Where are the stationary points on $\cot x-8\cos x$?

Where are the stationary points on $\cot x-8\cos x$?

Question

The equation of a curve is $y = \cot x - 8 \cos x, \quad (0 < x < \pi).$ Find the coordinates of the points on the curve where $\dfrac{dy}{dx} = 0$.

Sketch the curve.

Suggestion

Should we include $\dfrac{\pi}{2}$ in the domain of this function?

If we think of $\cot x$ as $\dfrac{1}{\tan x}$, then as $\tan \dfrac{\pi}{2}$ is not defined, neither is $\cot x$ at this value.

But what if we define $\cot x=\dfrac{\cos x}{\sin x}$ instead?

As well as the stationary points, we should consider the $x$-intercepts and any asymptotes.

Solution

Using $\cot x=\dfrac{1}{\tan x}$ will give us problems at $x=\dfrac{\pi}{2}$, so we’ll use the alternative, $\cot x = \frac{\cos x}{\sin x}.$

We can differentiate $y$: $\frac{dy}{dx} = \frac{d}{dx}(\cot x) + 8\sin x.$ What’s the derivative of $\cot x$? It’s $-\cos ec^2 x$.

Finding the derivative of $\cot x$

Let $u = \cos x$ and $v = \sin x$, so that $\cot x=\dfrac{u}{v}$, allowing us to use the quotient rule.

Noting that

$\dfrac{du}{dx}=-\sin x$

and

$\dfrac{dv}{dx}=\cos x$

, we find

$$\begin{align*} \frac{d}{dx}(\cot x) &= \frac{\sin x(-\sin x)-\cos x(\cos x)}{(\sin x)^2}\\ &= \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}\\ &= \frac{-1}{\sin^2 x}\\ &= -\cos ec^2 x. \end{align*}$$

So we have that $\frac{dy}{dx} = -\cos ec^2 x + 8\sin x.$

Now

$$\begin{align*} -\cos ec^2 x + 8\sin x &=0\\ \iff 8 \sin x &= \cos ec^2 x\\ \iff (\sin x)^3 &= \dfrac{1}{8}\\ \iff \sin x &= \dfrac{1}{2}.\\ \end{align*}$$

We’re allowed to multiply both sides by $\sin^2 x$ in the above, since $\sin x$ is not zero for $0 < x < \pi$.

Solutions for $x$ in the given range are $\dfrac{\pi}{6}$ and $\pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$.

We were asked, however, to find the coordinates of the stationary points. The first is $\left( \frac{\pi}{6}, \frac{\cos \tfrac{\pi}{6}}{\sin \tfrac{\pi}{6}} - 8 \cos \frac{\pi}{6} \right) = \Biggl( \frac{\pi}{6}, \frac{\tfrac{\sqrt{3}}{2}}{\tfrac{1}{2}} - 8 \frac{\sqrt{3}}{2} \Biggr) = \left( \frac{\pi}{6}, -3\sqrt{3} \right)$ and the second is $\left( \frac{5\pi}{6}, \frac{\cos \tfrac{5\pi}{6}}{\sin \tfrac{5\pi}{6}} - 8 \cos \frac{5\pi}{6} \right) = \Biggl( \frac{5\pi}{6}, \frac{-\tfrac{\sqrt{3}}{2}}{\tfrac{1}{2}} + 8 \frac{\sqrt{3}}{2} \Biggr) = \left( \frac{5\pi}{6}, 3\sqrt{3} \right).$

Now we have the coordinates of the stationary points, we can think about the sketch. First we should think about what happens as $x \to 0$ and $x \to \pi$:

As $x \to 0$, $\dfrac{\cos x}{\sin x} \to + \infty$, and $- 8 \cos x \to -8$, so $y \to + \infty$

As $x \to \pi$, $\dfrac{\cos x}{\sin x} \to - \infty$, and $- 8 \cos x \to 8$, so $y \to - \infty$

So since there are only two stationary points, and we have their coordinates, there is really only one shape the curve can have. So we have the beginnings of our sketch:

Now all we need to do is find the points where the graph crosses the $x$-axis, in the range given. We can factorise $y$ as

$$y = \cos x \left( \dfrac{1}{\sin x} - 8 \right),$$

so $y = 0$ if and only if either $\cos x = 0$ or $\sin x = \dfrac{1}{8}$, which in turn happens only when $x$ takes one of the three values

$$\dfrac{\pi}{2}, \quad \sin^{-1}{\tfrac{1}{8}}, \quad \pi - \sin^{-1}{\tfrac{1}{8}}$$

We can see $\dfrac{\pi}{2}$ is in the middle of the range, and if we calculate $\sin^{-1}{\tfrac{1}{8}}$ we can see that it is less than $\dfrac{\pi}{6}$ which is where our first turning point occurs. So the first $x$-intercept does indeed occur where we thought it did, looking back at our preliminary sketch. And this in turn implies that $\pi - \sin^{-1}{\tfrac{1}{8}} > \dfrac{5 \pi}{6}$

So we can mark on our intercepts, and our sketch is done!

A sketch of our function.

Note that the curve has rotational symmetry around the point $\left(\dfrac{\pi}{2}, 0\right)$. This is to be expected since both $\cos x$ and $\cot x$ have the same symmetry.