Pre-Calc Homework Solutions 109

Pre-Calc Homework Solutions 109 - Section 3.7 d (4 dt 109...

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Unformatted text preview: Section 3.7 d (4 dt 109 44. v(t) s (t) 9(4 6t)3/2 3 (4 2 6t)1/2(6) (b) The tangent is horizontal when x2 x2 0, or y . 3x 3 x2 Substituting for y in the original equation, we have: 3 dy dx 3y y2 6t)1/2 d [9(4 dt a(t) v (t) 9 (4 2 6t)1/2] 27(4 6t) 1/2 6t) 1/2 (6) x3 x3 x 3 2 3 y3 9x x6 27 x3 3 (x 27 9xy x 3 2 0 0 0 0 0 or x 02 3 3 At t 2, the velocity is v(2) 36 m/sec and the acceleration is a(2) 45. Acceleration dv dt 27 m/sec2. 4 d [8(s dt x3 t) dt 1/2 3x 3 54) x 1] 4(s 4(s 4(s 32(s t) t) t) t) 1/2 ds 1/2 1/2 1 1) t)1/2 t) 1/2 (v [(8(s (s 54 3 3 2 1) 1] At x 0, we have y 1 3 2 (3 2) 3 0, which gives the 3 3 1/2 point (0, 0), which is the origin. At x y 1 3 (9 4) 3 3 2, we have 32 ft/sec2 4 2 4 2 3 3 3 4, so the point other than the origin is (3 2, 3 4) or approximately 46. d 4 (y ) dx y 4y x 9x (3.780, 4.762). d (9x 2) dx d (4y 2) dx dy dy 8y 4y 3 dx dx dy dx d 4 (x ) dx (c) The equation x 3 the line y 9x 4y y3 9xy is not affected by 4x 3 4x 3 4y 3 9(3) 4(2) 18x 18x 8y 27 8 9( 3) 4(2) 9( 3) 4( 2) 9(3) 4( 2) 27 8 27 8 27 8 2x 3 2y 3 interchanging x and y, so its graph is symmetric about x and we may find the desired point by 3 3 interchanging the x-value and the y-value in the answer to part (b). The desired point is (3 4, 3 approximately (4.762, 3.780). 48. d 2 (x ) dx 2(3)3 Slope at (3, 2): 2(2)3 2) or 2( 3)3 Slope at ( 3, 2): 2(2)3 Slope at ( 3, Slope at (3, 2): 2): 2( 3)3 2( 2)3 x2 2xy 3y2 0 2(3)3 2( 2)3 2x 47. (a) x3 d 3 d 3 (x ) (y ) dx dx dy dy 3x 2 3y 2 9x dx dx y3 d dx 9xy 0 d (0) dx 9 (xy) 9(y)(1) 9x) dy dx dy dx d d d 2 (xy) (3y2) (0) dx dx dx dy dy 2x 2(y)(1) 6y 0 dx dx dy (2x 6y) 2x 2y dx dy 2x 2y x y dx 2x 6y 3y x 1 1 3(1) 1 2 2 0 9y 3x 2 3y y2 x2 3x At (1, 1) the curve has slope normal line is y Substituting 2 1, so the x 2. (3y 2 1(x 1) + 1 or y 3(2) Slope at (4, 2): 2 (2) 3(4) Slope at (2, 4): 2 (4) (4)2 3(4) (2)2 3(2) 9y 3x 2 3y 2 9x 10 5 8 4 4 8 5 10 x 2 for y in the original equation, we have: x 2 2xy 3y 2 0 x 2x( x 2) 3( x 2)2 0 2 x 2x 2 4x 3(x 2 4x 4) 0 4x 2 16x 12 0 4(x 1)(x 3) 0 x 1 or x 3 Since the given point (1, 1) had x 1, we choose x and so y (3) 2 1. The desired point is (3, 3 1). ...
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