A hollow sphere of radius r is being filled with water. The surface area of a hemisphere is 3pi*r^2. Question: When the water is at height r, and filling at a rate of 4cm^3s^-1, what is dS/dT?
By the chain rule ds/dt = ds/dr * dr/dv * dv/t. At a height of r, the water fills a hemisphere. So ds/dr = 6pir. dr/dv = 1/(dv/dr), so we need to find dv/dr. Students should have the formula for the volume of a sphere, which is (4/3)pir3 so the volume of a hemisphere is (2/3)pir3, which makes dv/dr = 2pir2. Now we know dv/dt=4, so ds/dt = 6pir * 1/(2pir2) * 4 = 24pir/2pir2 = 12/r.
HB
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