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Pre-Calc Homework Solutions 100

Pre-Calc Homework Solutions 100 - 100 Section 3.6 d dt 1 2...

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Unformatted text preview: 100 Section 3.6 d dt 1 2 1 4 2 1 4t 2 1 4(6) 1 2 4t 2 m/sec 5 d (1 4t dt dT du dT dL dL du d 2 dL 1 L (kL) g d L (kL) dL g L g kT 2 62. Velocity: s (t) 1 4t 4t) 66. 2 2 L g At t 6, the velocity is L g 1 (kL) g k Acceleration: s (t) d dt ( 1 2 1 4t 4t) (2) ( 2 1 2 1 d dt 2 d dt 1 4t 1 4t)2 4t) 67. No, this does not contradict the Chain Rule. The Chain Rule states that if two functions are differentiable at the appropriate points, then their composite must also be differentiable. It does not say: If a composite is differentiable, then the functions which make up the composite must all be differentiable. 68. Yes. Note that y d f(g(x)) dx d (1 4t dt f (g(x))g (x). If the graph of 1, then 0. This 1 4 1 4t 4t 4 4t)3/2 4 m/sec 2 125 d (k ds f(g(x)) has a horizontal tangent at x 0, so either g (1) f (g(1))g (1) 0 or f (g(1)) 1 4t (1 means that either the graph of y At t 6, the acceleration is dv dt dv ds ds dt [1 4 4(6)]3/2 g(x) has a horizontal f(u) has a horizontal tangent at x tangent at u 1, or the graph of y g(1). 63. Acceleration k 2 s dv (v) ds s) (k s) 69. For h 1: (k s) k2 , a constant. 2 64. Note that this Exercise concerns itself with the slowing down caused by the earth's atmosphere, not the acceleration caused by gravity. Given: v Acceleration k s dv dt k dv ds ds dt d k s ds s s (k) k s k ds d [ 2, 3.5] by [ 3, 3] For h 0.5: dv (v) ds (v) dv ds [ 2, 3.5] by [ 3, 3] k d ds For h s 0.2: ( s)2 (2 s) k s k2 2s 2 s [ 2, 3.5] by [ 3, 3] As h 0, the second curve (the difference quotient) ,s 0 Thus, the acceleration is inversely proportional to s 2. 65. Acceleration dv dt df (x) dt df (x) dx dx dt approaches the first (y 2 cos 2x). This is because 2 cos 2x is the derivative of sin 2x, and the second curve is the difference quotient used to define the derivative of sin 2x. As h 0, the difference quotient expression should be approaching the derivative. f (x) f(x) ...
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