A man walks along a straight path at a speed of 5 ft/s. A searchlight is located

Question

A man walks along a straight path at a speed of 5 ft/s. A searchlight is located on the ground 6 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 8 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 A be the angle between the beam of the searchlight and the perpendicular to the path.

We are given that dx/dt = 5 ft/s and are asked to find dA/dt when x = 8. The equation that relates x and A can be written from the figure:'

x/ _________ = tan A
x =__________ tan A

Differentiating each side with respect to t, we get

dx/dt = 6 ___________ * _________/________

so

dA/dt = 1/6 _____________ · dx/dt = 1/6 ______________ ·(5)

= ____________ cos(A)2

When x = 8, the length of the beam is 10, so cos(A) = 6/10 = ___________ and

dAdt = ___________ (____________ )^2 = ________________

The searchlight is rotating at a rate of _____________ rad/s.

Explanation / Answer

x/6ft = tan A
x = 6ft tan A

Differentiating each side with respect to t, we get

dx/dt = 6 * (1/cosA)2 * dA/dt

so

dA/dt = 1/6 * (cosA)2 · dx/dt = 1/6 * (cosA)2 * 5

= (0.833) * cos(A)2

When x = 8, the length of the beam is 10, so cos(A) = 6/10 = 0.6 and

dA/dt = (0.833) * (0.6)^2 = 0.29988

The searchlight is rotating at a rate of 0.29988 rad/s.

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