Parametric points

Excerpt: Students are asked to match up parametric equations with locations of stationary points and with their graphs. This requires use of the Chain Rule in the context of parametrics. We encounter some interesting looking curves.

Warm-up

A curve in the $x-y$ plane is defined parametrically in terms of a parameter $t$ as $x=2 \cos t,\quad y=3 \sin t.$

Find an expression for $\dfrac{dy}{dx}$ in terms of $t$ .

Answer

\begin{align*} x=2\cos t \quad\implies\quad \frac{dx}{dt}&=-2\sin t \\ y=3\sin t \quad\implies\quad \frac{dy}{dt}&=3\cos t \end{align*}

Using the chain rule, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = -\frac{3\cos t}{2\sin t}.$

Hence find the stationary points on the curve.

Are there any values of $t$ for which $\dfrac{dy}{dx}$ is undefined? What would that mean?

Answer

At the stationary points,

$\frac{dy}{dx} = 0$

which means

\begin{equation*} -\frac{3\cos t}{2\sin t} = 0 \quad\implies\quad \cos t=0. \end{equation*}

We don’t need to know the values of $t$ ; to find $x$ and $y$ it is sufficient to know $\cos t$ and $\sin t$ . When $\cos t=0$ , $\sin t = 1$ or $-1$ , so substituting we find the stationary points are at $(0,3)\text{ and }(0,-3).$

$\dfrac{dy}{dx}$ is undefined when $\sin t=0$ which happens at $(-2,0) \text{ and } (2,0).$ A sketch of the curve should make it clear what is happening here.

Problem

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*}

Using the interactive below or the printable cards, match up each equation (blue card) with

the location of a stationary point on the curve (green card),
the location of a critical point on the curve where the tangent is vertical (pink card),
its graph (white card).

(Note that some of the curves have other stationary points and critical points that do not appear on cards.)

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

\begin{align*} x &= t^3-t \\ y &= t^3-3t \end{align*}

\begin{align*} x &= t^2+t+1 \\ y &= 2t^2-2 \end{align*}

\begin{align*} x &= t^2 \\ y &= t^3-4t \end{align*}

\begin{align*} x &= 2 \cos t \\ y &= \sin 2t \end{align*}

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

Stationary point at $x=1$

Stationary point at $x=0$

Stationary point at $x=-\frac{1}{3}$

Stationary point at $x=\sqrt{2}$

Stationary point at $x=2$

Stationary point at $x=\frac{4}{3}$

Critical point at $(0,0)$

Critical point at $(5,4)$

Critical point at $\left(\frac{2}{3\sqrt{3}},\frac{8}{3\sqrt{3}}\right)$

Critical point at $(2,0)$

Critical point at $(5,0)$

Critical point at $\left(\frac{3}{4},-\frac{3}{2}\right)$

Solution

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.

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).

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).

Set of matched cards

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