Question

A snowball melts at a rate of 0.05 [cm3 / s]. At what speed does the surface area decrease when the sphere is 10 [cm] in diameter?

Answer 1

`(dA)/dt= - 0.02 """"" "cm^2"/"sec"`

Explanation:

The figure is a sphere
Volume decreases
`(dV)/dt=-0.05""""" "cm^3"/"sec"`

`V=(4/3)pir^3`
Differentiate with respect to time

`(dV)/dt=(4/3)*pi*3r^2*(dr)/dt`

Substitute `(dV)/dt=-0.05` and `r=D/2=10/2=5`

`-0.05=(4/3)*pi*3(5)^2*(dr)/dt`

Solve for `(dr)/dt`

`(dr)/dt=-0.0005/pi""""" "cm"/"sec"`

Now, the surface Area

`A=4pir^2`
Differentiate with respect to time

`(dA)/dt=4pi*2r*(dr)/dt`

Substitute now `r=5` and `(dr)/dt=-0.0005/pi`

`(dA)/dt=4pi*2r*(dr)/dt`

`(dA)/dt=4pi*2*5*(-0.0005/pi)`

`(dA)/dt=-0.02""""" "cm^2"/"sec"`

Negative sign means decreasing surface area.

I hope the explanation is useful...God bless