Question

There is a square pyramid with its vertex on bottom. Its base is `12` `meters` on each side. Its height is `4` `meters`. It is being filled with water at a rate of `9` `m^3`/`min`. How fast is the depth of the water growing when the depth is `2` `meters`?

First, I set up the primary equation:
`V=1/3bh`
(Volume of

Then, I found its derivative:
`(dV)/dt=1/3[(b)((dh)/dt)+(h)((db)/dt)]`

We know the base and height from the given information, so I plug that in:
`9=1/3[(36)((dh)/dt)+(2)((db)/dt)]`

We are trying to solve for `(dh)/dt`, but the `(db)/dt` is in the way. How do I solve this from here?

Answer 1

`(dh)/dt=1/4` meters/minute

Explanation:

The trick to these types of problems is to try to reduce the amount of different variables you have. You have a square pyramid, and you notice there is a direct relationship between the side and the height. The side is three times the height. We can write this as...

`3h=s`

In our primary equation:
`V=1/3bh`

We can rewrite it as:
`V=1/3(s^2)h`

And then plug in `3h`:
`V=1/3((3h)^2)h`

Simplify:
`V=3h^3`

Now you can just take the derivative and plug in the given information:
`(dV)/dt=9h^2*(dh)/dt`

`9=9*2^2*(dh)/dt`

`(dh)/dt=1/4`