Question

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

Answer 1

Not quite.

Answer 2

Well i keep getting an answer of -80 which I don't think is right

Answer 3

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 ....?

Answer 4

wouldn't dr/dt = -2/pi

Answer 5

why not?

Answer 6

I guess I am confused on where to go from -2/pi to -2pi

Answer 7

the radius is continuing shorten until it get 20 ft. The rate of changing is -2pi ft/ second
That means dr/dt = -2pi

Answer 8

And as above , \(\dfrac{dA}{dt}=\dfrac{dA}{dr}\dfrac{dr}{dt}\)
So that, \(\dfrac{dA}{dt}= 2\pi r*(-2pi)\)

Answer 9

and you plug r = 20 into it, you get : -80pi^2 ft^2/second

Answer 10

\[dA/dt = 2pir * dr/dt\]

and \[dr/dt = -2\pi\]
so
\[2\pi(20)*(-2\pi) = -80 \pi^2\]

Answer 11

Oh you beat me to the punch!