Calculus: Related Rates - Inverted Pyramid Filling w/ Water | Yahoo Answers

[MarkFL]

Here is the question:

How do I do this calculus related rates problem? Full solution please.?

The Louvre museum in Paris features an inverted square pyramid with a height of 10 m. The side of the square is 13 m. Supervillains decide to fill the pyramid with water, and do so at a rate of 10 m3 per second. How quickly is the water level rising when it reaches the top? *Please note that the pyramid is inverted* Thanks guys

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Temitope,

Let's begin with the formula for the volume of a pyramid:

\(\displaystyle V=\frac{1}{3}bh\)

Now, the base $b$ is a square, and so let's let $s$ be the measure of the sides of the square, and so our volume formula becomes:

\(\displaystyle V=\frac{1}{3}s^2h\)

At any point in time, the volume of water will be a square pyramid that is similar to the container, and so we know we will always have:

\(\displaystyle \frac{s}{h}=\frac{13}{10}\implies s=\frac{13}{10}h\)

Now we may express the volume in terms of one variable $h$:

\(\displaystyle V=\frac{1}{3}\left(\frac{13}{10}h\right)^2h=\frac{169}{300}h^3\)

Now, if we differentiate with respect to time $t$, we obtain:

\(\displaystyle \frac{dV}{dt}=\frac{169}{100}h^2\frac{dh}{dt}\)

Now, we are told:

\(\displaystyle \frac{dV}{dt}=10\frac{\text{m}^3}{\text{s}}\)

And so (given that our units of length are in meters and our units time in seconds) we may write:

\(\displaystyle 10=\frac{169}{100}h^2\frac{dh}{dt}\)

Solving for \(\displaystyle \frac{dh}{dt}\), we obtain:

\(\displaystyle \frac{dh}{dt}=\frac{1000}{169h^2}\)

And so we find that when the pyramid is full, or when $h=10$, we have:

\(\displaystyle \bbox[7px,border:2px solid #207498]{\left.\frac{dh}{dt}\right|_{h=10}=\frac{1000}{169(10)^2}\frac{\text{m}}{\text{s}}=\frac{10}{169}\frac{\text{m}}{\text{s}}}\)