EXAMPLE 5 A man walks along a straight path at a speed of 6 ft/s. A searchlight
ID: 2888145 • Letter: E
Question
EXAMPLE 5 A man walks along a straight path at a speed of 6 ft/s. A searchlight is located on the ground 10 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 24 ft from the point on the path closest to the searchlight? SOLUTION We draw the figure to the left and let x be the distance from the man to the point on the path closest to the searchlight. we let be the angle between the beam of the searchlight and the perpendicular to the path. we are given that dx/dt = 6 ft/s and are asked to find deat when x = 24, The equation that relates x and can be written from the figure: -tan(0) x- tan(8). Video Example Differentiating each side with respect to t, we get dx dt dt So de dt dx dt when x = 24, the length of the bearm is 26, so cos(0) = 10 26 and dt The searchlight is rotating at a rate of rad/sExplanation / Answer
x/10= tan theta
x=10 tan theta
Differentiating both sides
dx/dt=10 sec2theta(d theta/dt)
d theta / dt= (1/10) cos2theta (dx/dt)
d theta / dt= (1/10) cos2theta(6) = (3/5)cos2theta
cos(theta) =10/26
Therefore d theta/dt= (3/5)(10/26)2=0.09
The searchlight is rotating at a rate of 0.09 rad/s
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