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

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The discussion revolves around calculating the depth of water in a trapezoidal trough when it is half-full, specifically using the trapezium rule. The problem is simplified by focusing on the areas of two trapezia formed by the water line, rather than employing integration. The width of the trough at the water line is determined as a function of the water depth, leading to the formulation of the areas below and above the water line. By equating these areas and solving the resulting quadratic equation, the depth of water when the trough is half-full is calculated to be approximately 27.19 cm. This approach provides a clear method for solving similar problems involving trapezoidal cross-sections.
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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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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:

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

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

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

$$w=\frac{3}{5}d+70$$

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

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

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

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

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

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

Equating the two areas, we have:

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

Multiplying through by $10$ we get:

$$d(3d+700)=(50-d)(3d+850)$$

Distributing and arranging in standard quadratic form, we obtain:

$$3d^2+700d-21250=0$$

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

$$d=\frac{25}{3}\left(\sqrt{298}-14 \right)\approx27.1889708469339$$
 
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 $$k$$ full, where $0\le k\le1$.

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

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

and so the slope of the linear width function is:

$$m=\frac{w_2-w_1}{h}$$

and the point-slope formula gives us:

$$w=\frac{w_2-w_1}{h}d+w_1$$

Now, in order for the trough to be $$k$$ full, we require:

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

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

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

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

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