FILL IN THE BLANK!!!!!!!!!!!!!!!!!!!!!!! A man walks along a straight path at a
ID: 2837165 • Letter: F
Question
FILL IN THE BLANK!!!!!!!!!!!!!!!!!!!!!!!
A man walks along a straight path at a speed of 2 ft/s. A searchlight is located on the ground 4 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 3 ft from the point on the path closest to the searchlight?
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 = 2 ft/s and are asked to find dA/dt when x = 3. The equation that relates x and A can be written from the figure:
x/4 = tanA
x = 4 tanA
Differentiating each side with respect to t, we get dx/dt = 4 sec^2(A) * (dA)/(dt)
dA/dt = 1/4______ * dA/dt = 1/4_______ * 2 = ______ cos^2 (A)
When x=3, the length of the beam is 5, so cos(A) = 4/5 = ________
dA/dt = ______(______)^2 = ________
The searchlight is rotating at the rate of ________ rad/s.
Explanation / Answer
dA/dt = 1/4 cos^(A) dx/dt means here i fill cos^(A) dx/dt
= 1/4 *cos^(A) * 2
cos^(A) = 2 * dA/dt
when x = 3, cos (A) = 4/5 = cos(36.87)
dA/dt = 1/2 {cos(A)}^2 = 1/2 * cos^2 (36.87) = 1/2 * 16/25 = 16/50= 0.32
hence searchline rotates at 0.32 rad/s
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