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 \(t\).
\[\begin{align*}
x=2\cos t \quad\implies\quad \frac{dx}{dt}&=-2\sin t \\
y=3\sin t \quad\implies\quad \frac{dy}{dt}&=3\cos t
\end{align*}\]
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?
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.
