Review question

# When will the arclength $QR$ take a maximum value? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8936

## Solution

$P$ is a point on the circumference of a circle $X$ of diameter $\quantity{3}{in.}$ . 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:

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, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\text{length of PQ}}{\text{length of PP'}} \implies \text{length of PQ} = 3 \cos \theta.$

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 $PQ \times 2\theta = 6\theta\cos\theta.$

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

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.