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정은's question at Yahoo! Answers regarding depth of water in trough when half full

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MarkFL

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Feb 24, 2012
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Here is the question:

Integration using trapezium rule?

The diagram shows the cross section of a water-trough. One side is vertical and the other side slopes. Calculate the depth of water in the trough when it is exactly half-full. (Hint: this means the areas of two trapezia must be equal.)

https://plus.google.com/u/0/118370402069852386878/posts/Ei4LooTa5ze

Above link is shows the diagram of this question.

The answer to this question= 27.19cm

Please explain to me how to get this answer.
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello 정은,

This problem does not require integration, or the trapezium rule to approximate a definite integral. While it could be used, it is simpler to just use the given hint.

To compute the areas of the trapezia in the cross-section of the trough above and below the water line, we need to know the width $w$ of the trough at the water line as a function of the depth of the water, which we are calling $d$. All linear measures are in cm.

We know this width increases linearly, and we know two points of the form $(d,w)$:

$(0,70)$ and $(50,100)$

and so the slope of the linear function is:

\(\displaystyle m=\frac{\Delta w}{\Delta h}=\frac{100-70}{50-0}=\frac{3}{5}\)

Thus, using the slope, and the first point in the point-slope formula, we get:

\(\displaystyle w-70=\frac{3}{5}(d-0)\)

\(\displaystyle w=\frac{3}{5}d+70\)

Now, using the formula for the area of a trapezium:

\(\displaystyle A=\frac{h}{2}(B+b)\)

We find the area $A_1$ of the trapezium below the water line is:

\(\displaystyle A_1=\frac{d}{2}\left(\frac{3}{5}d+70+70 \right)=\frac{d}{2}\left(\frac{3}{5}d+140 \right)\)

We find the area $A_2$ of the trapezium above the water line is:

\(\displaystyle A_2=\frac{50-d}{2}\left(100+\frac{3}{5}d+70 \right)=\frac{50-d}{2}\left(\frac{3}{5}d+170 \right)\)

Equating the two areas, we have:

\(\displaystyle \frac{d}{2}\left(\frac{3}{5}d+140 \right)=\frac{50-d}{2}\left(\frac{3}{5}d+170 \right)\)

Multiplying through by $10$ we get:

\(\displaystyle d(3d+700)=(50-d)(3d+850)\)

Distributing and arranging in standard quadratic form, we obtain:

\(\displaystyle 3d^2+700d-21250=0\)

Application of the quadratic formula, and discarding the negative root, there results:

\(\displaystyle d=\frac{25}{3}\left(\sqrt{298}-14 \right)\approx27.1889708469339\)
 
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MarkFL

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Feb 24, 2012
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Suppose we wish to generalize a bit, and let the width of the trapezoidal cross-section of a trough at the base and top be $w_1$ and $w_2$ respectively, where $w_1<w_2$. The depth of the trough we can call $h$, and we will, as before, let $d$ be the depth when the trough is \(\displaystyle k\) full, where $0\le k\le1$.

To find the width $w$ of the trough at $d$, we note we have the points:

\(\displaystyle \left(0,w_1 \right),\,\left(h,w_2 \right)\)

and so the slope of the linear width function is:

\(\displaystyle m=\frac{w_2-w_1}{h}\)

and the point-slope formula gives us:

\(\displaystyle w=\frac{w_2-w_1}{h}d+w_1\)

Now, in order for the trough to be \(\displaystyle k\) full, we require:

\(\displaystyle \frac{d}{2}\left(\frac{w_2-w_1}{h}d+2w_1 \right)=\frac{kh}{2}\left(w_1+w_2 \right)\)

Multiplying through by $2h$, and arranging in standard quadratic form in $d$, we obtain:

\(\displaystyle \left(w_2-w_1 \right)d^2+2hw_1d-kh^2\left(w_1+w_2 \right)=0\)

Applying the quadratic formula, and discarding the negative root, we find:

\(\displaystyle d=\frac{h\left(\sqrt{w_1^2+k\left(w_2^2-w_1^2 \right)}-w_1 \right)}{w_2-w_1}\)