If a cylindrical tank with radius 5 meters is being filled with water at a rate of 3 cubic meters per minute, how fast is the height of the water increasing?

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The answer is ##(dh)/(dt)=3/(25 pi)m/(min)##.

With related rates, we need a function to relate the 2 variables, in this case it is clearly volume and height. The formula is: ##V=pi r^2 h##

There is radius in the formula, but in this problem, radius is constant so it is not a variable. We can substitute the value in: ##V=pi (5m)^2 h##

Since the rate in this problem is time related, we need to implicitly differentiate wrt (with respect to) time: ##(dV)/(dt)=(25 m^2) pi (dh)/(dt)##

In the problem, we are given ##3(m^3)/min## which is ##(dV)/(dt)##. So we substitute this in: ##(dh)/(dt)=(3m^3)/(min (25m^2) pi)=3/(25 pi)m/(min)##

In general – find a formula to relate the 2 variables – substitute values to remove the constant variables – implicitly differentiate wrt time (most often the case) – substitute the given rate – and solve for the desired rate.

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