Parametric equations and modelling – A-level Maths exam questions with mark scheme

1CoreCalculator3 marks

A curve $C$ has parametric equations

$$x=3t-1,\qquad y=t^{2}+2,\qquad 0\leqslant t\leqslant 4$$

(a) Find a Cartesian equation of $C$ in the form $y=\mathrm{f}(x)$. [2]

(b) State the domain of $\mathrm{f}$. [1]

(a) Rearranges $x=3t-1$ to $t=\dfrac{x+1}{3}$ and substitutes into $y=t^{2}+2$M1
(a) $y=\left(\dfrac{x+1}{3}\right)^{2}+2$ o.e., e.g. $y=\dfrac{x^{2}+2x+19}{9}$. Must be in the form $y=\ldots$A1
(b) $-1\leqslant x\leqslant 11$ (accept $[-1,11]$). Strict inequalities B0B1
Answer: (a) $y=\left(\dfrac{x+1}{3}\right)^{2}+2$ (b) $-1\leqslant x\leqslant 11$
Examiner tip: The domain comes from the range of $t$: substitute the end values $t=0$ and $t=4$ into $x$, not into $y$.

2CoreCalculator5 marks

A curve $C$ has parametric equations

$$x=4+5\cos\theta,\qquad y=-1+5\sin\theta,\qquad 0\leqslant\theta<2\pi$$

(a) Show that $C$ is a circle, stating its centre and radius. [3]

(b) Find the values of $\theta$ at the points where $C$ crosses the $y$-axis, giving your answers to 3 significant figures, and state the coordinates of these points. [2]

(a) Isolates $\cos\theta=\dfrac{x-4}{5}$, $\sin\theta=\dfrac{y+1}{5}$ and uses $\cos^{2}\theta+\sin^{2}\theta=1$M1
(a) $(x-4)^{2}+(y+1)^{2}=25$ o.e.A1
(a) Centre $(4,-1)$ and radius $5$ (both required)A1
(b) Sets $4+5\cos\theta=0\Rightarrow\cos\theta=-\tfrac{4}{5}$ and finds one correct value, $\theta=2.50$ or $3.79$M1
(b) Both $\theta=2.50$ and $\theta=3.79$ with points $(0,2)$ and $(0,-4)$. Degrees ($143.1^{\circ}$, $216.9^{\circ}$) A0A1
Answer: (a) $(x-4)^{2}+(y+1)^{2}=25$: centre $(4,-1)$, radius $5$ (b) $\theta=2.50$ giving $(0,2)$; $\theta=3.79$ giving $(0,-4)$
Examiner tip: $\theta$ is in radians here – a calculator left in degrees gives $143^{\circ}$ and loses the accuracy mark.

3StandardCalculator4 marks

A curve $C$ has parametric equations

$$x=2\sec t,\qquad y=3\tan t,\qquad 0\leqslant t<\frac{\pi}{2}$$

Show that a Cartesian equation of $C$ is

$$y=\frac{3}{2}\sqrt{x^{2}-4},\qquad x\geqslant k$$

where $k$ is a constant to be found. [4]

Uses the identity $1+\tan^{2}t=\sec^{2}t$ (o.e.) with $\sec t=\dfrac{x}{2}$ and $\tan t=\dfrac{y}{3}$M1
$\dfrac{y^{2}}{9}=\dfrac{x^{2}}{4}-1$ o.e., e.g. $y^{2}=\dfrac{9}{4}(x^{2}-4)$A1
Reaches $y=\dfrac{3}{2}\sqrt{x^{2}-4}$ with a reason for taking the positive root, e.g. $\tan t\geqslant 0$ for $0\leqslant t<\frac{\pi}{2}$ so $y\geqslant 0$. Given answer: no errors seenA1
$k=2$ (from $\sec 0=1$, so $x\geqslant 2$)B1
Answer: $y=\dfrac{3}{2}\sqrt{x^{2}-4}$, $x\geqslant 2$ (so $k=2$)
Examiner tip: For a 'show that' you must justify dropping the $\pm$: say why $y$ cannot be negative for this range of $t$.

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4StandardCalculator4 marks

A curve $C$ has parametric equations

$$x=3\sin t,\qquad y=4\cos 2t,\qquad -\frac{\pi}{2}\leqslant t\leqslant\frac{\pi}{2}$$

(a) Show that the Cartesian equation of $C$ can be written as $y=a+bx^{2}$, where $a$ and $b$ are constants to be found. [2]

(b) State the domain and range of the function whose graph is $C$. [2]

(a) Uses $\cos 2t=1-2\sin^{2}t$ with $\sin t=\dfrac{x}{3}$M1
(a) $y=4-\dfrac{8}{9}x^{2}$, i.e. $a=4$, $b=-\dfrac{8}{9}$A1
(b) Domain $-3\leqslant x\leqslant 3$B1
(b) Range $-4\leqslant y\leqslant 4$ (accept $-4\leqslant \mathrm{f}(x)\leqslant 4$)B1
Answer: (a) $y=4-\dfrac{8}{9}x^{2}$ (b) domain $-3\leqslant x\leqslant 3$; range $-4\leqslant y\leqslant 4$
Examiner tip: Choose the version of $\cos 2t$ that matches the other parameter: here $x$ involves $\sin t$, so use $1-2\sin^{2}t$.

5StandardCalculator5 marks

A curve $C$ has parametric equations

$$x=\ln(t+2),\qquad y=\frac{4}{t+1},\qquad t>-1$$

(a) Find a Cartesian equation of $C$ in the form $y=\mathrm{f}(x)$, stating the domain of $\mathrm{f}$. [3]

The point $P$ on $C$ has $y$-coordinate 2.

(b) Find the exact coordinates of $P$. [2]

(a) Rearranges to $t=\mathrm{e}^{x}-2$ and substitutes into $y$M1
(a) $y=\dfrac{4}{\mathrm{e}^{x}-1}$A1
(a) Domain $x>0$ (from $t>-1\Rightarrow t+2>1\Rightarrow \ln(t+2)>0$)B1
(b) Solves $\dfrac{4}{t+1}=2$ to get $t=1$ (or solves $\dfrac{4}{\mathrm{e}^{x}-1}=2$ to get $\mathrm{e}^{x}=3$)M1
(b) $P(\ln 3,\,2)$. Decimal $1.10$ only is A0 (exact required)A1
Answer: (a) $y=\dfrac{4}{\mathrm{e}^{x}-1}$, $x>0$ (b) $P(\ln 3,\,2)$
Examiner tip: 'Exact' means leave $\ln 3$ – writing $1.0986$ loses the final mark.

6HardCalculator7 marks

A ball is thrown from the point $A$ and moves in a vertical plane until it hits horizontal ground at $B$, as shown in the diagram. The position of the ball $t$ seconds after it is thrown is modelled by

$$x=12t,\qquad y=1.5+9t-4.9t^{2}$$

where $x$ metres is the horizontal distance from $O$ and $y$ metres is the height above the ground.

(a) Write down the height of $A$ above the ground. [1]

(b) Find, according to the model, the horizontal distance $OB$. [3]

(d) State one limitation of the model. [1]

(a) $1.5$ mB1
(b) Sets $1.5+9t-4.9t^{2}=0$ and solves the quadratic (formula, calculator or completing the square)M1
(b) $t=1.99$ (awrt), rejecting the negative root $t=-0.154$A1
(b) $OB=12\times 1.9905\ldots=23.9$ m (awrt 23.9)A1
(c) Substitutes $t=\dfrac{x}{12}$ into $y$M1
(c) $y=1.5+\dfrac{9x}{12}-\dfrac{4.9x^{2}}{144}$ simplified to the given answer with no errorsA1
(d) Any one sensible limitation, e.g. ignores air resistance; ball modelled as a particle; ignores spin/wind; ground assumed horizontal; $g$ assumed constant. Ignore extra commentsB1
Answer: (a) $1.5$ m (b) $23.9$ m (c) shown (d) e.g. air resistance is ignored
Examiner tip: Don't stop at the time: the question asks for a horizontal distance, so feed $t$ back into $x=12t$.

7HardCalculator8 marks

A passenger on an observation wheel has position modelled by

$$x=20\sin\left(\frac{\pi t}{10}\right),\qquad y=22-20\cos\left(\frac{\pi t}{10}\right),\qquad t\geqslant 0$$

where $t$ is the time in minutes after the passenger boards, $y$ metres is the passenger's height above the ground and $x$ metres is the horizontal displacement from the vertical line through the centre of the wheel.

(a) Find the height of the passenger when they board. [1]

(b) Show that the passenger moves on a circle, giving its centre and radius. [2]

(d) Find the length of time, during the first revolution, for which the passenger is more than 32 m above the ground. Give your answer in minutes and seconds. [4]

(a) $2$ mB1
(b) Uses $\sin^{2}+\cos^{2}=1$ with $\sin\left(\frac{\pi t}{10}\right)=\dfrac{x}{20}$, $\cos\left(\frac{\pi t}{10}\right)=\dfrac{22-y}{20}$M1
(b) $x^{2}+(y-22)^{2}=400$; centre $(0,22)$, radius $20$A1
(c) $\dfrac{\pi t}{10}=2\pi\Rightarrow t=20$ minutesB1
(d) Sets $22-20\cos\left(\frac{\pi t}{10}\right)=32$ (or $>32$) to obtain $\cos\left(\frac{\pi t}{10}\right)=-\tfrac{1}{2}$M1
(d) $\dfrac{\pi t}{10}=\dfrac{2\pi}{3}$ or $\dfrac{4\pi}{3}$A1
(d) $t=\dfrac{20}{3}$ and $t=\dfrac{40}{3}$ (awrt 6.67 and 13.3) and subtractsM1
(d) $\dfrac{20}{3}$ minutes $=6$ minutes $40$ secondsA1
Answer: (a) $2$ m (b) $x^{2}+(y-22)^{2}=400$, centre $(0,22)$, radius $20$ m (c) $20$ minutes (d) $6$ min $40$ s
Examiner tip: There are two solutions in one revolution – the passenger rises past 32 m and later falls past it; the time above is the gap between them.

8StandardCalculator6 marks

The diagram shows the curve $C$ with parametric equations

$$x=t^{2},\qquad y=2t-1,\qquad t\in\mathbb{R}$$

and the line $l$ with equation $x-y=4$. The line $l$ meets $C$ at the points $A$ and $B$.

(a) Find the coordinates of $A$ and $B$. [4]

(b) Find the exact length of $AB$, giving your answer in the form $p\sqrt{2}$. [2]

(a) Substitutes the parametric equations into the line: $t^{2}-(2t-1)=4$M1
(a) Forms and solves a 3TQ: $t^{2}-2t-3=0\Rightarrow (t-3)(t+1)=0$M1
(a) $t=3$ and $t=-1$A1
(a) $A(9,5)$ and $B(1,-3)$ (either labelling accepted). Alternative via Cartesian $(y+1)^{2}=4x$ and $x=y+4$ scores equivalentlyA1
(b) Uses Pythagoras: $\sqrt{(9-1)^{2}+(5-(-3))^{2}}$M1
(b) $8\sqrt{2}$ (do not accept $\sqrt{128}$ or $11.3$ alone)A1
Answer: (a) $(9,5)$ and $(1,-3)$ (b) $8\sqrt{2}$
Examiner tip: Substitute the parametric expressions straight into the line – you get a quadratic in $t$ without needing the Cartesian equation.

9HardCalculator6 marks

A curve $C$ has parametric equations

$$x=1+\cos 2t,\qquad y=3\sin^{2}t,\qquad 0\leqslant t\leqslant\frac{\pi}{2}$$

(a) Show that $C$ is part of a straight line and find the equation of this line in the form $ax+by=c$, where $a$, $b$ and $c$ are integers. [3]

(b) Find the exact length of $C$. [2]

A student claims that $C$ is the whole of the line found in part (a).

(a) Uses $\cos 2t=1-2\sin^{2}t$ to write $x=2-2\sin^{2}t$M1
(a) Substitutes $\sin^{2}t=\dfrac{y}{3}$ to get a linear relation, e.g. $x=2-\dfrac{2y}{3}$M1
(a) $3x+2y=6$ (or any integer multiple)A1
(b) Finds the end points $(2,0)$ at $t=0$ and $(0,3)$ at $t=\frac{\pi}{2}$, and applies PythagorasM1
(b) $\sqrt{13}$A1
(c) Explains that $0\leqslant\sin^{2}t\leqslant 1$ (or $-1\leqslant\cos 2t\leqslant 1$), so $0\leqslant x\leqslant 2$ and $0\leqslant y\leqslant 3$: $C$ is only the line segment joining $(2,0)$ and $(0,3)$B1
Answer: (a) $3x+2y=6$ (b) $\sqrt{13}$ (c) $x$ and $y$ are bounded ($0\leqslant x\leqslant 2$), so $C$ is only a segment
Examiner tip: Eliminating the parameter can hide the domain: always check what values $x$ and $y$ can actually take.

10HardCalculator6 marks

A curve $C$ has parametric equations

$$x=2t+1,\qquad y=t^{2}-3,\qquad t\in\mathbb{R}$$

The line $l$ has equation $y=x+k$, where $k$ is a constant.

(a) Show that the $t$-values of any points where $l$ meets $C$ satisfy $t^{2}-2t-(4+k)=0$. [1]

(b) Given that $l$ is a tangent to $C$, find the value of $k$ and the coordinates of the point of contact. [4]

(a) Substitutes: $t^{2}-3=2t+1+k$ and rearranges to the given equation with no errorsB1
(b) Uses discriminant $=0$ for a tangent: $(-2)^{2}-4(1)(-(4+k))=0$M1
(b) $k=-5$A1
(b) Substitutes $k=-5$ to obtain the repeated root $t=1$M1
(b) Point of contact $(3,-2)$A1
(c) $k<-5$ (accept $\{k:k<-5\}$). Ft their $k$ from (b) with strict inequalityB1ft
Answer: (a) shown (b) $k=-5$, point $(3,-2)$ (c) $k<-5$
Examiner tip: A tangent gives a repeated root in $t$ – set $b^{2}-4ac=0$ in the parameter, then remember to convert $t$ back to $(x,y)$.

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