SA = 4 pi R^2
dSA/dt = 8 pi R dR/dt
7 = 8 pi (12/2) dR/dt
7 = 48 pi dR/dt
dR/dt = 7 / (48pi) cm/min
Change in Diameter with respect to time is 14 / (48pi) cm per min
Kayla S.
asked • 03/23/22Calculus Question
If a snowball melts so that its surface area decreases at a rate of 7 cm2/min, find the rate at which the diameter decreases when the diameter is 12 cm.
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