Question 4
The position of a particle can be modelled using $\mathbf{r}=\cos(t)\,\hat{\mathbf{i}}-2\sin(t)\,\hat{\mathbf{j}}$, $t\ge0$.
Which curve best represents the path of the particle?
Not quite. This is a hyperbola, which comes from secant and tangent. Sine and cosine with the same angle give a closed curve. In 2023, 10.7% of Queensland students chose (A).
Not quite. This is also a hyperbola, which comes from secant and tangent. Sine and cosine with the same angle give a closed curve. In 2023, 8.2% of Queensland students chose (B).
Not quite. It is an ellipse, but a wide one. The coefficient $2$ is on the $\hat{\mathbf{j}}$ part, so the path reaches $\pm2$ up and down and only $\pm1$ across. In 2023, 13.2% of Queensland students chose (C).
That’s it. Step through the hints below to see why. In 2023, 67.2% of Queensland students chose (D).
Cosine and sine with coefficients $1$ and $-2$: a closed path that reaches $\pm1$ across and $\pm2$ up and down. This is option (D).
Swap the coefficients and the ellipse lies on its side, as in option (C).
Secant and tangent give a hyperbola. The path splits into two branches that run off and never close, as in options (A) and (B).
Name the curve by sight before any algebra, as in game plan 4.1. Look at the trig pair first.
$$\mathbf{r}=\textcolor{#1D5FB8}{\cos}(t)\,\hat{\mathbf{i}}-2\textcolor{#1D5FB8}{\sin}(t)\,\hat{\mathbf{j}}$$
Sine and cosine with the same angle, $t$, give a circle or an ellipse. A hyperbola needs secant and tangent. That rules out (A) and (B) straight away.
Now compare the coefficients in front of the cosine and the sine.
$$\mathbf{r}=\textcolor{#1D5FB8}{1}\cos(t)\,\hat{\mathbf{i}}\textcolor{#1D5FB8}{-2}\sin(t)\,\hat{\mathbf{j}}$$
They are different, so the path is an ellipse. The minus sign only changes the direction of travel. The $\hat{\mathbf{i}}$ part reaches $\pm1$ and the $\hat{\mathbf{j}}$ part reaches $\pm2$, so the ellipse is taller than it is wide. That rules out (C).
If you want to confirm it, make each trigonometric function the subject, then square and add.
$$x^2+\frac{y^2}{4}=\cos^2t+\sin^2t=1$$
Move the slider to trace the path. Then try the other paths, which give the shapes in the wrong options.
Sine and cosine with the same angle give a circle or an ellipse, which rules out (A) and (B). The coefficients $1$ and $-2$ are different sizes, so it is an ellipse that reaches $\pm1$ across and $\pm2$ up and down.
$$x^2+\frac{y^2}{4}=1$$
This tall ellipse passes through $(\pm1,0)$ and $(0,\pm2)$.
How Queensland students answered in 2023
From the QCAA Specialist Mathematics subject report for 2023. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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