Thread: Related Rates 6

A kite 116 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle between the string and the horizontal decreasing when 232 ft of string have been let out?

Hello, camherokid!

This is a messy one . . . no wonder no one wants to take it on.

A kite 116 ft above the ground moves horizontally at a speed of 11 ft/s.
At what rate is the angle between the string and the horizontal decreasing
when 232 ft of string have been let out?

Code:

                  x
      * - - - - - - - - - - - *
      :                    *  :
      :                 *     :
      :           L  *        :
  116 :           *           : 116
      :        *              :
      :     *                 :
      :  * θ                  :
      * - - - - - - - - - - - *
                  x

From the right triangle: .L² .= .x² + 116²

Differentiate with respect to time: .2L(dL/dt) .= .2x(dx/dt)

. . Hence: .dL/dt .= .(x/L)(dx/dt)

When L = 232, θ = π/6. .Hence: x = 116√3

. . So: .dL/dt .= .[(116√3)/232](11) .= .(11√3)/2 . [1]


We have: .cosθ .= .x/L . → . L·cosθ .= .x

Differentiate with respect to time: .cosθ(dL/dt) + L(-sinθ)(dθ/dt) .= .dx/dt . [2]


When L = 232, θ = π/6. .Hence: sinθ = 1/2, cosθ = √3/2


Substitute into [2]: .(√3/2)(11√3/2) + 116(-1/2)(dθ/dt) .= .11

And we have: .dθ/dt .= .-11/464 .≈ .-0.0237 radians/sec