SOLUTION: Water exits a conical tank at a constant rate of o.2 m3/minutes. If the surface of the water has radius r: a) find V(r), the volume of the water remaining b) find the rate at whi

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Question 1106186: Water exits a conical tank at a constant rate of o.2 m3/minutes. If the surface of the water has radius r:
a) find V(r), the volume of the water remaining
b) find the rate at which the surface radius is changing at the instant when the water is 5m deep.
The diameter is 6m. Height 8m.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The conical tank has a radius of 3m and a height (depth) of 8m. The ratio of the radius of the surface of the water to the depth of the water will always be 3:8. So

r+=+%283%2F8%29h; h+=+%288%2F3%29r

(a) The volume remaining at any time is


V%28r%29+=+%288%2F9%29%28pi%29%28r%5E3%29
(b) We want to find the rate at which the radius of the surface of the water is changing (dr/dt) when the depth is 5m. We are given that the volume of water in the tank is decreasing at a rate of 0.2 m^3/minute: dV/dt = -0.2.

dV%2Fdt+=+%28dV%2Fdr%29%2A%28dr%2Fdt%29
dr%2Fdt+=+%28dV%2Fdt%29%2F%28dV%2Fdr%29

We know dV/dt; and from part (a) we can find dV/dr to be %288%2F3%29%28pi%29%28r%5E2%29

So

dr%2Fdt+=+%28-0.2%29%2F%28%288%2F3%29%28pi%29%28r%5E2%29%29

When the depth is 5m, the radius is (3/8)*5 = (15/8)m. So dr/dt when the depth is 5 is

dr%2Fdt+=+%28-0.2%29%2F%28%288%2F3%29%28pi%29%28%2815%2F8%29%5E2%29%29
dr%2Fdt+=+%28-1%2F5%29%2F%28%2875%2F8%29%28pi%29%29
dr%2Fdt+=+%28-8%29%2F%28375%28pi%29%29