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

`(1)/(32\pi) \ cm/min`

Step-by-step explanation:

-Assume the snowball is a spherically perfect:

`A=4\pi r^2`#

`A=4\pi((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\\\\(dA)/(dt)=4\pi(2r)(dr)/(dt)=8\pi r (dr)/(dt)\\\\-1.5=8\pi(6)(dr)/(dt)=48\pi (dr)/(dt)\\\\(dr)/(dt)=-(1)/(32\pi)`

Hence, the diameter is decreasing at a rate of
`(1)/(32\pi) \ cm/min`