Explanation
This is a classic "Related Rates" problem where we need to find the rate of change of the water level (h) with respect to time (t).
1. Given Data:
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Radius of the tank (R) = 5 ft
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Height of the tank (H) = 10 ft
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Rate of change of volume (dtdV) = 2 ft3/min
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Current depth of water (h) = 6 ft
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To find: Rate of change of height (dtdh)
2. Establish the Relationship:
The volume V of a cone is given by:
Using similar triangles (since the water forms a smaller cone inside the tank):
3. Substitute r into the Volume Formula:
Substitute r=2h into the volume equation to get V in terms of h only:
V=31π(2h)2h=31π(4h2)h
4. Differentiate with respect to time (t):
5. Solve for dtdh:
Plug in the known values dtdV=2 and h=6:
Final Answer:
The water level is rising at a rate of 9π2 ft/min. The correct option is (b).
