Tom flies a kite at a height of 300 ft, and the wind is carrying the height horizontally away at a rate of 25 ft/s. How fast is the angle of elevation changing when y=500ft?

Answer 1

The rate, u, is equal to dx/dt.

tan ∅ = h/x

Taking the derivative:

d∅/dt tan ∅ = d/dx h/x = -h/x²dx/dt

One of the trig identities is d/d∅ tan∅ = sec²∅

sec²∅ d∅/dt = -h/x^{2 (}dx/dt)

Isolate the derivative:

   

d∅/dt = -1/sec²-h/x^{2 (}dx/dt)

sec ∅ = y/x

d∅/dt = -(x²/y²⁾h/x^{2 (}dx/dt) After simplifying:

d∅/dt = -h/y^{2 (}dx/dt)

When we substitute the values in we have the following:

d∅/dt = -300ft/(500ft)² (25 ft/s) = -3/100 radians per sec.