Question

A spherical block of ice melts so that its surface area decreases at a constant rate: ds/dt = - 8 pi cm^2/s. calculate how fast the radius is decreasing when the radius is 3cm. (recall that s = 4 pi r^2.)

Answer 1

The rate a which the surface area, S, decreases is

`(dS)/(dt) =-8 \pi \, cm^(2)/s`

The surface area is
S = 4πr²
where r = the radius at time t.

Therefore

`(dS)/(dt) = (dS)/(dr) (dr)/(dt) =8 \pi r (dr)/(dt) \\\\ -8 \pi = 8 \pi r (dr)/(dt) \\\\ (dr)/(dt) =- (1)/(r)`

When r = 3 cm, obtain

`(dr)/(dt)]_(r=3) = - (1)/(3) \, cm/s`

Answer: -1/3 cm/s (or -0.333 cm/s)