Fluency exercise

## Solution

Here is a set of equations defined parametrically in terms of $t$.

\begin{align*} x &= 1+ 4 \cos t \\ y &= t+ 3 \sin t \end{align*}
\begin{align*} x &= 2 \cos t \\ y &= \sin 2t \end{align*}
\begin{align*} x &= t^2 \\ y &= t^3-4t \end{align*}
\begin{align*} x &= t^2+t+1 \\ y &= 2t^2-2 \end{align*}
\begin{align*} x &= t^3-t \\ y &= t^3-3t \end{align*}
\begin{align*} x &= 2+ 3 \cos t \\ y &= 4+ 5 \sin t \end{align*}

The first thing you might notice is that three of the equations are based on trigonometric functions while the other three are based on powers of $t$.

What might that tell us about the kind of curves they represent?

Match up each equation (blue card) with

• the location of a stationary point on the curve (green card),

To find the stationary points on a curve, we want to find $\frac{dy}{dx}$ and when it is equal to zero.

Taking equation (E1), we have \begin{align*} x = 1+ 4 \cos t&, \quad y = t+ 3 \sin t \\ \implies\quad \frac{dx}{dt} = -4\sin t&, \quad \frac{dy}{dt} = 1+3\cos t. \end{align*} We can combine these using the chain rule. $\begin{equation*} \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1+3\cos t}{-4\sin t} \end{equation*}$ This is zero when $\begin{equation*} 1+3\cos t = 0 \quad\implies\quad \cos t=-\frac{1}{3}. \end{equation*}$ Substituting this back into the equation tells us the curve is stationary when $\begin{equation*} x = 1+4\left(-\frac{1}{3}\right) = -\frac{1}{3}. \end{equation*}$

This therefore goes with card (S4).

Note that we quietly ignored the $\sin t$ in the denominator of $\frac{dy}{dx}$ when setting it equal to zero. We should really check that the denominator is not zero at the same point as the numerator is zero, because if it were the gradient would be undefined. In this case, that doesn’t happen ($\sin t\ne0$ when $\cos t=-1/3$) so we are OK.

• the location of a critical point on the curve where the tangent is vertical (pink card),

At a critical point the tangent is vertical and $\frac{dy}{dx}$ is undefined. This will happen when our expression for $\frac{dy}{dx}$ is a fraction with zero in the denominator and non-zero in the numerator.

Taking equation (E5), we have \begin{align*} x = t^3-t&, \quad y = t^3-3t \\ \implies\quad \frac{dx}{dt} = 3t^2-1&, \quad \frac{dy}{dt} = 3t^2-3. \end{align*} We could combine these to write $\frac{dy}{dx}$ as a fraction, but in fact we are only interested in when $\frac{dx}{dt}=0$.

As above, we should also check that $\frac{dy}{dt}\ne 0$ at the same value of $t$. In this case, they are quadratics in $t$ differing only by a constant, so they can never be zero at the same time.

So the critical points happen at \begin{align*} 3t^2-1&=0 \quad\implies\quad t=\pm\frac{1}{\sqrt{3}} \\ \implies\quad x&=\pm\left(\frac{1}{3\sqrt{3}}-\frac{1}{\sqrt{3}}\right) = \pm\frac{2}{3\sqrt{3}} \\ \text{and}\quad y&=\pm\left(\frac{1}{3\sqrt{3}}-\frac{3}{\sqrt{3}}\right) = \pm\frac{8}{3\sqrt{3}} \end{align*}

This therefore goes with card (V4).

• its graph (white card).

We can use the above features of the curve to work out which of the graphs it goes with. Taking equation (E2) as an example, $x=2\cos t, \: y=\sin 2t,$ we find it has

• stationary points at $x=\pm\sqrt{2}$,
• critical points at $(2,0)$ and $(-2,0)$.
Also, since we know the range of the $\sin$ and $\cos$ functions, we can deduce that $\begin{equation*} -2\le x\le 2 \quad\text{and}\quad -1\le y\le 1 \end{equation*}$

so the whole curve must be contained within a rectangle. From these clues, the only graph card it can match is (C2).

If we wanted an extra clue to help identify the graph, we could find the $x$-intercepts and/or $y$-intercepts. This curve, for instance, has $x$-intercepts at $x=0$ and $x=\pm2$, which confirms the match with (C2).