Pre-Calc Homework Solutions 166

Pre-Calc Homework - 166 Section 4.4 0 0 a 3 6 a 3 the relationship is 3 53(a According to the graph y(0(b According to the graph y L(c y(0 0 so d 0

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Unformatted text preview: 166 Section 4.4 0. 0. a 3 6 a 3 , the relationship is 3 53. (a) According to the graph, y (0) (b) According to the graph, y ( L) (c) y(0) 0, so d 0. 2bx (c) Since r r h and h 2. Now y (x) c 3ax 2 c, so y (0) implies that ax 3 bx 2 and aL 3 bL 2 H 55. (a) Let x0 represent the fixed value of x at point P, so that P has coordinates (x0, a), and let m f (x0) be the slope of line RT. Then the 0. Therefore, y(x) 3ax 2 y (x) 2bx. Then y( L) 3aL 2 2bL and y ( L) 0, so we have two linear equation of line RT is y m(0 m(x x0) x0) a a. The y-intercept of this line is a mx0, mx0 m a equations in the two unknowns a and b. The second 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 x 3 x 2 y(x) 2 3 x 3 3 2 x 2, or y(x) H 2 3 . L L L L equation gives b and the x-intercept is the solution of m(x x0) a 0, or x . Let O designate the origin. Then (Area of triangle RST) 2(Area of triangle ORT) 2 2 m m 1 (x-intercept of line RT)(y-intercept of line RT) 2 1 mx0 a 2 mx0 m mx0 m a 2 m a 2 m a 54. (a) The base radius of the cone is r height is h Therefore, V(x) 3 a2 r2 2 a 2 x 2 a2 a2 2 a x and so the 2 2 a x 2 . 2 2 a 2 x 2 r 2h 3 . (a mx0) a (b) To simplify the calculations, we shall consider the volume as a function of r: volume f (r) = f(r) d 2 (r 3 dr 3 3 2a 2r 3 r(2a a2 2 2 mx0 m 3 r2 a2 r 2) r 2, where 0 r a. m x0 a2 1 Substituting x for x0, f (x) for m, and f (x) for a, we ( 2r) ( a2 r 2)(2r) have A(x) f (x) x f(x) 2 . f (x) r2 2 r 3 a2 2r(a a 2 2 r2 r 2 r 2) (b) The domain is the open interval (0, 10). To graph, let y1 y2 2 3r 3 r2 3r 2) r 2 f (x) 5 5 1 x2 , 100 f (x) A(x) NDER(y1), and y2 x y1 2 y2 3 a The critical point occurs when r r h r a 3 2 a 3 a 6 . Then 3 2a 2 , which gives 3 y3 . A(x) is shown The graph of the area function y3 below. a2 6 r2 and h a2 a 3 , 3 2a 3 2 a 3 2 a 3 3 . Using we may now find the values of r and h for the given values of a. [0, 10] by [ 100, 1000] When a when a when a when a 4: r 5: r 6: r 8: r 4 6 ,h 3 5 6 ,h 3 4 3 5 3 3 3 ; ; 2 8 3 6, h 6 2 3; 8 3 3 The vertical asymptotes at x 0 and x 10 correspond to horizontal or vertical tangent lines, which do not form triangles. ,h ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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