# Related rates of change problem

Water is pumped into an empty trough which is $200{\rm{ cm}}$ long at the rate of $33000{\rm{ c}}{{\rm{m}}^3}/s$. The uniform cross section of the trough is an isosceles trapezium with the dimensions shown:

Find the rate at which the depth of the water is increasing at the instant when the depth is 20cm.

I dont know how to even begin to approach this question. If I was to guess.. The question says that the trapezium is 200 cm long and I'm given the cross sectional dimensions of the shape, does this mean I should figure out an expression for the volume of the trapezium? And differentiate this? However I must further connect this to the rate that the water is being pumped into the trapezium.. How do I do that? How do I form an expression for the water flow? Or as the water is already given as a rate all I need to do is find the derivative of the volume and multiply the two? This is all very confusing.. Essentially I'd like an answer that provides a framework of sorts that will allow me to intuitively approach similar problems in the future

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Consider that every layer of water in the trough is rectangular, so it has an infinitesimal volume of $dV \ = \ A(y) \ dy$ , with $y$ being the "level" in the trough (you may choose to start from the bottom up or the top down) and $A(y)$ is the surface area. We'll measure from the bottom as $y = 0$ in what follows. The layer at $y = 0$ thus a surface area of $\ 80 · 200 \$ sq.cm.

We must consider how the surface area changes as $y$ increases. Since the trapezoid has straight (though inclined) sides, the width increases linearly from $w(y) = 80 \$cm at $y = 0 \$cm to $w(y) = 140 \$cm at $\ y = Y \$, this being the height of the trough. They were not quite so nice to you here, since they didn't tell you the full depth of the trough, so you will need to use a little trig to find $Y$. Then construct a "width function" $w(y)$ which varies linearly over the range from $y = 0$ to $y = Y$; you can then produce a "surface area function" $A(y) = w(y) \cdot 200 \$sq.cm.

You will now have a volume function with respect to $y$ that you can use in your related rates problem.

EDIT: In response to your other question, water is entering the trough at the rate $\frac{dV}{dt} = 33,000 \frac{\text{cu.cm.}}{\text{sec}}$ . So find $\frac{dy}{dt}$ at $y = 20$ .

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Really thorough answer, thank you! –  Assad Apr 30 '13 at 23:48
You said the volume is DV= A(y) dy, I don't understand this..why is this? –  Assad Apr 30 '13 at 23:49
The infinitesimal volume of a layer of water with a "thickness" $dy$ is $dV = A(y) dy$; that is the infinitesimal amount of water that would need to be added when the trough is filled up to a height $y$ in order to increase the level of water by $dy$. This gives you the related rate $\frac{dV}{dt} = A(y) \cdot \frac{dy}{dt}$ after differentiating implicitly with respect to time. –  RecklessReckoner May 1 '13 at 1:50
could you help me in making the width function please? I'm at a loss to do it... it would 80 + ?, the function would increase as h the height of water increases yes? as the trapezium will get wider and wider, but i don't know how to express this. Help would be appreciated, thank you! –  Assad May 1 '13 at 23:54
You will be setting up a "slope-intercept" equation for the line representing the function $w(y) = my + b$. We have arranged that $w(0) = 80$, so this is the value of $b$. Have you figured out how tall the trough is? We have $w(Y) = 140$, with $Y$ being the trough's height. The slope will then be $m = \frac{w(Y) - w(0)}{Y - 0} = \frac{140 - 80}{Y}$. –  RecklessReckoner May 2 '13 at 0:37

The trough is a prism so we can translate the 3D problem into a 2D problem. Instead of a 3D trough filling at a rate of $33000\,\text{cm}^3/\text{s}$, we can consider the 2D cross-section filling at a rate of $165\,\mathrm{cm}^2/\mathrm{s}$. (I divided by the length of the trough $200\,\mathrm{cm}$ to go from $33000\,\text{cm}^3/\text{s}$ to $165\,\mathrm{cm}^2/\mathrm{s}$.) This isn;t necessary, it just makes the numbers smaller.

First, we need to find an expression that connects the area, $A\,\mathrm{cm}^2$, covered by water of height $h\,\mathrm{cm}$ from the bottom of the cross-section. Looking at the diagram we see that $|AB|=80\,\text{cm}$ and $|CF|=140\,\mathrm{cm}$ and so it follows that $|CD|=|EF|=30\,\mathrm{cm}$. The question tells us that $|AC|=|BF|=50\,\mathrm{cm}$ and so Pythagoras tells us that the height is $|AD|=|BE|=40\,\mathrm{cm}$.

If the height of the water above the bottom is $h\,\mathrm{cm}$, then $|PQ|=80+12h$, where $0\le h \le 5$. It follows that the area of the trapezium $ABQP$ is given by $$A = \frac{1}{2}((80)+(80+12h))h = 80h+6h^2$$

The area $A$ and the height $h$ are both functions of time and so we can differentiate with respect to $t$ $$\frac{\mathrm{d}A}{\mathrm{d}t} = 80\frac{\mathrm{d}h}{\mathrm{d}t}+12h\frac{\mathrm{d}h}{\mathrm{d}t}=(80+12h)\frac{\mathrm{d}h}{\mathrm{d}t}$$

We know that the area increases at a rate of $165\,\text{cm}^2/\mathrm{s}$ and hence: $$\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{165}{80+12h}$$

When the height of the water is $20\,\text{cm}$ we have $$\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{165}{80+240} = \frac{33}{64} \approx 0.52\,\mathrm{cm}/\mathrm{s}$$

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