anonymous
  • anonymous
The radius of a circle is changing at the rate of -2/pi ft/s. At what rate is the circle's area when radius is 20 ft. I get dA/dt = 2pi*r dr/dt
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Not quite.
anonymous
  • anonymous
Well i keep getting an answer of -80 which I don't think is right
anonymous
  • anonymous
A = pi r^2 dA/dr = 2pi r but \(\dfrac{dA}{dr}\dfrac{dr}{dt}= \dfrac{dA}{dt}\) and \(\dfrac{dA}{dr}= 2\pi r\\\dfrac{dr}{dt}= -2\pi\) hence ....?

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anonymous
  • anonymous
wouldn't dr/dt = -2/pi
anonymous
  • anonymous
why not?
anonymous
  • anonymous
I guess I am confused on where to go from -2/pi to -2pi
anonymous
  • anonymous
|dw:1450379463078:dw|
anonymous
  • anonymous
|dw:1450379543013:dw|
anonymous
  • anonymous
the radius is continuing shorten until it get 20 ft. The rate of changing is -2pi ft/ second That means dr/dt = -2pi
anonymous
  • anonymous
And as above , \(\dfrac{dA}{dt}=\dfrac{dA}{dr}\dfrac{dr}{dt}\) So that, \(\dfrac{dA}{dt}= 2\pi r*(-2pi)\)
anonymous
  • anonymous
and you plug r = 20 into it, you get : -80pi^2 ft^2/second
anonymous
  • anonymous
\[dA/dt = 2pir * dr/dt\] and \[dr/dt = -2\pi\] so \[2\pi(20)*(-2\pi) = -80 \pi^2\]
anonymous
  • anonymous
Oh you beat me to the punch!
anonymous
  • anonymous
|dw:1450380048644:dw|
anonymous
  • anonymous
Thank you very much for the help!

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