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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 43e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 43e

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# Trajectory high point A stone is launched vertically

ISBN: 9780321570567 2

## Solution for problem 43E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Problem 43E

Trajectory high point A stone is launched vertically upward from a cliff 192 ft above the ground at a speed of 64 ft/s. Its height above the ground ?t seconds after the launch is g?iven by? ?s =?? 6?t2+ 64?t + 192? for ? ? 6. When does the stone reach its maximum height?

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STEP_BY_STEP SOLUTION Step-1 Let f be a continuous function defined on an open interval containing a number ‘c’.The number ‘c’ is critical value ( or critical number ). If f (c) 1 1 = 0 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Step-2 When an output value of a function is a maximum or a minimum over the entire domain of the function, the value is called the absolute maximum or the absolute minimum. Let f be a function with domain D and let c be a fixed constant in D . Then the output value f ) is the 1. Absolute maximum value of f on D if and only if f(x) f(c) , for all x in D. 2. Absolute minimum value of f on D if and only if f(c) f(x) , for all x in D. Step_3 Given is; A stone is launched vertically upward from the cliff 192 ft above the ground at a speed of 64 ft/s , and also given that its height above the ground t seconds after the launch is given by 2 S = -16t +64t+192 for 0 t 6. Now , we have to find out the value of stone reach its maximum height . For that we have to evaluate the function at critical points.so, the critical points satisfies the equation . 2 Given function is; S = -16t +64t+192 . For the critical points we have to differentiate the function with respect to t .[ Since from the step-1] d d 2 dt(s) = dt( -16t +64t+192 ) s = d ( -16t ) + d (64t)+ d (192 ),since d (u+v) = d (u)+ dt dt dt dx dx = -16 d ( t ) + 64 d (t)+ d (192 ), since d (Cx) = C d dt dt dt dx dx = -16 (2t) + 64(1)+0 , since dx(x ) = n x n1 , dx(c)=0, C is constant. = -32t+64 1 Since , from the definition f (c) = 0 = -32t+64 -32(t -2) = 0. Therefore , (t -2) = 0 t= 2. Therefore , t = 2 is clearly lies in the given interval [0,6]. 2 Therefore , at t =2 then S(2) = -16(2) +64(2)+192 = -64 +128+192 = 256. Hence, the function attains the critical value at t =2 , and the critical point is (2,256). Step-4 Now , we have to determine the absolute extreme values of f on the given interval ; Here the critical value is 2 , and the given interval is [0,6] so the endpoints are 0,6 since from the step -2, the absolute extreme values are ; Therefore, at x =0 , then s(0) = -16(0) +64(0)+192 = 0+ 0+192 = 192. 2 At t =2 then S(2) = -16(2) +64(2)+19 = -64 +128+192 = 256. At t =6 then S(6) = -16(6) +64(6)+192 = - 576+ 384+192 = 0. Therefore , s(0) = 192 , s( 2) = 256 , and s(6) = 0 Hence , the largest value of s(t) is 256 ,this attains at t=2. Therefore , the absolute maximum is S(2) = 256 Hence , the smallest value of s(t) is 0,this attains at t=6 Therefore , the absolute minimum is s(6) =0. Therefore , the maximum value occurs at t =2. Therefore, at t =2 the stone reach its maximum height.

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