Question 1
The position of a particle is given by $\mathbf{r}=(t+2)\,\hat{\mathbf{i}}+t^2\,\hat{\mathbf{j}}$ for $t\ge0$.
Determine the corresponding Cartesian equation.
Not quite. At $t=0$ the particle is at $(2,0)$, which fits, but at $t=1$ it is at $(3,1)$ and $3^2-4=5$. Squaring $x-2$ needs the middle term: $(x-2)^2=x^2-4x+4$. In 2023, 4.4% of Queensland students chose (A).
Not quite. At $t=0$ the particle is at $(2,0)$, but this option gives $y=8$. Expand $(x-2)^2$ term by term. In 2023, 8.8% of Queensland students chose (B).
That’s it. Step through the hints below to see why. In 2023, 76.0% of Queensland students chose (C).
Not quite. That is $(x+2)^2$, which comes from $t=x+2$. Solving $x=t+2$ for $t$ gives $t=x-2$. In 2023, 10.3% of Queensland students chose (D).
Solid: the path for $t\ge0$. Dashed: the rest of the parabola, which the particle never reaches.
At this point both (A) and (C) give the right value of $y$. One point isn’t enough to decide, so move $t$ and check again.
Only (C) gives the right value of $y$ at this point. Move $t$ and check that it keeps working.
Read the two components off the position vector. The $\hat{\mathbf{i}}$ part is $x$ and the $\hat{\mathbf{j}}$ part is $y$.
$$x=t+2\qquad y=t^2$$
Make $t$ the subject of the equation for $x$.
$$t=x-2$$
Substitute into the equation for $y$ and expand.
$$y=(x-2)^2=x^2-4x+4$$
Because $t\ge0$, the path is only the part with $x\ge2$.
Eliminate the parameter $t$.
$$x=t+2\ \Rightarrow\ t=x-2$$
$$y=t^2=(x-2)^2=x^2-4x+4$$
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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