Business Calc Homework w answers_Part_34

Business Calc Homework w answers_Part_34 - 166 Section 4.4...

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Unformatted text preview: 166 Section 4.4 53. (a) According to the graph, y (0) 0. (b) According to the graph, y ( L) (c) y(0) 0, so d c 2bx r h 3ax 2 bx 2 and 3aL 2 and y ( L) 2bL bL 2 H 0, so we have two linear 2. and let m y 3aL . Substituting into the first 2 3aL 3 aL 3 equation, we have aL 3 H, or H, so 2 2 H H a 2 3 . Therefore, b 3 2 and the equation for y is L L H H x3 x2 y(x) 2 3 x 3 3 2 x 2, or y(x) H 2 3 . L L L L m(0 equation gives b 2a x and so the 2 2a x2 . 2 a2 height is h r2 a2 r 2h 3 x2 2a 2 3 x2 2a 2 a2 m(x x0) x0) m(x x0) a a 3 d2 (r 3 dr f (r) = a2 a2 2 r 3 3 a2 2a 2r 3 a2 r(2a 2 2 2 a2 h r a 6 3 r 2)(2r) mx0 m x0 have A(x) y2 2 2a 2 , which gives 3 2 a6 . Then 3 a2 mx0 a m a m a2 m a2 m f(x) 2 . f (x) f (x) x To graph, let y1 3r 2) r2 mx0 mx0) (b) The domain is the open interval (0, 10). 2 The critical point occurs when r r ( a2 r2 r (a m m a. 3r 3 2 . 1 (x-intercept of line RT)(y-intercept of line RT) 2 1 mx0 a 2 2 r 2) r 2 3a 2 a 3 r a m Substituting x for x0, f (x) for m, and f (x) for a, we ( 2r) r2 2r(a a r 2, where 0 mx0 0, or x 2(Area of triangle ORT) . r 2) 1 r2 3 r2 mx0, (Area of triangle RST) m f(r) a Let O designate the origin. Then (b) To simplify the calculations, we shall consider the volume as a function of r: volume a. The y-intercept of this line is and the x-intercept is the solution of Therefore, V(x) f (x0) be the slope of line RT. Then the equation of line RT is equations in the two unknowns a and b. The second 54. (a) The base radius of the cone is r a3 , the relationship is 3 and h P has coordinates (x0, a), aL 3 2bx. Then y( L) 6 3 55. (a) Let x0 represent the fixed value of x at point P, so that c, so y (0) implies that ax 3 0. Therefore, y(x) y (x) 0. 0. 3ax 2 Now y (x) a (c) Since r f (x) y3 A(x) f (x) 5 51 x2 , 100 NDER(y1), and y2 x y1 2 y2 . The graph of the area function y3 A(x) is shown below. 2a 3 2 2 a 3 a 3 3 . Using a3 , 3 and h we may now find the values of r and h for the given values of a. [0, 10] by [ 100, 1000] When a 4: r when a 5: r when a 6: r when a 8: r 46 ,h 3 56 ,h 3 2 4 3 3 5 3 3 ; ; 6, h 6 8 3 2 3; 8 ,h 3 3 The vertical asymptotes at x 0 and x 10 correspond to horizontal or vertical tangent lines, which do not form triangles. Section 4.4 167 (c) Using our expression for the y-intercept of the tangent line, the height of the triangle is a mx f (x) f (x) x 1 2 1 2 5 5 x x2 100 2 100 x2 2 100 x2 100 x2 x x2 We may use graphing methods or the analytic method in part (d) to find that the minimum value of A(x) occurs at x 8.66. Substituting this value into the expression above, the height of the triangle is 15. This is 3 times the y-coordinate of the center of the ellipse. (d) Part (a) remains unchanged. The domain is (0, C). To graph, note that f (x) f (x) x2 C2 B1 B B C2 1 C2 x2 B C B C2 x 2 and Bx ( 2x) C C2 . x2 Therefore, we have A(x) f (x) 2 f (x) f (x) x C2 C B C B Bx x x2 Bx C C2 Bx C C2 x2 1 BCx C2 BCx C2 BCx C2 x2 1 x2 1 C2 x2) Bx [Bx 2 (BC B C2 x 2)( [Bx 2 BC C 2 x2 x2 x C2 C2 [BC(C C2 x2 2 B(C 2 C2 BC(C x2 C2 x 2)]2 x 2)]2 x 2)]2 x 2)2 x2 (x C 2 A (x) B (BC x x2 2 C2 x 2)(2)(C C2 x x 2) C2 BC x 2 (C 2 C 2 x 2) BC(C 22 x (C x 2) 2x 2 C 2 x 2) BC(C 22 x (C x 2) [ BC(C C 2 x 2) 22 x (C x 2) ( C2 (C Cx2 2x 2 C2 Cx 2 C2 BC(C x 2(C 2 C 2 x 2) [Cx 2 x 2)3/2 BC 2(C x 2(C 2 C 2 x 2) (2x 2 x 2)3/2 x2 x2 C C(C 2 C2 C2 x 2) C C2 x 2) C2 (C x2 ( x2 C 2 C2 x2 x2 ) C2 C2 C2 x 2) x C2 x2 x 2) C C2 x2 x 2 )2 x x 2] x2 (C 2 ) x2 x 2) C2 x 2(1) 168 Section 4.5 55. continued To find the critical points for 0 2x 2 4x 4 C2 C 4C 2x 2 C4 C4 4x 4 3C 2x 2 3C 2) x2 C 2x 2 0 The minimum value of A(x) for 0 corresponding triangle height is a C, we solve: 0 x 2(4x 2 C2 x mx f (x) C occurs at the critical point x C3 , or x 2 2 3C 2 . The 4 f (x) x B C B x C2 x2 Bx2 C C2 x2 2 B 3C B B C B BC C2 B B 2 C2 3C 2 4 4 CC 2 3C 2 4 3BC 2 4 C2 2 3B 2 3B This shows that the triangle has minimum area when its height is 3B. s Section 4.5 Linearization and Newton’s Method (pp. 220–232) Exploration 1 1. f (x) Approximating with Tangent Lines 2x, f (1) 2, so an equation of the tangent line is y 1 2(x 1) or y 2x 1. 3. Since (y1 y2)(1) y1(1) y2(1) 1 1 0, this view shifts the action from the point (1, 1) to the point (1, 0). Also (y1 y2) (1) y1 (1) y2 (1) 2 2 0. Thus the tangent line to y1 y2 at x 1 is horizontal (the x-axis). The measure of how well y2 fits y1 at (1, 1) is the same as the measure of how well the x-axis fits y1 y2 at (1, 0). 4. These tables show that the values of y1 y2 near x x 1. Here are two tables with Table 0.0001. 1 are close to 0 so that y2 is a good approximation to y1 near Section 4.5 Exploration 2 Grapher Using Newton’s Method on the 1–3. Here are the first 11 computations. 7. (a) x x 1 169 0 1 (b) 2ex 2ex x e e 1 (e 1 0 1) 0.684 2e 8. f (x) 3x 2 4 f (1) 3(1)2 4 1 Since f (1) 2 and f (1) 1, the graph of g(x) passes through (1, 2) and has slope 1. Its equation is g(x) 1(x 1) ( 2), or g(x) x 1. x 2. dy dx (x 1)(1 1) sin x) (x (x 1)2 x x sin x 1 (x sin x 1)2 1 x 2x cos (x 2 2 2.069 2.1 2.072 2.2 2.003 2.3 9. f (x) cos x f (1.5) cos 1.5 Since f (1.5) sin 1.5 and f (1.5) cos 1.5, the tangent line passes through (1.5, sin 1.5) and has slope cos 1.5. Its equation is y (cos 1.5)(x 1.5) sin 1.5, or approximately y 0.071x 0.891 Quick Review 4.5 d2 (x dx 1.9 2 1.3 1) 1.871 1.2 cos (x 2 1.8 1.1 dy dx 1.7 1.688 1.0 1. 1.457 0.8 0.9 2. g(x) 0.7 4. Answers will vary. Here is what happens for x1 f (x) 1) cos x)(1) cos x cos x (x 1) sin x (x 1)2 [0, ] by [ 0.2, 1.3] 10. For x , and so f (4) 2 1 . Since 2 x3 1 , the tangent line passes through 2 1 1 (4, 1) and has slope . Its equation is y (x 4) 1, or 2 2 1 y x 1. 2 f (4) 3. 1 3, f (x) 1 and f (4) [ 2, 6] by [ 3, 3] x 0.567 4. [ 1, 7] by [ 2, 2] [ 4, 4] by [ 10, 10] x 0.322 5. f (x) (x)( e x) (e x)(1) e x xe x f (0) 1 The lines passes through (0, 1) and has slope 1. Its equation is y x 1. 6. f (x) (x)( e x) ( e x)(1) e x xe x f ( 1) e 1 ( e 1) 2e The lines passes through ( 1, e 1) and has slope 2e. Its equation is y 2e(x 1) ( e 1), or y 2ex e 1. Section 4.5 Exercises 1. (a) f (x) 3x 2 2 We have f (2) 7 and f (2) 10. L(x) f (2) f (2)(x 2) 7 10(x 2) 10x 13 (b) Since f (2.1) 8.061 and L(2.1) 8, the approximation differs from the true value in absolute value by less than 10 1. 170 Section 4.5 1 2. (a) f (x) 2 x 9 We have f ( 4) L(x) x 8. (a) f (x) 2 9 4 . 5 5 and f ( 4) f ( 4) 4 x 5 x)6 (1 2 (b) f (x) 1 2 9 5 (b) Since f ( 3.9) 4.9204 and L( 3.9) 4.92, the approximation differs from the true value by less than 10 3. 1 2[1 ( x)] 1 1/2 x (1 ( 4)) 4) ( x)]6 [1 1 6( x) 1 x 2 2[1 1 6x ( 1)( x)] 2x (c) f (x) f ( 4)(x 4 (x 5 5 x (2x) 2 (d) f (x) x) x2 2 21 x 2 1/2 2 x2 4 21 1 x2 22 x 2 1 21 2 3. (a) f (x) 1 x We have f (1) 2 and f (1) L(x) f (1) f (1)(x 1) 2 0(x 1) 2 0. (e) f (x) 41/3 1 (b) Since f (1.1) 2.009 and L(1.1) 2, the approximation differs from the true value by less than 10 2. 4. (a) f (x) (f) f (x) x 1 2 3 1 2x We have f (0) L(x) (b) Since f (0.1) 0.0953 and L(0.1) 0.1 the approximation differs from the true value by less than 10 2. 5. (a) f (x) sec2 x We have f ( ) 0 and f ( ) L(x) f ( ) f ( )(x ) 0 1(x ) x f (0) f (0)(x ( 1)(x 2 x x 3 3 and f (0) 2 x 2 1 x 1. 0) 0) (b) Since f (0.1) 1.47063 and L(0.1) 1.47080, the approximation differs from the true value in absolute value by less than 10 3. x 6 3x cos x 1 1 and f (0) f (0)(x 3 2 0) 3 x 2 1 x 2 1. 0.002)100 1 0.021 10 0.009)1/3 1 1.009 (1 1.009 1.003 9 10 6 (100)(0.002) 1 1 (0.009) 3 10 f (x) 5 0(x 1) 5 1 2/3 x 3 We have f (8) L(x) 1.003; 8 2 and f (8) k1 k. 2 2 11. Center 1 f (x) 4x 4 We have f ( 1) 5 and f ( 1) 0 L(x) f ( 1) f ( 1)(x ( 1)) 5 12. Center 2 7. f (x) k(1 x) We have f (0) 1 and f (0) L(x) f (0) f (0)(x 0) 1 k(x 0) 1 kx 2/3 1 1 1 10. (a) (1.002)100 (1 1.2; 1.002100 1.2 x2 2 2/3 1 The linearization is the sum of the two individual for 1 We have f (0) f (0) 1 (b) 1 41/3 1 linearizations, which are x for sin x and 1 1. (b) Since f ( 0.1) 0.10033 and L( 0.1) 0.1, the approximation differs from the true value in absolute value by less than 10 3. 6. (a) f (x) 1 3x 34 3x 1/3 4 x 4 41/3 1 1 9. f (x) 1. 3x)1/3 1 1 We have f (0) 0 and f (0) L(x) f (0) f (0)(x 0) 0 1x x L(x) (4 f (8) f (8)(x 8) 1 . 12 2 1 (x 12 8) x 12 4 3 ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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