Question

Water leaking onto a floor creates a circular pool with an area that increases at the rate of 3 square centimeters per minute. How fast is the radius of the pool increasing when the radius r is 3 centimeters?

Answer 1

So, I tried doing\[dA/dt = 3\] and \[dA/dt = 2\pi r(dr/dt)\] to get \[(dr/dt) =1/2\pi \] but that didn't work. help please?

Answer 2

\[\Large A'=3\]And they want us to find \(\Large r'\) when \(\Large r=3\).
Your work looks good.\[\Large A=\pi r^2\]\[\Large A'=2\pi r r'\]\[\Large r=\frac{A'}{r' 2 \pi}=\frac{1}{2\pi}\]

Answer 3

Are you entering this online?
Does it require you to enter it like this maybe? D:

1/(2pi)

Answer 4

Woops I solved for r instead of r' lol :) silly mistake there.
Should work out the same though.

Answer 5

Ah yeah. I see. I was supposed to enter it in as litterally 1/2pi rather than 1.57. Thanks.