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A conical tank with radius 6 ft. and height 14 ft that was initially full of water is being drained at a rate of (1/8)√(h). Find a formula for the depth and the amount of water in the tank at any time t.
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A conical tank with radius 6 ft. and height 14 ft that was initially full of water is being drained at a rate of (1/8)√(h). Find a formula for the depth and the amount of water in the tank at any time t.
If you know the formula for volume of a cone (V= (1/3) pi r^2 h), there is no integration in this problem: it is a "related rates" problem that requires differentiation. Another complication is that you don't say whether the vertex of the cone or the base is downward. I am going to asume that the vertex is downward. Draw a picture: The tank, as seen from the side, should be an isosceles triangle which base 2r and height h. Now draw in the water level. Do you see that you have "similar triangles"? From that you can get a relationship between "h" and "r" for the triangle formed by the water and so write the volume of the water as a function of h only. Now differentiate both sides of that equation by t to get a relation between the rates of change.
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