Fluency exercise

## 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 $\left(\frac{3}{4},-\frac{3}{2}\right)$