Question

A spherical balloon whose initial radius was 3 m is expanded such that its surface area is increasing at a constant rate of 72pi m2/sec. What is the speed of a point on the surface in the radially outward direction when radius becomes 9m

Answer 1

Hey Dear,

◆ Answer -

dr/dt = 1 m/s

● Explanation -

Surface area of the balloon is given by -

A = 4πr²

Differentiating on both sides -

dA/dt = 8πr.dr/dt

72 π = 8π × 9 × dr/dt

dr/dt = 72π / 72π

dr/dt = 1 m/s

Therefore, the speed of a point on the surface in the radially outward direction is 1 m/s.

Thanks for asking..

Answer 2

Answer:

Explanation:

by differenciating the value and equating them to 72py u get the ans 1sec