# C x23 y 23 a23 cos2 sin2 a23 y 42 y y 1 50 50 0 3 2 1

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Unformatted text preview: at a = 0 and b = 0; eliminate the parameter to get (x − h)2 /a2 + (y − k )2 /b2 = 1. If |a| = |b| the curve is a circle with center (h, k ) and radius |a|; if |a| = |b| the curve is an ellipse with center (h, k ) and major axis parallel to the x-axis when |a| > |b|, or major axis parallel to the y -axis when |a| < |b|. (a) ellipses with a ﬁxed center and varying axes of symmetry (b) (assume a = 0 and b = 0) ellipses with varying center and ﬁxed axes of symmetry (c) circles of radius 1 with centers on the line y = x − 1 40. Refer to the diagram to get bθ = aφ, θ = aφ/b but θ − α = φ + π/2 so α = θ − φ − π/2 = (a/b − 1)φ − π/2 x = (a − b) cos φ − b sin α = (a − b) sin φ − b sin 41. a−b a−b b a−b b θ φ, bθ aφ φ. = (a − b) cos φ + b cos y = (a − b) sin φ − b cos α y φ α x y (a) a -a a x -a (b) Use b = a/4 in the equations of Exercise 40 to get 1 3 1 3 x = a cos φ + a cos 3φ, y = a sin φ − a sin 3φ; 4 4 4 4 but trigonometric identities yield cos 3φ = 4 cos3 φ − 3 cos φ, sin 3φ = 3 sin φ − 4 sin3 φ, so x = a cos3 φ, y = a sin3 φ. (c) x2/3 + y 2/3 = a2/3 (cos2 φ + sin2 φ) = a2/3 y 42. y y 1 50 50 0 -3 -2 -1 x x -2 -1 -1 1 -50 -50 -1 a = −1 a = −2 a=0 y y 50 50 x 1 0 2 1 2 -50 -50 a=1 a=2 3 x x Supplementary Exercises 1 38 CHAPTER 1 SUPPLEMENTARY EXERCISES 1. 1940-45; the greatest ﬁve-year slope 2. (a) (c) f (−1) = 3.3, g (3) = 2 x < −2, x > 3 (b) (d) (e) the domain is −4 ≤ x ≤ 4.1, the range is −3 ≤ y ≤ 5 (f ) 3. 4. T x = −3, 3 the domain is −5 ≤ x ≤ 5 and the range is −5 ≤ y ≤ 4 f (x) = 0 at x = −3, 5; g (x) = 0 at x = −3, 2 x 70 60 50 40 0 2 4 6 t t 5 8 13 5. If the side has length x and height h, then V = 8 = x2 h, so h = 8/x2 . Then the cost C = 5x2 + 2(4)(xh) = 5x2 + 64/x. 6. Assume that the paint is applied in a thin veneer of uniform thickness, so that the quantity of paint to be used is proportional to the area covered. If P is the amount of paint to be used, P = kπr2 . The constant k depends on physical factors, such as the thickness of the paint, absorption of the wood, etc. 7. y 5 -5 -1 5 x 8. Suppose the radius of the uncoated ball is r and that of the coated ball is r + h. Then the plastic has 4 4 4 volume equal to the diﬀerence of the volumes, i.e. V = π (r + h)3 − πr3 = πh[3r2 + 3rh + h2 ] in3 . 3 3 3 9. (a) The base has sides (10 − 2x)/2 and 6 − 2x, and the height is x, so V = (6 − 2x)(5 − x)x ft3 . (b) From the picture we see that x < 5 and 2x < 6, so 0 < x < 3. (c) 3.57 ft ×3.79 ft ×1.21 ft 10. {x = 0} and ∅ (the empty set) 11. impossible; we would have to solve 2(3x − 2) − 5 = 3(2x − 5) − 2, or −9 = −17 12. (a) 13. 1/(2 − x2 ) 15. (3 − x)/x x −4 f (x) 0 g (x) 3 (f ◦ g )(x) 4 (g ◦ f )(x) −1 (b) no; f (g (x)) can be deﬁned at x = 1, whereas g , and therefore f ◦ g , requires x=1 14. g (x) = x2 + 2x −3 −2 −1 0 1 2 3 4 −1 2 1 3 −2 −3 4 −4 2 1 −3 −1 −4 4 −2 0 −3 −2 −1 1 0 −4 2 3 −3 4 −4 −2 1 2 0 3 39 16. Chapter 1 (a) y = |x − 1|, y = |(−x) − 1| = |x + 1|, y = 2|x + 1|, y = 2|x + 1| − 3, y = −2|x + 1| + 3 y (b) 3 1 -3 -1 2 x -1 17. 18. 19. (a) even × odd = odd (c) even + odd is neither (b) a square is even (d) odd × odd = even π 3π π 5π 7π 11π (a) y = cos x − 2 sin x cos x = (1 − 2 sin x) cos x, so x = ± , ± , , ,− ,− 2 266 6 6 π 3π π√ 5π √ 7π √ 11π √ (b) (± , 0), (± , 0), ( , 3/2), ( , − 3/2), (− , − 3/2), (− , 3/2) 2 2 6 6 6 6 If x denotes the distance from A to the base of the tower, and y the distance from B to the base, then x2 + d2 = y 2 . Moreover h = x tan α = y tan β , so d2 = y 2 − x2 = h2 (cot2 β − cot2 α), d2 d2 sin2 α sin2 β . The trigonometric identity h2 = 2 2 β − cot2 α = cot sin α cos2 β − cos2 α sin2 β d sin α sin β sin(α + β ) sin(α − β ) = sin2 α cos2 β − cos2 α sin2 β yields h = . sin(α + β ) sin(α − β ) (b) 20. (a) 295.72 ft. (a) 3π 2π (t − 101) = , or t = 374.75, when 365 2 which is the same date as t = 9.75, so during the night of January 10th-11th (b) y 60 40 20 t 100 200 300 -20 (c) from t = 0 to t = 70.58 and from t = 313.92 to t = 365 (the same date as t = 0) , for a total of about 122 days 21. C is the highest nearby point on the graph; zoom to ﬁnd that the coordinates of C are (2.0944, 1.9132). Similarly, D is the lowest nearby point, and its coordinates are (4.1888, 1.2284). Since f (x) = 1 x − sin x 2 is an odd function, the coordinates of B are (−2.0944, −1.9132) and those of A are (−4.1888, −1.2284). 22. Let y = A + B sin(at + b). Since the maximum and minimum values of y are 35 and 5, A + B = 35 and A − B = 5, so A = 20, B = 15. The period is 12 hours, so 12a = 2π and a = π/6. The maximum occurs at t = 2, so 1 = sin(2a + b) = sin(π/3 + b), π/3 + b = π/2, b = π/2 − π/3 = π/6 and y = 20 + 15 sin(πt/6 + π/6). 23. (a) T...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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