SOLUTION: Good day! Please help me with my problem solving. => Water is flowing into a conical reservoir 20 ft deep and 10 ft deep across the top at the rate of 15 ft³ per minute. Find ho

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Question 1146721: Good day!
Please help me with my problem solving.
=> Water is flowing into a conical reservoir 20 ft deep and 10 ft deep across the top at the rate of 15 ft³ per minute. Find how fast the surface is rising when the water is 8 ft deep.
Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


(1) Formula for the volume of a cone: V+=+%281%2F3%29%28pi%29%28r%5E2%29%28h%29

(2) Use the given dimensions of the cone to get the volume formula in terms of a single variable. Since the problem asks for the rate of change of the depth (height), we want a volume formula in terms of h.

The cone has a depth of 20 and a diameter of 10, so a radius of 5. So at all times as the cone is filling, the radius is 1/4 of the depth: r = h/4.

V+=+%281%2F3%29%28pi%29%28%28h%2F4%29%5E2%29%28h%29+=+%281%2F48%29%28pi%29%28h%5E3%29

(3) Find the derivative with respect to time:



dV/dt is given; solve for dh/dt when h=8:

15+=+%281%2F16%29%28pi%29%288%5E2%29%28dh%2Fdt%29
15+=+4pi%2A%28dh%2Fdt%29
dh%2Fdt+=+15%2F%284pi%29

ANSWER: the depth of the water in the tank is changing at a rate of 15/(4pi) feet per minute when the depth is 8 feet.