Given a water tank shaped like a right circular cone with radius 15 feet at its base and altitude 10 feet. Water leaks from the tank at a constant rate of 1 cubic foot per second, and water is added to the tank at a constant rate of cubic feet per second. Find the value of so that the water level will be rising at a rate of 4 feet per second when the depth of the water is 2 feet.
The following diagram illustrates the situation:
First, we compute the radius of the water level in terms of the height of the water,
Furthermore, the radius at the water line is 3 feet when the height of the water is 2 feet. So, computing the volume of water in terms of and we have,
(As in the previous exercise, we are careful to use the product rule when evaluating this derivative.)
The problem then gives us the rate of change of volume of the water is
(Since we are adding cubic feet per second, and 1 cubic foot per second is leaking out.) So we set and solve for ,
Alternative solution: «Forget» the formula for the volume of a cone and start with integrating it from scratch.
I guess the radium is 10/15 of the height isnt it?
or change the numbers of the image