Specialist Mathematics · Mock exam 2026 All mock exam questions
Mock exam · Paper 2 · Simple familiar

Question 13

[4 marks] Graphics calculator
The question

The position vector of a particle at time $t$ is given by

$$\begin{aligned}\boldsymbol{r}(t)&=\left(2+4\cos(t)\right)\hat{\boldsymbol{\imath}}\\&\quad+\left(3\sin(t)-1\right)\hat{\boldsymbol{\jmath}},\quad t\geq0\end{aligned}$$

a)

Show that the particle moves along the ellipse $\dfrac{(x-2)^{2}}{16}+\dfrac{(y+1)^{2}}{9}=1$.

[2 marks] b)

Determine the velocity of the particle at time $t$.

[1 mark] c)

Determine the speed of the particle when $t=2$.

[1 mark]
Watch the situation first

The coordinates trace an ellipse, and the velocity points along it

Watch the particle trace its path. Measured from the centre, it is always 4cos(t) across and 3sin(t) up. Then follow the orange velocity arrow to t = 2.

As t increases, the particle moves around a closed path Its coordinates are x = 2 + 4cos(t) and y = 3sin(t) − 1 From the centre (2, −1) it is 4cos(t) across and 3sin(t) up cos²(t) + sin²(t) = 1 for every t, so the path is the ellipse The velocity v(t) = −4sin(t) î + 3cos(t) ĵ points along the path The speed is the length of the velocity arrow At t = 2 (radians), v(2) ≈ −3.64 î − 1.25 ĵ Speed: |v(2)| = √(16sin²(2) + 9cos²(2)) ≈ 3.85 −4 −2 2 4 6 −4 −2 2 x y (2, −1) 4cos(t) 3sin(t) −3.64−1.25 3.85 x = 2 + 4cos(t) y = 3sin(t) − 1 cos(t) =x − 24 sin(t) =y + 13 cos²(t) + sin²(t) = 1 (x − 2)²16+(y + 1)²9= 1 v(t) = −4sin(t) î v(t) + 3cos(t) ĵ speed = |v| = arrow length v(2) = −4sin(2) î v(2) + 3cos(2) ĵ v(2) ≈ −3.64 î − 1.25 ĵ |v(2)| = √(16sin²(2) + 9cos²(2)) ≈ 3.85

Drawn to scale. The velocity arrow uses the same scale as the axes, so its length is the speed.

Now for the mathematics

Move the particle and read off its velocity

Drag $t$. The blue legs are $x-2$ and $y+1$, so they give $\cos(t)$ and $\sin(t)$. The arrow is the velocity at that instant. Set $t=2$, then compare radian mode with degree mode.

−4 −2 2 4 6 −4 −2 2 x y (2, −1) t
Legs $x-2=4\cos(t)$ and $y+1=3\sin(t)$ Velocity, to scale
Calculator angle mode

Whatever $t$ is, the legs give $\cos(t)=\frac{x-2}{4}$ and $\sin(t)=\frac{y+1}{3}$, so the particle stays on the ellipse. The arrow always points along the path.

At $t=2$ the velocity is about $-3.64\,\hat{\boldsymbol{\imath}}-1.25\,\hat{\boldsymbol{\jmath}}$. The question asks for speed, so give its length: $\sqrt{16\sin^2(2)+9\cos^2(2)}\approx3.85$. That is part c).

Degree mode gives 3.00. The calculator worked out $\sin(2^\circ)$ and $\cos(2^\circ)$, not $\sin(2)$ and $\cos(2)$. Here $t$ is in radians, so switch to radian mode.

In degree mode the calculator reads $t$ as degrees. The velocity formula assumes radians, so this number is not the speed.

Now set $t=2$ in radian mode for part c).

You have part c). Try degree mode at $t=2$ to see the most common slip.

You have seen both. Only the radian value, 3.85, gains the mark.

Marking guide · 4 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Choose a method

Step 1 · part a)1 mark

Make $\cos(t)$ and $\sin(t)$ the subjects of the two coordinate equations.

$$\begin{aligned}x=2+4\cos(t)&\Rightarrow\cos(t)=\frac{x-2}{4}\\y=3\sin(t)-1&\Rightarrow\sin(t)=\frac{y+1}{3}\end{aligned}$$

Why?

You need to eliminate $t$. Rather than solving for $t$ itself, isolate $\cos(t)$ and $\sin(t)$, because an identity links them with no $t$ left over.

Marker: correctly expresses $\cos(t)$ and $\sin(t)$ in terms of $x$ and $y$

Step 2 · part a)1 mark

Substitute both results into the Pythagorean identity and square each fraction.

$$\begin{aligned}\cos^2(t)+\sin^2(t)&=1\\\frac{(x-2)^2}{16}+\frac{(y+1)^2}{9}&=1\end{aligned}$$

Every position $(x,y)$ of the particle satisfies this, so the particle moves along the ellipse.

Marker: correctly uses $\cos^2(t)+\sin^2(t)=1$ to show the required result

Step 3 · part b)1 mark

Velocity is the derivative of position. Differentiate each component.

$$\begin{aligned}\boldsymbol{v}(t)&=\frac{d\boldsymbol{r}}{dt}\\&=-4\sin(t)\,\hat{\boldsymbol{\imath}}+3\cos(t)\,\hat{\boldsymbol{\jmath}}\end{aligned}$$

Why?

The constants differentiate to 0, and $\dfrac{d}{dt}\cos(t)=-\sin(t)$. The minus sign is the one most often lost.

Marker: correctly determines the velocity

Step 4 · part c)1 mark

Speed is the magnitude of the velocity. Substitute $t=2$, then evaluate on your calculator in radian mode.

$$\left|\boldsymbol{v}(2)\right|=\sqrt{16\sin^2(2)+9\cos^2(2)}$$

$\left|\boldsymbol{v}(2)\right|\approx3.85$ units of length per unit of time
Calculator use

Set Radian mode. TI-84 Plus CE: mode, highlight Radian, enter. Casio fx-CG50: SHIFT MENU (SET UP), Angle, F2 (Rad). Enter √(16(sin(2))² + 9(cos(2))²). Output: 3.84548. On the Casio, sin 2² is read as $\sin(4)$, so keep the brackets.

Marking

Your calculator must be in radian mode. The degree-mode value, 3.00, does not gain this mark. Follow-through applies from an incorrect velocity in b).

Marker: correctly determines the speed

Putting it all together

Each phrase of the question gave you something

“The position vector of a particle at time $t$ is given by $\boldsymbol{r}(t)$”
The coordinates separately: $x=2+4\cos(t)$ and $y=3\sin(t)-1$.
“Show that the particle moves along the ellipse”
A target with no $t$ in it. Make $\cos(t)$ and $\sin(t)$ the subjects, then use $\cos^2(t)+\sin^2(t)=1$ to eliminate $t$.
“Determine the velocity of the particle at time $t$”
Differentiate each component: $\boldsymbol{v}(t)=\dfrac{d\boldsymbol{r}}{dt}=-4\sin(t)\,\hat{\boldsymbol{\imath}}+3\cos(t)\,\hat{\boldsymbol{\jmath}}$.
“Determine the speed of the particle”
The magnitude of the velocity, $\left|\boldsymbol{v}\right|=\sqrt{v_x^2+v_y^2}$, a single number, not a vector.
“when $t=2$”
2 radians, so your calculator must be in radian mode: $\sqrt{16\sin^2(2)+9\cos^2(2)}\approx3.85$.
Common errors

If I had to guess where you might lose marks…

  • Degree mode in c). You get $\sqrt{16\sin^2(2^\circ)+9\cos^2(2^\circ)}$ $=3.00142\approx3.00$. This does not gain the mark.
  • Forgetting to square the trig values in c). $\sqrt{16\sin(2)+9\cos(2)}$ $=\sqrt{10.8034}$ $=3.28686\approx3.29$, which is wrong.
  • Not squaring the coefficients in c). $\sqrt{4\sin^2(2)+3\cos^2(2)}$ $=1.95623\approx1.96$, which is wrong.
  • Using position instead of velocity in c). $\boldsymbol{r}(2)=0.335413\,\hat{\boldsymbol{\imath}}$ $+1.72789\,\hat{\boldsymbol{\jmath}}$, so $\left|\boldsymbol{r}(2)\right|$ $=1.76015\approx1.76$. That is a distance from the origin, not a speed.
  • Sign slip in b) or giving a vector in c). Writing $+4\sin(t)$ for the $\hat{\boldsymbol{\imath}}$ component loses the b) mark, though c) still gives 3.85 because squaring removes the sign (and follow-through applies). Stopping at $\boldsymbol{v}(2)\approx-3.64\,\hat{\boldsymbol{\imath}}-1.25\,\hat{\boldsymbol{\jmath}}$ answers velocity, not speed. In a), writing $\frac{x-2}{4}+\frac{y+1}{3}=1$ (adding instead of squaring) is wrong: it equals $\cos(t)+\sin(t)$, which is not always 1.

Only mark this done when you could do it without help. Reading the solution does not count.

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A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.

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