Question 11
An aerial view of the surface of a dam, 6 km in length, is symmetrically positioned on a Cartesian plane as shown. A dam wall is located along the $y$-axis.
The surrounding edge of the dam can be modelled by the ellipse $\dfrac{(x-2)^2}{16}+\dfrac{y^2}{9}=1$, for $0\le x\le6$.
Use Simpson’s rule with four strips to determine an approximate area of the surface of the dam.
[4 marks] b)Evaluate the reasonableness of this approximation.
[2 marks]Work out half the dam, then double it
The dam is symmetric about the $x$-axis. Simpson’s rule needs the heights of the top half at five equally spaced values of $x$.
The height at $x=6$ is zero because the ellipse crosses the $x$-axis there. That term adds nothing to the sum.
Change the number of strips and compare
Simpson’s rule fits a parabola, dashed in orange, over each pair of strips. Change the number of strips, and choose whether to measure the top half or the full width of the dam.
This is part a), with four strips. Look at the right-hand end: the edge of the dam turns vertical at $x=6$, and a parabola can’t follow that. Almost all of the shortfall comes from the last two strips.
With two strips, one parabola covers the whole dam, so this estimate is the furthest from the exact area.
More strips bring the estimate closer, but slowly. The vertical edge at $x=6$ keeps causing most of the error.
The full-width heights are double the top-half heights, so you skip the doubling at the end. The QCAA accepts either set of values.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow the QCAA marking guide, which works with the top half of the dam.
The dam runs from the wall at $x=0$ to the end of the ellipse at $x=6$. Four strips across 6 km gives the width of each one.
$$w=\frac{b-a}{n}=\frac{6-0}{4}=1.5$$
This mark can be implied by your later working.
Marker: correctly determines the width of each interval
Rearrange the ellipse for $y$ and take the positive root, which is the top half of the dam.
$$\begin{aligned}&\frac{(x-2)^2}{16}+\frac{y^2}{9}=1\\&\Rightarrow\ y=\sqrt{9\left(1-\frac{(x-2)^2}{16}\right)}\end{aligned}$$
Store this as a function on your calculator and evaluate it at $x=0,\ 1.5,\ 3,\ 4.5,\ 6$.
$$\begin{array}{c|c}x&y\\\hline0&2.598\\1.5&2.976\\3&2.905\\4.5&2.342\\6&0\end{array}$$
Values to at least one decimal place get the mark. Heights of the full width, double these ($5.20,\ 5.95,\ 5.81,\ 4.68,\ 0$), are also accepted. You don’t need a table, and this mark can be implied by your substitution.
Marker: correctly determines the 5 required $y$-values
Write out Simpson’s rule from the formula sheet, then substitute. The values at $x_1$ and $x_3$ take the 4s.
$$\begin{aligned}A&\approx\tfrac{w}{3}\Big[f(x_0)+f(x_4)\\&\quad+4\big(f(x_1)+f(x_3)\big)\\&\quad+2f(x_2)\Big]\\[4pt]&\approx\tfrac{1.5}{3}\Big[2.598+0\\&\quad+4(2.976+2.342)\\&\quad+2\times2.905\Big]\end{aligned}$$
With the full-width values, the QCAA also accept this substitution.
$$\begin{aligned}A&=\tfrac{1.5}{3}\Big[5.2+0\\&\quad+4(5.95+4.68)\\&\quad+2\times5.81\Big]\end{aligned}$$
If you made an error earlier, you can still earn follow-through marks from here on.
Marker: substitutes $y$-values into Simpson’s rule
The bracket comes to $29.68$, and $\frac{1.5}{3}\times29.68\approx14.84$, so the top half is about $14.84$ km². The dam is symmetric about the $x$-axis, so double it.
$$A\approx2\times14.84$$
With the full-width values you get the whole area straight away, with no doubling. The two-decimal values give about $29.67$ km², and unrounded values give $29.68$ km². Either is fine.
Marker: determines an approximate area
To judge the estimate you need something better to compare it with. On Paper 2 your calculator can find the exact area, so write the integral down first. That line communicates your approach.
$$\begin{aligned}&\text{Area}\\&=2\int_0^6\sqrt{9\left(1-\tfrac{(x-2)^2}{16}\right)}\,dx\end{aligned}$$
The QCAA also accept other ways of writing it, such as
$$2\int_0^6\sqrt{\frac{144-9(x-2)^2}{16}}\,dx$$
or integrating the top half to get $15.16$ and then doubling. Another accepted strategy compares the estimate with a rectangle of about the same size, $l\times w\approx6\times5.2$.
The marking guide prints this alternative with a denominator of $144$ under the square root. That is a typo for $16$: $9\left(1-\frac{(x-2)^2}{16}\right)=\frac{144-9(x-2)^2}{16}$.
Marker: uses a suitable mathematical representation to communicate the approach
Evaluate the integral on your calculator, then compare it with your estimate.
$$\begin{aligned}\text{Area}&\approx30.33\ \text{km}^2\\\text{difference}&\approx30.33-29.68\\&=0.65\ \text{km}^2\\\text{error}&\approx\frac{0.65}{30.33}\times100\%\\&\approx2.1\%\end{aligned}$$
The QCAA sample response gives $30.32$, which is $2\times15.16$ after rounding the half first. The rectangle strategy from Step 5 gives about $31.2$ km², which is also accepted. The estimate is a little low because the parabolas cut inside the curve near $x=6$, where the edge of the dam turns vertical.
Marker: determines area and evaluates the reasonableness of the approximation
Each phrase of the question gave you something
Simpson’s rule with a given number of strips is a routine you have practised. The calculator evaluates the square roots and the exact integral for you, and part b) is the usual check of an estimate against a calculated value.
If I had to guess where you might lose marks, it would be forgetting to double the top half and giving $14.84$ km² as the area of the dam. In part b), a bare “it is reasonable” earns nothing. The mark needs a value to compare with and how close the two are.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2022 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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