A man walks along a straight path at a speed of 3 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

0.48 rad/sec

Step-by-step explanation:

dx/dt = 3ft

to find d∅/dt when x = 3

Using the right angled triangle

ΔABC

Now perpendicular (BC) = X FT

Base ( AC) 4 ft

so Hypotenuse(AB) = √ (BC² + AC²) = √ ( x² + 4²) = √ ( x² + 16)

and x/4 = tan∅ ; x = 4tan∅

now we differentiate each side

dx/dt = d/dt(4tan∅) = d/d∅(4tan∅) d∅/dt

⇒ dx/dt = (4sec²∅) d∅/dt  = ( 1/4 cos²∅) dx/dt

⇒ d∅/dt = cos²∅/4 × 3 = 3/4 cos²∅

now when x = 3, the length of the beam is

{from our initial equation (AB) = √ (BC² + AC²)}

AB = √(3² + 4²) = √(9 + 16) = 5ft

therefore cos∅ = 4/5

so we substitute in  (d∅/dt = ( 1/4 cos²∅) dx/dt)

d∅/dt = 1/4 ( 4/5)² × 3

d∅/dt = 12/15 = 0.48 rad/sec

the answer is su.k.my? ? co.kc