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n v = rω has units length We need to talk about units here. The units of v are length time , the units of r are length only, and the units of ω are radians time , whereas time = length·radians the right hand side has units length · radians . The supposed contradiction in units is resolved by remembering that radians ar... |
ne and Sine 717 10.2 The Unit Circle: Cosine and Sine In Section 10.1.1, we introduced circular motion and derived a formula which describes the linear velocity of an object moving on a circular path at a constant angular velocity. One of the goals of this section is describe the position of such an object. To that end... |
e √ = 3 2 . 2 = − 2 and sin − 5π 3 measures more than 2π = 6π y 1 π 4 1 x θ = − 5π 4 y 1 θ = 7π 3 π 3 1 x Finding cos − 5π 4 and sin − 5π 4 Finding cos 7π 3 and sin 7π 3 724 Foundations of Trigonometry 6 as a reference angle, those with a denominator of 4 have π The reader may have noticed that when expressed in radian... |
4 x Q(4, −2) −2 −4 y 3960 Q (x, y) 41.628◦ 3960 x The terminal side of θ contains Q(4, −2) A point on the Earth at 41.628◦N Using Theorem 10.3, we get x = 3960 cos (41.628◦). Using a calculator in ‘degree’ mode, we find 3960 cos (41.628◦) ≈ 2960. Hence, the radius of the Earth at North Latitude 41.628◦ is approximately ... |
ng is the side opposite θ? 63. If θ = 5◦ and the hypotenuse has length 10, how long is the side adjacent to θ? 64. If θ = 37.5◦ and the side opposite θ has length 306, how long is the side adjacent to θ? In Exercises 65 - 68, let θ be the angle in standard position whose terminal side contains the given point then comp... |
provided cos(θ) = 0; if cos(θ) = 0, sec(θ) is undefined. csc(θ) = 1 sin(θ) , provided sin(θ) = 0; if sin(θ) = 0, csc(θ) is undefined. tan(θ) = sin(θ) cos(θ) , provided cos(θ) = 0; if cos(θ) = 0, tan(θ) is undefined. cot(θ) = cos(θ) sin(θ) , provided sin(θ) = 0; if sin(θ) = 0, cot(θ) is undefined. It is high time for an exa... |
d side and multiplying the numerator and denominator by the quantity 1 + cos(θ): sin(θ) 1 − cos(θ) = = = sin(θ) (1 − cos(θ)) · (1 + cos(θ)) (1 + cos(θ)) = sin(θ)(1 + cos(θ)) (1 − cos(θ))(1 + cos(θ)) sin(θ)(1 + cos(θ)) 1 − cos2(θ) sin(θ)(1 + cos(θ)) sin(θ) sin(θ) = = sin(θ)(1 + cos(θ)) sin2(θ) 1 + cos(θ) sin(θ) In Examp... |
imit ∞, but rather, ranges through all of the natural numbers, the index k in the union ∞ (2k + 1)π 2 , (2k + 3)π 2 k=−∞ can never actually be ∞ or −∞, but rather, this conveys the idea that k ranges through all of the integers. Now that we have painstakingly determined the domain of F (t) = sec(t), it is time to discu... |
- 128, verify the identity. Assume that all quantities are defined. 82. cos(θ) sec(θ) = 1 84. sin(θ) csc(θ) = 1 86. csc(θ) cos(θ) = cot(θ) 88. 90. 92. cos(θ) sin2(θ) = csc(θ) cot(θ) 1 − cos(θ) sin(θ) = csc(θ) − cot(θ) sin(θ) 1 − cos2(θ) = csc(θ) 83. tan(θ) cos(θ) = sin(θ) 85. tan(θ) cot(θ) = 1 87. 89. 91. 93. sin(θ) cos... |
erminal with θ0, there is some integer k so that θ = θ0 + 2π · k. Therefore, −θ = −θ0 − 2π · k = −θ0 + 2π · (−k). Since k is an integer, so is (−k), which means −θ is coterminal with −θ0. Hence, cos(−θ) = cos(−θ0) and sin(−θ) = sin(−θ0). Let P and Q denote the points on the terminal sides of θ0 and −θ0, respectively, w... |
pplicable angles θ, cos(2θ) = cos2(θ) − sin2(θ) 2 cos2(θ) − 1 1 − 2 sin2(θ) sin(2θ) = 2 sin(θ) cos(θ) tan(2θ) = 2 tan(θ) 1 − tan2(θ) The three different forms for cos(2θ) can be explained by our ability to ‘exchange’ squares of cosine and sine via the Pythagorean Identity cos2(θ) + sin2(θ) = 1 and we leave the d... |
, find (a) cos(α + β) (b) sin(α + β) (c) tan(α + β) (d) cos(α − β) (e) sin(α − β) (f) tan(α − β) 23. If csc(α) = 3, where 0 < α < π 2 , and β is a Quadrant II angle with tan(β) = −7, find (a) cos(α + β) (b) sin(α + β) (c) tan(α + β) (d) cos(α − β) (e) sin(α − β) (f) tan(α − β) 24. If sin(α) = 3 5 , where 0 < α < π 2 , an... |
= cos(t) for any integer k. To determine the period of f , we need to find the smallest real number p so that f (t + p) = f (t) for all real numbers t or, said differently, the smallest positive real number p such that cos(t + p) = cos(t) for all real numbers t. We know that cos(t + 2π) = cos(t) for all real numbers t bu... |
the form prescribed in Theorem 10.23, ω induced by φ. f (x) = 3 cos πx − π 2 + 1 = 3 cos π 2 x + − π 2 + 1, 7Try using the formulas in Theorem 10.23 applied to C(x) = cos(−x + π) to see why we need ω > 0. 796 Foundations of Trigonometry 2 , φ = − π π/2 = 4, the amplitude is |A| = |3| = 3, the phase shift is − φ so tha... |
nce, sec(x) inherits its period from cos(x). 12In Section 10.3.1, we argued the range of F (x) = sec(x) is (−∞, −1] ∪ [1, ∞). We can now see this graphically. 10.5 Graphs of the Trigonometric Functions 801 As one would expect, to graph y = csc(x) we begin with y = sin(x) and take reciprocals of the corresponding y-valu... |
- 24, graph one cycle of the given function. State the period of the function. 13. y = tan 16. y = sec x − x − π 3 π 2 14. y = 2 tan 17. y = − csc x − 3 1 4 x + π 3 15. y = 1 3 18. y = − 19. y = csc(2x − π) 20. y = sec(3x − 2π) + 4 21. y = csc −x − tan(−2x − π) + 1 1 3 sec 22. y = cot x + π 6 23. y = −11 cot x 1 5 24.... |
ons Properties of F (x) = arccos(x) – Domain: [−1, 1] – Range: [0, π] – arccos(x) = t if and only if 0 ≤ t ≤ π and cos(t) = x – cos(arccos(x)) = x provided −1 ≤ x ≤ 1 – arccos(cos(x)) = x provided 0 ≤ x ≤ π Properties of G(x) = arcsin(x) – Domain: [−1, 1] – Range – arcsin(x) = t if and only if − π – sin(arcsin(x)) = x ... |
4 , which 2 . Returning ∪ − π means the values of t under consideration are − to arctan(2t), we note the double angle identity tan(2t) = 2 tan(t) 1−tan2(t) , is valid for all the values of t under consideration, hence we get tan(2 arctan(x)) = tan(2t) = 2 tan(t) 1 − tan2(t) = 2x 1 − x2 4It’s always a good idea to make... |
ot (arccsc (−3)) 2. Rewrite the following as algebraic expressions of x and state the domain on which the equiv- alence is valid. (a) tan(arcsec(x)) (b) cos(arccsc(4x)) 832 Solution. Foundations of Trigonometry 1. 2. (a) Since 2 ≥ 1, we may invoke Theorem 10.29 to get arcsec(2) = arccos 1 2 (b) Unfortunately, −2 is not... |
input and the result below. arctan 2 x arctan 2 x + 2π, π, + π, if x < 0 if x = 0 if x > 0 y = g(x) in terms of arctangent y = g(x) = arccot x 2 + π The inverse trigonometric functions are typically found in applications whenever the measure of an angle is required. One such scenario is presented in the following examp... |
r state that it is undefined. 131. sin arccos − 1 2 132. sin arccos 3 5 134. sin arccot √ 5 137. cos arctan √ 7 135. sin (arccsc (−3)) 140. tan arcsin − √ 2 5 5 141. tan arccos − 1 2 143. tan (arccot (12)) 144. cot arcsin 12 13 138. cos (arccot (3)) 139. cos (arcsec (5)) 133. sin (arctan (−2)) 136. cos arcsin − 5 13 142... |
arccsc 124. arccsc 126. arccsc π 6 2π 3 11π 6 9π 8 3 5 134. sin arccot √ 130. arccsc 132. sin arccos csc 136. cos arcsin − 5 = = 4 5 √ 6 6 5 13 = 12 13 √ 3 10 138. cos (arccot (3)) = 140. tan arcsin − √ 2 5 = −2 10 5 = 4 3 141. tan arccos − √ 3 = − 1 2 143. tan (arccot (12)) = 1 12 142. tan arcsec 144. cot arcsin 5 3 1... |
s, we obtain six answers, corresponding to k = 0 through k = 5, which lie in [0, 2π): x = π 2 and 11π 6 . To confirm these graphically, we must be careful. On many calculators, there is no function button for cotangent. We choose4 to use sin(3x) . Graphing y = cos(3x) the quotient identity cot(3x) = cos(3x) sin(3x) and ... |
x). cos(3x) = 2 − cos(x) 4 cos3(x) − 3 cos(x) = 2 − cos(x) 2 cos3(x) − 2 cos(x) − 2 = 0 4u3 − 2u − 2 = 0 Let u = cos(x). 864 Foundations of Trigonometry To solve 4u3 − 2u − 2 = 0, we need the techniques in Chapter 3 to factor 4u3 − 2u − 2 into (u − 1) 4u2 + 4u + 2. We get either u − 1 = 0 or 4u2 + 2u + 2 = 0, and since... |
x : x = − π 2 k for integers k. To help visualize the domain, we follow the 6 , . . ., old mantra ‘When in doubt, write it out!’ We get x : x = − π where we have kept the denominators 6 throughout to help see the pattern. Graphing the situation on a numberline, we have = 0, we get x = − π 6 + π 6 , − 7π 6 , − 4π 6 + π ... |
1 2 14. tan (2x − π) = 1 17. cos2 (x. sin (−2x) = 6. sec (3x) = 9. sin = x 3 12. 2 cos x + √ 3 = 7π 4 15. tan2 (x) = 3 18. sin2 (x) = 3 4 In Exercises 19 - 42, solve the equation, giving the exact solutions which lie in [0, 2π) 19. sin (x) = cos (x) 21. sin (2x) = cos (x) 23. cos (2x) = cos (x) 20. sin (2x) = sin (x) ... |
r Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height 136 feet. It completes two revolutions in 2 minutes and 7 seconds. Assuming that the riders are at the edge of the circle, find a sinusoid which describes the height of the passengers above the ground t seconds aft... |
t (ft.) to measure displacement. If we second2 and 1 Newton = 1 kg meter second2 . are in the SI system, we measure displacement in meters (m). Time is always measured in seconds (s). 11.1 Applications of Sinusoids 887 object and initial velocity v0 of the object. As with x(t), x0 = 0 means the object is released from ... |
ncept as well.17 y = 5 sin(6x) − 5 sin(8x) over [0, π] y = 5 sin(6x) − 5 sin(8x) and y = ±10 sin(x) over [0, 2π] 16The reader is invited to investigate the destructive implications of resonance. 17A good place to start is this article on beats. 11.1 Applications of Sinusoids 891 11.1.2 Exercises 1. The sounds we hear a... |
he side opposite α, b is the side opposite β and c is the side opposite γ. Taken together, the pairs (α, a), (β, b) and (γ, c) are called angle-side opposite pairs. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not alwa... |
a) and (γ, c) are intended to be angle-side pairs in a triangle where α, a and c are given. Let h = c sin(α) If a < h, then no triangle exists which satisfies the given criteria. If a = h, then γ = 90◦ so exactly one (right) triangle exists which satisfies the criteria. If h < a < c, then two distinct triangles exist wh... |
in standard position with 0◦ ≤ θ < 360◦ which corresponds to each of the bearings given below. (a) due west (b) S83◦E (c) N5.5◦E (d) due south 14The word ‘plumb’ here means that the tree is perpendicular to the horizontal. 906 Applications of Trigonometry (e) N31.25◦W (f) S72◦4112W15 (g) N45◦E (h) S45◦W 26. The Colonel... |
tionship c2 = a2 + b2. What this means is that in the larger mathematical sense, the Law of Cosines and the Pythagorean Theorem amount to pretty much the same thing.2 Example 11.3.1. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. β = 50◦,... |
The Law of Cosines 917 22. A naturalist sets off on a hike from a lodge on a bearing of S80◦W. After 1.5 miles, she changes her bearing to S17◦W and continues hiking for 3 miles. Find her distance from the lodge at this point. Round your answer to the nearest hundredth of a mile. What bearing should she follow to return... |
− 5π 2 Since P is 117 units from the pole, any representation (r, θ) for P satisfies r = ±117. For the r = 117 case, we can take θ to be any angle coterminal with − 5π 2 . In this case, we choose as one answer. For the r = −117 case, we visualize moving left 117 θ = 3π units from the pole and then rotating through an an... |
0 Subtract 9 from both sides. Since cos2(θ) + sin2(θ) = 1 Factor. r2 − 6r cos(θ) = 0 r(r − 6 cos(θ)) = 0 We get r = 0 or r = 6 cos(θ). From Section 7.2 we know the equation (x − 3)2 + y2 = 9 describes a circle, and since r = 0 describes just a point (namely the pole/origin), we choose r = 6 cos(θ) for our final answer.... |
s into rectangular coordinates. 77. r = 7 81. θ = 2π 3 78. r = −3 82. θ = π 79. r = 83. θ = √ 2 3π 2 80. θ = π 4 84. r = 4 cos(θ) 85. 5r = cos(θ) 86. r = 3 sin(θ) 87. r = −2 sin(θ) 88. r = 7 sec(θ) 89. 12r = csc(θ) 90. r = −2 sec(θ) √ 91. r = − 5 csc(θ) 92. r = 2 sec(θ) tan(θ) 93. r = − csc(θ) cot(θ) 94. r2 = sin(2θ) 9... |
e left θ runs from π 2 to 3π 2 2π θ , we see that r increases from 4 to 6. On the xy-plane, the curve Over the interval π, 3π 2 sweeps out away from the origin as it travels from the negative x-axis to the negative y-axis 3π 2 2π θ θ runs from π to 3π 2 Finally, as θ takes on values from 3π xy-plane pulls in from the n... |
π order for a point P to be on the graph of r = 2 sin(θ), it must have a representation P (r, θ) which satisfies r = 2 sin(θ). If P is also on the graph of r = 2−2 sin(θ), then P has a (possibly different) representation P (r, θ) which satisfies r = 2 sin(θ). We first try to see if we can find any points which have a singl... |
θ) which satisfy both equations. Keep in mind that k is an integer. Substitute (−r) for r and (θ + (2k + 1)π) for θ in either one of E1 or E2 (but not both) and solve for pairs (r, θ) which satisfy both equations. Keep in mind that k is an integer. 956 Applications of Trigonometry Our last example ties together graphin... |
in(2θ) is indeed symmetric about the origin. (See Example 11.5.2 number 3.) (b) Show that f (θ) = 3 cos θ 2 is not odd, yet the graph of r = 3 cos θ 2 is symmetric about the origin. (See Example 11.5.3 number 4.) 64. Show that if f (π − θ) = f (θ) for all θ in the domain of f then the graph of r = f (θ) is symmetric ab... |
φ to get x and y in terms of x and y, but that seems like cheating. The matrix A introduced here is revisited in the Exercises. 11.6 Hooked on Conics Again 975 Example 11.6.1. Suppose the x- and y- axes are both rotated counter-clockwise through the angle θ = π 3 to produce the x- and y- axes, respectively. 1. Let P (x... |
and C in the original equation. To that end, we make the usual substitutions x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) into Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. We leave it to the reader to show that, after gathering like terms, the coefficient A on (x)2 and the coefficient C on (y)2 are A = A cos2(θ) + B cos(θ) sin(θ)... |
s (−3, 0) and (6, 0), so these are our 2 , 0.8 vertices. The center of the ellipse is the midpoint of the vertices, which in this case is 3 2 , 0 and this allows us to find the We know one focus is (0, 0), which is 3 1−e2 = (2)(4) 2 from the center 3 8As a quick check, we have from Theorem 11.12 the major axis should ha... |
ins down the value of |z| to one and only one number. Thus the modulus is well-defined in this case, too.2 Even with the requirement r ≥ 0, there are infinitely many angles θ which can be used in a polar representation of a point (r, θ). If z = 0 then the point in question is not the origin, so all of these angles θ are ... |
sin − 3π 4 . From this, we find (b) Expanding, we get z = 2cis − 3π √ 4 2 = Im(z). 2, so Re(zc) We get z = 3cis(0) = 3 [cos(0) + i sin(0)] = 3. Writing 3 = 3 + 0i, we get Re(z) = 3 and (d) Lastly, we have z = cis π 2 Im(z) = 0, which makes sense seeing as 3 is a real number. = cos π 2 = i. Since i = 0 + 1i, we get Re(z)... |
ity. Since we have found three distinct zeros, we know these are all of the zeros, so there are exactly three distinct cube roots of 8. Let us now solve this same problem using the machinery developed in this section. To do so, we express z = 8 in polar form. Since z = 8 lies 8 units away on the positive real axis, we ... |
together, thus pairing up the complex conjugate pairs of zeros Theorem 3.15 told us we’d get, we have that p(x) = (x2 − 2x + 2)(x2 + 2x + 2). Use the 12 complex 12th roots of 4096 to factor p(x) = x12 − 4096 into a product of linear and irreducible quadratic factors. 1006 Applications of Trigonometry 80. Complete the p... |
ardless of its initial point. In the case of our vector v above, any vector which moves three units to the right and four up3 from its initial point to arrive at its terminal point is considered the same vector as v. The notation we use to capture this idea is the component form of the vector, v = 3, 4, where the first ... |
Property: For every vector v and scalars k and r, (kr)v = k(rv). Identity Property: For all vectors v, 1v = v. Additive Inverse Property: For all vectors v, −v = (−1)v. Distributive Property of Scalar Multiplication over Scalar Addition: For every vector v and scalars k and r, (k + r)v = kv + rv Distributive Property ... |
50◦) + 35 sin(−30◦)). Since both of these coordinates are positive,14 we know θ is a Quadrant I angle, as depicted below. Furthermore, tan(θ) = y x = 175 sin(50◦) + 35 sin(−30◦) 175 cos(50◦) + 35 cos(−30◦) , so using the arctangent function, we get θ ≈ 39◦. Since, for the purposes of bearing, we need the angle between ... |
d heading should she set to take into account the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. HINT: If v denotes the velocity of the HMS Sasquatch and w denotes the velocity of the current, what does v + w need to be to reac... |
ourtesy of Definition 11.8. The following example puts Theorem 11.22 to good use. As in Example 11.8.3, we work out the problem in great detail and encourage the reader to supply the justification for each step. 2 = v2, 1 + v2 Example 11.9.1. Prove the identity: v − w2 = v2 − 2(v · w) + w2. Solution. We begin by rewritin... |
, the point R coincides with the point O, so p = −→ OR = −−→ OO = 0. 1040 Applications of Trigonometry 3 −2 −1 1 Suppose we wanted to verify that our answer p in Example 11.9.4 is indeed the orthogonal projection of v onto w. We first note that since p is a scalar multiple of w, it has the correct direction, so what rem... |
w(v) = √ 3 √ q = 2+ 8 √ √ 2− 8 6 √ √ 2+ 8 6 , √ 6 2 20. v = 1 and 19. v = √ 3 2 , 1 2 √ and w = √ v · w = − θ = 165◦ 2 6+ 4 √ proj w(v) = √ q = 3−1 4 , 1− 4 3+1 4 √ 3 √ , 3++ 4 v · w = θ = 15◦ √ proj w(v) = 1− 4 q = √ 3 , 1− 4 3+1 4 √ 3 √ 3+1 4 , − 21. (1500 pounds)(300 feet) cos (0◦) = 450, 000 foot-pounds 22. (10 pou... |
at the point (2, 1) and heads towards, but never reaches,6 (0, 0). = x 2 , and since x = 2e−t means e− = 2e−t, t ≥ 0 y = e−2t, t ≥ 0 x = 2e−t, y = e−2t , t ≥ 0 6 gives the point 1 3. For the system {x = sin(t), y = csc(t) for 0 < t < π, we start by graphing x = sin(t) and y = csc(t) over the interval (0, π). We find tha... |
the points generated by the parametric equations {x = 2 cos(t), y = 3 sin(t) lie on the ellipse x2 9 = 1. Employing 2 ≤ t ≤ 3π the techniques demonstrated in Example 11.10.2, we find that the restriction π 2 generates the left half of the ellipse, as required. 2 . Our final answer is {x = 2 cos(t), y = 3 sin(t) for π 4 +... |
g utility. Be sure to indicate the orientation imparted on the curve by the parametrization. 21. 23. x = t3 − 3t y = t2 − 4 x = et + e−t y = et − e−t for − 2 ≤ t ≤ 2 for − 2 ≤ t ≤ 2 22. 24. x = 4 cos3(t) y = 4 sin3(t) x = cos(3t) y = sin(4t) for 0 ≤ t ≤ 2π for 0 ≤ t ≤ 2π 1060 Applications of Trigonometry In Exercises 2... |
e opposite pairs, 896 angular frequency, 708 annuity annuity-due, 667 ordinary definition of, 666 future value, 667 applied domain of a function, 60 arccosecant calculus friendly definition of, 831 graph of, 830 properties of, 831 1069 Index 1070 trigonometry friendly definition of, 828 graph of, 827 properties of, 828 ar... |
n, 225 linear function, 156 of best fit, 225 parallel, 166 1076 Index perpendicular, 167 point-slope form, 155 slope of, 151 slope-intercept form, 155 vertical, 23 linear equation n variables, 554 two variables, 549 linear function, 156 local maximum formal definition of, 102 intuitive definition of, 101 local minimum for... |
rty of, 1034 geometric interpretation, 1035 properties of, 1034 relation to magnitude, 1034 relation to orthogonality, 1037 work, 1042 head, 1012 initial point, 1012 magnitude definition of, 1020 properties of, 1020 relation to dot product, 1034 normalization, 1024 orthogonal projection, 1038 1082 Index multiplicity of,... |
ferent times and see how they differ (i.e., how they have changed). The two times of comparison are usually called the final time and the initial time. We really need to be precise about this, so here is what we mean: ∆ quantity = (value of quantity at final time) − (value of quantity at initial time) ∆ time = (final time... |
pm and the profit is a constant rate of $11 on 10 orders. When will phone order profit exceed $1,000? Problem 1.10. Aleko’s Pizza has delivered a beautiful 16 inch diameter pie to Lee’s dorm room. The pie is sliced into 8 equal sized pieces, but Lee is such a non-conformist he cuts off an edge as pictured. John then tak... |
ne unit on the horizontal axis . Figure 2.5(a) has aspect ratio 1, whereas Figure 2.5(b) has aspect ratio 1 2. In problem solving, you will often need to make a rough assumption about the relative axis scaling. This scaling will depend entirely on the 14 CHAPTER 2. IMPOSING COORDINATES y-axis 1.0 0.8 0.6 0.4 0.2 −1.0 −... |
e two axes) is given by the formula (∆x)2 + (∆y)2 (x2 − x1)2 + (y2 − y1)2. d = = p p (2.3) y-axis |∆y| y1 y2 P = (x1, y1) Beginning point. d Q = (x2, y2) Ending point. x2 x1 x-axis |∆x| Figure 2.10: A different direction. If your algebra is a little rusty, a very common mistake may crop up when you are using the distan... |
rst want to become acquainted with the sorts of pictures that will arise. This is surprisingly easy to accomplish: Impose an xy-coordinate system on a blank sheet of paper. Take a sharp pencil and begin moving it around on the paper. The resulting picture is what we will call a curve. For example, here is a sample of t... |
is (x − 20)2 + (y − 40)2 = 3600. To find the circular disc intersection with the y-axis, we have a system of two equations to work with: (x − 20)2 + (y − 40)2 = 3,600; x = 0. To find the intersection points we simultaneously solve both equations. To do this, we replace x = 0 in the first equation (i.e., we impose the 30 ... |
ear 30000 25000 20000 15000 10000 5000 y y-axis (dollars) S = (1987,28313) (x,y) on the line: This means men earn y dollars in year x. R = (1970,9521) x-axis (yeara) Linear model for women. (b) Linear model for men. Figure 4.2: Linear models of earning power. 4.2. RELATING LINES AND EQUATIONS ywomen = = ymen = = 18,531... |
lue of the house is changing at a rate of 10,000 dollars/year. ∆y ∆x At the other extreme, if the units for both x and y are the same, then the units cancel out in the rate of change calculation; in other words, the slope is a unit-less quantity, simply a number. This sort of thing will come up in the mathematics you s... |
2 crop duster P 1.5 miles South Figure 4.11: The flight path of a crop duster. Example 4.10.2. A crop dusting airplane flying a constant speed of 120 mph is spotted 2 miles South and 1.5 miles East of the center of a circular irrigated field. The irrigated field has a radius of 1 mile. Impose a coordinate system as pictur... |
rice? (f) When will the average sales price in Port Townsend be $15,000 less than the Seattle sales price? What are the two sales prices at this time? (g) When will the Port Townsend sales price be $15,000 more than the Seattle sales price? What are the two sales prices at this time? (h) When will the Seattle sales pri... |
ulated data. This same equation produces ALL of the data points for the other two plots, using 1/2 second and 1/4 second time intervals. (Granted, we have swept under the rug the issue of “...where the heck the equation comes from...”; that is a consequence of mathematically modeling the motion of this gull. Right now,... |
a single unique 62 CHAPTER 5. FUNCTIONS AND GRAPHS amount, call it W(z). So, T = W(z) is a function, where the independent variable is z. The domain could be taken to be 0 1,000,000, which would cover all items costing up to one-million dollars. The range of the function would be the set of all values W(z), as z range... |
n graphically relate three 0 x ≤ ≤ y-axis (dollars) y-axis (dollars) 1500 1000 500 x, 15x) P x-axis (units sold) 20 40 60 80 100 e s n e p x e 1500 1000 500 (x, e(x)) Q x-axis (units sold) 20 40 60 80 100 (a) Gross income graph. (b) Expenses graph. Figure 5.12: Visualizing income and expenses. things: x on the horizont... |
In that discussion, we looked at the graphs of the relevant functions and used these as visual aids to help us answer the questions posed. This was a concrete illustration of what is typically called “graphical analysis of a function.” This is a fundamental technique we want to carry forward throughout the course. Let’... |
origin. With this coordinate system, the graph of the equation x2 + y2 = 152 = 225 will be the circular cross-section of the tunnel. In the case of Deck A, we basically need to determine how close to each edge of the tunnel a 6 foot high person can stand without hitting his or her head on the tunnel; a similar remark ... |
s the range of g? (b) Where is the function increasing? Where is the function decreasing? (c) Find the multipart formula for y = g(x). (d) If we restrict the function to the smaller domain −5 x ≤ ≤ 0, what is the range? (e) If we restrict the function to the smaller domain 0 x ≤ ≤ 4, what is the range? 3D−view of ditch... |
bols of an equation mean? y-axis 20 15 10 5 x-axis −6 −4 −2 2 4 6 −5 −10 −15 (b) What does the equation look like? Figure 7.8: Interpreting an equation. Solution. To begin with, we can make some initial conclusions about the specific shifts, reflections and dilations involved, based on looking at the vertex form of the e... |
ormation is given. Can you see why? !!! CAUTION !!! 2. The function y(t) is defined for all t; however, in the context of the problem, there is no physical meaning when t < 0. 96 CHAPTER 7. QUADRATIC MODELING The next example illustrates how we must be very careful to link the question being asked with an appropriate fu... |
quadratic function is one of the form • f(x) = ax2 + bx + c. where a = 0. 6 102 CHAPTER 7. QUADRATIC MODELING • The graph of a quadratic function is a parabola which is symmetric about the vertical line through the highest (or lowest) point on the graph. This highest (or lowest) point is known as the vertex of the gra... |
n) to build the model we want (the new function)? We will return in Exam- 107 108 CHAPTER 8. COMPOSITION ple 8.2.4 to see the answer is yes; first, we need to develop the tool of function composition. 8.1 The Formula for a Composition The basic idea is to start with two functions f and g and produce a new function calle... |
-negative real numbers, z. We graph z = g(x) in the xz-plane, mark the desired range denoted 0 0 z on the vertical z-axis, then determine which x-values would lead to points on the graph with second coordinates in this zone. We find that the domain of all x-values greater or equal to −1 (denoted −1 x) leads to the desir... |
an always find a “reverse process” for the function. To find it, you must solve the equation y = f(x) for x in terms of y: y = mx + b y − b = mx 1 m (y − b) = x So, if m = 3 and b = −1, we just have the first example above. For another example, suppose y = −0.8x + 2; then m = −0.8 and b = 2. In this case, the reverse proc... |
ber y in the domain of f−1(y), y = f(x) for some x in the domain of f(x); i.e., we are using the fact that the domain of f−1 equals the range of f. The function f−1(y) takes the number f(x) and sends it to x, by Fact 9.3.2. So when f(x) is the input value, x becomes the output value. Conclude a point on the graph of f−... |
s a fixed positive integer exponent. What happens if we turn this around, interchanging x and b, defining a new rule: y = f(x) = bx. (10.1) 10.1. FUNCTIONS OF EXPONENTIAL TYPE 135 We refer to x as the power and b the base. An expression of this sort is called a function of exponential type. Actually, if your algebra is a... |
lack key. A piano keyboard is commonly tuned according to a rule requiring that each key (white and black) has a frequency 21/12 times the frequency of the key to its immediate left. This makes the ratio of adjacent keys always the same (21/12), and it means that keys 12 keys apart have a ratio of frequencies exactly e... |
welfth of a year. Arguing as before, paying special attention that the periodic rate is now r 12 , we have n = 0.08 P(1/12) = P0 + (periodic rate)P0 = P0 1 + .08 12 = $1,000(1 + 0.006667) = $1,006.67. 148 CHAPTER 11. EXPONENTIAL MODELING After two compounding periods, the value is P(2/12), P(2/12) = P(1/12) + (periodic... |
es the theory of inverse functions. Assuming we can find the inverse function of f(t) = et, we can apply f−1(t) to each side of the equation and solve for t: f−1(5) = f−1(e0.08t) = 0.08t (12.5)f−1(5) = t The goal in this section is to describe the function f−1, which is usually denoted by the symbol f−1(t) = ln(t) and c... |
ted on the eardrum, which we refer to as the intensity of the sound. We can try to measure the intensity using some sort of scale. This becomes challenging, since the human ear is an amazing instrument, capable of hearing a large range of sound intensities. For that reason, a logarithmic scale becomes most useful. The ... |
he population of Abnarca to double? (b) When will Abnarca’s population equal that of Bonipto? Chapter 13 Three Construction Tools Sometimes the composition of two functions can be understood by graphical manipulation. When we discussed quadratic functions and parabolas in the previous section, certain key graphical man... |
ained by horizontally shifting the graph of f(x) by h. If h is positive, the picture shifts to the right h units; if h is negative, the picture shifts to the left h units. If the domain of f(x) is an interval a b, then the domain ≤ of f(x − h) is a x − h b. The range remains unchanged under ≤ horizontal shifting. ≤ ≤ x... |
e right c units. 2 Replace x with (x + c). y = f(x + c) A shift to the left c units. Replace with (f(x) + c). f(x) y = f(x) + c A shift up c units. 2 4 2 −4 −2 4 2 −2 2 Replace with (f(x) − c). f(x) y = f(x) − c A shift down c units. −2 2 2 Table 13.2: Shifting y = f(x). 13.6. SUMMARY OF RULES 177 Dilation Symbolic Cha... |
gets larger in the negative direction (i.e., it gets more and more negative). The curve gets closer and closer to the negative y-axis as y becomes more and more negative. Again, we say that the y − axis is a vertical asymptote for the curve y = 1/x. It turns out that every linear-to-linear rational function has a grap... |
Sketch a graph and indicate any vertical or horizontal asymptotes. Give equations for the asymptotes. (a) f(x) = 2x x−1 (c) h(x) = x+1 x−2 (e) k(x) = 8x+16 5x− 1 2 (b) g(x) = 3x+2 2x−5 (d) j(x) = 4x−12 x+8 (f) m(x) = 9x+24 35x−100 Problem 14.2. Oscar is hunting magnetic fields with his gauss meter, a device for measurin... |
s of these wedges as one-degree arcs. Why 360 equal sized arcs? The reason for doing so is historically tied to the fact that the ancient Babylonians did so as they developed their study of astronomy. (There is actually an alternate system based on dividing the circular region into 400 equal sized wedges.) Any central ... |
ie) 1 12 = (82π) = 16π 3 . (fraction of pie) × B Rθ θ O A Cr Figure 15.12: Finding the area of a “pie shaped wedge”. Let’s apply the same reasoning to find the area of a circular sector. We know the area of the circular disc bounded by a circle of radius r is πr2. Let Rθ be the “pie shaped wedge” cut out by an angle ∠AO... |
traveled? (d) Suppose bug B lands on the end of the wiper blade closest to the pivot. Assume the wiper turns through an angle of 110◦. In one cycle of the wiper blade, how far has the bug traveled? (e) Suppose bug C lands on an intermediate location of the wiper blade. Assume the wiper turns through an angle of 110◦. ... |
eady speed of 40 MPH (miles per hour). If the rear wheel is 28 inches in diameter, determine the angular speed of a location on the rear tire. A pebble becomes stuck to the tread of the rear tire. Describe the location of the pebble after 1 second and 0.1 second. * pebble sticks to tread here Figure 16.3: Where is the ... |
r speeds of each of the wheels A, B, and C, respectively. In addition, the chain connecting the wheels B and C will have a linear speed, which we will denote by vchain. The strategy is broken into a sequence of steps which leads us from the known linear speed vA to the angular speed ωC of wheel C: • • • • Step 1: Given... |
2, cos(θ) = √3 13, tan(θ) = 5 12. , cos(θ) = 1 √2 2 , tan(θ) = 1 √3 For example, we have three right triangles in Figure 17.4; you can verify that the Pythagorean Theorem holds in each of the cases. In the left-hand triangle, sin(θ) = 5 13, cos(θ) = 12 In the middle triangle, sin(θ) = 1 , tan(θ) = 1. In the right-hand... |
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