If a snowball melts so that its surface area decreases at a rate of 4 cm^2/min, find the rate at which the diameter decreases when the diameter is 12 cm.
Please Help!!!! Thanks.
Please Help!!!! Thanks.
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S = 4πr^2
dS/dt = 8πr dr/dt
d = 2r
dd/dt = 2 dr/dt
-4 cm^2/min = 8π(6 cm) dr/dt
-1/(12π) cm/min = dr/dt
The diameter decreases at a rate of 1/(6π) cm/min ≈ 0.053 cm/min.
dS/dt = 8πr dr/dt
d = 2r
dd/dt = 2 dr/dt
-4 cm^2/min = 8π(6 cm) dr/dt
-1/(12π) cm/min = dr/dt
The diameter decreases at a rate of 1/(6π) cm/min ≈ 0.053 cm/min.