I was working through my calculus textbook's practice problems when I came across a problem I couldn't figure out. I really suck at rate of change word problems and this one stumped me. Unfortunately, the textbook only gives answers to even-numbered problems and this is an odd question. Anyway it goes something like this:

An empty oil container is 10 meters long. A cross-section of the container is in the shape of an isosceles trapezoid that is 30cm wide at the bottom and 80cm wide at the top and has a height of 50cm. The container is filled with oil at a rate of 2 meters cubed per minute. How fast is the oil level rising when it is 30cm deep?

I would give my work but I have little to no idea if it's right and I'm afraid I would be just wasting my time as it's probably wrong. How do you do problems like these? I feel like I was never properly taught how and I would like to have a solid understanding by the time I'm actually being tested on this material.

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    $\begingroup$ The problem is about volume of an isosceles trapezoid prism. Start with the formula for that. $\endgroup$ – randomgirl Oct 25 '18 at 21:05
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    $\begingroup$ An "isosceles trapezoid that is 30cm wide at the top and 80cm wide at the top"? Is the 30cm or the 80cm supposed to be the width at the bottom? $\endgroup$ – Kurt Schwanda Oct 25 '18 at 21:12
  • $\begingroup$ @KurtSchwanda My bad, fixed that typo. $\endgroup$ – Bob Smith Oct 25 '18 at 23:03

Refer to the figure:

enter image description here

The volume of the container filled with oil is: $$V=\frac{(0.3+2x)+0.3}{2}\cdot h\cdot 10.$$

From the similarity of the two triangles on the left we get: $$\frac{x}{0.25}=\frac{h}{0.5} \Rightarrow x=\frac{h}{2}.$$

Substitute this into the volume formula: $$V=\frac{(0.3+2\cdot \frac{h}{2})+0.3}{2}\cdot h\cdot 10=\frac{(h+0.6)h}{2}\cdot 10=5h^2+3h.$$ Take the derivative of the volume function with respect to time: $$\frac{dV}{dt}=\frac{dV}{dh}\cdot \frac{dh}{dt}=2 \ \frac{m^3}{\text{min}} \Rightarrow \\ (10h+3)\cdot \frac{dh}{dt}=2 \Rightarrow \\ (10\cdot 0.3+3)\cdot \frac{dh}{dt}=2 \Rightarrow \\ \frac{dh}{dt}=\frac{2}{6}=\frac13.$$

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    $\begingroup$ Thank you for the very clear explanation! I obviously underestimated the geometric component to these questions. I'm curious, how would you do a problem like this if the given shape was a cone or a sphere and you were given the radius and height? How would you approach your first step where you put x in terms of h in a cone? For example, say you had a cone on it's tip with an arbitrary radius and height, as well as the height the cone is filled with fluid. How would you find the volume of the cone filled with fluid in terms of its height? $\endgroup$ – Bob Smith Oct 26 '18 at 19:39
  • $\begingroup$ Sorry forgot to @ you in the last comment and now I can't edit it. $\endgroup$ – Bob Smith Oct 28 '18 at 2:08
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    $\begingroup$ Search MSE. E.g.: this, this, this, this. Good luck. $\endgroup$ – farruhota Oct 28 '18 at 2:29
  • $\begingroup$ Thank you that answers my question perfectly. $\endgroup$ – Bob Smith Oct 28 '18 at 5:30

Let v(h) be the volume of that shape at height of h.
That is a formula you'll need to create.
You'll also have to calculate dv/dh.

Rate of increase of volume = dv/dh × dh/dt.
Solve for dh/dt knowing the volume increases
2 cubic meters/sec. Finally set h = 0.3 meters.


First note that the trough is of constant cross-section, so we can easily convert this from a volume problem to an area problem. If the trough is filling at $2 \mathrm{m^3/min}$, then the filled cross-sectional area is increasing at a rate of $\frac{2}{10} = 0.2 \mathrm{m^2/min}$. Also, work in consistent units, so convert everything in $\mathrm{cm}$ to $\mathrm{m}$.

Now look at the geometry of the problem. Similarity is often very useful here. If you extend (produce) the slant sides of the trapezoid to meet at a point, you'll get a larger inverted triangle with height $0.8 \mathrm{m}$ and base $0.8 \mathrm{m}$. If you consider fluid level height $H$ from this (imaginary) vertex instead, the problem is made rather simple. The base of this triangle is always equal to the height, so its area $A = \frac 12 H^2$. At the specified instant, $H = 0.3 + 0.3 = 0.6$ (the actual fluid depth in the trough plus the "imaginary" height of the produced triangle).

By chain rule, $\frac{dA}{dt} = \frac{dA}{dH} \cdot \frac{dH}{dt}$

Since $\frac{dA}{dH} = H$, you get $\frac{dH}{dt} = \frac{\frac{dA}{dt}}H = \frac{0.2}{0.6} = \frac 13 \mathrm{m/min}$, which is the required answer.

Note that introducing the "imaginary" triangular portion below the base of the trapezoid doesn't affect the analysis because when you take the derivative of the area, the constant area of that portion vanishes.

If you find the above hard to imagine, you can work out the actual area of the fluid filled cross-section of the trapezoid. It's only a little harder. If you wanted to employ this method, you need a relationship between the top of the filled cross-sectional trapezoid (call this $p$) to the actual fluid level (let's call this $h$). You'll find (again, by similarity) that this is a simple linear relationship, namely $p = h + 0.3$.

So the area $a$ of the filled trapezoid will be given by $a = \frac 12(p+0.3)(h) = \frac 12 h(h+0.6) = \frac 12(h^2 + 0.6h)$. This gives $\frac{da}{dh} = \frac 12(2h + 0.6) = h + 0.3$.

Following what we did before, $\frac{da}{dt} = \frac{da}{dh} \cdot \frac{dh}{dt}$, so $0.2 = (h+0.3)\cdot \frac{dh}{dt}$ and $\frac{dh}{dt} = \frac{0.2}{0.3+0.3} = \frac 13 \mathrm{m/min}$, as before.


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