Question

If the volume of a solid hemisphere increases at a uniform rate, prove that its surface area varies inversely as its radius.

Answer 1

V = `2/3pir^3`

S = 3πr²

`(dV)/dt=k`

`(dV)/dt=d/dt(2/3pir^3)`

k = `2/3pi(3r^2(dr)/dt)`

k = `2pir^2(dr)/dt`

`(dr)/dt=k/(2pir^2)`

Differentiate Surface Area with Respect to Time (t):

`(dS)/dt=d/dt(3pir^2)`

`(dS)/dt=6pir(dr)/dt`

Substitute `(dr)/dt` into the surface area equation:

Substitute `(dr)/dtk/(2pir^2)` into `(dS)/dt=6pir(k/(2pir^2))`

`(dS)/dt=(3k)/r`

`(dS)/dt∝1/r`