SA = 4πr^2
d(SA)/dt = 8πr dr/dt
when d(SA)/dt = -1 , and r = 5
-1 = 8π(5) dr/dt
dr/dt = -1/(40π)
dD/dt = -1/(20π) or appr -.016 cm/min
(the negative sign indicates that the diameter is decreasing)
if a snowball melts so that its surface area decreases at a rate f 1cm^2/min, find the rate at which the diameter decrease when the diameter is 10cm?
1 answer
1 answer