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