0
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.