2020 Paper 2, Q19
Consider the following information when completing this question.
The length of a curve $y=f(x)$ over the interval $[a,b]=\displaystyle\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$
You are driving along a road with a vertical distance above sea level $D$ (in metres) given by the function $D(x)=300+\ln(x^2-3x+e)$ where $x$ is the horizontal distance from an initial point of measurement (in kilometres) at sea level.
Assume that if $x$ is positive you are east of the initial point of measurement and if $x$ is negative you are west of it. You start your drive at a horizontal distance of 10 kilometres west of the initial point and drive until you are 10 kilometres east of it.
Determine the time you spend driving downhill, if you drive downhill at an average speed of 40 km/h.
A road that rises and falls over 20 km
Drive west to east. The road falls gently for 11.5 km, bottoms out at $x=1.5$, then climbs for the last 8.5 km. The vertical scale below is stretched about 900 times. In reality this road is almost dead flat.
twenty kilometres in twelve seconds, then a short hold · only the orange stretch is paid for at 40 km/h
Where the slope is negative, and how long that road is
$D'(x)=\frac{2x-3}{x^2-3x+e}$. The denominator can never be zero (its discriminant is $9-4e<0$), so the sign depends only on the numerator, which is negative until $x=1.5$.
downhill is the shaded half
$\sqrt{1+\left(D'(x)\right)^2}$
Why is a length formula handed to you at all?
QCAA marking guide · 7 marks
Driving eastwards, downhill is where the altitude is decreasing: $D'(x)<0$.
Marker: correctly identifies values of $x$ associated with the downhill drive.
$$D'(x)=\frac{2x-3}{x^2-3x+e}$$
Graph $D'$ on the GDC, or graph $D$ and find the negative-sloping section. Either earns the mark.
Marker: correctly uses an appropriate mathematical representation.
$$\frac{2x-3}{x^2-3x+e}<0\ \Rightarrow\ 2x-3<0\ \Rightarrow\ -10\le x<1.5$$
Marker: determines the decreasing interval. The denominator never changes sign, because $x^2-3x+e$ has discriminant $9-4e\approx-1.87$, so it is positive for every $x$ and can be divided out safely.
$$\text{distance}=\int_{-10}^{1.5}\sqrt{1+\left(D'(x)\right)^2}\,dx$$
Marker: establishes the integral expression for the total distance travelled downhill. The limits are not given anywhere. They come out of step 3.
By GDC, $11.5$ km once $D$ is converted to kilometres. You get $13.25$ km if you leave $D$ in metres.
Marker: determines the distance travelled downhill.
$$t=\frac{11.5}{40}=0.29\ \text{hours}\qquad\left(\text{or}\ \frac{13.25}{40}=0.33\ \text{hours}\right)$$
About 0.29 hours, a little over 17 minutes
Marker: determines the time travelling downhill. Equivalent times accepted, e.g. 19.88 minutes for the unconverted version.
Integral notation written properly with its limits, the inequality solved in steps, the units named, and a closing sentence that answers in hours or minutes.
Marker: shows logical organisation communicating key steps.
QCAA gave full marks for both answers
11.50 km · 17.25 min
$D$ converted to kilometres makes $\frac{dy}{dx}$ about a thousandth, so the integrand is 1.0000 and the arc length is just the horizontal span.
13.25 km · 19.88 min
Metres against kilometres inflates the slope a thousandfold, so the integrand climbs to 1.77 and the road looks 15% longer.
QCAA’s own note: the 0.33-hour answer “was awarded full marks as the length of a curve formula was unfamiliar to students.” You should still do the conversion. The marks were for your reasoning, and not for the arithmetic that came out of it.
The QCAA hand you the formula, which tells you the horizontal distance is not the answer. What they do not tell you is the limits. You have to find them yourself by differentiating, solving an inequality, and discovering that the drive changes from downhill to uphill at $x=1.5$, a number that appears nowhere in the question. I would expect this to be the step most students miss. Then you have to watch the units, because the height is in metres and the distance is in kilometres.
Question wording and marking-guide steps are from the 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.