Question

A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at 2 ft/sec, how fast is the area changing when the radius is 5 ft? 10 ft?

Answer 1

Given:

Rate of increase in ripple radius,
`(dr)/(dt)` = 2 ft/sec

Solution:

Area of circle =
`\pi r^(2)`

Differentiating w.r.t 'r', we get:

`(dA)/(dr)` = 2
`\pi r`

Rate of change in area is given by:

`(dA)/(dt)` =
`(dA)/(dr)`.
`(dr)/(dt)`

(by chain rule)

`(dA)/(dt)` = 2
`\pi r` .
`(dr)/(dt)`

when radius, r = 5ft

`(dA)/(dt)` = 2
`\pi 5` . 2 = 20
`\pi`

Now,

when r = 10 ft

`(dA)/(dt)` = 2
`\pi r` .
`(dr)/(dt)`

`(dA)/(dt)` = 2
`\pi 10` . 2 = 40
`\pi`