2024 H2 Maths Paper 1 Q9 — A Parametric Curve

The curve, livethe tangent

Walk along the curve,

t

0.60

The curve C The tangent The area P and Q

Part (a)

2 marks

Show that $\frac{dy}{dx}=kt$ for the curve $x=3t^{2}+2t$ , $y=t^{2}+2t^{3}$ , $t\geq0$ , finding $k$ .

Approach

For a parametric curve the gradient is the ratio of the two rates, $\frac{dy/dt}{dx/dt}$ . Differentiate each equation separately and divide — there is no need to eliminate $t$ , and here trying to would be painful for no gain.

Differentiating each with respect to $t$ :

\frac{dx}{dt}=6t+2,\qquad\frac{dy}{dt}=2t+6t^{2}

so the gradient of the curve is

\begin{aligned}\frac{dy}{dx}=\frac{dy/dt}{dx/dt}&=\frac{2t+6t^{2}}{6t+2}\\[2pt]&=\frac{2t(1+3t)}{2(3t+1)}\\[2pt]&=t\ \ (\text{shown})\end{aligned}

To check your answer:

At $t=1$ both rates come to $8$ , so the gradient there is $1$ — which is $t$ , as claimed.

$\therefore\ \dfrac{dy}{dx}=t$ , so $k=1$ .

Note — what part (a) just handed you

The gradient is the parameter. Every question about the steepness of this curve now becomes a question about $t$ directly, with no solving to do. Walk the point along the curve in the model and watch the gradient readout track the slider exactly.

Part (b)

3 marks

The tangent at a point $P$ makes an angle of $60^\circ$ with the $x$ -axis. Find the exact coordinates of $P$ .

A line at $60^\circ$ to the $x$ -axis has gradient $\tan60^\circ=\sqrt{3}$ , and by part (a) the gradient is $t$ :

\frac{dy}{dx}=t=\sqrt{3}

Substituting $t=\sqrt{3}$ into each parametric equation:

\begin{aligned}x&=3\left(\sqrt{3}\right)^{2}+2\sqrt{3}=9+2\sqrt{3}\\[2pt]y&=\left(\sqrt{3}\right)^{2}+2\left(\sqrt{3}\right)^{3}=3+2\left(3\sqrt{3}\right)=3+6\sqrt{3}\end{aligned}

To check your answer:

$t=\sqrt{3}\approx1.732$ , comfortably inside $t\geq0$ , and the point $(12.46,\ 13.39)$ does sit on the curve between the origin and $Q$ . Snap to it in the model and the tangent's angle reads $60.0^\circ$ .

$\therefore\ P=\left(9+2\sqrt{3},\ 3+6\sqrt{3}\right)$ .

Where the marks go

"Exact" rules out $(12.5,\ 13.4)$ — the surds must survive to the answer line. And $\left(\sqrt{3}\right)^{3}=3\sqrt{3}$ , not $3$ : cubing a surd is where this part quietly goes wrong.

Part (c)

5 marks

The point $Q(16,20)$ lies on $C$ . Using calculus, find the exact area of the region bounded by $C$ , the $x$ -axis and the line through $Q$ parallel to the $y$ -axis.

Thought process

The area under a curve is $\int y\,dx$ , but $y$ is not a function of $x$ here. Convert the whole integral to the parameter: $dx=\frac{dx}{dt}\,dt$ , and the limits move from values of $x$ to values of $t$ . That means finding the parameter at $Q$ first.

At $Q$ the $x$ -coordinate is 16:

\begin{aligned}3t^{2}+2t&=16\\3t^{2}+2t-16&=0\\(3t+8)(t-2)&=0\end{aligned}

so $t=2$ or $t=-\frac{8}{3}$ , and the second is rejected since $t\geq0$ .

To check your answer:

At $t=2$ , $y=2^{2}+2\left(2^{3}\right)=4+16=20$ , which is the $y$ -coordinate of $Q$ . The parameter is right.

Before integrating, note the shape: $\frac{dx}{dt}=6t+2\gt0$ and $\frac{dy}{dt}=2t+6t^{2}\geq0$ for $t\geq0$ , so the curve climbs steadily from the origin to $Q$ and stays above the $x$ -axis throughout. The area is therefore

\text{Area}=\int_{0}^{16}y\,dx=\int_{0}^{2}y\,\frac{dx}{dt}\,dt=\int_{0}^{2}\left(t^{2}+2t^{3}\right)(6t+2)\,dt

Expanding first makes the integration routine:

\left(t^{2}+2t^{3}\right)(6t+2)=6t^{3}+2t^{2}+12t^{4}+4t^{3}=12t^{4}+10t^{3}+2t^{2}

\begin{aligned}\text{Area}&=\int_{0}^{2}\left(12t^{4}+10t^{3}+2t^{2}\right)dt\\[2pt]&=\left[\frac{12}{5}t^{5}+\frac{5}{2}t^{4}+\frac{2}{3}t^{3}\right]_{0}^{2}\\[2pt]&=\frac{12}{5}(32)+\frac{5}{2}(16)+\frac{2}{3}(8)\\[2pt]&=\frac{384}{5}+40+\frac{16}{3}\\[2pt]&=\frac{1152+600+80}{15}=\frac{1832}{15}\end{aligned}

To check your answer:

$\frac{1832}{15}=122.13\ldots$ The region sits inside the rectangle from $(0,0)$ to $(16,20)$ , area 320, and the curve bulges below the diagonal, so something around a third of the rectangle is the right size. Fill the region in the model and the running total ends on the same value.

$\therefore$ the exact area is $\dfrac{1832}{15}$ units $^{2}$ .

The two ways this part is lost

Forgetting the $\frac{dx}{dt}$ factor — $\int y\,dt$ is not an area — and leaving the limits as 0 and 16 after changing the variable to $t$ . Both give a tidy-looking number that is simply not the area.

What this question was really testing

The parametric gradient. $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ — two separate differentiations and a division, never an attempt to eliminate the parameter.
Spotting the gift. Once $\frac{dy}{dx}=t$ , the tangent question is a single line. Part (a) is not a warm-up; it is the tool the rest of the question uses.
Angle to gradient. "Makes an angle of $60^\circ$ with the $x$ -axis" means gradient $\tan60^\circ$ . The wording is the equation.
Area in a parameter. $\int y\,\frac{dx}{dt}\,dt$ , with the limits converted too. Justify that the curve stays above the axis before you integrate, not after.

The curve whose gradient is its own parameter

Part (a)

Part (b)

Part (c)

What this question was really testing

This is how we mark, and how we teach.