A kite 50 ft above the ground moves horizontally at a speed of 6 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?

Answer 1

Draw a right triangle. The hypotenuse (of length y at time t) represents the string, the vertical side has length 50(constant), let the horizontal side be x at time t, and let θ be the angle opposite the vertical side at time t.

Given: dx/dt = 6. Find: dθ/dt when y = 200.

When y = 200, we have x² + 50² = 200². So, x²= 37500

tanθ = 50 / x

sec²θ (dθ/dt) = (-50 / x²)(dx/dt)

dθ/dt = cos²θ(-50/37500)(6) = (37500/40000)(-50/37500)(6) = -300/40000) = -3/400 radians/second.