Question

If a snowball melts so that its surface area decreases at a rate of 3 cm2/min, how do you find the rate at which the diameter decreases when the diameter is 12 cm?

Answer 1

Answer:

$\frac{1}{32\pi} \ cm/min$

Explanation:

-Assume the snowball is a spherically perfect:

$A=4\pi r^2$#

$A=4\pi(\frac{D}{2})^2\\\\A=\pi D^2$

Therefore, at the time of interest, we plug the value of dA/dt and r(r=6cm):

$D=2r\\\\\frac{dA}{dt}=4\pi(2r)\frac{dr}{dt}=8\pi r \frac{dr}{dt}\\\\-1.5=8\pi(6)\frac{dr}{dt}=48\pi \frac{dr}{dt}\\\\\frac{dr}{dt}=-\frac{1}{32\pi}$

Hence, the diameter is decreasing at a rate of  $\frac{1}{32\pi} \ cm/min$