Question

(Calculus) Related Rates:
A pebble is dropped into a pond which causes a circular ripple. Find the rate at which the radius of the ripple is growing at the time the radius is 1 foot, and the area enclosed by the ripple is increasing at a rate of 4 ft^ 2 / s .

Answer 1

My work so far:

dA/dt = 4 ft^2 / s
4 = dA/dr * dr/dt
4 = 2pi*r*dr/dt

Answer 2

\[A=\pi r^2\]
\[A'=2\pi r r'\] replace \(A'\) by 4, \(r\) by 2, solve for \(r'\)

Answer 3

Where did you get that 2?

Answer 4

the derivative of \(r^2\) with respect to \(t\) is \(2rr'\) by the chain rule

Answer 5

the derivative of \(r^2(t)\) is \(2r(t)\frac{dr}{dt}\)

Answer 6

r=1.

Answer 7

put the numbers in at the end, not at the beginning

Answer 8

\[\frac{dA}{dt}=2\pi r \frac{dr}{dt}\]
\[4=2\pi \times 1\times \frac{dr}{dt}\]
\[\frac{dr}{dt}=\frac{4}{2\pi}=\frac{2}{\pi}\]

Answer 9

"replace A′ by 4, r by 2, solve for r′"
This is what I was referring to. Looks like you fixed it.

Answer 10

ooooooooooh thanks. that was a typo on my part