Not sure how to approach the question, tried a few different angles but came out

Question

Not sure how to approach the question, tried a few different angles but came out with the wrong answer multiple times, thank you in advance for the help!

Suppose we pump water into an inverted right-circular cone tank at the rate of 2 cubic feet per minute. The tank has the height 9 ft and radius on the top is 7ft. What is the rate at which the water level is rising when the water is 5 ft deep? Leave the answer approximate. You may use the formula for the volume of the right-circular cone of radius r and height h: V=(1/3) pi(r^2)h. Not sure how to approach the question, tried a few different angles but came out with the wrong answer multiple times, thank you in advance for the help!

Explanation / Answer

Let's start with some variables R= 9ft maximum radius

r(h) - radius as a function of the height

The radius of a cone varies like r(h) = R/H*h = 9/7 *h

the volume at the height h is V=1/3 * pi * r(h)^2 * h

We can plug in r(h) V=1/3 * pi * (9/7)^2 h^2 * h = 1/3 * (9/7)^2 * pi * h^3

The change in volume of the fluid in the cone is found by taking the derivative of that formula using the chain rule

dV/dt = 1/3*pi*(9/7)^2*(3*h^2)*dh/dt

Now we know dV/dt which is the given volumetric flow rate. By plugging in h=4ft for the height we want, we can solve the above equation for dh/dt which is the rate of change of the height of the fluid.

so we got dh/dt = 49/(1296 pi)*dV/dt

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