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Given a cross-section of an object that is parabolic in shape, how do you find the depth of the object when it is "half full".

A full example given in an exam:

A long trough whose cross-section is parabolic is $1\frac{1}{2}$ metres wide at the top and $2$ metres deep. Find the depth of water when it is half-full.

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  • $\begingroup$ Best guess for now (while writing longer reply) is that 1/2 full is 1/2 the area. Otherwise, the problem is 1/2 the depth = 2/2 = 1 m. Easy, huh? ;-) $\endgroup$ – FundThmCalculus Jul 29 '14 at 1:38
  • $\begingroup$ 1m isn't the answer. That was my first thought after graphing the shape out onto axes. $\endgroup$ – jduncanator Jul 29 '14 at 1:39
  • $\begingroup$ I know, I was showing the trivial solution for being a smart-alec on the test. Not saying you were dumb or anything, just a joke. :-) $\endgroup$ – FundThmCalculus Jul 29 '14 at 1:43
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Put the parabola in standard position. It has equation $b^2y=x^2$ for some $b$ which we won't bother to evaluate.

The area of the parabolic segment up to depth $d$ is given by $$\int_{y=0}^d 2x\,dy.$$ Here $x=by^{1/2}$. So the area is $$\frac{2}{3}bd^{3/2},$$ or more simply $kd^{3/2}$ for some constant $k$. If the full depth is $2$, and $m$ is the depth at half the volume, we have $$km^{3/2}=\frac{1}{2}k\, 2^{3/2}.$$ Solve. We get $m^{3/2}=2^{1/2}$ and therefore $m=2^{1/3}$.

Remark: Note that the width at the top is irrelevant to the calculation. Archimedes already knew this, one and a half millenia "before calculus." All the material needed to solve this problem was in his Quadrature of the Parabola. For details, please see the Wikipedia article.

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Hint. First find the equation of the parabola. Take the origin to be at the vertex of the parabola. We can probably assume (even though the question didn't say so) that the axis of the parabola is vertical. Then the equation is $$y=ax^2\ ,$$ where the value of the constant $a$ can be found from the given dimensions.

Now if the trough is filled out to the $x$ coordinate $x=x_0$ to give half the volume, then $$\int_{-x_0}^{x_0} ax_0^2-ax^2\,dx=\frac{1}{2}\int_{-3/4}^{3/4}2-ax^2\,dx\ .$$ From this you can find $x_0$, and the corresponding $y$ is the depth you are looking for.

Comment. You don't actually need to know $a$ in order to find $x_0$, but you do need it for the final step.

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  • $\begingroup$ Just a note. I think that gives the area under the parabola, which is not inside the trough, based upon my understanding of the problem. $\endgroup$ – FundThmCalculus Jul 29 '14 at 1:50
  • $\begingroup$ Ooooops yes of course. Fixed now. (I hope.) $\endgroup$ – David Jul 29 '14 at 5:42
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Model the cross-section as a parabola centred at the origin. Then we know that: $$ y = ax^2 $$ for some constant $a$. But from the given dimensions, we also know that the parabola contains the point $(3/4, 2)$. Substituting, we find that: $$ 2 = a(3/4)^2 \iff 2 = \frac{9a}{16} \iff a = \frac{32}{9} $$ Now the area of the cross-section when full is given by: $$ A = 2\int_0^{3/4} (2 - \tfrac{32}{9}x^2) \, dx = 2 \left[2x - \frac{32}{27}x^3 \right]_0^{3/4} = 2\left[\frac{3}{2} - \frac{1}{2} \right] = 2 $$ Now suppose that the half-width of the cross-section when half-full is $x = k$ so that its depth when half-full is $\frac{32}{9}k^2$. Then it follows that: \begin{align*} \frac{A}{2} &= 2\int_0^{k} (\tfrac{32}{9}k^2 - \tfrac{32}{9}x^2) \, dx \\ \frac{9A}{128} &= \int_0^{k} (k^2 - x^2) \, dx \\ \frac{9(2)}{128} &= \left[k^2x - \frac{x^3}{3} \right]_0^k \\ \frac{9}{64} &= \frac{2k^3}{3} - 0 \\ \frac{27}{128} &= k^3 \\ k &= \frac{3}{4\sqrt[3]{2}} \end{align*}

Hence, we conclude that the desired depth is: $$ \frac{32}{9}k^2 = \frac{32}{9}\cdot \frac{9}{16\sqrt[3]{4}} = \frac{2}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{2}}{\sqrt[3]{2}} = \sqrt[3]{2} $$

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  • $\begingroup$ Congrats. Nice work. You beat me to it, but since we ran it both ways and it worked, that's nice confirmation. $\endgroup$ – FundThmCalculus Jul 29 '14 at 1:58
  • $\begingroup$ Hahah thanks, you too. It's cool that you wrote $x$ as a function of $y$ instead and got everything to work out nicely. $\endgroup$ – Adriano Jul 29 '14 at 2:01
  • $\begingroup$ Just remember this: I'm lazy. I don't like to have to remember the capping on the top of the integral, so I rotated the direction of gravity... :D $\endgroup$ – FundThmCalculus Jul 29 '14 at 2:04
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The way to do this is to find the area of the parabola as a function of $x$. Let's assume for simplicity that function is: $x = c*y^2$. This is equivalent to $y = \sqrt{\frac{x}{c}}$ Since the parabola is symmetric, we can double this to get the total width at any point $x$. $$ w(x) = 2 \sqrt{\frac{x}{c}}$$ Now we can integrate with respect to $x$ from $0$ to $2$. But first, we need to find the constant $c$ that gives us the particular parabola that we have in this case. To do this, at the point $(0.75,2)$ solve for $c$. $$ y= \sqrt{\frac{x}{c}} = 2 = \sqrt{\frac{3/4}{c}} \rightarrow 4 = \frac{3}{4c} \rightarrow c = \frac{3}{16} $$ Now we can do the actual integration that will give the area. $$A(u) = \int_0^u w(x) dx = \int_0^u 2 \sqrt {\frac{16x}{3}} dx = \frac{8}{\sqrt 3} \int_0^u \sqrt x dx = \frac{8}{\sqrt 3} [\frac{2}{3} x^{\frac{3}{2}}]_0 ^ u = \frac{16}{3 \sqrt 3} u^{\frac{3}{2}}$$. Thus, you now have the area as a function of integration distance along the length. Just plug in $2$ to get the total area, and then solve it for $1/2$ the total area. $$A_t = \frac{16}{3 \sqrt 3} 2^{\frac{3}{2}} = \frac{32}{3}\sqrt{\frac{2}{3}}$$ Solve: $$\frac{\frac{32}{3}\sqrt{\frac{2}{3}}}{2} = \frac{16}{3 \sqrt 3} u^{\frac{3}{2}}$$ $$\frac{16}{3}\sqrt{\frac{2}{3}} = \frac{16}{3 \sqrt 3} u^{\frac{3}{2}}$$ $$\sqrt{\frac{2}{3}} = \frac{1}{\sqrt 3} u^{\frac{3}{2}}$$ $$\sqrt{2} = u^{\frac{3}{2}}$$ $$\sqrt{2} = u^{\frac{3}{2}}$$ $$u = \sqrt[3]{2}$$

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