I don't know how to do related rates with the correct "derivative with respect to time" when the variables are not constant.

A girl flies a kite at a height of 300 feet, the wind carrying the kite horizontally away from her at a rate of 25 ft/sec. How fast must she let out the string when the kite is 500 feet from her?

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    $\begingroup$ when the variables are not constant. that cracked me up! $\endgroup$ – user31280 Nov 16 '12 at 12:22
  • $\begingroup$ ... I blame my teacher. Ha, I mean when the variables move so I can't plug them into the basic equation. Does that sound smarter now? $\endgroup$ – Courtney Nov 16 '12 at 12:24
  • $\begingroup$ $300^2+y^2=r^2$ where $y$ is horizontal distance and $r$ the length of the string. So $2yy'=2rr'$ and plug in what you know. $\endgroup$ – coffeemath Nov 16 '12 at 12:39

let $x$ be the horizontal component and $y$ the vertical component of the kite of length $k$, then from Pythagoras' $$k^2=x^2+y^2$$ Apply implicit differentiation with respect to time and you get $$2k\cdot\cfrac{dk}{dt}= 2x\cdot\cfrac{dx}{dt}+ 2y\cdot\cfrac{dy}{dt} $$ The kite flies only horizontally, thus there is no variation of $y$ with respect to time and $\cfrac{dy}{dt}=0$.

Find $x$ using Pythagras', the goal is to look for $\cfrac{dk}{dt}$, so with the values you were given $$\cfrac{dx}{dt}=25ft\cdot s^{-1},\ \ k=500ft$$

You should be able to complete it.


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