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…

Question

asked Jan 4, 2020 64.5k views

1 Answer

Ajoseps · Answer 1 · 2020-01-11T19:27:34+0000

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
$\pi$