A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep. radius, r height, h Need to know the volume, V

The volume of a cone is given by: Product Rule There are 3 variables, V, r, h

We want to find the “rate of change of the depth of the water”. We want dh/dt This is still the volume of a cone

To find dh/dt, we need to know how many things? 4 From the problem: We still need the radius, r, and its rate of change

We have to review Geometry and similar triangles: 5 12 r Water level h = 8 From the problem, we know the radius of the tank is 5 feet, the height of the tank is 12 feet and the water is 8 feet deep. From this we can find the radius of the water in the tank

The radius of the water, r, is 10/3 feet. Now, we need to find dr/dt. Again, we need to use similar triangles. No matter what the level of the water, we can relate the radius, r, to the depth, h

Now, take the derivative of each side with respect to t This gives the last thing we need since we know dh/dt

Make the substitution Plug in the known numbers and solve for dh/dt

Of course, this is left for the student to do.