Question 18
Polar curves are defined by points that are a variable distance of $r$ units from the origin and dependent on the angle $\theta$ (in radians) measured from the positive $x$-axis.
Consider the polar curve $r=1+\cos(\theta)$.
A table of four polar coordinates on this curve is shown.
| $\theta$ | $r$ |
|---|---|
| $0$ | $2$ |
| $\frac{\pi}{6}$ | $1+\frac{\sqrt{3}}{2}$ |
| $\frac{\pi}{3}$ | $1.5$ |
| $\frac{\pi}{2}$ | $1$ |
The graph shows the polar curve $r=1+\cos(\theta)$ for $0\le\theta\le2\pi$ on a Cartesian plane. The polar coordinates from the table have been plotted on the curve.
The length of a polar curve, $L$, from $\theta=a$ to $\theta=b$ can be determined using the rule
$$L=\int_{a}^{b}\sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}}\,d\theta$$
Use a complete algebraic method to determine the length of the section of the given polar curve that lies above the $x$-axis.
Only the top half of the heart counts towards the length
Watch the point trace the curve from $\theta=0$ to $\theta=\pi$. You will see the length build up to exactly 4.
The white dots are the four points from the table, so you can check the curve against them.
Keep going past π and see what goes wrong
Drag $\theta$ to trace the curve. You can compare the true length with $4\sin\left(\frac{\theta}{2}\right)$ and see where the two stop agreeing.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The marking guide has two methods, and they share the first four marks.
You start by choosing the limits. The curve leaves the $x$-axis at $(2,\ 0)$ when $\theta=0$ and comes back to it at the origin when $\theta=\pi$, because $r=1+\cos(\pi)=0$.
$$a=0,\qquad b=\pi$$
This mark can be implied by your later working.
Marker: correctly determines the values of $a$ and $b$ by considering the curve length above the $x$-axis
The rule needs $\frac{dr}{d\theta}$, so you differentiate $r$ and then square it.
$$\frac{dr}{d\theta}=-\sin(\theta)\qquad\left(\frac{dr}{d\theta}\right)^{2}=\sin^{2}(\theta)$$
This mark can be implied by your later working.
Marker: correctly determines expressions for $\frac{dr}{d\theta}$
Put both pieces into the rule and expand the bracket.
$$\begin{aligned}L&=\int_{0}^{\pi}\sqrt{\left(1+\cos(\theta)\right)^{2}+\sin^{2}(\theta)}\,d\theta\\&=\int_{0}^{\pi}\sqrt{1+2\cos(\theta)+\cos^{2}(\theta)+\sin^{2}(\theta)}\,d\theta\end{aligned}$$
Follow-through marks are allowed here if you made a slip in step 2.
Marker: determines an expression for the integrand in expanded form
You should spot $\cos^{2}(\theta)+\sin^{2}(\theta)=1$ straight away.
$$L=\int_{0}^{\pi}\sqrt{2+2\cos(\theta)}\,d\theta$$
You can write the integrand as $\sqrt{2\left(1+\cos(\theta)\right)}$ instead, and it is just as good. This is where the two methods split.
Marker: uses suitable identity to determine a simplified integrand expression
You cannot integrate a square root like this directly. The trick is the double angle identity written with half angles, $\cos(\theta)=2\cos^{2}\left(\frac{\theta}{2}\right)-1$.
$$L=\int_{0}^{\pi}\sqrt{2\left(2\cos^{2}\left(\tfrac{\theta}{2}\right)-1\right)+2}\,d\theta$$
Using it in the form $1+\cos(\theta)=2\cos^{2}\left(\frac{\theta}{2}\right)$ is also accepted.
Marker: uses suitable double angle identity within integrand
The square root now comes off cleanly, and you can integrate.
$$\begin{aligned}L&=\int_{0}^{\pi}\sqrt{4\cos^{2}\left(\tfrac{\theta}{2}\right)}\,d\theta=\int_{0}^{\pi}2\cos\left(\tfrac{\theta}{2}\right)d\theta\\&=\left[4\sin\left(\tfrac{\theta}{2}\right)\right]_{0}^{\pi}\end{aligned}$$
Be careful here. $\sqrt{4\cos^{2}\left(\frac{\theta}{2}\right)}$ is really $2\left|\cos\left(\frac{\theta}{2}\right)\right|$. It only equals $2\cos\left(\frac{\theta}{2}\right)$ because $\cos\left(\frac{\theta}{2}\right)\ge0$ for every $\theta$ between 0 and $\pi$.
Marker: uses suitable integration method to determine an expression for $L$
Substitute the limits.
$$L=4\left(\sin\left(\tfrac{\pi}{2}\right)-\sin(0)\right)$$
$L=4$ units
The interactive above shows you what happens if you ignore the absolute value and integrate all the way to $2\pi$. You get 0, which cannot be a length.
Marker: determines value for the required length
Instead of a half angle, you substitute $u=\cos(\theta)$. You need $\sin(\theta)=\sqrt{1-u^{2}}$ to swap $d\theta$ for $du$, and that works because $\sin(\theta)\ge0$ between 0 and $\pi$.
$$u=\cos(\theta),\qquad du=-\sin(\theta)\,d\theta,\qquad d\theta=\frac{-du}{\sqrt{1-u^{2}}}$$
$$\theta=0\Rightarrow u=1\qquad\theta=\pi\Rightarrow u=-1$$
$$L=-\int_{1}^{-1}\frac{\sqrt{2}\sqrt{1+u}}{\sqrt{1-u^{2}}}\,du$$
Remember to change the limits to $u$-values when you substitute.
Marker: uses suitable substitution within integrand including the $du$ term
Factorise $1-u^{2}=(1-u)(1+u)$ and cancel the $\sqrt{1+u}$. What is left is a power of $1-u$ that you can integrate.
$$\begin{aligned}L&=-\sqrt{2}\int_{1}^{-1}\frac{1}{\sqrt{1-u}}\,du\\&=-\sqrt{2}\left[-2\sqrt{1-u}\right]_{1}^{-1}\end{aligned}$$
Marker: uses suitable integration method to determine an expression for $L$
Substitute the $u$ limits.
$$L=-\sqrt{2}\left(-2\sqrt{2}+2\sqrt{0}\right)=4$$
$L=4$ units
This route avoids the half-angle identity, but you have to be comfortable swapping $\sin(\theta)$ for $\sqrt{1-u^{2}}$.
Marker: determines value for the required length
Each phrase of the question gave you something
You have probably never seen the arc length rule for a polar curve before. You are given it, so what you are really being tested on is choosing the limits and then simplifying an integrand that does not look like anything you know how to integrate.
The slip I would watch for is the limits and the square root together. If you go from 0 to $2\pi$, $\sqrt{4\cos^{2}\left(\frac{\theta}{2}\right)}$ stops being $2\cos\left(\frac{\theta}{2}\right)$ halfway round, and you get an answer of 0. The QCAA love to ask for part of a curve so that you have to think about which values of $\theta$ you actually need.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2025 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.