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?

1 Answer

(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