A kite is released from a spot 50 feet from where you are on a calm day. It rises at a rate of 3 ft/s. How fast is the angle of elevation changing when it is $\frac{\pi}{6}$ radians above the line of sight from where you are? What about when the angle is $\frac{\pi}{4}$?

By creating a triangle, I can tell that the kite will be 25 feet in the air but I'm unsure how to incorporate the rate. Can someone walk me through the problem?

Just a heads up, this isn't a homework problem; rather it is a problem I got wrong on an exam.


Let $D$ be the horizontal distance to the kite, and $H$ the height of the kite above the ground. Then $\tan \theta = H/D$, where $\theta$ is the angle of elevation from where you're standing. Now take $d / dt$ of this equation, noting that $D$ is constant: $$ \frac{d}{dt}\tan \theta = \frac{d}{dt} \left(\frac{H}{D}\right) \Rightarrow \sec^2 \theta \frac{d\theta}{dt} = \frac{1}{D} \frac{dH}{dt} \Rightarrow \frac{d\theta}{dt} = \frac{\cos^2 \theta}{D} \frac{dH}{dt} $$ We know $D = 50$ ft and $dH / dt = 3$ ft/s, so you just have to substitute $\theta$ to get $d\theta / dt$.

  • $\begingroup$ So that would make $\frac{\pi}{6}$ at .045 degrees per second? I wish I had known it was this simple to set up the equation :/ I missed the obvious $\endgroup$ – StrugglingWithMath Dec 5 '12 at 19:32
  • $\begingroup$ 0.045 radians per second. Multiply by 180/$\pi$ to get degrees per second. $\endgroup$ – Eric Angle Dec 5 '12 at 22:09

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