2023 Paper 2, Q19
Over a suitable domain, a hill has a cross-sectional area given by $\displaystyle\int h(x)\,dx=\frac{a}{b}e^{bx}+c$, where:
- $a$, $b$ and $c$ are constants,
- $h(x)$ represents vertical distance (m), $x$ represents horizontal distance (m).
It is known that $h(0)=1.22$ and $h(40)=25$.
Where the gradient of the hill is 0.86 there is a tree stump. A second tree stump is located further up the hill. The difference in hill gradient between the two tree stumps is 0.44.
The vertical distance separating the two tree stumps is between 7.5 m and 8.5 m.
Evaluate the reasonableness of this prediction.
Walk up the hill and watch it get steeper
The stumps are not marked by where they are. They are marked by how steep the hill is there. The gauge finds them: 0.86 at the first, 1.3 at the second.
one walk up the hill, then a short hold · the gauge crosses each white line exactly as the walker passes that stump
Use two gradients to find two heights
Each stated gradient is an equation in $x$. Solve it for $x$, then put that $x$ back into $h$. The answer is the difference of the two heights, not of the two $x$ values.
$h(x)=1.22e^{0.0755x}$
$h'(x)=0.0921e^{0.0755x}$
QCAA marking guide · 6 marks
What the stem hands you is the area, i.e. an antiderivative. Differentiate it to get the hill itself:
$$\int h(x)\,dx=\frac{a}{b}e^{bx}+c\ \Rightarrow\ h(x)=ae^{bx}$$
Then $h(0)=1.22$ gives $a=1.22$, and $h(40)=25$ gives $25=1.22e^{40b}$, so
$$b=\frac{\ln\!\left(\frac{25}{1.22}\right)}{40}=0.0755\ \Rightarrow\ h(x)=1.22e^{0.0755x}$$
Marker: correctly determines the model for the hill with constants $a$ and $b$ found.
$$h'(x)=1.22\times0.0755\,e^{0.0755x}=0.09211e^{0.0755x}$$
Marker: differentiates $h(x)$ to determine the gradient of the hill formula. Follow-through marks are allowed for errors in prior working.
Set the gradient to 0.86 and solve for the position, then feed that position back into $h$ for the height:
$$0.09211e^{0.0755x}=0.86\ \Rightarrow\ x=29.5887,\qquad h(29.5887)=11.3907$$
Marker: determines the $y$-coordinate location of the first tree stump where the hill gradient is 0.86.
“The difference in hill gradient between the two stumps is 0.44” means the second gradient is $0.86+0.44=1.3$. A gradient, not a distance:
$$0.09211e^{0.0755x}=1.3\ \Rightarrow\ x=35.0614,\qquad h(35.0614)=17.2185$$
Marker: determines the $y$-coordinate of the second tree stump.
$$h(35.0614)-h(29.5887)=17.2185-11.3907=5.8278\ \mathrm{m}$$
Marker: determines the vertical distance between the tree stumps. Equivalent decimal values accepted.
5.83 m, the prediction is not reasonable
The vertical separation of 5.8278 m is not between 7.5 m and 8.5 m.
Marker: provides appropriate statement of reasonableness. Alternative statements accepted.
Three different “distances” you could report
$h(x_2)-h(x_1)$. This is what “vertical distance separating” asks for, and it is the answer.
$x_2-x_1$. It is very close to the vertical answer, but it is wrong. It would also fall outside the prediction, so the verdict survives but the marks do not.
The straight-line distance between the stumps. This one does land inside 7.5 to 8.5. So misreading the question flips the verdict to “reasonable”.
That third column is why the word “vertical” is doing so much work. The surveyor’s 7.5 to 8.5 m window is close to the slope distance, which is exactly the misreading a hurried student makes.
The function you are given is not the hill. It is the hill’s cross-sectional area, so everything starts with differentiating it. If you treat $\frac{a}{b}e^{bx}+c$ as $h(x)$, you have the wrong model from the first line. This is an inversion. The gradients are then used as inputs, so you solve each one for a position and substitute that position back into $h$. Nothing tells you to do that second substitution, and I see a lot of students stop at $x_2-x_1$. That is a horizontal distance, and the question wants a vertical one.
Question wording and marking-guide steps are from the 2023 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.