When the radius of the balloon reaches 2 cm the radius is increasing at a rate of 0.3 cm/min. At this moment, how fast is the surface area of the balloon increasing?

Answer 1

A = 4πr²

Want dA/dt when r=2 cm and dr/dt = 0.3 cm/min

dA/dt = 8πr dr/dt = 8π(2)(0.3) = 4.8π cm²/min ≈ 15.1 cm²/min

Answer 2

This is called a related rates problem (related variables are changing with respect to time).

The surface area of a sphere is given by S = 4πr², where r is the radius of the sphere.

Differentiating with respect to time:

dS/dt = 8πr dr/dt

We are given that when r = 2, dr/dt = 0.3 cm/min

dS/dt = 8π(2)(0.3)

= 4.8π cm²/min