2020 Paper 1, Q20
At the end of the first stage of its growth cycle, a species of tree has a height of 5 metres and a trunk radius of 15 cm.
In the second stage of its growth cycle, the tree stays at this height for the next 10 years. However, the growth rate of the trunk radius (in cm per year) varies over the 10 years and is given by the function $r(t)=1.5+\sin\left(\frac{\pi t}{5}\right)$.
Assume the density (mass per unit volume) of the tree trunk is approximately 1 g/cm³ and the tree trunk is in the shape of a cylinder.
Determine the ratio of the trunk’s mass at the end of the second stage to its mass at the end of the first stage.
Ten years of rings, and the rate never sits still
The trunk widens between 0.5 and 2.5 cm a year. The fast years and the slow years are mirror images, so over a full ten-year cycle the wobble cancels and the radius gains a clean 15 cm.
ten years in twelve seconds, then a short hold · the dashed inner ring is where the trunk started
The area under the rate is the extra radius
Drag the slider. The shaded area on the left is exactly the number of centimetres the radius has gained. And by year 10 the sine has contributed nothing at all.
$r(t)=1.5+\sin\left(\frac{\pi t}{5}\right)$
$R(t)=15+1.5t+\frac{5}{\pi}\left(1-\cos\frac{\pi t}{5}\right)$
What is $r(t)$?
QCAA marking guide · 6 marks
$$R(t)=\int 1.5+\sin\left(\frac{\pi t}{5}\right)dt=1.5t-\frac{5}{\pi}\cos\left(\frac{\pi t}{5}\right)+c$$
Marker: correctly establishes an integrated expression for the radius of the tree. A definite integral is equally acceptable. QCAA accepts either.
$$15=-\frac{5}{\pi}+c\ \Rightarrow\ c=15+\frac{5}{\pi}$$
$$R(10)=15-\frac{5}{\pi}\cos(2\pi)+15+\frac{5}{\pi}=30\ \text{cm}$$
Marker: determines the radius of the tree. The two $\frac{5}{\pi}$ terms cancel because $\cos 2\pi=\cos 0$. The sine completes exactly one full cycle in the ten years.
$$V=\pi R^2 h=\pi R^2\times 500\ \text{cm}$$
Marker: identifies use of the formula and the radius to determine the volume of the tree trunk. Note the height must be converted: 5 m is 500 cm, because the radius is in centimetres.
$$V_{10}=\pi\times 30^2\times 500=450\,000\pi\ \text{cm}^3$$
$$V_{0}=\pi\times 15^2\times 500=112\,500\pi\ \text{cm}^3$$
Marker: determines volumes of the tree trunk initially and at 10 years. Leaving $\pi$ in is the smart move on a technology-free paper, because it cancels next.
$$\frac{\text{mass at 10}}{\text{mass at 0}}=\frac{1\times 450\,000\pi}{1\times 112\,500\pi}=4$$
A ratio of 4 : 1, so the trunk has quadrupled in mass
Marker: determines the ratio of mass. Density 1 g/cm³ means mass and volume are numerically equal, so the densities cancel too.
Define what you are integrating (“let $R(t)$ be the radius of the trunk”), keep the units visible, state the initial condition before you use it, and finish with a sentence. Do not just write 4.
Marker: shows logical organisation communicating key steps.
Doubling the radius makes the trunk four times heavier
The height never changes, so the whole answer lives in the cross-section: mass is proportional to $R^2$, and the radius exactly doubled. Four identical starting trunks would fit inside the finished one.
In kilograms, the trunk is about 353 kg at the end of the first stage and about 1414 kg at the end of the second. The question only asks for the ratio, and the ratio is the part that works for any trunk that grows like this one.
It is worth noticing that the sine term does not change the answer. Over a whole number of ten-year cycles it integrates to zero, so only the constant 1.5 cm/yr is left. If you spot that, you get $R(10)=15+15$ in one line.
There are three separate steps here, and the question does not point you to any of them. The function is called $r(t)$, but it is a rate, so you have to antidifferentiate it before it means anything. The 15 cm is not a number to substitute at the end. It is the initial condition you use to find $c$. The 5 metres also has to become 500 centimetres, or your volumes will be out by a factor of a hundred. The mass and the density then cancel, which is why the answer is a clean whole number on a technology-free paper. The QCAA like numbers that cancel this neatly on Paper 1, so I would treat a clean answer like this as a good sign you are on the right track.
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.