SA = 4πr^2
d(SA)/dt = 8πr dr/dt
subbing in our values
d(SA)/dt = 8π(3)(-2) = -48π
you differentiated volume, it asked to find the rate of change of the surface area. So you need to differentiate surface area
A snowball is melting at the rate of 2 inches per hour. How fast is the surface area of the snowball changing at the instant the snowball has a radius of 3 inches
i tried (d/dt) = 4pir^2 (dr/dt)
its multiple choice and -72pi isnt there so im just trying to figure this out
2 answers
it's -48pi. ur welcome flvs fam
2 answers