# Thread: Calculus help, need help starting it and explaining

1. ## Calculus help, need help starting it and explaining

Word Problem

The base of a pyramid-shaped tank is a square with sides of length 12 feet, and vertex of the pyramid is 10 feet above the base. The tank is filled to a depth of 4 feet, and water is flowing into the tank at the rate of 2 cubic feet per minute. Find the rate of change of the depth of water in the tank.

I need help starting this problem and explaination

2. You want $\displaystyle \frac{dh}{dt}$
$\displaystyle \frac{dh}{dt} = \frac{dh}{dV} \frac{dV}{dt}$

You are given \frac{dV}{dt} as 2 cubic feet per minute
$\displaystyle \frac{dh}{dV} = \frac{1}{\frac{dV}{dh}}$

and $\displaystyle \frac{dV}{dh}$ is just the area of the top of the water.

Consider the triangle formed by the top of the pyramid, the midpoint of one of the sides of the base and the center of the base.

We know the length of the bottom is 6 as this is half of 12.

We also know that the smaller triangle formed with the surface of the water is symmetrical to the larger triangle, as 2 angles are the same.
So if the length of the side on the surface of the water is x,
$\displaystyle \frac{x}{6} = \frac{6}{10}$
x=3.6

This means that the side lengths of the surface of the water is 2*3.6 = 7.2 feet and the area is $\displaystyle 7.2^2$= 51.84 square feet
So $\displaystyle \frac{dh}{dt} = \frac{2}{51.84}$
=$\displaystyle \frac {50}{1296}$