Radius of a spherical balloon increases at the rate of 2 cm per second. Then, what would be the rate of increase of its curved surface area when the radius reaches 30 cm?

Question

Question

This question was previously asked in

OSSC CGLRE (Mains) Exam (Mathematics) Official Paper (Held On: 23 July, 2023)

Answer 1

Given:

The radius of the spherical balloon increases at the rate of 2 cm/s.

Radius when calculating the rate of increase of curved surface area = 30 cm.

Formula Used:

Curved Surface Area (CSA) of a sphere = 4πr²

Rate of change of CSA with respect to time (d(CSA)/dt) = d(4πr²)/dt

Calculation:

Given, dr/dt = 2 cm/s

First, find the derivative of the CSA with respect to the radius, r:

d(CSA)/dr = d(4πr²)/dr

⇒ d(CSA)/dr = 8πr

Now, use the chain rule to find d(CSA)/dt:

d(CSA)/dt = d(CSA)/dr × dr/dt

⇒ d(CSA)/dt = 8πr × 2

Substitute r = 30 cm:

⇒ d(CSA)/dt = 8π × 30 × 2

⇒ d(CSA)/dt = 480π cm²/s

The rate of increase of the curved surface area when the radius reaches 30 cm is 480π cm²/s.