底面为菱形，∠BAD=1200，AB=2.平面PCD⊥平面ABCD，PC=PD，E，F分别是BC，PD的中点。
（1）求证：EF//平面PAB;
（2）若直线PB与平面ABCD所成的角为450，求直线DE与平面PBC所成角的正弦值。
本文只分析第（1）小题，第（2）小题略。
分析1：借助线面平行的判定定理构造相交平面找线线平行→线面平行
线线平行两个常见模型（构造平行四边形或三角形）
（一）平行四边形的对边互相平行（需要能证明平行四边的条件）
（二）三角形的中位线平行于第三边（需要构造三角形的两个中点）
策略一、构造平行四边形——线线平行→线面平行
取PA的中点M构造平行四边形EFMB，得EF//BM，又EF在平面PAB外，BM在平面PAB内，所以EF//平面PAB。
【点评】构造平行四边形需包含要证的直线并能够证明要证直线的一组邻边平行且相等。此题涉及中点E、F又涉及平行且相等线段AD、BC，符合条件构造平行四边形。
策略二、构造中位线——线线平行→线面平行
连接DE并延长交AB的延长线于点N，则点E是DN的中点，在ΔDPN中，EF是中位线，则EF//PN，又EF在平面PAB外，PN在平面PAB内，所以EF//平面PAB。
【点评】构造中位线时，需想办法将待证明线段及中点所在线段纳入同一个三角形，利用中位线的平行关系找到平面内的直线。此题平面PDE与平面PAB已有一个公共点P，由公理3“如果两个不重合的平面有一个公共点，那么它们有且只有一条过该点的公共直线”顺藤摸瓜找到公共直线PN。
策略三、构造平行平面——线线平行→线面平行→面面平行→线线平行
取AD中点Q，连接EQ，FQ，可证平面QEF//平面PAB，又EF在平面QEF内，所以EF//平面PAB。
【点评】在构造平行线和平行平面两个方法上，一般构造平行平面更容易些，思维量相对比较小，但书写过程中只能依据面面平行的判定定理：一个平面两条相交直线与另一个平面平行，则两平面平行。而学生易犯的错误是：直接由线线平行→面面平行。
分析2：向量是一个具有代数与几何双重属性的量，为我们用代数方法研究几何问题提供了强有力的工具，灵活使用直线的方向向量和平面的法向量来证明线面平行，有时也会起到意想不到的效果。
策略四、向量法——建系、点坐标、向量坐标、法向量、套公式、下结论
取CD的中点R，由PC=PD得PR⊥CD，又平面PCD⊥平面ABCD，平面PCD∩平面ABCD=CD
所以PR⊥平面ABCD，设AC∩BD=O，以OC，OD所在直线分别为x轴，y轴，过点O作直线与PR平行为z轴建立空间直角坐标系
平面PAB的一个法向量为
又EF在平面PAB外，所以EF//平面PAB。
【点评】由于几何体结构的特殊性，且明确了线段的长度，为我们建立空间直角坐标系提供了方便。利用向量数量积运算解决线面平行问题，思路简单，程式化的流程，降低了立体几何的空间难度，给学生一个比较低的门槛。
策略五、基底法——以为基底：
又EF在平面PAB外，所以EF//平面PAB
【点评】只要能在几何体中寻找到合适的基向量，就可以很容易地说明向量共面。而且，有時可以省略求相关坐标的过程，使证明过程更加简洁。对于不方便建立坐标系的几何体，这种方法就更具有优越性。
变式题1：四棱锥P-ABCD的底面为菱形，∠BAD=1200，AB=2.平面PCD⊥平面ABCD，PC=PD，E是BC的中点，试问在棱PD上是否存在点F使EF//平面PAB？
思路一、先猜后证，如何猜才能准确呢？回顾前面种种策略，构造平行平面。
（一）从中点E出发，取线段AD的中点Q，则EQ//AB;
（二）从中点Q出发，取线段PD的中点F，则QF//PA;
可得平面QEF，再证平面QEF//平面PAB，又EF在平面QEF内，所以EF//平面PAB。
【点评】构造平行平面法的基本步骤：①选点——在所证直线上任取一点记作A;②选线——在所证平面内任取一线记作a;③优化组合——使点A作所选线a的平行线，交棱于一点，记作B;再由点B出发，选线，重复上述过程。
思路二、向量法，建系与前面例题中的策略四一样
可得平面PAB的一个法向量为
点F在PD上，一般设点F的向量坐标式：即设
所以
【点评】向量法的基本步骤：先假设存在——建系设点、列式解方程，利用方程是否有解来判断所探索的问题是否有解。其中设探究中的点是关键：
1.如果点F在AB上有两种设法：
①坐标式：直接设点F（x，y，z），利用已知条件寻找x，y，z的关系;
②向量式：假设存在常数λ，使得，然后用λ表示点F的坐标
2.如果点F在平面ABC上，一般存在常数λ，μ，使得（当然也可选用其他基底向量），然后用λ，μ来表示点F坐标。
变式题2：如图，四棱锥P-ABCD的底面为菱形，∠BAD=1200，AB=2.平面PCD⊥平面ABCD，PC=PD，点E是BC的中点，点M在四棱锥P-ABCD内部或表面上，且EM//平面PAB，则动点M的轨迹。
思路分析：已知动直线平行定平面，寻找定平面包含动直线，从而面面平行得线面平行。
回想变式题1中的构造平行平面法
（1）从中点E出发，取线段AD的中点Q，则EQ//AB;
（2）从中点Q出发，取线段PD的中点F，则QF//PA;
可得平面QEF，再证平面QEF//平面PAB，若EM在平面QEF内，则EM//平面PAB。
所以点M的轨迹是三角形QEF内部以及它的边界。
【点评】觅得规律明轨迹——动中有静，动态立体几何体在变化过程中总蕴含着某些不变的因素，用定平面包含动直线，通过作截面寻找动点M的轨迹，从而找到解决问题的突破口。
变式题3、如图，四棱锥P-ABCD的底面为菱形，∠BAD=1200，AB=2.平面PCD⊥平面ABCD，PC=PD，E，Q分别是BC，AD的中点，点F是直线PD上的一动点，求证：AB//平面QEF。
思路分析：证明定直线与动平面平行，几何法？向量法？一般会在动平面内找一定直线与已知的定直线平行。由中点E、Q易想到AB//EQ，由线面平行的判定定理可知AB//平面QEF。
【点评】在解决立体几何中的“动态”问题时，探寻变化过程中的不变关系，是解决动态问题的常用手段。利用动平面内一定直线与已知定直线平行的不变性证得线面平行。
三、提炼思想方法
证明直线与平面平行的基本思路：
思路一、证线面平行，由线面平行的判定定理可知需先找线线平行，要找线线平行，往往要找面面交线，要找面面交线，需构造辅助线（构造平行四边形，或三角形中位线，或等比例线段等加以证明）。
思路二、證明线面平行，也可先找面面平行再由面面平行的性质得以证明。
思路三、证明线面平行，还可利用空间向量的线性关系或数量积关系，通过代数运算加以证明。而空间向量中最主要的两个手段就是选基底与建立空间直角坐标系。
总的来说证明线面平行有两种方法：几何法与向量法。几何法是一种定性的研究方法，表现为识图与作图（常添加辅助线与面），利用所学的定义、公理、定理去分析，推理论证。从而能培养我们的空间想象能力，逻辑思维能力，在探索性问题或动态平行问题中更能显示它的威力。而向量法是一种定量的研究方法，把形的问题转化为数的计算，体现了几何解题的一种通用方法。在高考中，特别是解答题中，试题一般采取几何法和向量法的“双轨制”，不管选用哪种方法，各有千秋，各有优缺点，只要做出来，书写规范，结果正确，哪种方法都可以，都是满分。
参考文献
鲍启静.线面平行之常见题型[N].中学生数理化.2008（2）.
向鸿.空间向量在立体几何中的运用[J]凯里学院学报，2008年6月.
作者简介：林巧红，（1980—），女，福建晋江人，汉，晋江养正中学，中学一级教师，大学本科。高中数学教学。
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1395 | https://www.geeksforgeeks.org/engineering-mathematics/walks-trails-paths-cycles-and-circuits-in-graph/ | Walks, Trails, Paths, Cycles and Circuits in Graph
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Walks, trails, paths, cycles, and circuits in a graph are sequences of vertices and edges with different properties. Some allow repetition of vertices and edges, while others do not. In this article, we will explore these concepts with examples.
What is Walk?
A walk in a graph is a sequence of vertices and edges where both edges and vertices can be repeated. The length of the walk refers to the number of edges covered in the sequence. A graph can contain multiple walks.
There are two key points to note about a walk:
Edges can be repeated.
Vertices can be repeated.
Here, 1-> 2-> 3-> 4-> 2-> 1-> 3 is a walk.
Walk can be open or closed.
Types of Walks
There are two types of walks:
Open Walk
Closed Walk
Open Walk:
An open walk is a walk in which the starting and ending vertices are different. In other words, for a walk to be considered open, the origin and terminal vertices must not be the same. Additionally, the length of the walk must be greater than 0.
Closed Walk:
A closed walk occurs when the starting and ending vertices are identical, meaning the walk starts and ends at the same vertex. For a walk to be classified as closed, the origin and terminal vertices must be the same. Similar to an open walk, the length of the walk must be greater than 0.
In the above diagram:
1-> 2-> 3-> 4-> 5-> 3 is an open walk.
1-> 2-> 3-> 4-> 5-> 3-> 1 is a closed walk.
What is Trail?
A trail is an open walk in which no edge is repeated, though vertices may be repeated.
There are two types of trails:
Open Trail: A trail is considered an open trail if the starting and ending vertices are different.
Closed Trail: A trail is a closed trail if the starting and ending vertices are the same.
In a trail, the key point to remember is that: Edges cannot be repeated, but vertices can be repeated.
Here 1-> 3-> 8-> 6-> 3-> 2 is trail
Also 1-> 3-> 8-> 6-> 3-> 2-> 1 will be a closed trail
What is Circuit?
A circuit can be described as a closed trail in graph theory, where no edge is repeated, but vertices can be repeated.
Thus, the key characteristics of a circuit are:
Edges cannot be repeated.
Vertices can be repeated.
In other words, a circuit is a closed traversal of a graph where each edge is used exactly once, but a vertex may appear more than once.
Here 1-> 2-> 4-> 3-> 6-> 8-> 3-> 1 is a circuit.
Circuit is a closed trail.
What is Path?
A path is a trail in which neither vertices nor edges are repeated.
In other words, when traversing a graph along a path, each vertex and each edge is visited exactly once. Since a path is also a trail, it is inherently an open walk unless stated otherwise.
Another definition of a path is a walk with no repeated vertices. This automatically implies that no edges are repeated, making it unnecessary to explicitly mention edge repetition in the definition.
Key characteristics of a path:
Vertices are not repeated.
Edges are not repeated.
Here 6->8->3->1->2->4 is a Path
What is Cycle?
A cycle in graph is a closed path, meaning that it starts and ends at the same vertex while ensuring that no other vertices or edges are repeated.
In other words, a cycle is formed by traversing a graph such that: No vertex is repeated, except for the starting and ending vertex, which must be the same and No edge is repeated.
Key characteristics of a cycle:
Edges cannot be repeated.
Vertices cannot be repeated, except for the first and last vertex, which must be the same.
Here 1->2->4->3->1 is a cycle.
Cycle is a closed path.
Table for Walk, Trail and Path
The table below represents the repetition of edges and vertices in walk, trail and path.
| Category | Edges | Vertices |
---
| Walk | Can be repeated | Can be repeated |
| Trail | Cannot be repeated | Can be repeated |
| Path | Cannot be repeated | Cannot be repeated |
Solved Examples on Walks, Trails, Paths, Cycles and Circuits in Graph
1. Find a trail from A to D for below graph
A - B
| |
C - D
One possible trail is: A → B → D → C → A → B → D.
2. Find the shortest path from A to C.
A - B - C
| |
D - E
The shortest path is: A → B → C.
3. Finding All Possible Paths
A - B - C
| |
D - E
Following are paths
A → B → C
A → E → C
A → B → E → C
Practice Problems - Walks, Trails, Paths, Cycles and Circuits in Graph
Graph for Reference:
A
/ | \
B C D
\ | /
E
|
F
Find a path from A to F.
Find a simple path from B to F.
Find a trail that uses every edge exactly once.
Identify a cycle in the graph.
Find the shortest path from A to E.
List all possible paths from E to F.
Find a trail from C to F.
Find the longest simple path in the graph.
Count the number of distinct cycles in the graph.
Find a path from A to F that consists of exactly 3 edges.
Related Articles:
Euler and Hamiltonian Paths | Engineering Mathematics
Eulerian path and circuit for undirected graph
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Exercise 1
Calculate the mean for the following set of numbers: 5, 3, 6, 5, 4, 5, 2, 8, 6, 5, 4, 8, 3, 4, 5, 4, 8, 2, 5, 4.
Exercise 2
Find the mean for the following set of numbers:
3, 5, 2, 6, 5, 9, 5, 2, 8, 6.
Exercise 3
Find the mean for the following series:
3, 5, 2, 7, 6, 4, 9.
3, 5, 2, 7, 6, 4, 9, 1.
Exercise 4
Given the statistical distribution of the table.
| | | | | | |
--- --- --- |
| xi | 61 | 64 | 67 | 70 | 73 |
| fi | 5 | 18 | 42 | 27 | 8 |
Calculate the mean.
Exercise 5
A statistical distribution is given by the following table:
| | | | | | |
--- --- --- |
| | [10, 15) | [15, 20) | [20, 25) | [25, 30) | [30, 35) |
| fi | 3 | 5 | 7 | 4 | 2 |
Calculate the mean.
Exercise 6
Given the statistical distribution:
| | | | | | | |
--- --- ---
| | [0, 5) | [5, 10) | [10, 15) | [15, 20) | [20, 25) | [25, ∞) |
| fi | 3 | 5 | 7 | 8 | 2 | 6 |
Calculate the mean.
Exercise 1
Calculate the mean for the following set of numbers: 5, 3, 6, 5, 4, 5, 2, 8, 6, 5, 4, 8, 3, 4, 5, 4, 8, 2, 5, 4.
Exercise 2
Find the mean for the following set of numbers:
3, 5, 2, 6, 5, 9, 5, 2, 8, 6.
Exercise 3
Find the mean for the following series:
3, 5, 2, 7, 6, 4, 9.
3, 5, 2, 7, 6, 4, 9, 1.
Exercise 4
Given the statistical distribution of the table.
| | | | | | |
--- --- --- |
| xi | 61 | 64 | 67 | 70 | 73 |
| fi | 5 | 18 | 42 | 27 | 8 |
Calculate the mean.
Exercise 5
A statistical distribution is given by the following table:
| | | | | | |
--- --- --- |
| | [10, 15) | [15, 20) | [20, 25) | [25, 30) | [30, 35) |
| fi | 3 | 5 | 7 | 4 | 2 |
Calculate the mean.
Exercise 6
Given the statistical distribution:
| | | | | | | |
--- --- ---
| | [0, 5) | [5, 10) | [10, 15) | [15, 20) | [20, 25) | [25, ∞) |
| fi | 3 | 5 | 7 | 8 | 2 | 6 |
Calculate the mean.
Exercise 2
Find the mean for the following set of numbers:
3, 5, 2, 6, 5, 9, 5, 2, 8, 6.
Exercise 3
Find the mean for the following series:
3, 5, 2, 7, 6, 4, 9.
3, 5, 2, 7, 6, 4, 9, 1.
Exercise 4
Given the statistical distribution of the table.
| | | | | | |
--- --- --- |
| xi | 61 | 64 | 67 | 70 | 73 |
| fi | 5 | 18 | 42 | 27 | 8 |
Calculate the mean.
Exercise 5
A statistical distribution is given by the following table:
| | | | | | |
--- --- --- |
| | [10, 15) | [15, 20) | [20, 25) | [25, 30) | [30, 35) |
| fi | 3 | 5 | 7 | 4 | 2 |
Calculate the mean.
Exercise 6
Given the statistical distribution:
| | | | | | | |
--- --- ---
| | [0, 5) | [5, 10) | [10, 15) | [15, 20) | [20, 25) | [25, ∞) |
| fi | 3 | 5 | 7 | 8 | 2 | 6 |
Calculate the mean.
Exercise 3
Find the mean for the following series:
3, 5, 2, 7, 6, 4, 9.
3, 5, 2, 7, 6, 4, 9, 1.
Exercise 4
Given the statistical distribution of the table.
| | | | | | |
--- --- --- |
| xi | 61 | 64 | 67 | 70 | 73 |
| fi | 5 | 18 | 42 | 27 | 8 |
Calculate the mean.
Exercise 5
A statistical distribution is given by the following table:
| | | | | | |
--- --- --- |
| | [10, 15) | [15, 20) | [20, 25) | [25, 30) | [30, 35) |
| fi | 3 | 5 | 7 | 4 | 2 |
Calculate the mean.
Exercise 6
Given the statistical distribution:
| | | | | | | |
--- --- ---
| | [0, 5) | [5, 10) | [10, 15) | [15, 20) | [20, 25) | [25, ∞) |
| fi | 3 | 5 | 7 | 8 | 2 | 6 |
Calculate the mean.
Exercise 4
Given the statistical distribution of the table.
| | | | | | |
--- --- --- |
| xi | 61 | 64 | 67 | 70 | 73 |
| fi | 5 | 18 | 42 | 27 | 8 |
Calculate the mean.
Exercise 5
A statistical distribution is given by the following table:
| | | | | | |
--- --- --- |
| | [10, 15) | [15, 20) | [20, 25) | [25, 30) | [30, 35) |
| fi | 3 | 5 | 7 | 4 | 2 |
Calculate the mean.
Exercise 6
Given the statistical distribution:
| | | | | | | |
--- --- ---
| | [0, 5) | [5, 10) | [10, 15) | [15, 20) | [20, 25) | [25, ∞) |
| fi | 3 | 5 | 7 | 8 | 2 | 6 |
Calculate the mean.
Exercise 5
A statistical distribution is given by the following table:
| | | | | | |
--- --- --- |
| | [10, 15) | [15, 20) | [20, 25) | [25, 30) | [30, 35) |
| fi | 3 | 5 | 7 | 4 | 2 |
Calculate the mean.
Exercise 6
Given the statistical distribution:
| | | | | | | |
--- --- ---
| | [0, 5) | [5, 10) | [10, 15) | [15, 20) | [20, 25) | [25, ∞) |
| fi | 3 | 5 | 7 | 8 | 2 | 6 |
Calculate the mean.
Exercise 6
Given the statistical distribution:
| | | | | | | |
--- --- ---
| | [0, 5) | [5, 10) | [10, 15) | [15, 20) | [20, 25) | [25, ∞) |
| fi | 3 | 5 | 7 | 8 | 2 | 6 |
Calculate the mean.
Solution of exercise 1
Calculate the mean for the following set of numbers: 5, 3, 6, 5, 4, 5, 2, 8, 6, 5, 4, 8, 3, 4, 5, 4, 8, 2, 5, 4.
| | | |
---
| xi | fi | xi · fi |
| 2 | 2 | 4 |
| 3 | 2 | 6 |
| 4 | 5 | 20 |
| 5 | 6 | 30 |
| 6 | 2 | 12 |
| 8 | 3 | 24 |
| | 20 | 96 |
Mean
Solution of exercise 2
Find the mean for the following set of numbers:
3, 5, 2, 6, 5, 9, 5, 2, 8, 6.
2, 2, 3, 5, 5, 5, 6, 6, 8, 9.
Mean
Solution of exercise 3
Find the mean for the series:
3, 5, 2, 7, 6, 4, 9.
Mean
3, 5, 2, 7, 6, 4, 9, 1.
Mean
Solution of exercise 4
Given the statistical distribution of the table.
| | | | | | |
--- --- --- |
| xi | 61 | 64 | 67 | 70 | 73 |
| fi | 5 | 18 | 42 | 27 | 8 |
Calculate the mean.
| | | |
---
| xi | fi | xi · fi |
| 61 | 5 | 305 |
| 64 | 18 | 1152 |
| 67 | 42 | 2814 |
| 71 | 27 | 1890 |
| 73 | 8 | 584 |
| | 100 | 6745 |
Mean
Solution of exercise 5
A statistical distribution is given by the following table:
| | | | | | |
--- --- --- |
| | [10, 15) | [15, 20) | [20, 25) | [25, 30) | [30, 35) |
| fi | 3 | 5 | 7 | 4 | 2 |
Calculate the mean:
| | | | |
--- --- |
| | xi | fi | xi · fi |
| [10, 15) | 12.5 | 3 | 37.5 |
| [15, 20) | 17.5 | 5 | 87.5 |
| [20, 25) | 22.5 | 7 | 157.5 |
| [25, 30) | 27.5 | 4 | 110 |
| [30, 35) | 32.5 | 2 | 65 |
| | | 21 | 457.5 |
Mean
Solution of exercise 6
Given the statistical distribution:
| | | | | | | |
--- --- ---
| | [0, 5) | [5, 10) | [10, 15) | [15, 20) | [20, 25) | [25, ∞) |
| fi | 3 | 5 | 7 | 8 | 2 | 6 |
Calculate the mean
| | | | |
--- --- |
| | xi | fi | xi · fi |
| [10, 15) | 12.5 | 3 | 37.5 |
| [15, 20) | 17.5 | 5 | 87.5 |
| [20, 25) | 22.5 | 7 | 157.5 |
| [25, 30) | 27.5 | 4 | 110 |
| [30, 35) | 32.5 | 2 | 65 |
| | | 21 | 457.5 |
Mean
The mean cannot be calcuated, because the class mark of the last interval cannot be found.
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Emma
I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.
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Iam unable to understand
I specified that I wanted 7th-grader stuff but these are not even in my whole textbook
Someone relook at the examples. What z table are you referring to? Once you solve for the Z the rest of the answer is gibberish. Maybe sh o w the table, label the axes and red line the values you call out so the student can figure out what you are trying to convey.
Using facebook account,conduct a survey on the number of sport related activities your friends are involvedin.construct a probability distribution andbcompute the mean variance and standard deviation.indicate the number of your friends you surveyed
this page has a lot of advantage, those student who are going to be statitian
I’m a junior high school,500 students were randomly selected.240 liked ice cream,200 liked milk tea and 180 liked both ice cream and milktea
A box of Ping pong balls has many different colors in it. There is a 22% chance of getting a blue colored ball. What is the probability that exactly 6 balls are blue out of 15?
Where is the answer??
A box of Ping pong balls has many different colors in it. There is a 22% chance of getting a blue colored ball. What is the probability that exactly 6 balls are blue out of 15?
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The Lord of the Rings Wiki
7,045
pages
Contents
1 Biography
1.1 Childhood
1.2 Coming of age & flight from the Shire
1.2.1 To Bree
1.2.2 Strider
1.2.3 Weathertop
1.3 Quest of the Ring
1.3.1 Council of Elrond
1.3.2 Moria
1.3.3 Lothlórien
1.3.4 The Breaking of the Fellowship
1.3.5 Emyn Muil
1.3.6 The Dead Marshes
1.3.7 Ithilien
1.3.8 Minas Morgul and Shelob's Lair
1.3.9 Cirith Ungol
1.3.10 Mordor and Mount Doom
1.4 End of the War
1.4.1 The Scouring of the Shire
1.5 Close of the Third Age
1.6 Fourth Age
2 Characteristics
3 Attire
4 In other versions
5 In adaptations
5.1 Ralph Bakshi's The Lord of the Rings
5.2 Rankin/Bass
5.3 The Lord of the Rings film trilogy
5.4 Video games
5.5 Radio
5.6 Parodies
5.7 Voice dubbing actors
6 Gallery
7 Appearances
7.1 Books
7.2 Films
8 Trivia
9 Translations
10 References
in:
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Bearers of the One Ring
Elf friends
Fellowship members
The Hobbit: An Unexpected Journey characters
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Ring bearers
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Frodo Baggins
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: "I will take the Ring, though I do not know the way."
: —Frodo Baggins, at the Council of Elrond, in The Fellowship of the Ring
Frodo Baggins was a hobbit of the Shire in the late Third Age. He was a key figure in the Quest of the Ring, in which he bore the One Ring to Mount Doom, where it was destroyed. He was a Ring-bearer, best friend to his gardener, Samwise Gamgee, and was one of three hobbits who sailed from Middle-earth to the Uttermost West at the end of the Third Age.
Contents
1 Biography
1.1 Childhood
1.2 Coming of age & flight from the Shire
1.2.1 To Bree
1.2.2 Strider
1.2.3 Weathertop
1.3 Quest of the Ring
1.3.1 Council of Elrond
1.3.2 Moria
1.3.3 Lothlórien
1.3.4 The Breaking of the Fellowship
1.3.5 Emyn Muil
1.3.6 The Dead Marshes
1.3.7 Ithilien
1.3.8 Minas Morgul and Shelob's Lair
1.3.9 Cirith Ungol
1.3.10 Mordor and Mount Doom
1.4 End of the War
1.4.1 The Scouring of the Shire
1.5 Close of the Third Age
1.6 Fourth Age
2 Characteristics
3 Attire
4 In other versions
5 In adaptations
5.1 Ralph Bakshi's The Lord of the Rings
5.2 Rankin/Bass
5.3 The Lord of the Rings film trilogy
5.4 Video games
5.5 Radio
5.6 Parodies
5.7 Voice dubbing actors
6 Gallery
7 Appearances
7.1 Books
7.2 Films
8 Trivia
9 Translations
10 References
Biography
Childhood
Much of Frodo's youth was spent at Brandy Hall in Buckland, the ancestral home of the Brandybuck family, including his mother (Primula Brandybuck). Frodo was known as something of a rascal, befriending Meriadoc (Merry) Brandybuck and Peregrin (Pippin) Took and causing trouble wherever they went. They would often steal mushrooms from Farmer Maggot's farm Bamfurlong.
In TA 2980, when Frodo was only 12 years old, his parents drowned in a boating accident on the Brandywine River. An only child, Frodo stayed in Brandy Hall until his 99-year-old "uncle" Bilbo, his father's second cousin, adopted him in TA 2989. Bilbo took Frodo to live with him in his home at Bag End and made him his heir.
The two grew very close in the following years; Frodo, treated by his adopted guardian as he were his own son, learned much of the Elvish language during his time with Bilbo, as well as much of the lore of Middle-earth. They also had a weekly tradition of taking long walking trips together from Hobbiton to Michel Delving and Buckland; all throughout the Shire. The two shared the same birthday, September 22 by Shire Reckoning (around September 12–14 of our calendar), and a party of special magnificence was held when Frodo came of age of thirty-three and Bilbo achieved the notable age of 111.
Bilbo gave a memorable birthday speech before playing a joke on his fellow hobbits by using the One Ring to disappear, at which Gandalf quickly reacted and used his staff to create a blinding flash where Bilbo had been standing. The hobbits at the party were left confused and disgruntled, and Bilbo was never again seen in the Shire.
Before departing for his journey to Rivendell, Bilbo was persuaded by Gandalf to voluntarily surrender the One Ring. Bilbo left it on the fireplace mantel with a note for Frodo, who would now become the next Ring-bearer.
Coming of age & flight from the Shire
After the party finished, Frodo returned home and discovered that he was now the master of Bag End and the recipient of Bilbo's magic Ring. Gandalf, ever more curious about the Ring's origin, power, and purpose (but not yet positive it was the One Ring), advised the young hobbit against the using the Ring. For the next seventeen years, Frodo complied with the wizard's request and hid the Ring in a safe place. However, on April 12, 3018, Gandalf returned to Bag End and warned Frodo that the Ring was actually the One Ring, which the Dark Lord Sauron needed to rule over Middle-earth. Realizing that Sauron would be looking for the Ring, Gandalf advised the Hobbit to secretly follow Bilbo's journey to Rivendell.
After Frodo's discussion with Gandalf, a rumor started that he was running out of money. This rumor, though not begun by Frodo, was encouraged by him. Merry helped Frodo to purchase a small house at Crickhollow. With the exception of his gardener Sam Gamgee, who had agreed to accompany him to Rivendell, Frodo told the other Hobbits of the Shire that he intended to move to Buckland. He sold his home to the Sackville-Bagginses, and, on the September 23, 3018, the day after his fiftieth birthday, Frodo left from Bag End, taking with him Sam and Pippin. They left in the early morning for Bree, and just in time, as Sauron's most powerful servants, the nine Nazgûl, had entered the Shire dressed as Black Riders searching for a hobbit with the name of Baggins.
To Bree
Frodo was nearly captured by a Black Rider on the road, but was saved by Gildor Inglorion, whom he asked for advice. Leaving the roads to cut across country, Frodo and Sam reached Farmer Maggot's farm, who helped them to evade the Riders. Meeting Merry at Bucklebury Ferry, they saw a Rider tracking them from the bank they'd departed from. On arrival at Crickhollow, Frodo found that Merry and Pippin already knew about his "secret" journey. Frodo was left with no alternative but to bring the two youngsters with him. They cut through the Old Forest and the Barrow-downs in hopes of losing the Black Riders, which did succeed. They met other troubles in those places though, at the hands of Old Man Willow and the Barrow-wights, but were rescued twice by Tom Bombadil, a mysterious being who dwelt in a glade in the middle of the Old Forest.
Strider
In Bree, the hobbits stayed at The Prancing Pony, an old inn. Frodo went by the name of Mr Underhill, attempting to raise as little suspicion as possible. When he noticed a mysteriously cloaked Man sitting in the shadows and smoking a long-stemmed pipe, Frodo asked the innkeeper, Barliman Butterbur, who the man was. The innkeeper referred to the man, a Ranger, as Strider.
That night, Black Riders arrived in Bree and attacked the inn in search of Frodo and the One Ring, but Strider had managed to hide the Hobbits from them in time. Having gained their trust, Strider introduced himself as Aragorn to Frodo and the others, to whom he also revealed the backstory of the black riders, also called Nazgûl or Ringwraiths. With a pony named Bill that the Hobbits had acquired at Bree, Strider led Frodo and his companions into the Wild. Aragorn would be their guide to Rivendell, and he would lead them through the Midgewater Marshes and to the hill of Weathertop.
Weathertop
On the night of October the sixth, the Hobbits were attacked by five of the nine Ringwraiths at Weathertop. In the presence of the Nazgûl, Frodo made the mistake of putting on the Ring. He was able to resist their attempt to take him by drawing his sword and invoking the name of one of the Valar, Elbereth Gilthoniel. Unfortunately, the leader of the Nazgûl, the Witch-king of Angmar, stabbed Frodo in the shoulder (he would have stabbed his heart) with a Morgul-knife. If it had caught him in the heart, Frodo would have become like the Nazgûl, only weaker and under their control. The Ringwraiths were driven away by the appearance of Aragorn and his martial skill; also because he had torches, one of their few weaknesses.
Though Aragorn was a skilled healer, he could not heal Frodo's wound. A fragment of the Ringwraith's blade remained in Frodo's flesh, where it continued to move towards his heart. Near death (or worse), Frodo was rescued by Glorfindel, an Elf-lord, who put the injured Hobbit upon his horse Asfaloth. They were found and pursued by the Nazgûl, as Glorfindel rode and bore Frodo swiftly to the Ford of Bruinen, at the entrance to the valley of Rivendell. Once they had crossed the River Bruinen, the Nine Ringwraiths behind them demanded Frodo give up the Ring, but Frodo refused. Subsequently, the Ringwraiths entered the river and were washed away in a flood called up by Elrond.
Frodo was soon healed in Rivendell by Elrond, who knew the wound would not ultimately leave him, as it was both spiritual and physical. On the 24th of October 3018, Frodo awoke in Rivendell and was reunited with Bilbo, Gandalf, Aragorn, Sam, Merry, and Pippin. Although Elrond had healed his wound, it continued to ail him yearly for as long as he lived in Middle-earth.
Quest of the Ring
Council of Elrond
After his healing, Frodo was summoned to a great Council that Elrond had organized. Representatives of all the Free Peoples of Middle-earth discussed the history of the Rings of Power and decided that the One Ring must be destroyed. As the Ring was shown and tempers flared, argument broke out as to who should carry the Ring on this mission, until Frodo bravely volunteered to take the Ring to Mordor and cast it into the fires of Mount Doom. A member of each of the Free Peoples offered to join Frodo in his quest, thus forming the Fellowship of the Ring.
The Fellowship consisted of Frodo, Samwise, Merry, Pippin, Aragorn, Gandalf, Boromir of Gondor, Legolas of the Woodland Realm, and Gimli of the Lonely Mountain. Before leaving Rivendell, Bilbo gave Frodo his dwarf-made coat of mithril mail and his elven blade Sting. The mithril coat had been given to Bilbo by Thorin Oakenshield during the events of The Hobbit, and Sting had been taken by Bilbo from a troll den. On December 25, the Fellowship of the Ring departed from Rivendell and headed south.
Moria
After an attack by White Wolves in Eregion, on January 11, 3019, the Fellowship attempted to cross the Misty Mountains (specifically the Pass of Caradhras), but were unable to due to a snowstorm. They instead traveled through the underground city of Moria at the urging of Gimli.
Some days in into their dangerous trek through Moria, the Fellowship entered the Chamber of Mazarbul, and was attacked by Orcs and a cave-troll. Frodo helped to defeat the troll before he was stabbed by an Orc captain, his mithril-shirt saving him from a deadly blow. The Fellowship then ran from there to the Bridge of Khazad-dum, at which Gandalf dueled Durin's Bane and fell. Once outside Moria, while the Fellowship was grieving, Gimli took Frodo and Sam to look upon the Mirrormere, even in their great hurry.
Lothlórien
Deeply grieved by their loss, the Fellowship journeyed to the Elven kingdom of Lothlórien, where they met the Lady Galadriel and Lord Celeborn. Galadriel showed Frodo a vision of the future in her Mirror. Frodo offered her the One Ring, but she resisted the temptation to take it, passing the test that was laid before her, and accepting the diminishing of the power of the Elves. Before the Fellowship departed from Lothlórien, Galadriel gave each member a gift. To Frodo, she gave a phial with the light of the star Eärendil captured inside; this gift would prove hugely important later on in the quest. They were also provided with elven way-bread, other supplies, and boats for their voyage down the Anduin River.
The Breaking of the Fellowship
The Fellowship continued their journey south to Parth Galen. There, Boromir, a Man of Gondor and a member of the Fellowship, attempted to convince Frodo to bring the Ring to Minas Tirith and regroup from there. When the hobbit asked for an hour alone to consider his options, Boromir followed him. Seeing that Frodo did not intend to take the suggested course of action, Boromir tried to take the Ring from him by force.
Frodo put on the Ring and escaped to the Seat of Seeing, where he watched as war brewed across Middle-earth and the Eye of Sauron searching for him as he traveled. He was nearly spotted, but the Dark Lord's attention was drawn away by a resurrected Gandalf.
Taking off the Ring, he decided to take the item to Mordor alone, without telling the other members of the Fellowship. However, he was joined by his friend Samwise Gamgee, who felt it was necessary that he should protect and guide Frodo. Frodo gave in to Sam's protests, and although reluctant to lead anyone else to his fate, was glad to have Sam's company.
The two hobbits continued toward Mordor, dividing the Fellowship. Meanwhile, Boromir was killed by Uruk-archers while defending Merry and Pippin; the two young hobbits were then captured by Uruk-hai, and were to be taken to Isengard. Instead of following Samwise and Frodo to Mordor, the Three Hunters, Aragorn, Legolas and Gimli, decided it more important to rescue Merry and Pippin from their captors. The breaking of the Fellowship was now complete.
Emyn Muil
After leaving what remained of the Fellowship at Amon Hen, Frodo and Sam tried to navigate through the winding paths and razor sharp rocks of the Emyn Muil. After getting lost several times, they were found by Gollum, who at first tried to take the One Ring, but was captured by Sam (with Frodo's help) and tied up with the Elven rope. Frodo, now pitying the creature, decided not to slay Gollum, but forced him to swear an oath of servitude to the master of the precious. Gollum then led them out of the maze and into the Dead Marshes.
The Dead Marshes
The Dead Marshes lay to the south of Emyn Muil, and were just as disorienting, if not more so. There was thought to be no route through the marshes, as Orcs marched for miles around, although Gollum had secretly discovered a path when traveling. He led Frodo and Sam on a safe pathway through the marshes, warning them not to follow what seemed like small torches in the water.
Soon after Frodo, Samwise, and Gollum entered the Dead Marshes, a Nazgûl flew over them on a Fellbeast, looking for the Ring. Gollum had led the two hobbits to a safe place that they could hide in.
Ithilien
Gollum led the two Hobbits to the Black Gate of Mordor, as Frodo had desired, but stopped the Hobbits from passing its doors, as the danger was too great. He then explained about a secret way into Mordor, 'Up the stairs and through the tunnel'. The Hobbits once again found themselves being led by Gollum. After venturing into Ithilien, and witnessing a skirmish between a company of Haradrim warriors (along with Oliphaunts) and Rangers from Gondor, they were apprehended by the Ranger's captain, Faramir. When the skirmish had ended, Faramir blindfolded the ring-bearer and his companions and led them to Henneth Annûn, the Window on the West. Upon much interrogation, Sam foolishly misspoke, and gave away that Frodo was indeed carrying the One Ring. Realizing the importance of the quest, Faramir proved his quality, unlike his brother, Boromir, and let the Ring-bearer go free, resupplying the hobbits with food for their journey. Later, Gollum was captured in the Forbidden Pool and forcibly taken into the hidden lair. Frodo begged for his safety, and he was not killed, although the rift between master and servant had once again begun to open.
Minas Morgul and Shelob's Lair
Gollum led the hobbits past the lair of the Witch-king of Angmar, Minas Morgul, and up the Stairs of Cirith Ungol into 'The Tunnel'. When they arrived at the top though, they were abandoned by Gollum, causing them to realize that Gollum had betrayed them. They cautiously traveled through the tunnel, and managed to get to the end only to find their way barred by Shelob's great web. Whilst attempting to cut through the webbing, Frodo bravely stood up to Shelob and forced her back further into the tunnels giving him and Sam time enough to hack through the threads and escape. Upon escaping the tunnels, Frodo thought himself safe; however, Shelob, through one of her many tunnels, managed to sneak out and jab him with her stinger. As he was being encased in Shelob's webbing, Sam was able to draw her into single combat wherein he, using Sting and the Phial of Galadriel, was able to mortally wound her and drive her back into her caves. Sam took the Ring from around Frodo's neck upon hearing Orcish voices, and hid behind some nearby rocks. He overheard the Orcs speaking of Frodo, and Sam realized that his master was not dead, but merely paralyzed. Frodo was then taken to the Tower of Cirith Ungol to await further torture and questioning.
Cirith Ungol
Frodo was imprisoned on the highest floor of the Tower. He was stripped of all his clothes and possessions. Fighting broke out amongst the two lead Orcs and their battalions from squabbling over claim of the mithril vest - in this violence almost all Orcs and Uruks in the tower died. Sam soon arrived at the gate of the Tower, only to find his way blocked by the Two Watchers; he eventually overcame them, entered the tower, finding and rescuing his master. They descended and fled the tower, having to pass the Watchers again (this time, destroying them), and entered Mordor.
Mordor and Mount Doom
Frodo and Sam crawled onward through the plains of Mordor, which lay vacant as many hosts of Orcs were sent to the Black Gate to meet the Men of the West's army, and, after falling in and out of one such company of Orcs, started to climb Mount Doom. They journeyed on for days with little food or water, and Frodo became progressively weaker as the Ring's power over him grew the closer they drew to Orodruin. Frodo was eventually unable to go on, and Sam was forced to carry him a fair distance. It was then that Gollum reappeared, and, after a brief struggle, Sam cut Gollum in the stomach while Frodo fled up the mountain to its entrance.
Inside the Cracks of Doom, Frodo finally had the chance to destroy the Ring and relieve his burden, but the Ring's power was at its strongest due to the surroundings where it had been created. Frodo finally yielded to the temptation and power of the Ring. Sam yelled for Frodo to destroy it, but Frodo, overcome by its power, claimed the Ring for himself aloud. Gollum attacked Sam, who fell and hit his head on a rock, temporarily knocking him unconscious. When he regained consciousness, he saw Gollum fighting with an invisible Frodo. Gollum bit off Frodo's finger that wore the Ring, causing Frodo great pain, and he was reunited with his treasure for a short time, until, dancing with joy, he toppled off the brink and fell into the lava, destroying himself and the One Ring.
The two hobbits tried to escape as Mount Doom erupted. In the very moment that they believed themselves doomed, Gwaihir the Lord of Eagles saw them, and, with his Eagle companions Landroval and Meneldor, rescued Sam and Frodo, and flew them westward, out of Mordor to safety.
End of the War
The Scouring of the Shire
After recovering at the Field of Cormallen and witnessing the crowning of Aragorn as King Elessar, Frodo, Sam, Merry, and Pippin all returned to the Shire. Arriving, they found it under the control of one Sharkey (later revealed to be Saruman) and his forces. Saruman was ruling the Shire from Bag End, although he was later murdered by Gríma Wormtongue. Frodo, Sam, Merry, and Pippin, however, started to gather all Shirriffs and townsfolk of the Shire, and they successfully defeated of Saruman's Ruffians at the Battle of Bywater.
Frodo was not directly involved in the fighting at the Battle of Bywater; instead, he made sure that no hobbits were harmed (saying that no Hobbit had ever intentionally harmed another in the Shire, and that this would stay true), and also that any Ruffians that surrendered were not harmed.
Close of the Third Age
Following the Scouring of the Shire and end of the War of the Ring in November, Frodo went on to serve as Deputy Mayor of the Shire. During his brief tenure of six months, he helped lead the rebuilding of the Shire, but soon realized that he still bore the wounds of his quest. On May 1st, TA 3020, Frodo attended the wedding of his best friend Sam and his wife Rosie Cotton and they moved in with him at Bag End. On mid-years day later that year, Frodo retired from his post as Deputy Mayor, letting the office revert to Will Whitfoot, who Frodo held place for as he recovered from his wounds and imprisonment by Saruman.
During his last days in middle-earth, he tried to live as peacefully as possible. Frodo was still troubled by his shoulder wound, which pained him on each anniversary of the attack at Weathertop, in addition to falling sick on each anniversary of being stung by Shelob. As such, October 6th of that year, he would suffer from his wounds again. March 13th, TA 3021 brought similar ailments, Frodo being ill from his stinging. While living with Sam and Rosie he would witness the birth of their oldest daughter Elanor twelve days later. Prior to departing Middle-earth, he also wrote his own story "The Lord of the Rings" along with Bilbo's "There and Back Again" in the Red Book of Westmarch.
On September 22, TA 3021 (Third Age), at the age of 53, Frodo joined Bilbo, Gandalf, Elrond, and Galadriel aboard the White Ship at the Grey Havens. He was allowed passage across the sea to the Undying Lands, as he was a Ring-bearer, with the hope of healing the damage to his spirit that bearing the Ring had caused.
Fourth Age
During the Fourth Age, Frodo spent his last days on Tol Eressëa in "a period of reflection and peace", attempting to understand his place in Arda, having been given the opportunity as he healed from the trauma afflicted upon him due to his quest during the War of the Ring.
In the year FA 61, fellow Ring-bearer Samwise Gamgee, sailed to the Undying Lands, possibly reuniting with Frodo, assuming he was still alive.
Characteristics
Frodo, as described by Gandalf, was "taller than some and fairer than most," with "a cleft in his chin: perky chap with a bright eye." He had thick, curly brown hair like most other hobbits, and had lighter-than-usual skin due to his Fallohide ancestry through his Brandybuck mother. He could be described as fairly good looking for a hobbit. Frodo is described as appearing thirty-three, even when he is fifty, due to the influence of the Ring.
Bilbo and Frodo shared a common birthday on September 22, but Bilbo was 78 years Frodo's senior. At the opening of The Fellowship of the Ring, Frodo and Bilbo were celebrating their thirty-third and eleventy-first (111th) birthdays, respectively.
Frodo, like Bilbo, was considered by many others in Hobbiton to be a little odd. His curiosity of the outside world, fascination with Elves and faraway places (like those to which Bilbo traveled in The Hobbit) did not fit the general content personality of most Hobbits. This curiosity was also attributed to his Took ancestry. He was very kind and compassionate, pitying Gollum and allowing him to guide him and Sam to Mordor despite Sam's distrust of the creature. This act of kindness later proved to be a factor in the quest's success in destroying the Ring.
Attire
Frodo was dressed in typical Hobbit-fashion when he left the Shire: knee breeches, shirt, waistcoat, jacket, and cloak. Colors such as bright green and yellow were typical for Shire-folk. He was unarmed, save for a pocketknife.
When his little group was waylaid by Barrow-wights, he lost his summer-weight clothing and was wearing a burial shrift when rescued by Tom Bombadil. When their pack-ponies returned, he was forced to put on heavier woollen clothing intended for colder weather. The Hobbits found several long Dúnedain daggers in the wight's treasure. These served as short-swords for the Hobbits, but Frodo's was broken when he resisted the Witch-king at Weathertop. At Imladris, he removed his Hobbit clothing upon finding new Elven clothes that fitted him perfectly. Therefore, throughout his quest, he wore a green silk tunic, trousers together with cloaks made of fur for the first stages of the quest, and then towards Mordor he shed them to wear his tunic and trousers in the warmer weather. Later, his uncle Bilbo gave him both Sting, a magic Elven dagger, and a coat of mithril chain mail. The mail saved his life twice, first when it deflected a spear-point in the Mines of Moria, and second when it turned aside the dagger that Saruman used to try and kill him.
As with the other members of the Fellowship of the Ring, Frodo received a special cloak from Galadriel in Lórien, which allowed him to blend in with natural surroundings. Upon being betrayed by Gollum and captured by Orcs at Cirith Ungol, Frodo lost all of his clothing and most of his possessions. Sam Gamgee saved Sting, however. After the two tribes of Orcs had slain each other in the tower of Cirith Ungol, Frodo dressed himself in Orc-garb. This successfully fooled the Mordor-Orcs they encountered, but he dropped the Orc mail and helmet as he and Sam approached Mount Doom.
In other versions
When J.R.R. Tolkien first conceived of the character, he christened him Bingo Bolger-Baggins; for he thought that Lord of the Rings was going to be a children's book, but then soon decided that it was far too 'silly' of a name for The Lord of the Rings, which was an increasingly more serious topic.
In adaptations
Ralph Bakshi's The Lord of the Rings
Christopher Guard voiced Frodo in Ralph Bakshi's 1978 The Lord of the Rings film, with Sharon Baird doing the modeling.
Rankin/Bass
Orson Bean voiced Frodo in the Rankin/Bass The Return of the King film, the same actor who portrayed Bilbo in The Hobbit and in The Return of the King.
The Lord of the Rings film trilogy
Frodo is portrayed in The Lord of the Rings film trilogy by Elijah Wood. There are several differences between Peter Jackson's film trilogy and the original story. In the movies, Frodo seems to have owned the Ring for only a few days or perhaps a few months before Gandalf returned, as opposed to the seventeen years of the book. Frodo never sells Bag End, but sets out early next morning with Sam. Merry and Pippin run into the pair at the farm of Farmer Maggot and are pulled into the journey. The Hobbits are pursued by the Black Riders all the way to Bucklebury Ferry on the borders of Buckland. There the Black Riders are forced to ride to the Brandywine Bridge while the Hobbits make for Bree. The movies remove several parts of the journey as well. These include their encounters with the High Elves, Farmer Maggot, and Tom Bombadil, as well as their visit to Buckland, the Old Forest, and the Barrow-downs.
Arwen, Elrond's daughter, leads Frodo to Rivendell instead of Glorfindel. The Cave-troll inflicts the wound on Frodo in Moria, instead of the Orc captain. In the novel, Faramir declared right from the first that he wanted no part of the One Ring, but in the film Faramir at first follows what he believes is his duty to bring the Ring back to Minas Tirith. But while travelling with Frodo, Sam, and Gollum through the city of Osgiliath, the city is attacked by a Nazgûl and the forces of Mordor, and Faramir realizes he should not take the Ring after he sees the effect it has on Frodo.
Video games
Frodo appears in many video games:
J.R.R. Tolkien's The Lord of the Rings, Volume 1 (1994) SNES by Interplay. The main playable character.
The Lord of the Rings: The Fellowship of the Ring (2002) by WXP and Sierra (voiced by Steve Staley, playable)
The Lord of the Rings: The Two Towers (2002) by Stormfront Studios and Electronic Arts (voiced by Elijah Wood)
The Lord of the Rings: The Return of the King (2003) by EA Redwood Shores and EA (voiced by Elijah Wood, playable)
The Lord of the Rings: War in the North (2011) by Snowblind Studios and Warner Bros. Interactive (cameo appearance)
LEGO The Lord of the Rings: The Video Game
Frodo is a primary character in EA's The Lord of the Rings: The Battle for Middle-earth (2004), and in The Lord of the Rings: The Battle for Middle-earth II (2005), he can only be temporarily summoned with the Summon Hobbits power, available to factions of good in Skirmish mode.
In The Lord of the Rings Online, (2007), Frodo is first met in Rivendell, preparing for departure. Later, he is found on Cerin Amroth in Lothlórien, weary from the loss of Gandalf. From Amon Hen onwards, player experiences Frodo's journey in a series of Session Plays, alternatively playing as either Frodo, Sam or Gollum. The player meets Frodo again at the Field of Cormallen, he later gives a speech to Aragorn and Arwen. A Hobbit actor portrays Frodo Baggins in a Hobbit-made theater play "The Disappearance of Mad Baggins". Notably, the player is not told about Frodo's mission for a very long time, with Elrond, Gandalf, Aragorn and others only saying that it is "of great importance".
Radio
Oliver Burt voiced the character in the 1956 BBC Radio radio adaptation of The Lord of the Rings.
James Arrington voiced the character in the 1979 The Mind's Eye radio adaptation of The Lord of the Rings.
Ian Holm (who later portrayed Bilbo in The Lord of the Rings film trilogy) voiced the character in the 1981 BBC Radio serial of The Lord of the Rings.
Nigel Planer voiced the character in Tales from the Perilous Realm (1992 radio series).
Matthias Haase voiced the character in the 1991-1992 German radio serial adaptation of The Lord of the Rings.
Dušan Cinkota voiced the character in the 2001 first season and Ľuboš Kostelný in the 2002 second season and 2003 third season of the 2001-2003 three-season Slovak radio serial adaptation of The Lord of the Rings.
Parodies
Frodo was portrayed by Lauren Lopez in a parody of The Lord of the Rings by the group Team Starkid.
Elijah Wood reprised his role in an AT&T streaming app commercial, wherein, shortly after a request was made to see The Lord of the Rings, Frodo requests the streaming app to play his music from the Shire playlist. As with the other people in the ad, only his mouth is actually seen.
Voice dubbing actors
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| Foreign Language | Voice dubbing artist |
| Japanese | Daisuke Namikawa (浪川大輔) |
| Korean (SBS TV Edition) | Kim Yeong Seon (김영선) |
| French (Québec) | Guillaume Champoux |
| French (France) | Alexandre Gillet |
| Spanish (Latin America) | Enzo Fortuny / Edson Matus (The Hobbit trilogy) |
| Spanish (Spain) | Óscar Muñoz |
| Catalan | ???? |
| German | Timmo Niesner |
| Italian | Davide Perino |
| Portuguese (Brazil) (Television/DVD) | Vagner Fagundes / Sérgio Cantú (The Hobbit trilogy) |
| Turkish | Emrah Özertem |
| Polish | Łukasz Lewandowski (1978) Józef Pawłowski (The Hobbit trilogy) |
| Czech | Jan Maxián |
| Slovak | Michal Hallon |
| Hungarian | Gábor Csőre |
| Tagalog (Flipino) | Ryan Ang |
| Russian | Aleksey Elistratov (Елистратов, Алексей Анатольевич) |
| Ukrainian | ???? |
| Mandarian Chinese (China / Taiwan) | Jiang Guangtao (姜广涛) |
| Cantonese Chinese (Hong Kong) | Bosco Tang (鄧肇基) |
| Thai | Achita Pramote (อชิตะ ปราโมช ณ อยุธยา) (Kabook) Tonsak Unon (ธนศักดิ์ อุ่นอ่อน) (Channel 7) |
| Hindi | ???? |
| Tamil | ???? |
| Telugu | ???? |
| Arabic (MBC TV Edition) | Safi Mohammed (محمد صافي) |
| Persian | Afshin Zinoori (افشین زینوری) |
| Punjabi | ???? |
| Urdu | ???? |
Gallery
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Frodo standing off against the Nazgul
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Frodo claiming the Ring in The Return of the King
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Frodo in The Lord of the Rings: The Card Game - Shadow in the East Expansion
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Frodo in The Lord of the Rings: The Card Game - Conflict at the Carrock Adventure Pack
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Frodo in The Lord of the Rings: The Card Game - The Black Riders Expansion
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Frodo in The Lord of the Rings: The Card Game - The Road Darkens Expansion
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Frodo in The Lord of the Rings: The Card Game - The Land of Shadow Expansion
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Frodo in The Lord of the Rings: The Card Game - The Mountain of Fire Expansion
Appearances
Books
The Lord of the Rings
The Fellowship of the Ring - First appearance
The Two Towers
The Return of the King
The Silmarillion
Of the Rings of Power and the Third Age
Films
The Lord of the Rings (1978 film)
The Return of the King (1980 film)
The Lord of the Rings: The Fellowship of the Ring
The Lord of the Rings: The Two Towers
The Lord of the Rings: The Return of the King
The Hobbit: An Unexpected Journey
Trivia
| Trivia that should be moved |
| Lists of miscellaneous information should be avoided. Please relocate any relevant information into appropriate sections or articles. |
There is a deleted scene in The Lord of the Rings: The Two Towers of Faramir envisioning Frodo becoming a creature like Gollum.
Frodo has the most mentions of any character throughout the entire The Lord of the Rings trilogy, with 2028 mentions.
Before Elijah Wood was chosen to play Frodo, he would originally have been played by Jake Gyllenhaal.
Frodo has a street named after him in the Dutch town of Geldrop.
There is another street named after him in West Jordan, Utah, next to another street named after Frodo's uncle.
He has been determined by some fans to be a member of the INFP personality type.
Translations
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| Foreign Language | Translated name |
| Arabic | فرودو باجنز |
| Aragonese | Frodo Bolsón |
| Armenian | Ֆրոդո Բեգինս |
| Asturian | Frodo Bolsona |
| Basque | Frodo Zorrozabal |
| Belarusian Cyrillic | Фрода Бэгінс |
| Bengali | ফ্রোডো ব্যাগিন্স |
| Breton | Frodo Sac'heg |
| Bulgarian Cyrillic | Фродо Бегинс |
| Catalan | Frodo Saquet |
| Chinese (China) | 弗罗多 |
| Chinese (Hong Kong) | 佛羅多·巴金斯 |
| Colognian | Frodo Büggels |
| Czech | Frodo Pytlík |
| Danish | Frodo Sækker |
| Dutch | Frodo Balings |
| Esperanto | Frodo Baginzo |
| Estonian | Frodo Paunaste |
| Faroese | Fróði Pjøkin |
| Finnish | Frodo Reppuli |
| French | Frodon Sacquet (first translation) Frodo Bessac (second translation) |
| Galician | Frodo Bolseiro |
| Georgian | ფროდო ბეგინსი |
| German | Frodo Beutlin |
| Greek | Φρόντο Μπάγκινς |
| Hebrew | פרודו בגינס |
| Hindi | फ्रोडो बेगिन्स |
| Hungarian | Zsákos Frodó |
| Icelandic | Fróði Baggi |
| Japanese | フロド・バギンズ |
| Kannada | ಫ್ರೋಡೊ ಬ್ಯಾಗ್ಗಿನ್ಸ್ |
| Kazakh | Фродо Бэггинс (Cyrillic) Frodo Béggïns (Latin) |
| Korean | 프로도 배긴스 |
| Kurdish | Firodo Torbik (Kurmanji) |
| Kyrgyz Cyrillic | Фродо Баггинс |
| Latvian | Frodo Baginss |
| Lithuanian | Frodas Beginsas |
| Macedonian Cyrillic | Фродо Багинс |
| Marathi | फ्राडो बॅगिन्स |
| Mongolian Cyrillic | Фродо Баггинз |
| Norwegian | Frodo Lommelun |
| Persian | فرودو بگینز |
| Polish | Frodo Baggins (Skibniewska tr.) Frodo Bagosz (Łoziński tr.) |
| Portuguese | Frodo Bolseiro (Brazil) |
| Punjabi | ਫ੍ਰੋਡੋ ਬੈਗੇਂਸ |
| Russian | Фродо Бэггинс |
| Serbian | Фродо Багинс (Cyrillic) Frodo Bagins (Latin) |
| Sinhalese | ෆ්රෝඩෝ බැගින්ස් |
| Slovak | Frodo Bublík |
| Slovenian | Frodo Bisagin |
| Spanish (Spain and Latin America) | Frodo Bolsón |
| Swedish | Frodo Bagger (First translation) Frodo Secker (Second translation) |
| Tajik Cyrillic | Фродо Баггинс |
| Tamil | புரோடோ பாகின்சு |
| Telugu | ఫ్రోడో బాగ్గినస్ |
| Thai | โฟรโด แบ๊กกิ้นส์ |
| Ukrainian Cyrillic | Фродо Торбин |
| Urdu | فروڈو بیگنز |
| Yiddish | פראָדאָ באַגגינס |
| Ring-bearers | |
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| Sauron • Isildur • Déagol • Sméagol (Gollum) • Bilbo Baggins • Frodo Baggins • Samwise Gamgee |
| The Fellowship of the Ring | | | | | | | | |
--- --- --- ---
| | | | | | | | | |
| Frodo · Sam · Merry · Pippin · Gandalf · Aragorn · Legolas · Gimli · Boromir | | | | | | | | |
| |
| The Lord of the Rings Wiki Featured articles |
| People: Faramir · Sauron · Witch-king of Angmar · Gollum · Elrond · Frodo Baggins · Samwise Gamgee · Meriadoc Brandybuck · Peregrin Took · Gandalf · Aragorn II · Legolas · Gimli · Boromir · Galadriel · Elves · Hobbits Locations: Middle-earth · Gondor · Mordor · Rohan Other: Mithril · Middle-earth Strategy Battle Game · The Fellowship of the Ring: Being the First Part of The Lord of the Rings · Works inspired by J. R. R. Tolkien · The Lord of the Rings · The Lord of the Rings (1978 film) · Ainulindalë · Tolkien vs. Jackson · Tengwar · Quenya |
References
↑ The Lord of the Rings, The Fellowship of the Ring, Book One, Ch. IV: "A Short Cut to Mushrooms"
↑ The Lord of the Rings, The Fellowship of the Ring, Book One, Ch. I: "A Long-expected Party"
↑ Jump up to: 3.0 3.1 The Letters of J. R. R. Tolkien, Letter 246, "a gaining of a truer understanding of his position..."
↑ The Lord of the Rings, The Fellowship of the Ring, Chapter X: "Strider"
↑ Der Herr der Ringe (hörspiel). (German: "The Lord of the Rings (radio play)". Ardapedia.org (German-language wiki of Tolkien's Legendarium). Retrieved/cited 30 May 2021.
↑ Pán prsteňov. (Slovak: "The Lord of the Rings) Slovak 2001-2003 radio play. Tolkien Gateway.net (English-language wiki of Tolkien's Legendarium). Retrieved/cited 30 May 2021.
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1398 | http://www1.us.elsevierhealth.com/studentconsult/Kumar_Robbins/virtual_microscope_slides/?srsltid=AfmBOor1l7KkqOnerLS_7zcqj1wtkkDhRtIXgdCW0AFSajgAy-QIMxg2 | STUDENT CONSULT: Robbins & Cotran Pathologic Basis of Disease, 8th edition: Virtual Microscope
Blood Vessels
Vein - organizing and recanalizing thrombus
The Heart
Heart - glycogen storage disease
Heart - myocardial hypertrophy
Aorta - aortic arteriosclerosis
Heart and coronary artery - arteriosclerosis with organizing thrombus
Heart - acute transmural infarct (5 days old) with rupture
Heart - healing transmural infarct
Heart - healed transmural infarct
Heart - acute rheumatic carditis
Heart with mitral valve - infective endocarditis
Aorta - cystic medionecrosis
Heart - acute myocardial infarct
Heart - normal
Aorta - normal
White Blood Cells, Lymph Nodes, Spleen and Thymus
Spleen - Gaucher
Lymph Node - metastatic carcinoma of the breast
Lymph Node - sarcoidosis
Lymph Node - normal
Spleen - normal
The Lung
Lung - chronic passive congestion
Lung - acute passive congestion and edema
Lung - oxygen toxicity
Lung - pseudomonal bronchopneumonia
Lung - viral pneumonia
Lung - abscesses
Lung - pulmonary tuberculosis with cavity
Lung - emphysema
Lung - interstitial pulmonary fibrosis
Lung, bronchus - asthma
Lung, bronchus - squamous cell carcinoma
Lung, bronchus - small cell undifferentiated carcinoma
Lung, bronchus - adenocarinoma
Lung - normal
The Kidney
Kidney - acute infarct
Kidney - systemic lupus erythematosus
Kidney - arteriolar nephrosclerosis
Kidney - malignant nephrosclerosis
Kidney - acute glomerulonephritis
Kidney - membranous glomerulopathy
Kidney - crescentic glomerulonephritis
Kidney - acute pyelonephritis with necrotizing papillitis
Kidney - Renal cell carcinoma
Kidney - diabetic glomerulosclerosis
Kidney - normal
The Gastrointestinal Tract
Small intestine, jejunum - organizing peritonitis
Stomach - chronic peptic ulcer
Colon - chronic ulcerative colitis
Colon - villous adenoma
Colon - adenocarcinoma
Colon - amebiasis
Appendix - fibrous obliteration of the tip
Appendix - acute appendicitis
Stomach, fundus - normal
Appendix - normal
Stomach, pylorus - normal
Colon - normal
Small Intestine - normal
Esophagus - normal
Liver and Biliary Tract
Liver - fatty change (steatosis)/ alcoholic fatty liver
Liver - chronic passive congestion with centrilobular necrosis
Liver - metastatic carcinoma of pancreas
Liver - miliary tuberculosis
Liver - viral hepatitis
Liver - alcoholic hepatitis with early cirrhosis
Liver - alcoholic (portal) cirrhosis
Liver - cirrhosis secondary to chronic viral hepatitis
Liver - biliary cirrhosis
Liver - primary carcinoma of the liver (hepatoma)
Gallbladder - acute and chronic cholecystitis
Liver - normal
Gallbladder - normal
The Pancreas
Pancreas - acute interstitial pancreatitis
Pancreas - chronic (alcoholic) pancreatitis
Pancreas - adenocarcinoma
Pancreas - cystic fibrosis
Pancreas - normal
The Lower Urinary Tract and Male Genital System
Skin, genitalia - chancre of the prepuce
Prostate - adenocarcinoma
Testis - seminoma
Bladder - squamous cell carcinoma
Bladder - frozen section diagnosis: squamous cell carcinoma
Bladder - normal
Testis - normal
Epididymus - normal
Prostate - normal
The Female Genital Tract
Fallopian Tube - suppurative salpingitis
Uterus, cervix - chronic cervicitis
Uterus, cervix - carcinoma in situ
Uterus, cervix - squamous cell carcinoma
Uterus, Endometrium - endometrial carcinoma
Uterus - leiomyoma
Uterus - leiomyosarcoma
Ovary - serous cystadenocarcinoma
Ovary - benign teratoma (dermoid cyst)
Fallopian tube - normal
Ovary - normal
Endometrium - normal
Uterus - normal
Uterus, cervix - normal
The Breast
Breast - fibrocystic change
Breast - fibroadenoma
Breast - invasive ductal carcinoma
Breast - normal
The Endocrine System
Thyroid - Graves disease - hypertrophy and hyperplasia
Thyroid - follicular adenoma
Thyroid - papillary carcinoma
Thyroid - Hashimoto thyroiditis (autoimmune thyroiditis)
Adrenal gland - nodular hyperplasia
Adrenal gland - pediatric tumor (neuroblastoma)
Adrenal gland - pheochromocytoma
Pituitary - pituitary adenoma
Parathyroid - secondary parathyroid hyperplasia
Thyroid - normal
Adrenal gland - normal
Pituitary - normal
Bones, Joints and Soft Tissue Tumors
Joint - rheumatoid arthritis
Joint - degenerative joint disease
Bone - giant cell tumor
Bone - osteosarcoma
Bone - chondrosarcoma
Bone - normal
Peripheral Nerve and Skeletal Muscle
Skeletal muscle - progressive muscular dystrophy
Skeletal muscle - denervation atrophy (motor neuron disease)
Peripheral Nerve and Skeletal Muscle, Skeletal muscle - normal
The Skin
Skin - healing surgical incision with foreign-body granuloma
Skin - normal
The Central Nervous System
Brain - cerebral edema
Brain - anoxic necrosis
Brain - acute bacterial meningitis
Brain - Alzheimer disease
Brain - organizing cerebral infarct
Brain - old cerebral infarct
Brain, cerebellum - astrocytoma
Brain, cerebrum - glioblastoma multiforme
Brain, cerebellum - medulloblastoma
Brain, meninges, and cerebrum - meningioma
Brain - herpes simplex virus encephalitis
Brain, cerebrum - multiple sclerosis
Spinal cord - amyotrophic lateral sclerosis
Brain, cerebrum - leukodystrophy (myelin stain)
Brain - normal
Brain, cerebellum - normal
Head and Neck
Larynx - squamous cell
Larynx, right vocal cord - normal |
1399 | https://emedicine.medscape.com/article/212127-treatment | Anthrax Treatment & Management
Updated: Aug 08, 2023
Author: David J Cennimo, MD, FAAP, FACP, FIDSA, AAHIVS; Chief Editor: Michael Stuart Bronze, MD more...;)
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Sections
Anthrax
Sections Anthrax
Overview
Practice Essentials
Background
Pathophysiology
Etiology
Epidemiology
Prognosis
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Presentation
History
Physical Examination
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DDx
Workup
Approach Considerations
Gram Stain and Blood Culture
Enzyme-Linked Immunosorbent Assay
Chest Radiography and Computed Tomography
Lumbar Puncture
Histologic Findings
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Treatment
Approach Considerations
Cutaneous Anthrax
Prehospital Care
Emergency Department Care
Consultations
Deterrence/Prevention
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Medication
Medication Summary
Antibiotics, Other
Corticosteroids
Antidotes, Other
Vaccines
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Media Gallery;)
Tables)
References;)
Treatment
Approach Considerations
With doxycycline, a loading regimen should be used (200 mg PO/IV every 12 hours for 72 hours). In severely ill patients, 200 mg IV/PO every 12 hours may be continued (without toxicity) for the duration of therapy. Oral doxycycline and quinolones have excellent bioavailability; the same blood/tissue levels are obtained with PO and IV therapy. Use any quinolone in patients who are unable to take penicillin or doxycycline.
The preferred agent used to treat nonbioterrorist anthrax is penicillin. Penicillin is the preferred agent to treat inhalational anthrax and anthrax meningitis. Use meningeal doses for inhalational anthrax because meningitis is often also present.
For bioterrorist anthrax, use any quinolone or doxycycline for 1-2 weeks. Clindamycin may be added for its anti-exotoxin effect. Use doxycycline or any quinolone (eg, ciprofloxacin, levofloxacin) for postexposure prophylaxis (PEP) to prevent inhalational anthrax. PEP to prevent inhalational anthrax should be continued for 60 days.
Raxibacumab is a human IgG1 gamma monoclonal antibody directed at the protective antigen of Bacillus anthracis. It is produced by recombinant DNA technology in a murine cell expression system. This agent was approved by the FDA in December 2012 for treatment of inhalational anthrax or for prevention when alternative therapies are not available or appropriate. It is used as part of a combination regimen with appropriate antibiotic drugs. The efficacy of raxibacumab as a prophylactic agent and after disease onset was assessed in 4 randomized controlled animal model trials to provide surrogate endpoints applicable to human use.
Obiltoxaximab is another monoclonal antibody directed at the protective antigen of B anthracis that was approved by the FDA in March 2016. It is a chimeric IgG1 kappa monoclonal antibody.
Human anthrax immune globulin (Anthrasil) is indicated for treatment of inhalational anthrax in adults and children in combination with antibiotic therapy.
The indication for anthrax vaccine adsorbed (BioThrax) was expanded in November 2015 to include postexposure use following suspected or confirmed B anthracis exposure in combination with antimicrobial therapy. It was originally approved for pre-exposure prophylaxis in high-risk individuals.
Antimicrobial therapy renders lesions culture-negative within hours, but the lethal effects of anthrax are related to the effects of the organism's toxin. Patients with septic and hemorrhagic shock, which is the final common pathway for end-stage anthrax infection, should be admitted to the intensive care unit for hemodynamic monitoring and management. In addition, progressive respiratory insufficiency may necessitate the use of ventilatory support.
Despite early treatment, persons infected with inhalational, gastrointestinal, or meningeal anthrax have a very poor prognosis. Although prophylaxis and vaccinations confer almost complete protection, adequately providing immunity to a potentially exposed community is extremely difficult.
The Infectious Diseases Society of America (IDSA) published guidelines for the treatment of both naturally acquired and bioterrorism-related cases of cutaneous anthrax (see Practice Guidelines for the Diagnosis and Management of Skin and Soft Tissue Infections: 2014 Update by the Infectious Diseases Society of America).
The CDC has issued updated guidelines on anthrax postexposure prophylaxis (PEP) and treatment in nonpregnant and pregnant adults. Recommendations include the following: [14, 15]
All individuals exposed to aerosolized B anthracis spores should receive a full 60 days of PEP antimicrobial drugs, regardless of their vaccination status
Compared with monotherapy, antimicrobial combination therapy is more likely to result in a cure; combined bactericidal and protein synthesis inhibitor agents may be beneficial
Ciprofloxacin, levofloxacin, and doxycycline are approved by the FDA for PEP for inhalation anthrax in adults aged 18 years or older; ciprofloxacin and doxycycline are first-line treatments
Treatment for anthrax meningitis should include at least 3 antimicrobial drugs with activity against B anthracis, at least 1 of which should have bactericidal activity, and at least 1 of which should be a protein synthesis inhibitor
For patients suspected to have systemic anthrax, antitoxin should be added to combination antimicrobial drug treatment
Uncomplicated cutaneous anthrax can be treated with a single oral agent; fluoroquinolones (ciprofloxacin, levofloxacin, and moxifloxacin) and doxycycline are equivalent first-line drugs
Treatments for pregnant, postpartum, and lactating women are generally the same as those for nonpregnant patients
Next: Cutaneous Anthrax
Cutaneous Anthrax
Patients with isolated cutaneous anthrax without systemic involvement (ie, without edema, fever, cough, headache, etc) or complications may be treated on an outpatient basis with antibiotic monotherapy. The treatment of choice has been outlined as recommended by the CDC, IDSA, and AAP.
Table 3. CDC Expert Panel Recommendations for Treatment of Cutaneous Anthrax (Open Table in a new window))
| Nonpregnant adults | Pregnant/lactating women | Children |
---
| Recommended therapy : Treatment duration, 7-10 days | | |
| Ciprofloxacin 500 mg every 12 hours | Ciprofloxacin 500 mg every 12 hours | Ciprofloxacin 30 mg/kg/day divided every 12 hours (max dose, 500 mg/dose) |
| Doxycycline 100 mg every 12 hours | | Amoxicillin 75 mg/kg/day divided every 8 hours (max dose, 1 g/dose) |
| Levofloxacin 750 mg every 12 hours | | |
| Moxifloxacin 400 mg every 24 hours | | |
| Alternative therapy | | |
| Clindamycin 600 mg every 8 hours | Levofloxacin 750 mg every 12 hours | Doxycycline < 45 kg: 4.4 mg/kg/day divided every 12 hours (max dose, 100 mg/dose) >45 kg: 100 mg every 12 hours |
| Amoxicillin 1 g every 8 hours (susceptible strain only) | Amoxicillin 1 g every 8 hours (susceptible strain only) | Clindamycin 30 mg/kg/day divided every 8 hours (max dose, 600 mg/dose) |
| | | Levofloxacin < 50 kg: 16 mg/kg/day divided every 12 hours (max dose, 250 mg/dose) >50 kg: 500 mg every 24 hours |
Systemic anthrax without meningitis
All patients with suspected systemic anthrax should be begun immediately on IV antibiotics with co-administration of an antitoxin (raxibacumab or anthrax IgG) and should receive aggressive supportive care. All patients with suspected systemic illness should be admitted to inpatient for treatment. Early diagnosis and clinical suspicion are critical to improving outcomes. Workup should include standard fever workup pending the clinical situation, often including blood cultures and urine samples.
For adults with systemic anthrax (inhalational, intestinal, meningitis, injection), the CDC expert panel recommends the following:
Table 4. CDC Expert Panel Recommendations for Treatment of Systemic Anthrax Without Meningitis (Open Table in a new window))
| Adults | Children |
--- |
| 1. Bactericidal agent | |
| Ciprofloxacin 400 mg every 8 hours | Ciprofloxacin 30 mg/kg/day divided every 8 hours (max dose, 400 mg/dose) |
| Alternative | |
| Levofloxacin 750 mg every 24 hours -OR- | Meropenem 60 mg/kg/day divided every 8 hours (max dose, 2 g/dose) -OR- |
| Moxifloxacin 400 mg every 24 hours -OR- | Levofloxacin < 50 kg: 20 mg/kg/day divided every 12 hours (max dose, 250 mg/dose) >50 kg: 500 mg every 24 hours -OR- |
| Meropenem 2 g every 8 hours -OR- | Imipenem 100 mg/kg/day divided every 6 hours (max dose, 1 g/dose) -OR- |
| Imipenem 1 g every 6 hours -OR- | Vancomycin 60 mg/kg/day divided every 8 hours (max dose, 2 g/dose); trough target, 15-20 mcg/mL |
| Vancomycin 60 mg/kg/day divided every 8 hours (max dose, 2 g/dose); trough target, 15-20 mcg/mL | |
| PLUS | |
| 2. Protein synthesis inhibitor | |
| Clindamycin 900 mg every 8 hours -OR- | Clindamycin 40 mg/kg/day divided every 8 hours (max dose, 900 mg/dose) |
| Linezolid 600 mg every 12 hours | |
| Alternative | |
| Doxycycline 200-mg loading dose followed by 100 mg every 12 hours -OR- | Linezolid < 12 years: 30 mg/kg/day divided every 8 hours >12 years: 30 mg/kg/day divided every 12 hours (max dose, 600 mg/dose) -OR- |
| Rifampin 600 mg every 12 hours | Doxycycline < 45 kg: 4.4 mg/kg loading dose (max 200 mg) followed by 4.4 mg/kg/day divided every 12 hours (max 100 mg/dose) >45 kg: 200 mg loading dose followed by 100 mg every 12 hours -OR- |
| | Rifampin 20 mg/kg/day divided every 12 hours (max dose, 300 mg/dose) |
Anthrax meningitis, suspected or confirmed
In patients with suspected or confirmed anthrax meningitis, if not already done, a lumbar puncture should be performed for CSF analysis. Immediate antibiotics are warranted, and lumbar puncture should never delay therapy. Two bactericidal agents plus a protein synthesis inhibitor are indicated. All must be dosed for CSF penetration.
Table 5. CDC Expert Panel Recommendations for Treatment of Anthrax Meningitis (Open Table in a new window))
| Adults | Children |
--- |
| 1. Bactericidal agent | |
| Ciprofloxacin 400 mg every 8 hours | Ciprofloxacin 30 mg/kg/day divided every 8 hours (max 400 mg/dose) |
| Alternative | |
| Levofloxacin 750 mg every 24 hours -OR- | Levofloxacin < 50 kg: 16 mg/kg/day divided every 12 hours (max 250 mg/dose) ≥50 kg: 500 mg every 24 hours -OR- |
| Moxifloxacin 400 mg every 24 hours | Moxifloxacin Age 3 months to < 2 years: 12 mg/kg/day divided every 12 hours Age 2-5 years: 10 mg/kg/day divided every 12 hours Age 6-11 years: 8 mg/kg/day divided every 12 hours Age 12-17 years, < 45 kg: 8 mg/kg/day divided every 12 hours (Max 200 mg/dose) Age 12-17 years, ≥45 kg: 400 mg every 24 hours |
| PLUS | |
| 2. Second bactericidal agent | |
| Meropenem 2 g every 8 hours | Meropenem 120 mg/kg/dose divided every 8 hours (max 2 g/dose) |
| Alternative | |
| Imipenem 1 g every 6 hours -OR- | Imipenem 100 mg/kg/day divided every 6 hours (max 1 g/dose) -OR- |
| Ampicillin 3 g every 6 hours -OR- | Vancomycin 60 mg/kg/day divided every 8 hours (max 2 g/dose); target trough, 15-20 mcg/mL -OR- |
| | Ampicillin 400 mg/kg/day divided every 6 hours (max 3 g/dose) |
| PLUS | |
| 3. Protein synthesis inhibitor | |
| Linezolid 600 mg every 12 hours | Linezolid < 12 years old: 30 mg/kg/day divided every 8 hours ≥12 years old: 30 mg/kg/day divided every 12 hours (Max 600 mg/dose) |
| Alternative | |
| Clindamycin 900 mg every 8 hours -OR- | Clindamycin 40 mg/kg/day divided every 8 hours (max 900 mg/dose) -OR- |
| Rifampin 600 mg every 12 hours -OR- | Rifampin 20 mg/kg per day divided every 12 hours (max 300 mg/dose) -OR- |
| Chloramphenicol 1 g every 6-8 hours | Chloramphenicol 100 mg/kg per day divided every 6 hours (max 1 g/dose) |
Anticipate a therapy duration of at least three weeks or until clinical improvement, whichever comes last, as clinical improvement may take several weeks. Thereafter, patients should complete a 60-day course of antibiotics with oral monotherapy to prevent relapse involving dormant endospores. Oral antibiotic should be dosed according to guidelines for postexposure prophylaxis.
Previous
Next: Cutaneous Anthrax
Prehospital Care
All suspected cases of inhalational anthrax should be considered a bioterror event until proven to be otherwise. As with any potential epidemic biologic exposure, patients should be decontaminated in the field when possible, and paramedical health care workers should wear masks and gloves. If protection is needed from exposure, responders are advised to use splash protection, gloves, and a full-face respirator with high-efficiency particulate air (HEPA) filters (level C) or self-contained breathing apparatus (SCBA) (level B).
Persons who are potentially contaminated should wash with soap and water, not bleach solutions. Clothing and evidence/materials should be placed in plastic bags (triple). If the contamination is confirmed, then a 1:10 dilution of household beach may be used to decontaminate any materials and surfaces not sufficiently cleaned by soap and water.
Chemoprophylaxis with antibiotics should be instituted only if exposure is confirmed. For persons not at risk for repeated exposures to aerosolized B. anthracis spores through their occupation, preexposure vaccination with anthrax vaccine is not recommended.
Previous
Next: Cutaneous Anthrax
Emergency Department Care
The emergency department workup includes rapid initiation of intravenous antibiotic therapy. If risk of exposure is considerable, initiate PEP.
During the 2001 bioterrorist attacks in the United States, the Centers for Disease Control and Prevention (CDC) recommendations for antimicrobial PEP included ciprofloxacin or doxycycline; the CDC recommended amoxicillin for children and pregnant or breastfeeding women exposed to strains susceptible to penicillin. The duration of postexposure antimicrobial prophylaxis should be 60 days if used alone for PEP of unvaccinated exposed persons.
There is a potential preventive benefit of using anthrax vaccine along with an antimicrobial drug for PEP, and the vaccine was made available for this use during the 2001 bioterrorism attacks; however, anthrax vaccine is not licensed for use in PEP.
Raxibacumab and obiltoxaximab are monoclonal antibodies available from the CDC for treatment of inhalational anthrax or as prophylaxis when other therapies are not available or appropriate. These monoclonal antibodies are used as part of a combination regimen with appropriate antibiotic drugs. [9, 10, 16]
Human anthrax immune globulin was approved by the FDA in March 2015. It provides passive immunity to adults and children exposed to inhalational anthrax. It is used in conjunction with appropriate antibiotic therapy.
The indication for anthrax vaccine adsorbed (BioThrax) was expanded in November 2015 to include postexposure use following suspected or confirmed B anthracis exposure in combination with antimicrobial therapy. It was originally approved for pre-exposure prophylaxis in high-risk individuals.
Patients can be admitted to a normal hospital room with barrier nursing procedures (ie, gown, gloves, mask) and secretion precautions (ie, special handing of potentially infectious dressings, drainage, and excretions).
Previous
Next: Cutaneous Anthrax
Consultations
Anthrax is a reportable disease; notify local health care authorities and the Centers for Disease Control and Prevention of suspected cases. In addition, consultation with an infectious disease specialist may be warranted, although treatment of patients in whom anthrax is suspected is straightforward. If biologic terrorism is a threat, consider contacting the Federal Bureau of Investigation (FBI) through the local police department.
Previous
Next: Cutaneous Anthrax
Deterrence/Prevention
For PEP in adults, the CDC recommends vaccination and the use of oral fluoroquinolones (ciprofloxacin, 500 mg bid; levofloxacin, 500 mg qd; or ofloxacin, 400 mg bid). Doxycycline is an acceptable alternative. Prophylaxis should continue until exposure to B anthracis is excluded or for a period of 4 weeks if exposure is confirmed.
The monoclonal antibodies raxibacumab and obiltoxaximab are indicated for prophylaxis of inhalational anthrax when alternative therapies are not available or not appropriate. They should be used as part of a combination regimen with appropriate antibiotic drugs. [9, 10, 16]
Three doses of vaccine should be administered during the 4-week period (at 0, 2, and 4 weeks post exposure). If a vaccine is not available, the antibiotic treatment should continue for at least 60 days. A second option is treatment for 100 days. A third option is 100 days of antibiotic prophylaxis with vaccine.
Anthrax vaccine
Vaccines exist, but are not readily available. They are approved for individuals aged 18-65 years. Preexposure vaccination is recommended only for populations at high risk for exposure to aerosolized B anthracis spores, according to the CDC's Advisory Committee on Immunization Practices (ACIP). The populations at high risk include the following:
Those who work directly with the organism in the laboratory
Those who work with imported animal hides or furs in areas where standards are insufficient to prevent exposure to anthrax spores
Those who handle potentially infected animal products in high-incidence areas; while incidence is low in the United States, veterinarians who travel to work in other countries where incidence is higher should consider being vaccinated
Those who work in the military and are deployed to areas with high risk for exposure to the organism
Anthrax vaccine adsorbed
Pre-exposure prophylaxis regimen is as follows:
3-dose primary series: 0.5 mL IM at 0, 1, and 6 months
Booster series: 0.5 mL IM at 6 and 12 months after primary series initiation and then at 1-year intervals thereafter for persons who remain at risk
In persons who are at risk for hematoma formation following IM injection, the vaccine may be administered SC as a 4-dose primary series at 0, 2, and 4 weeks and at 6 months
Note: Individuals are not considered protected until completion of the primary series
Administer in deltoid region
Postexposure prophylaxis regimen is as follows:
3-dose series: 0.5 mL SC at 0, 2, and 4 weeks postexposure combined with antimicrobial therapy
Administer over deltoid region
SC administration is recommended for postexposure prophylaxis as it results in higher antibody concentrations by week 4 than the IM route
Anthrax vaccine adsorbed, adjuvanted
Anthrax vaccine adsorbed adjuvanted is indicated for postexposure prophylaxis in individuals aged 18-65 years. Using an additional adjuvant, 2 doses administered over 14 days elicit protective levels of immune response, which can be especially important in response to a large-scale public health emergency involving anthrax.
Postexposure prophylaxis regimen is as follows:
2-dose series: 0.5 mL IM administered 2 weeks apart (0, 2 weeks) combined with antimicrobial therapy
Previous
Medication
References
Akbayram S, Dogan M, Akgün C, et al. Clinical findings in children with cutaneous anthrax in eastern Turkey. Pediatr Dermatol. 2010 Nov-Dec. 27(6):600-6. [QxMD MEDLINE Link].
Inglesby TV, O'Toole T, Henderson DA, et al. Anthrax as a biological weapon, 2002: updated recommendations for management. JAMA. 2002 May 1. 287(17):2236-52. [QxMD MEDLINE Link].
John TJ, Dandona L, Sharma VP, Kakkar M. Continuing challenge of infectious diseases in India. Lancet. 2011 Jan 15. 377(9761):252-69. [QxMD MEDLINE Link].
Doganay M, Metan G, Alp E. A review of cutaneous anthrax and its outcome. J Infect Public Health. 2010. 3 (3):98-105. [QxMD MEDLINE Link].
Beatty ME, Ashford DA, Griffin PM, Tauxe RV, Sobel J. Gastrointestinal anthrax: review of the literature. Arch Intern Med. 2003 Nov 10. 163 (20):2527-31. [QxMD MEDLINE Link].
Knox D, Murray G, Millar M, et al. Subcutaneous anthrax in three intravenous drug users: a new clinical diagnosis. J Bone Joint Surg Br. 2011 Mar. 93(3):414-7. [QxMD MEDLINE Link].
Hupert N, Person M, Hanfling D, Traxler RM, Bower WA, Hendricks K. Development and Performance of a Checklist for Initial Triage After an Anthrax Mass Exposure Event. Ann Intern Med. 2019 Apr 16. 170 (8):521-530. [QxMD MEDLINE Link].
US Food and Drug Administration (FDA). FDA approves raxibacumab to treat inhalational anthrax. December 14, 2012 (press announcement). Available at
Migone TS, Subramanian GM, Zhong J, Healey LM, Corey A, Devalaraja M, et al. Raxibacumab for the treatment of inhalational anthrax. N Engl J Med. 2009 Jul 9. 361(2):135-44. [QxMD MEDLINE Link]. [Full Text].
Anthim (obiltoxaximab) [package insert]. Pine Brook, NJ: Elusys Therapeutics, Inc. March 2016. Available at [Full Text].
Anthrasil (anthrax immune globulin intravenous [human]) [package insert]. Winnipeg, MB; Canada: Cangene Corp (Emergent BioSolutions). March 24, 2015. Available at [Full Text].
BioThrax (anthrax vaccine adsorbed) [package insert]. Lansing, MI: Emergent BioSolutions. November, 2015. Available at [Full Text].
[Guideline] Stevens DL, Bisno AL, Chambers HF, Dellinger EP, Goldstein EJ, Gorbach SL, et al. Practice guidelines for the diagnosis and management of skin and soft tissue infections: 2014 update by the infectious diseases society of america. Clin Infect Dis. 2014 Jul 15. 59(2):e10-52. [QxMD MEDLINE Link].
Barclay L. Anthrax Guidelines Address Nonpregnant, Pregnant Adults. Available at Accessed: January 19, 2014.
[Guideline] Hendricks KA, Wright ME, Shadomy SV, et al. Centers for Disease Control and Prevention expert panel meetings on prevention and treatment of anthrax in adults. Emerg Infect Dis 2014. Available at
Raxibacumab [package insert]. Research Triangle Park, NC: GlaxoSmithKline. December, 2012. Available at [Full Text].
Bower WA, Schiffer J, Atmar RL, and the, ACIP Anthrax Vaccine Work Group. Use of Anthrax Vaccine in the United States: Recommendations of the Advisory Committee on Immunization Practices, 2019. MMWR Recomm Rep. 2019 Dec 13. 68 (4):1-14. [QxMD MEDLINE Link]. [Full Text].
Food and Drug Administration. 17.5 FDA-Approved Medication Guide. Levaquin (levofloxacin). [Full Text].
Cyfendus (anthrax vaccine adsorbed, adjuvanted) [package insert]. Gaithersburg, MD: Emergent BioSolutions Inc. July 2023. Available at [Full Text].
WHO. Leading health agencies outline updated terminology for pathogens that transmit through the air. World Health Organization. Available at April 18, 2024; Accessed: April 25, 2024.
Media Gallery
Polychrome methylene blue stain of Bacillus anthracis. Image courtesy of Anthrax Vaccine Immunization Program Agency, Office of the Army Surgeon General, United States.
Histopathology of mediastinal lymph node showing a microcolony of Bacillus anthracis on Giemsa stain. Image courtesy of Marshall Fox, MD, Public Health Image Library, US Centers for Disease Control and Prevention, Atlanta, Georgia.
Cutaneous anthrax. Image courtesy of Anthrax Vaccine Immunization Program Agency, Office of the Army Surgeon General, United States.
Skin lesion of anthrax on face. Image courtesy of the Public Health Image Library, US Centers for Disease Control and Prevention, Atlanta, Georgia.
Skin lesions of anthrax on neck. Cutaneous anthrax showing the typical black eschar. Image courtesy of the Public Health Image Library, US Centers for Disease Control and Prevention, Atlanta, Georgia.
Histopathology of large intestine showing marked hemorrhage in the mucosa and submucosa. Image courtesy of Marshall Fox, MD, Public Health Image Library, US Centers for Disease Control and Prevention, Atlanta, Georgia.
Histopathology of the large intestine showing submucosal thrombosis and edema. Image courtesy of Marshall Fox, MD, Public Health Image Library, US Centers for Disease Control and Prevention, Atlanta, Georgia.
Inhalation anthrax. Chest radiograph with widened mediastinum 22 hours before death. Image courtesy of P.S. Brachman, MD, Public Health Image Library, US Centers for Disease Control and Prevention, Atlanta, Georgia.
Histopathology of mediastinal lymph node showing mediastinal necrosis. Image courtesy of Marshall Fox, MD, Public Health Image Library, US Centers for Disease Control and Prevention, Atlanta, Georgia.
Hemorrhagic meningitis resulting from inhalation anthrax. Image courtesy of the Public Health Image Library, US Centers for Disease Control and Prevention, Atlanta, Georgia.
Anthrax infection. Histopathology of hemorrhagic meningitis in anthrax. Image courtesy of Marshall Fox, MD, Public Health Image Library, US Centers for Disease Control and Prevention, Atlanta, Georgia.
Microscopic picture of anthrax showing gram-positive rods. Image courtesy of Ramon E. Moncada, MD.
Seven-month-old infant with anthrax. In this infant, the infection progressed rapidly with significant edema developing the day after exposure. This large hemorrhagic lesion developed within 3 more days. The infant was febrile and was admitted to the hospital on the second day after the symptoms appeared.On September 28, 2001, the infant had visited the mother's workplace. On September 29, nontender massive edema and a weeping erosion developed. On September 30, a 2-cm sore developed over the edematous area. (Note that edema preceded the primary lesion.) On October 2, an ulcer or eschar formed, and the lesion was diagnosed as a spider bite. Hemolytic anemia and thrombocytopenia developed, and the patient was hospitalized. Serum was drawn on October 2; the polymerase chain reaction results were positive for Bacillus anthracis. On October 13, skin biopsy results were positive with immunohistochemical testing for the cell wall antigen.Note that the initial working diagnosis was a Loxosceles reclusa spider bite with superimposed cellulitis. Courtesy of American Academy of Dermatology with permission of NEJM.
Fourth patient with cutaneous anthrax in New York City, October 2001. This dry ulcer was present. Photo used with permission of the patient. Courtesy of American Academy of Dermatology. Courtesy of Sharon Balter of the New York City Department of Health.
Note the hemorrhage that is associated with cutaneous anthrax lesions. The early ulcer has a moist base. Courtesy of American Academy of Dermatology.
Note the central ulcer and eschar. Courtesy of American Academy of Dermatology.
An example of a central ulcer and eschar with surrounding edema. Courtesy of American Academy of Dermatology with permission from Boni Elewski, MD.
Note the black eschar. Courtesy of American Academy of Dermatology. Courtesy of Gorgas Course in Clinical Tropical Medicine.
Anthrax with facial edema. Courtesy of American Academy of Dermatology.
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Tables
Table 1. Microbiological Differences Between B anthracis and Non– B anthracis Bacilli;)
Table 2. Toxins and Protein Toxins of Bacillus anthracis;)
Table 3. CDC Expert Panel Recommendations for Treatment of Cutaneous Anthrax;)
[Table 4. CDC Expert Panel Recommendations for Treatment of Systemic Anthrax Without Meningitis ;)15]
[Table 5. CDC Expert Panel Recommendations for Treatment of Anthrax Meningitis ;)15]
Table 1. Microbiological Differences Between B anthracis and Non– B anthracis Bacilli
| | |
--- |
| B anthracis | Non–B anthracis bacilli (pseudoanthrax bacilli) |
| Nonmotile long chains | Generally motile short chains |
| Capsule formation on bicarbonate agar | No capsule formation in bicarbonate |
| No growth on penicillin agar (10 mcg/mL) | Usually good growth on penicillin agar |
| Growth in gelatin resembles inverted fir tree | Growth in gelatin absent or resembles atypical fir tree |
| Gelatin liquefaction slow | Gelatin liquefaction usually rapid |
| No hemolysis of sheep RBCs | Hemolysis of sheep RBCs |
| Ferments salicin slowly or not at all | Usually ferments salicin rapidly |
| Pathogenic to laboratory animals | Nonpathogenic to laboratory animals |
| Adapted from Cunha CB. Anthrax: Ancient Plague, Persistent Problem. Infect Dis Pract. 1999;23(4):35-9. | |
Table 2. Toxins and Protein Toxins of Bacillus anthracis
| |
| Edema factor (EF) + lethal factor (LF) = Host cell penetration by B anthracis |
| EF + protective antigen (PA) = Edema toxin |
| LF + PA = Lethal toxin (primary virulence factor of B anthracis) |
| Edema toxin + lethal toxin = Inhibited PMN function and phagocytosis |
Table 3. CDC Expert Panel Recommendations for Treatment of Cutaneous Anthrax
| Nonpregnant adults | Pregnant/lactating women | Children |
---
| Recommended therapy : Treatment duration, 7-10 days | | |
| Ciprofloxacin 500 mg every 12 hours | Ciprofloxacin 500 mg every 12 hours | Ciprofloxacin 30 mg/kg/day divided every 12 hours (max dose, 500 mg/dose) |
| Doxycycline 100 mg every 12 hours | | Amoxicillin 75 mg/kg/day divided every 8 hours (max dose, 1 g/dose) |
| Levofloxacin 750 mg every 12 hours | | |
| Moxifloxacin 400 mg every 24 hours | | |
| Alternative therapy | | |
| Clindamycin 600 mg every 8 hours | Levofloxacin 750 mg every 12 hours | Doxycycline < 45 kg: 4.4 mg/kg/day divided every 12 hours (max dose, 100 mg/dose) >45 kg: 100 mg every 12 hours |
| Amoxicillin 1 g every 8 hours (susceptible strain only) | Amoxicillin 1 g every 8 hours (susceptible strain only) | Clindamycin 30 mg/kg/day divided every 8 hours (max dose, 600 mg/dose) |
| | | Levofloxacin < 50 kg: 16 mg/kg/day divided every 12 hours (max dose, 250 mg/dose) >50 kg: 500 mg every 24 hours |
Table 4. CDC Expert Panel Recommendations for Treatment of Systemic Anthrax Without Meningitis
| Adults | Children |
--- |
| 1. Bactericidal agent | |
| Ciprofloxacin 400 mg every 8 hours | Ciprofloxacin 30 mg/kg/day divided every 8 hours (max dose, 400 mg/dose) |
| Alternative | |
| Levofloxacin 750 mg every 24 hours -OR- | Meropenem 60 mg/kg/day divided every 8 hours (max dose, 2 g/dose) -OR- |
| Moxifloxacin 400 mg every 24 hours -OR- | Levofloxacin < 50 kg: 20 mg/kg/day divided every 12 hours (max dose, 250 mg/dose) >50 kg: 500 mg every 24 hours -OR- |
| Meropenem 2 g every 8 hours -OR- | Imipenem 100 mg/kg/day divided every 6 hours (max dose, 1 g/dose) -OR- |
| Imipenem 1 g every 6 hours -OR- | Vancomycin 60 mg/kg/day divided every 8 hours (max dose, 2 g/dose); trough target, 15-20 mcg/mL |
| Vancomycin 60 mg/kg/day divided every 8 hours (max dose, 2 g/dose); trough target, 15-20 mcg/mL | |
| PLUS | |
| 2. Protein synthesis inhibitor | |
| Clindamycin 900 mg every 8 hours -OR- | Clindamycin 40 mg/kg/day divided every 8 hours (max dose, 900 mg/dose) |
| Linezolid 600 mg every 12 hours | |
| Alternative | |
| Doxycycline 200-mg loading dose followed by 100 mg every 12 hours -OR- | Linezolid < 12 years: 30 mg/kg/day divided every 8 hours >12 years: 30 mg/kg/day divided every 12 hours (max dose, 600 mg/dose) -OR- |
| Rifampin 600 mg every 12 hours | Doxycycline < 45 kg: 4.4 mg/kg loading dose (max 200 mg) followed by 4.4 mg/kg/day divided every 12 hours (max 100 mg/dose) >45 kg: 200 mg loading dose followed by 100 mg every 12 hours -OR- |
| | Rifampin 20 mg/kg/day divided every 12 hours (max dose, 300 mg/dose) |
Table 5. CDC Expert Panel Recommendations for Treatment of Anthrax Meningitis
| Adults | Children |
--- |
| 1. Bactericidal agent | |
| Ciprofloxacin 400 mg every 8 hours | Ciprofloxacin 30 mg/kg/day divided every 8 hours (max 400 mg/dose) |
| Alternative | |
| Levofloxacin 750 mg every 24 hours -OR- | Levofloxacin < 50 kg: 16 mg/kg/day divided every 12 hours (max 250 mg/dose) ≥50 kg: 500 mg every 24 hours -OR- |
| Moxifloxacin 400 mg every 24 hours | Moxifloxacin Age 3 months to < 2 years: 12 mg/kg/day divided every 12 hours Age 2-5 years: 10 mg/kg/day divided every 12 hours Age 6-11 years: 8 mg/kg/day divided every 12 hours Age 12-17 years, < 45 kg: 8 mg/kg/day divided every 12 hours (Max 200 mg/dose) Age 12-17 years, ≥45 kg: 400 mg every 24 hours |
| PLUS | |
| 2. Second bactericidal agent | |
| Meropenem 2 g every 8 hours | Meropenem 120 mg/kg/dose divided every 8 hours (max 2 g/dose) |
| Alternative | |
| Imipenem 1 g every 6 hours -OR- | Imipenem 100 mg/kg/day divided every 6 hours (max 1 g/dose) -OR- |
| Ampicillin 3 g every 6 hours -OR- | Vancomycin 60 mg/kg/day divided every 8 hours (max 2 g/dose); target trough, 15-20 mcg/mL -OR- |
| | Ampicillin 400 mg/kg/day divided every 6 hours (max 3 g/dose) |
| PLUS | |
| 3. Protein synthesis inhibitor | |
| Linezolid 600 mg every 12 hours | Linezolid < 12 years old: 30 mg/kg/day divided every 8 hours ≥12 years old: 30 mg/kg/day divided every 12 hours (Max 600 mg/dose) |
| Alternative | |
| Clindamycin 900 mg every 8 hours -OR- | Clindamycin 40 mg/kg/day divided every 8 hours (max 900 mg/dose) -OR- |
| Rifampin 600 mg every 12 hours -OR- | Rifampin 20 mg/kg per day divided every 12 hours (max 300 mg/dose) -OR- |
| Chloramphenicol 1 g every 6-8 hours | Chloramphenicol 100 mg/kg per day divided every 6 hours (max 1 g/dose) |
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Contributor Information and Disclosures
Author
David J Cennimo, MD, FAAP, FACP, FIDSA, AAHIVS Associate Professor of Medicine and Pediatrics, Adult and Pediatric Infectious Diseases, Rutgers New Jersey Medical School
David J Cennimo, MD, FAAP, FACP, FIDSA, AAHIVS is a member of the following medical societies: American Academy of HIV Medicine, American Academy of Pediatrics, American College of Physicians, American Medical Association, HIV Medicine Association, Infectious Diseases Society of America, Medical Society of New Jersey, Pediatric Infectious Diseases Society
Disclosure: Nothing to disclose.
Coauthor(s)
Justin R Hofmann, MD Resident Physician, Departments of Internal Medicine and Pediatrics, Rutgers New Jersey Medical School
Justin R Hofmann, MD is a member of the following medical societies: American Medical Association, Infectious Diseases Society of America
Disclosure: Nothing to disclose.
Chief Editor
Michael Stuart Bronze, MD David Ross Boyd Professor and Chairman, Department of Medicine, Stewart G Wolf Endowed Chair in Internal Medicine, Department of Medicine, University of Oklahoma Health Science Center; Master of the American College of Physicians; Fellow, Infectious Diseases Society of America; Fellow of the Royal College of Physicians, London
Michael Stuart Bronze, MD is a member of the following medical societies: Alpha Omega Alpha, American College of Physicians, American Medical Association, Association of Professors of Medicine, Infectious Diseases Society of America, Oklahoma State Medical Association, Southern Society for Clinical Investigation
Disclosure: Nothing to disclose.
Additional Contributors
Burke A Cunha, MD Professor of Medicine, State University of New York School of Medicine at Stony Brook; Chief, Infectious Disease Division, Winthrop-University Hospital
Burke A Cunha, MD is a member of the following medical societies: American College of Chest Physicians, American College of Physicians, Infectious Diseases Society of America
Disclosure: Nothing to disclose.
Acknowledgements
Hilarie Cranmer, MD, MPH, FACEP Director, Global Women's Health Fellowship, Associate Director, Harvard International Emergency Medicine Fellowship, Department of Emergency Medicine, Brigham and Women's Hospital; Director, Humanitarian Studies Program, Harvard Humanitarian Initiative; Assistant Professor, Harvard University School of Medicine
Hilarie Cranmer, MD, MPH, FACEP is a member of the following medical societies: American College of Emergency Physicians, American Medical Association, Massachusetts Medical Society, Physicians for Human Rights, and Society for Academic Emergency Medicine
Disclosure: Nothing to disclose.
Robert G Darling, MD, FACEP Adjunct Clinical Assistant Professor of Military and Emergency Medicine, Uniformed Services University of the Health Sciences, F Edward Hebert School of Medicine; Associate Director, Center for Disaster and Humanitarian Assistance Medicine
Robert G Darling, MD, FACEP is a member of the following medical societies: American College of Emergency Physicians, American Medical Association, American Telemedicine Association, and Association of Military Surgeons of the US
Disclosure: Nothing to disclose.
Ronald A Greenfield, MD Professor, Department of Internal Medicine, University of Oklahoma College of Medicine
Ronald A Greenfield, MD is a member of the following medical societies: American College of Physicians, American Federation for Medical Research, American Society for Microbiology, Central Society for Clinical Research, Infectious Diseases Society of America, Medical Mycology Society of the Americas, Phi Beta Kappa, Southern Society for Clinical Investigation, and Southwestern Association of Clinical Microbiology
Disclosure: Pfizer Honoraria Speaking and teaching; Gilead Honoraria Speaking and teaching; Ortho McNeil Honoraria Speaking and teaching; Abbott Honoraria Speaking and teaching; Astellas Honoraria Speaking and teaching; Cubist Honoraria Speaking and teaching; Forest Pharmaceuticals Speaking and teaching
James Li, MD Former Assistant Professor, Division of Emergency Medicine, Harvard Medical School; Board of Directors, Remote Medicine
Disclosure: Nothing to disclose.
Mauricio Martinez, MD Assistant Medical Director, Department of Emergency Medicine, Winchester Medical Center
Mauricio Martinez, MD is a member of the following medical societies: American Academy of Emergency Medicine
Disclosure: Nothing to disclose.
Barry J Sheridan, DO Chief, Department of Emergency Medical Services, Brooke Army Medical Center
Barry J Sheridan, DO is a member of the following medical societies: American Academy of Emergency Medicine
Disclosure: Nothing to disclose.
Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference
Disclosure: Medscape Salary Employment
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