The-volume-of-a-sphere-is-increasing-at-the-constant-rate-of-10cm-3-sec-Calculate-the-rate-of-increase-of-the-surface-area-at-the-instant-when-the-radius-is-5cm-What-is-the-radius-of-the-sphere-whe

Question Number 141776 by 7770 last updated on 23/May/21

The volume of a sphere is increasing

at the constant rate of 10 {cm}^{3} / \sec .

Calculate the rate of increase of the

surface area at the instant when the

radius is 5 cm . What is the radius of the

sphere when the surface area is increasing

at 2 {cm}^{2} / \sec .

Any step by step solution pls .

Answered by physicstutes last updated on 23/May/21

V = \frac{4}{3} π r^{3}, \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}

10 = 4 π {(5)}^{2} \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{10}{25 (4 π)}

A = 4 π r^{2} \Rightarrow \frac{d A}{d t} = 8 π \frac{d r}{d t}