text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
A(l) is a segment of a parabola opening down. The vertex
of the parabola is contained in the segment. Thus the vertex is the maximum.
Step 5. Compute the derivative. In this case,
A�(l) = −2l + 10.
Step 6. Find all critical points, endpoints, discontinuity points, etc. In most cases, it
suffices to find all critical ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
The point is a meters parallel to the shore. If the swimmer
swims v1 meters per second and runs v2 meters per second, at what distance x from the closest
point on shore should she aim to minimize her time to the target? Mathematically, the swimmer
is at point (0, b1) and wants to reach point (a, −b2), where the shor... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2
.
26
18.01 Calculus
Thus the total time to reach the target is,
Jason Starr
Fall 2005
T = T1 + T2 = (x 2 + b2
1)1/2 +
1
v1
1
v2
((a − x)2 + b2
2)1/2
.
The derivative of T with respect to x is,
(x 2 + b2
1)−1/2(2x) +
((a − x)2 + b2
2)−1/2(−2(a − x))
�
.
dT
dx
=
This simplifies to,
�
1
1
v1 2
� ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
v1 in the first medium and with velocity v2 in the second medium, light rays will refract
upon crossing the boundary between media. Snell’s Law describes the angles of this refraction.
Lecture 11. October 4, 2005
Homework. Problem Set 3 Part I: (g) and (h).
Practice Problems. Course Reader: 2E4, 2E8, 2E9.
1. Rel... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(T ) = k(5)(5)cm 2/s =
25
kcm
2/s.
2. Method for solving relatedrates problems. Many of these steps apply to any word
problem in mathematics.
(i) Identify the independent variable. In the example, this is t.
(ii) Label all constants. In the example, k is a constant.
(iii) Label all dependent variables. In the exa... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Label a coordinate system with the trooper at the origin and the highway equal to the line y = a.
Label the position of the car along the highway as x, moving in the positive direction. Denote by
r the distance of the car from the trooper. Then x and r are dependent variables. The rateof
change dr/dt(T ) is given a... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
the rateofchange of
θ at t = T ?
The independent variable is time t. There is no constant. The dependent variables are the x
coordinate of the point, x(t), and the angle θ(t). The rateofchange dx/dt(T ) is given to be v. The
rateofchange dθ/dt is unknown.
The constraint is somewhat tricky. There are two tang... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
/
− (2
3).
Jason Starr
Fall 2005
3. Another applied max/min problem. As review for Exam 2, this is another applied max/min
problem. A trapezoid is inscribed inside the upper unit semicircle, x2 + y = 1, y ≥ 0. The base
of the trapezoid is the diameter of the semicircle lying on the xaxis. The top of the trapezoid... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
+ (x + 1)
−2x
1 − x
√
2
2
= √
1
1 − x
((1 − x 2) − (x + x)) =
2
2
Because the quadratic polynomial 2x2 + x − 1 factors as,
2x 2 + x − 1 = (2x − 1)(x + 1),
−(2x2 + x − 1)
.
1 − x
√
2
the critical points of A are x = −1 and x = 1/2. But x = −1 does not give a point in the first
quadrant. Thus A is maximized eith... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Sabri Kilic. Notes from the lecture will not be posted. As always,
please do the required reading in the course textbook.
Lecture 13. October 13, 2005
Homework. Problem Set 4 Part I: (a) and (b); Part II: Problem 3.
Practice Problems. Course Reader: 3A1, 3A2, 3A3.
1. Differentials. An alternative notation for derivat... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(x)
G(x)dF (x) + F (x)dG(x)
d(cF (x))
The chain rule has a particularly simple form,
d(F (u)) =
dF
du
du =
dF du
du dx
dx.
31
18.01 Calculus
Jason Starr
Fall 2005
Example. Using differential notation, the derivative of sin(
√
x2 + 1) is,
d sin((x2 + 1)1/2) = cos((x2 + 1)1/2)d(x2 + 1)1/2 = cos((x2 + 1)1/... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
to note, if F (x) is one antiderivative
of f (x), then for each constant C, F (x) + C is also an antiderivative of f (x). The constant C is
called a constant of integration.
In a sense that can be made precise, the problem of differentiation has a complete solution whenever
F (x) is a “simple expression”, i.e., a fu... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
= 1
3 x3 + x2 + x + E ,
32
18.01 Calculus
where E is any constant.
Jason Starr
Fall 2005
Guessandcheck is a game we can lose, as well as win. However, there are a few rules that better the
odds in this guessing game. In fact, they are basically the same rules for derivatives in differential
notation, simply ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
) = x2 . Also, the derivative u�(x) = 2x occurs in f (x) through x = 1/2(2x) =
u�(x)/2. Thus,
x sin(x ) = sin(u(x))u�(x)/2, u(x) = x 2 .
2
Applying integration by substitution,
�
x sin(x2)dx =
�
sin(u(x)) 1 u�(x)dx =
2
�
1
2 sin(u)du =
−1 cos(u) + C = −1 cos(x2) + C.
2
2
Here is a checklist for applying i... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
integrand is a function of sin(x). So substitute u(x) = sin(x). The differential of u is
du = cos(x)dx. The differential sin(x)3 cos(x)dx contains du = cos(x)dx as a factor. The remainder
of the integrand is sin(x)3 = u . So, according to integration by substitution,
3
�
sin(x)3 cos(x)dx =
�
u 3du =
1
4
u 4 + C. ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
a =
r cos(π/N ) and chord length b = 2r sin(π/N ). Thus the area of the polygon is,
A = N = N r 2 sin(π/N ) cos(π/N ) = r
ab
2
2
N
2
sin(2π/N ) = πr 2 sin(2π/N )
.
2π/N
As N increases, 2π/N decreases to 0. Because limt
the area of the polygon approaches,
0 sin(t)/t equals 1, as N approaches infinity,
→
lim
N ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
].
Since an interval is determined by its right and left endpoints, to specify a partition of [a, b], it is
equivalent to give an ordered sequence of increasing numbers,
a = x0 < x1 < x2 < · · · < xn−2 < xn−1 < xn = b.
The kth subinterval of the partition is the interval [xk−1, xk], having length,
Δxk = xk − xk−1. ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
· · ·
3. Riemann sums. Let f (x) be a function defined on an interval a ≤ x ≤ b. Given a partition
< xn = b of [a, b], and given a choice, for every k = 1, . . . , n, of element x∗ in the
a = x0
kth subinterval, xk−1 ≤ x∗ ≤ xk , the curvilinear region bounded by y = f (x) and the xaxis is
approximated by a union o... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
yk,min = min{f (x) xk−1 ≤ x ≤ xk+1}.
|
Similarly, the largest value f (x) takes on is denote by,
|
yk,max = max{f (x) xk−1 ≤ x ≤ xk+1}.
For every choice of x∗ in the kth interval, y∗ is trapped between these two values,
k
k
yk,min ≤ y∗
k ≤ yk,max.
Denoting,
the area ΔAk is trapped between these two values,
ΔAk,... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
is called the
Riemann integral,
� b
a
f (x)dx =
lim Amin =
mesh 0
→
lim Amax.
mesh 0
→
Also, f (x) is said to be Riemann integrable on the interval [a, b]. Another name for the Riemann
integral is the definite integral.
36
18.01 Calculus
Jason Starr
Fall 2005
Example. Consider the function f (x) = x on t... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
also gives,
k=1
n
�
k =
n(n + 1)
.
2
n
�
(k − 1) =
n−1
n−1
� �
l =
l =
k=1
l=0
l=1
(n − 1)n
2
,
by making the substitution l = k − 1. Substituting the formula gives,
Amin =
L2 n(n − 1)
2n
2
=
L2
2
(1 −
1
),
n
and,
Therefore,
Similarly,
Amin =
L2 n(n + 1) L2
n2
2
=
2
(1 + ).
1
n
lim Am... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
))dx =
a
�
b
a (r ·
=
f (x))dx
�
c
f (x)dx + b f (x)dx =
� b
a
� b
a
� b
f (x)dx + a g(x)dx,
� b
·
r
a f (x)dx,
�
f (x)dx.
c
a
Lecture 15. October 18, 2005
Homework. Problem Set 4 Part I: (d) and (e); Part II: Problem 2.
Practice Problems. Course Reader: 3B6, 3C2, 3C3, 3C4, 3C6.
1. The Riemann sum... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
=
n
�
k=1
ekb/n
b
.
n
To evaluate each of the sums, make the substitution c = eb/n . Then the lower sum is,
Amin =
The sum is a geometric sum,
n
b �
n
k=1
c k−1 =
n−1
b �
n
l=0
l
c .
Plugging this in gives,
(1 + c + c 2 + · · · + c n−2 + c n−1) =
n
c − 1
.
c − 1
Amin =
n
b c − 1
n c − 1
=
b
n... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
he − 1
lim
h 0 h
→
= 1.
Inverting gives,
h
lim
h→0 eh − 1
=
�
h
e − 1 �−1
lim
h 0 h
→
= (1)−1 = 1.
Also, because ex is continuous,
Putting this together gives,
lim e h = e 0 = 1.
h 0→
lim Amin = (e b − 1) lim
n→∞
h→0 eh
h
− 1
= (e b − 1)(1) = e b − 1.
Similarly,
lim Amax = (e b − 1)(lim e )(lim
n→... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(b1/n)n = b.
The length of the kth subinterval is,
Δxk = xk − xk−1 = q k − q k−1 = q k−1(q − 1).
Observe this increases as k increases. So this is not the partition of [1, b] into n equal subintervals.
The mesh size is,
mesh = max(Δx1, . . . , Δxn) = Δxn = (q − 1)b(n−1)/n ≤ q − 1.
As n tends to infinity, the mesh ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
to,
n
�
yk,maxΔxk =
n
�
q q(k−1)(q − 1).
kr
k=1
k=1
Amax = (q − 1)q r
n
�
q(k−1)(r+1)
.
To evaluate the sum, make the substitution c = qr+1 . Then the sum is,
k=1
n
�
c k−1 = 1 + c + c + · · · + c n−2 + c
2
n−1
.
k=1
This geometric sum equals,
n
c − 1
c − 1
Thus the upper and lower sums simplify t... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
=
d(xr+1)
dx
x=1 = ((r + 1)x r |x=1 = (r + 1).
|
41
18.01 Calculus
Jason Starr
Fall 2005
Also, since x is continuous,
r
Substituting this in gives,
lim q r = 1r = 1.
q→1
lim Amin = (br+1 − 1)
n→∞
lim Amax = (br+1 − 1)
n→∞
lim q r
q→1
�
qr+1 − 1 −1
�
lim
q→1 q − 1
� �
�
=
br+1
− 1
,
r + 1
qr... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
f (x) is a continuous function. Let
x = a be a fixed point where f (x) is defined. Form the function,
F (x) =
�
x
a
f (t)dt.
The function F (x) is defined whenever f (t) is defined on all of [a, x]. If f (x) is continuous, the
Fundamental Theorem of Calculus asserts F (x) is differentiable and,
dF
dx
(x) =
x
�
... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
k
k
n
�
ymaxΔtk .
k=1
n
�
yminΔtk = ymin
n
�
Δtk = yminΔx,
because the total length of the interval [x, x + Δx] is Δx. Similarly, the upper bound is,
k=1
k=1
n
�
ymaxΔtk = ymaxΔx.
Thus the Riemann sum is squeezed between,
k=1
yminΔx ≤
n
�
y
k=1
∗Δ ≤x
k
k
y
maxΔx.
Because the Riemann integral is a limi... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Calculus
Jason Starr
Fall 2005
4. Algorithm for computing Riemann integrals. The Fundamental Theorem of Calculus has many
important applications. The most obvious is to give us a simpler method for computing Riemann
integrals, under the hypothesis that we can compute the antiderivative. If f (x) is a continuous
f... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Observe, the variable x does not appear in this limit. It is very convenient to include the variable
x in the notation for the Riemann integral; for how else are we to express the function integrated?
But, since the definition of the Riemann integral does not involve x, x is really a dummy variable.
Any variable name... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
) is beyond our current techniques of integration (though soon we will have tech
niques to solve it). The simplest solution is indirect. Here, first, is the direct solution. The integral
f (θ) equals the area of 2 triangles and a circular sector. By highschool geometry, the area is,
f (θ) =
π − 2θ
2
+ 2( sin(θ) co... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
The argument above using the chain rule and the Fundamental Theorem of Calculus is quite general.
It gives the general equation,
d/dx
� v(x)
u(x) f (t)dt = f (v(x))v�(x) − f (u(x))u�(x).
3. Geometric area and algebraic area. The Riemann integral is the algebraic area,
� b
a
f (x)dx = Area above the xaxis − Area... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(cos(x) −π + (− cos(x) π = (1 − (−1)) + (−(−1) + 1) =
0
|
4.
Thus the geometric area does not equal the algebraic area. But computation of the geometric area
reduces to a straightforward Riemann integral.
46
18.01 Calculus
Jason Starr
Fall 2005
4. Estimates. For every pair of Riemann integrable functions f (x)... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2
1000
−
√
3
1
√
21
10000000
2
3
1
21
�
�
�
�
0
1
−
6
= 0.1210667926±10−10
.
0
Similarly,
1 + x
1/2dx =
�
x +
� 0.1
0
0.1
2
3
x
�
�
�
3/2
�
0
= 0.1 +
2
1000
√
3
= 0.1210818511 ± 10−10
.
Since these two integrals agree to within ±10−4, this gives the original integral,
� 0.1
�
1 +
0
sin(... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
�
π/3 sin(x)
π/4 cos(x)
dx.
In this form, the substitution u = cos(x) is natural,
�
x=π/3 sin(x)
x=π/4 cos(x) dx,
u = cos(x)
du = − sin(x)dx u(π/4) = cos(π/4) = 1/
u(π/3) = cos(π/3) = 1/2,
2.
√
�
u=1/2 1 (−du).
2 u
u=1/
√
The new integral can be computed by the Fundamental Theorem of Calculus, since 1/u is ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
comes so naturally for most calculus students, it barely warrants
mention. Technically, the Riemann integral,
� b
a
f (x)dx,
48
18.01 Calculus
Jason Starr
Fall 2005
is only defined if a ≤ b. What if a is larger than b? The only possible answer consistent with the
Fundamental Theorem of Calculus is the followin... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
trivial) solution,
y = sin(x 2).
Because the solution is unique, it depends on 0 parameters (and the order is 0).
(ii) The ordinary differential equation,
dy
dx
−
1
x + 1
= 0,
has order 1 because dy/dx occurs and no higher derivatives occur. Every solution is an antiderivative
of 1/x + 1,
�
y =
1
x + 1
dx... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
constant coefficient. The solution of constant
coefficient linear ordinary differential equations is a main focus of Math 18.03.
2. Separable differential equations. Many differential equations arising in applications are
examples of separable differential equation. A separable ordinary differential equation is a first
order... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(y)
⇒ h(y)dy = g(x)dx.
50
18.01 Calculus
In the example, this gives,
dy
dx
=
−x
y
⇒ ydy = −xdx.
Jason Starr
Fall 2005
(iii). Antidifferentiate both sides of the equation. In the example, the antiderivatives
�
�
ydy = −xdx,
give,
1 2 −1 2
y =
2
2
x + C.
(iv). If there is an inital value, use it to ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
a single separable ordinary differential
equation satisfied by every function,
y = (x − a)3 ,
where a is an arbitrary constant. Find this differential equation, and find all its solutions.
51
Jason Starr
Fall 2005
18.01 Calculus
The derivative of y is,
dy
dx
= 3(x − a)2 .
The constant a can be eliminated by wr... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
F (x, y) is both continuous and differentiable at (x0, y0).
In the previous example, F (x, y) = 3y2/3 is continuous at y0 = 0. But it is not differentiable at
y0 = 0. Ultimately, this is the reason for the extra solutions of the differential equation.
3. Applications. Separable differential equations come up often in ap... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
not too large.
A slightly more realistic model hypothesizes a constant, equilibrium population Nequi sustainable
indefinitely. The model is that the population grows at a rate proportional both to the population
N and the difference Nequi − N ,
dN
dt
= kN (Nequi − N ).
This is again a separable differential equatio... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
halflife is the length of time necessary
for the mass of radioactive isotope to decrease to onehalf the initial mass,
Solving in the formula gives,
m(Thalf) = m0/2.
Thalf =
ln(2)
/k.
Example. The halflife of a certain radioactive isotope is 20 years. How long is required for the
mass to decrease to 1% of the i... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
3G5.
1. Approximating Riemann integrals. Often, there is no simpler expression for the antideriva
tive than the expression given by the Fundamental Theorem of Calculus. In such cases, the simplest
method to compute a Riemann integral is to use the definition. However, this is not necessarily the
most efficient method.... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
1), (xk , 0) and (xk , yk ). In particular, if the graph of y = f (x) is a line, this
trapezoid is precisely the region between the graph and the xaxis over the interval [xk−1, xk ]. Thus,
the approximation above gives the exact integral for linear integrands.
Writing out the sum gives,
Itrap =
b − a
2n
((y0 + ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
the area under the
unique parabola containing the 3 points (x2l−2, y2l−2), (x2l−1, y2l−1), (x2l, y2l). For notation’s sake,
denote 2l − 1 by k. Thus the 3 points are (xk−1, yk−1), (xk , yk ), and (xk+1, yk+1) (this is slightly
more symmetric).
The first problem is to find the equation of this parabola. Since the para... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
k ) and (xk+1, yk+1) is,
y = A(x − xk )2 + B(x − xk )2 + yk ,
x)2 ,
yk−1 − 2yk + yk+1)/
A
yk+1 − yk−1)/
= (
2(Δ
2(Δ
x).
= (
B
The next problem is to compute the area under the parabola from x = xk−1 to x = xk+1. This is a
straightforward application of the Fundamental Theorem of Calculus,
�
xk+1
xk−1
A(x − x... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
, summing this contribution over each choice of l gives the Simpson’s rule approximation,
ISimpson =
Writing out the sum gives,
m
b − a �
6m
l=1
(y2l−2 + 4y2l−1 + y2l).
ISimpson =
b−a
6m
((y0 + 4y1 + y2) + (y2 + 4y3 + y4) + (y4 + 4y5 + y6)+
· · · + (y2m−4 + 4y2m−3 + y2m−2) + (y2m−2 + 4y2m−1 + y2m)).
Gathering ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2y2 + 2y3 + y4) = ( + 2 + 2 + 2 + ) =
1 4
8 4
4
5
4
6
4
7
4
8
1171
1680
≈
0.6970
b − a
2n
For Simpson’s Rule, because n equals 4, m equals 2. Thus,
ISimpson =
b − a
6m
(y0 + 4y1 + 2y2 + 4y3 + y4) =
1 4
12 4
4
( + 4 + 2 + 4 + ) =
6
4
8
4
5
4
7
1747
2520
≈
0.6933
57
18.01 Calculus
Jason St... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
x(t), the ycoordinate of the particle, y(t), and the distance L(t) from the particle to (0, 0). The
constant is the speed s = 3 of the particle. The constraints are that the point moves on the
parabola,
and the Pythagorean theorem,
Also, since the speed is constant,
2
y = x ,
L2 = x + y .
2
2
� �2 � �2
+
.
2... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
known rateofchange”, dx/dt at the moment when L equals
√
2
5. Implicitly differentiating the equation y = x2 gives,
dy
dt
= 2x
dx
.
dt
58
18.01 Calculus
Jason Starr
Fall 2005
Substituting this into the equation for s2 gives,
� �2 �
2 s =
+ 2x
dx
dt
�2
dx
dt
= (1 + 4x 2)
� �2
dx
dt
.
Since s is know... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
27/
85 .
Homework. Problem Set 5 Part I: (d) and (e); Part II: Problems 2 and 3.
Practice Problems. Course Reader: 4A1, 4A3, 4B1, 4B3, 4B6.
1. Differentials revisited. In a typical applied integration problem, the main difficulty is finding
the integrand and the limits of integration. An unknown quantity, for inst... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
between y = g(x) and the xaxis. Each of these is a
Riemann integral.
The differential method asks, what is the infinitesimal change in the area A as x varies from x
to x + dx? The infinitesimal region is a rectangle of base dx and height f (x) − g(x). Thus the
infinitesimal change in A is,
Integrating gives,
dA = he... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
= −1 is a solution. Thus the polynomial factors as,
3 x − 3x − 2 = (x + 1)(x 2 − x − 2) = (x + 1)2(x − 2).
The remaining intersection point is (2, 2). So the area bounded by the curve y = x(x2 − 3) and the
tangent line y = 2 is,
�
x=2
x=−1
2 − (x(x 2 − 3))dx =
� 2
−1
−x 3 + 3x + 2dx.
Using the Fundamental The... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
has
height H above the base.
The cone is the solid of revolution for the plane region bounded by x = 0, the xaxis, and the line
containing (0, R) and (H, 0). The equation of this line is,
Thus the area of an infinitesimal disk is,
y =
(H − x)R
.
H
dV = Area × width = π
(H − x)2R2
H 2
dx,
and the volume is, ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
01 Calculus
Jason Starr
Fall 2005
Evaluating gives the volume,
V =
πR
4
3/3.
4. The slice method. A generalization of the disk method is the slice method. The problem
is to find the volume of a region bounded by the planes x = a and x = b given the knowledge
of the area A(x) of the slice of the solid by the plane... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
x) = L − x.
Thus the slice area is
A(x) = base × altitude = (L − x)2 .
1
2
1
2
Thus the infinitesimal volume is,
dV = A(x)dx = (L − x)2dx,
1
2
giving the total volume,
�
V =
�
dV =
x=0
Make the substitution u = L − x, du = −dx to get,
x=L 1
2
(L − x)2dx.
Evaluating gives,
V
=
�
u=0
u=L
1 2
u
2
(−... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
.
Example. The main part of a dog dish is a solid of revolution whose radial crosssection is a
triangle of height H whose base has inner radius Ri and outer radius Ro. Find the volume of
material used to make the dog dish.
Note. This integral was only setup in lecture. The derivation will be completed in recitatio... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
ro + ri
2
�
(ro − ri).
Note the first factor is the circumference of the center of the annulus. And the second factor is the
radial width of the annulus. Thus the area of an annulus is the circumference of the center times
the radial width.
63
18.01 Calculus
Jason Starr
Fall 2005
Back to the problem, the cente... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
i )
2.H/
One reality check is that this is the same volume as a cylinder with the same center (Ri + Ro)/2
and height H as the dish, and whose (constant) radial width equals the average radial width of the
dish, (Ro − Ri)/2.
Lecture 20. November 1, 2005
Practice Problems. Course Reader: 4C2, 4C6, 4D1, 4D4, 4D8.... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Δx.
k
The sum is a Riemann sum. To get better approximations to the true average, increase the number
of points n at which f (x) is “sampled”. In the limit, this gives the true average,
Average =
n
1 �
lim
b − a n→∞
k=1
y∗Δx =
k
� b
a
f (x)dx/(b − a).
Example. Under ideal conditions, a wireproducing machine... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
likely. Typically, the probability
that x has value in the range from x0 to x0 + Δx is approximately,
Prob(x0 ≤ x ≤ x0 + Δx) ≈ ρ(x0)Δx,
65
18.01 Calculus
Jason Starr
Fall 2005
for some nonnegative continuous function ρ(x). The function ρ(x) is called a probability distribution.
Assuming this approximation beco... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
, the probability
distribution is empirically observed to be,
ρ(x) = Ce−x2/2σ2 ,
where σ is a constant determining the “width” of the probability distribution, and C is an unde
termined normalization constant. What is the average distance of the particle from the center of
the screen? Assume the particle lies in an... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Theorem of Calculus, this equals,
2Cσ2 (e −R2/2Σ2 = 2Cσ2(1 − e−R2/2Σ2 ).
u 0
|
As R becomes large, the quantity e−R2/2Σ2 becomes vanishingly small. Thus, in the limit as R
tends to ∞, the numerator equals,
� R
lim
R→∞ −R
|x|Ce−x2/2σ2 dx = 2Cσ2 .
Unfortunately, this is not an answer, because the normalization co... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
1 = 1/
2π.
Unfortunately, we cannot yet prove this. Taking it as true gives the final answer,
Average distance =
√
σ/
2
2π.
3. Volumes of solids of revolution: the shell method. An alternative to the disk and washer
method is the shell method. A shell is the region between 2 cylinders of the same height. If the
ave... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
The volume was computed using the washer method. This time it will be computed using the shell
method. The triangular region is the union of two regions. The first region is bounded by x = Ri,
x = (Ri + Ro)/2, the xaxis, and the line segment,
y =
2H
Ro − Ri
(x − Ri).
68
18.01 Calculus
Jason Starr
Fall 2005
... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
=
4πH
Ro − Ri
� x=Ro
x=(Ro+Ri )/2
x(Ro − x)dx.
Believe it or not, this will be simpler to deal with after the substitution u = x − (Ro + Ri)/2,
du = dx. The new integral is
V2 =
4πH
Ro − Ri
�
u=(Ro−Ri)/2
(u + (Ro + Ri)/2)(−u + (Ro − Ri)/2)du.
u=0
Notice how similar are the integrals for V1 and V2. They have... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Jason Starr
Fall 2005
By the Fundamental Theorem of Calculus, this equals,
4πH
Ro − Ri
(Ro + Ri)
�
−u
2
2
+
�
(Ro−Ri)/2
�
(Ro − Ri)u
�
�
2
0
.
This evaluates to,
This simplifies to give,
4πH
Ro − Ri
(Ro + Ri)
(Ro − Ri)2
.
8
V = πH(Ro − Ri)(Ro + Ri)/2 = π(R2
o − R2
i )
2.H/
This is precisely the sam... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
condition y > 0.
There is a third important way to specify a curve: using parametric equations. Given a parameter
t varying in an interval a ≤ t ≤ b and given functions f (t) and g(t) on this interval, the associated
parametric curve,
�
x = f (t),
y = g(t)
is simply the set of all pairs (x, y) = (f (t), g(t)) as ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
g/2)t2 + v0 sin(α)t + y0
v0 cos(α)t + x0,
0 ≤ t.
This is a parametric equation where time t is the parameter. Even when some other quantity is
the parameter, it is often useful to think of the parameter as time. Thus the curve is the trail left
by a point, or perhaps better, the tip of a pen, as it moves in the pl... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
⇔ t =
x − x0
.
v0 cos(α)
This can then be substituted into the equation for y to get an explicit equation for the curve,
y = −g(x − x0)2/(2v2
0 cos2(α
)) + tan(
α)(x − x0
) +
y0.
In going from a parametric equation to an implicit equation, there are 2 important warnings to
keep in mind:
71
18.01 Calculus
Ja... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
curve? Imagining the curve
made of some flexible extensible material like wire, what is the length when the wire is pulled taut?
The answer is called the arc length, s.
The method for expressing arc length is an integral is by now familiar. Break the interval a ≤ t ≤ b
into a large number n of subintervals with endp... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
k=1
Arc length =
� t=b
�
t=a
� �2 � �2
+
dt.
dx
dt
dy
dt
Example 1. For the parametric curve in Example A above,
dx
dθ
= −r sin(θ),
dy
dθ
= r cos(θ).
Therefore, the expression,
equals,
� �2 � �2 � �2
dx
dt
dy
dt
ds
dt
+
=
,
(−r sin(θ))2 + (r cos(θ))2 = r 2 sin2(θ) + r cos (θ) = r (sin2(θ) + cos 2(... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
,
x =
y = f (t)
a ≤ t ≤ b.
73
Jason Starr
Fall 2005
18.01 Calculus
Then,
Using this,
Thus the arc length is,
dx
dt
= 1,
dy
dt
= f �(t).
�
1 + (f �(t))2dt.
ds =
Arc length =
� t=b �
1 + (f �(t))2dt.
t=a
Since the parameter t in the Riemann integral is only a dummy variable anyway, it is allowed ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
= (x + x−1).
1
2
74
18.01 Calculus
Therefore,
ds =
1
2
(x + x−1)dx.
Integrating gives the arc length,
�
s =
ds =
�
x=b
x=a
1
2
(x + x−1)dx =
2
�
1 x
2
2
Evaluating gives the arc length,
Jason Starr
Fall 2005
.
b
�
�
�
+ ln(x)
�
a
Arc length =
(b2 − a2)/
/
4 + (1 2) ln(
b/a).
Lecture 22. Nove... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
for the area and circumference of a sector of a circle give the identity,
Area of sector = (Radius of sector) × (Circumference of sector).
1
2
Thus, the area of the cone equals,
A = (S)(2πR) =
1
2
πRS
.
In particular, the height H is involved only indirectly (as H depends on H).
Next, consider a conical band ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2 gives,
A = 2
.πrs
2. Surface area of a surface of revolution. Given a segment of a parametric curve,
�
x = x(t),
y = y(t)
a ≤ t ≤ b
the surface of revolution is the surface obtained by revolving the segment through xyzspace about
the yaxis. What is the area of this surface? The answer is called the surface a... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
2 ,
and the differential arc length of the line segment is,
S
ds = dx.
R
Thus the differential element of surface area is,
Integrating gives,
dA = 2πrds = 2πx dx.
S
R
�
A =
dA =
�
x=R
x=0
2πS
R
xdx =
2
�
2πS x
2
R
R
�
�
�
�
0
=
πRS
.
This is the same formula obtained above by more elementary means.... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
is to observe the surface area A and the volume V of a sphere of
radius R are related by,
A = 4πR2 =
= (4πR3/3).
dV
dr
d
dr
C. An astroid is a curve,
2/3 + y 2/3
x
= a
2/3
.
The part of the astroid in the first quadrant has parametric equation,
�
x = a cos3(t),
y = a sin3(t)
0 ≤ t ≤
π
.
2
The derivativ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
du = − sin(t)dt, u(0) = 1, u(π/2) = 1 to get,
A =
� u=0
u=1
6πa 2 u (−du) = 6πa 2
4
� u=1
u=0
u 4du.
78
18.01 Calculus
Jason Starr
Fall 2005
Thus the surface area of the surface of revolution is,
�
�
A = 6πa 2 u /5
5
�
1 =
0
πa
6
2/5.
3. Polar coordinate curves. After the explicit, Cartesian form of a cu... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
polar coordinates. But for a polar
coordinate curve these are simply parameters. They are very closely related to, but often different
from, the actual distance and angle. This is easiest to think about by imagining the point swerving
through the origin along the radius line to the opposite ray of the ray given by θ.... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
A similar curve occurs in Part II, Problem 2 of Problem
Set 6.
1. Find the range of θ. In almost every case, this will be given. In this case, the range is given
as −π/4 ≤ θ ≤ 7π/4. In some cases, the range must be determined. For instance, to sketch only
the “leaf” of the rose containing (1, 0), first the endpoints... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
occur at
such points, but also occur everytime the curve crosses the origin (so that |r| equals 0).
| |
| |
80
18.01 Calculus
Jason Starr
Fall 2005
In our example, the local minima are all points where r = 0, enumerated above. The derivative of
r is,
r�(θ) = −2 sin(2θ).
The critical points are θ = 0, π/2, π a... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
equation is y = a, which is a line
parallel to the xaxis, this is correct.
5. Find the tangent direction at important points. This will be discussed further next time. The
most important tangent directions are when the curve crosses the origin, and critical points of r.
If r(θ) = 0 and r�(θ) = 0, the tangent line ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
for
general parametric curves. It was discussed for polar curves, which are a special collection of
parametric curves.
81
�
�
18.01 Calculus
Given a parametric curve,
�
x = f (t),
y = g(t),
Jason Starr
Fall 2005
what is the slope of the tangent line at (f (a), g(a))? The relevant differentials are,
If g�(a) i... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
)) in the positive xdirection,
and the tangent line. To be precise, there are two such angles, differing by π. The defining equation
for α is,
tan(α) =
dy
.
dx
And, of course,
tan(θ) =
y
.
x
Define ψ to be the difference between α and θ,
The angle addition/subtraction formulas for tan(θ) are,
ψ = α − θ.
tan(φ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
xdx + ydy = r(θ)r�(θ)dθ.
To compute the numerator in the expression, differentiate both sides of,
tan(θ) =
y
,
x
to get,
1
= 2
x
Now substitute x = r cos(θ) in the denominator to get,
sec (θ)dθ =
ydx
x2
dy
x
−
2
(xdy − ydx).
2
sec (θ)dθ =
1
r2 cos2(θ)
(xdy − ydx) =
sec2(θ)
2r
(xdy − ydx).
Cancelling s... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
), this simplfies to,
2 cos2(θ/2)
−2 sin(θ/2) cos(θ/2)
= − cot(θ/2).
Of course there is an identity,
− cot(u) = tan(u − π/2).
Altogether, this gives,
Therefore,
tan(ψ) = − cot(θ/2) = tan(θ/2 − π/2).
Since α equals θ + ψ, this gives,
ψ =
(θ − π)/2.
α =
(3θ − π)/2.
In particular, the angle of the tangent lin... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
length for a polar curve,
ds = �
[r�(θ)]2 + [r(θ)]2dθ.
Example. For the cardioid,
the derivative is,
Thus,
r(θ) = a(1 + cos(θ)),
r�(θ) = −a sin(θ).
(r�)2 + r = a (1 + cos(θ))2 + (−a sin(θ))2 = a (1 + 2 cos(θ) + cos 2(θ)) + a 2 sin2(θ).
2
2
2
This simplifies to,
2a 2(1 + cos(θ)).
To simplify this further, writ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
only briefly
discussed in lecture. Here is a continuation of the previous problem.
Example. The top half of the cardioid,
r(θ) = a(1 + cos(θ)), 0 ≤ θ ≤ π,
is revolved about the xaxis to give a fairly good approximation of the surface of an apple. What
is the surface area of this apple?
85
18.01 Calculus
Jason Sta... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
integral is,
A = 16πa 2
�
u=0
u=1
u (−2du) = 32πa 2
4
�
u=1
u=0
u 4du = 32πa 2
�
5u
5
This evaluates to give the total surface area of the apple,
A = 32
πa
2/5.
1
�
�
�
�
0
.
5. Area of a region enclosed by a polar curve. What is the area of the planar region enclosed
by a cardioid? By the same sort of re... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
Calculus, this equals,
0
�
3θ + 4 sin(θ) +
1
2
2
a
4
�
�
sin(2θ)
�
�
2π
0
.
Evaluating gives,
A =
πa
3
2/2.
Lecture 24. November 15, 2005
Practice Problems. Course Reader: 5A1, 5A2, 5A3, 5A5, 5A6.
1. Inverse functions. Let a, b, s and t be constants. Let y = f (x) be a function defined on the
interv... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
)) = g(f (x2)) = x2,
i.e., x1 equals x2. In other words, two distinct inputs x1 and x2 give two distinct outputs f (x1) and
f (x2). A function satisfying this condition is called onetoone, because to every output, there is
at most one input. This is the first necessary condition: every invertible function is oneto... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
as the graph of f (y) = x. This
is simply the usual graph of y = f (x) with the roles of x and y reversed. What this translates to
is, the graph of f −1 is the same as the graph of f with the roles of the xaxis and yaxis reversed.
The simplest way to get the graph of f −1(x) is simply to reflect the graph of f (x) ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
graph is an increasing function that is asymptotic to the line y = −π/2 as x → −∞, and
asymptotic to the line y = +π/2 as x → +∞.
4. Derivatives of inverse functions. A particular simple formulation of the chain rule is the
differential formulation,
df (u) = f �(u)du.
88
18.01 Calculus
Jason Starr
Fall 2005
If... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(θ) + cos 2(θ) = 1.
cos(θ) =
�
1 − sin2(θ) =
√
1 − x .
2
This gives a very useful formula for the derivative of sin−1(x),
d
dx
(sin−1(x)) =
√
1/
1 − x2 .
There is a very similar derivation that,
d
dx
(cos−1(x)) =
√
−1/
1 − x2 .
89
18.01 Calculus
Jason Starr
Fall 2005
This looks remarkably similar to... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
ned on the entire real line.
6. Hyperbolic trigonometric functions. The trigonometric functions are very useful for dis
cussing point on the unit circle x2 + y2 = 1, because the circle is the parametric curve,
�
x = cos(θ),
sin(θ)
y =
Are there analogous continuous functions for the points on the hyperbola x2 − ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
t
2
,
et + e−t
2
,
et − e−t
=
=
sech(t) =
csch(t) =
and,
coth(t) =
1
tanh(t)
=
cosh(t)
sinh(t)
=
et + e−t
.
et − e−t
The first observation is that,
cosh2(t) =
1
4
1
(e t + e−t)2 = (e 2t + 2 + e−2t),
4
1
sinh2(t) = (e t − e−t)2 =
4
1
4
(e 2t − 2 + e−2t).
Taking the difference of these, most of the... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
(x)
(cosh(x) · cosh(x) − sinh(x) · sinh(x)) =
1
cosh2(x)
=
sech
2(x).
91
18.01 Calculus
Jason Starr
Fall 2005
Lecture 25. November 17, 2005
Homework. Problem Set 7 Part I: (a)–(e)
Practice Problems. Course Reader: 5D2, 5D6, 5D7, 5D10, 5D14
1. Inverse hyperbolic functions. There are a few other useful formul... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2005/400735651ecc95647e4c74067ad2fa3f_ocw01f05sums.pdf |
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