When will the arclength $QR$ take a maximum value?

When will the arclength $QR$ take a maximum value?

Question

$P$ is a point on the circumference of a circle $X$ of diameter $3\,\text{in.}$ An arc of another circle $Y$ is drawn with centre $P$, intersecting $X$ at $Q$ and $R$. Express the lengths of $PQ$ and the minor arc $QR$ of $Y$ in terms of $\theta$, where $\theta$ is the angle, in radians, between $PQ$ and the diameter of $X$ through $P$.

Prove that the length of the arc $QR$ has a stationary value when $\theta = \cot \theta$, and that this value is a maximum.

Suggestion

We ought to begin by drawing the configuration described in the question. How can we find the length of $PQ$? Could we try adding to the configuration?

A further suggestion

If we let $P'$ denote the other end of the diameter of $X$ that passes through $P$, what can we say about $\triangle P'QP$?

A final suggestion

Do we know anything about $\angle P'QP$?

When checking that a value is a maximum of a function, do we have to calculate the second derivative of the function?

Solution

$P$ is a point on the circumference of a circle $X$ of diameter $3\,\text{in.}$ . An arc of another circle $Y$ is drawn with centre $P$, intersecting $X$ at $Q$ and $R$. Express the lengths of $PQ$ and the minor arc $QR$ of $Y$ in terms of $\theta$, where $\theta$ is the angle, in radians, between $PQ$ and the diameter of $X$ through $P$.

Here’s a diagram of the two circles $X$ and $Y$.

If you drew your circle $Y$ larger than your circle $X$, your diagram will look quite different to this: there’s no need to worry, the mathematics works just the same.

You can explore different versions of the diagram with this interactive:

Varying $Y$

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/NFaPz3AU/width/700/height/450/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

Drag the point $R$ to change the size of $Y$.

We’ll calculate the two lengths in turn. Firstly, to find $PQ$, let’s look at the circle $X$.

Since the angle $PQP'$ lies in a semicircle, it must be a right angle. Now, from trigonometry, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\text{length of }PQ\text{}}{\text{length of }PP'\text{}} \implies \text{length of }PQ\text{} = 3 \cos \theta.$

Secondly, to find the length of the minor arc $QR$, let’s look at the circle $Y$.

The length of an arc of a circle is $r\phi$, where $\phi$ is the angle subtended at the centre of the circle by the arc.

By symmetry, $\angle QPR = \theta + \theta = 2\theta$, and so the length of the minor arc $QR$ is $PQ \times 2\theta = 6\theta\cos\theta.$

We know that $QR = 6 \theta \cos \theta.$ Using the product rule, we have $\dfrac{d(QR)}{d\theta}= 6\theta (-\sin \theta) + 6\cos \theta = 6 (\cos \theta - \theta \sin \theta).$

So the stationary points are those values of $\theta$ for which $\cos \theta = \theta \sin \theta,$ which is true if and only if $\theta = \cot \theta.$

We can divide through here by $\sin \theta$, since $\theta = 0$ is not going to be a solution to our equation.

We are only concerned with the interval $0 \le \theta \le \dfrac{\pi}{2}$.

Since $\theta$ increases from zero and $\cot \theta$ descends from infinity, as illustrated below, our equation has just one solution.

Is this a maximum? We know that $QR > 0$ when $0 < \theta < \dfrac{\pi}{2}$, that $QR = 0$ when $\theta = 0$ and $\dfrac{\pi}{2}$, and that there’s only one stationary point between $0$ and $\dfrac{\pi}{2}$; therefore, this point has to be a maximum.

We can see this in the changing values in the GeoGebra illustration above.

If we try dragging $R$, what happens to $QR$? For what value of $\theta$ is this a maximum? Does $\theta = \cot \theta$ here?

Alternatively, we could calculate the second derivative $f''(\theta)$ and check that it’s negative on this interval.