Water leaking onto a floor forms a circular pool. The radius of the pool increases at a rate of 8 cm/min. How fast is the area of the pool increasing when the radius is 13 cm?

Question

asked Sep 28, 2020 127k views

2 Answers

Very Very Cherry · Answer 1 · 2020-10-04T14:57:18+0000

Answer:

The area of the pool increasing at the rate of 653.12
cm^2/min when the radius is 13 cm

Explanation:

Given:

radius of the pool increases at a rate of 8 cm/min

To Find:

How fast is the area of the pool increasing when the radius is 13 cm ?

Solution:

we are given with the circular pool

hence the area of the circular pool =

A =
$\pi r^2$ -----------------------------(1)

The area of the pool os increasing at the rate of 8 cm/min, meaning that the arae of the pool is changing with respect to time t

so differentiating eq (1) with respect to t , we have

$(d A)/(d t)=\pi \cdot 2 r \cdot (d r)/(d t)$

we have to find
(d A)/(d t) with
(d r)/(d t) = 8 cm/min and r = 13cm

substituting the values

$(d A)/(d t)=\pi \cdot 2 (13) \cdot 8$

$(d A)/(d t)=\pi \cdot 26 \cdot 8$

$(d A)/(d t)=\pi \cdot 208$

$(d A)/(d t)= 208 \pi$

(d A)/(d t)=653.12