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# Watching an elevator An observer is 20 m above the ground ISBN: 9780321570567 2

## Solution for problem 41E Chapter 3.10

Calculus: Early Transcendentals | 1st Edition

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Problem 41E

Watching an elevator? An observer is 20 m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 20 m horizontally from the observer (see figure). The angle of elevation of the elevator is the angle that the observer’s line of sight makes with the horizontal (it may be positive or negative). Assuming that the elevator rises at a rate of 5 m/s, what is the rate of change of the angle of elevation when the elevator is 10 m above the ground? When the elevator is 40 m above the ground?

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Solution:- Step1 To find what is the rate of change of the angle of elevation when the elevator is 10 m above the ground When the elevator is 40 m above the ground Step2 Given that An observer is 20 m above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is 20 m horizontally from the observer (see figure). The angle of elevation of the elevator is the angle that the observer’s line of sight makes with the horizontal (it may be positive or negative). Assuming that the elevator rises at a rate of 5 m/s. Step3 Step 4 We have, tan(a)= (20-h)/20 = 1- h/20 Differentiate both side we get, sec^2(a) da/dt= -1/20 dh/dt ----------(1) We have h= 10 and dh/dt=5 m/s tan(a)= (20-10)/20 = 10/20= ½ sec^2(a) = 1+ tan^2(a) = 1+(½)*(½) = 1+¼ =( 4+1)/4= 5/4 Put values in equation (1) we get, sec^2(a) da/dt= -1/20 dh/dt 5/4(da/dt)= -1/20 *5 da/dt= -1/20 * 5* = - = radian/ sec (By ignoring negative) We have h= 40 and dh/dt=5 m/s tan(a)= (20-40)/20 = -20/20= -1 sec^2(a) = 1+ tan^2(a) = 1+(-1)*(-1) = 1+1 =2 Put values in equation (1) we get, sec^2(a) da/dt= -1/20 dh/dt 2*(da/dt)= -1/20 *5 da/dt= -1/20 * 5* 1/2 = -1/8 = radian/ sec (By ignoring negative)

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##### ISBN: 9780321570567

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