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The method used in this problem is correct, but the answer is wrong.

The conical tank has 10 feet across the top. Therefore, the radius of the tank is 5ft, hence r=5h/12

r^2=(25h^2)/(144)

Changing these values yields that the rate of change of the deph of the water is 9/10pi ft/s

Btw... the similar triangle equation should be r/h=5/12 not 10/12 because 10 is the diameter not the radius...

This answer is incorrect because the diameter is equal to 10, but the radius is equal to 10/2 = 5.

Therefore, r is expressed as h*(5/12).

The volume expression becomes v = 1/3 * pi * (5/12 * h)^2 * h... which is (25/432) * pi * h^3.

Derivative = (25/432) * pi * 3 * h^2 * dh/dt.

At h = 8 and dv/dt = 10 ft^3/min...

10 ft^3/min = (25/432) * pi * 3 * 8^2 * dh/dt.

Solve for dh/dt and you will get... (9/10) * pi feet/min.

This is incorrect. You have made the mistake of assuming that the radius is 10 when all the while it is the diameter that is 10.

Your steps are correct but your geometry is skew.

the r/h relationship is 5/12.

I was just curious, if the cone is 10 feet across the top, wouldn't the radius be 5ft not 10ft???

Your method is correct however your answer is incorrect. The step where comparing radius to height, the r/h should = 5/12. Therefore the answer is 9pi/10.

Calculus

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