Question

The surface of a spherical balloon is increasing at the rate of 4 cm²/seс. Find the rate of change of volume when its radius is 12 cm.

Answer 1

Differentiate the Surface Area S = 4πr²

`(ds)/dt = 8πr (dr)/(dt)`

4 = 8πr × `(dr)/(dt)`

4 = 96π `(dr)/(dt)`

`(dr)/(dt) = 4/(96π) = 1/(24π)`cm/sec

`(dv)/dt = 1/2πr`

V = `4/3πr^3`   ...[Differentiate the Volume]

`(dv)/dt = 4πr ^2 (dr)/dt`

Calculate the rate of change of volume

Substitute r = 12 and `(dr)/dt = 1/(24π):`

`(dv)/dt = 4π(12)^2 xx 1/(24π)`

`(dv)/(dt) = 4π × 144 × 1/(24π)`

`(dv)/(dt) = (4 xx 144)/24`

= `144/6`

= 2 × 12

= 24 cm³