Polar coordinates and curve sketchingEdexcel International A Level Further Maths: Revision notes

A point can be described by its distance $r\ge0$ from a fixed point $O$ (the pole) and the angle $\theta$ measured anticlockwise from a fixed half-line (the initial line), usually the positive $x$-axis. The pair $(r,\theta)$ gives the polar coordinates. Converting with the pole at the origin and the initial line along the positive $x$-axis: $x=r\cos\theta,\qquad y=r\sin\theta,\qquad r^2=x^2+y^2,\qquad\tan\theta=\frac yx.$ Example: $\left(4,\frac{\pi}{3}\right)$ has $x=4\cos\frac{\pi}{3}=2$ and $y=4\sin\frac{\pi}{3}=2\sqrt3$. For $\tan\theta=\frac yx$, use a sketch to choose the correct quadrant. To convert an equation, substitute $x=r\cos\theta$ and $y=r\sin\theta$. For example $x^2+y^2=6x$ becomes $r^2=6r\cos\theta$, so $r=6\cos\theta$.

• $\theta=\alpha$: a half-line from the pole at angle $\alpha$ (a full line if $r$ may be negative).
• $r=a$: a circle, centre the pole, radius $a$.
• $r=2a\cos\theta$: a circle of radius $a$ with centre $(a,0)$, passing through the pole, with diameter along the initial line ($-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$). In Cartesian form, $x^2+y^2=2ax$.
• $r=p\sec(\alpha-\theta)$, that is $r\cos(\theta-\alpha)=p$: a straight line whose perpendicular distance from the pole is $p$, the perpendicular making angle $\alpha$ with the initial line. Cartesian form: $x\cos\alpha+y\sin\alpha=p$. Example: $r=4\sec\left(\theta-\frac{\pi}{3}\right)$ is $x+\sqrt3y=8$.
• $r=k\theta$ ($\theta\ge0$): an Archimedean spiral starting at the pole. $r$ increases steadily with $\theta$, so each turn is a constant distance $2\pi k$ further out along any ray.

The curve $r=a(1+\cos\theta)$ is a cardioid, heart-shaped, with $r=2a$ at $\theta=0$ and $r=0$ at $\theta=\pi$, where it has a cusp at the pole. $r=a(1-\cos\theta)$ is the same shape pointing the other way. Both are symmetric about the initial line. The curve $r=a(3+2\cos\theta)$ is a limaçon with no cusp and no loop. Because $3>2$ the value of $r$ is never zero: it ranges from $5a$ at $\theta=0$ down to $a$ at $\theta=\pi$, giving a smooth closed curve around the pole, wider on the side $\theta=0$. To sketch: find $r$ at key angles ($0,\frac{\pi}{2},\pi,\frac{3\pi}{2}$), identify whether $r$ ever reaches $0$, use symmetry (cosine curves are symmetric about the initial line since $\cos(-\theta)=\cos\theta$), and join smoothly.

The curve $r=a\cos2\theta$ is a four-petalled rose. Petals along the $x$-axis occur where $\cos2\theta\ge0$, for example $-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}$, with $r=a$ at $\theta=0$ and $r=0$ at $\theta=\pm\frac{\pi}{4}$. Where $\cos2\theta<0$, a point $(r,\theta)$ with $r<0$ is plotted as $(|r|,\theta+\pi)$, giving the petals along the $y$-axis. The curve $r^2=a^2\cos2\theta$ is a lemniscate (figure of eight) with two loops along the $x$-axis. It exists only where $\cos2\theta\ge0$, that is $-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}$ and $\frac{3\pi}{4}\le\theta\le\frac{5\pi}{4}$, with greatest value $r=a$ at $\theta=0$ and $\theta=\pi$ and $r=0$ at $\theta=\pm\frac{\pi}{4}$. In an examination the range of $\theta$ is usually given, so check where $r\ge0$ before sketching. Always mark the intercepts with the axes and the value of $r$ at each key point.

On the curve $r=f(\theta)$, $x=r\cos\theta$ and $y=r\sin\theta$ are both functions of $\theta$.

• A tangent parallel to the initial line is horizontal, so $\frac{dy}{d\theta}=0$ where $y=r\sin\theta$.
• A tangent perpendicular to the initial line is vertical, so $\frac{dx}{d\theta}=0$ where $x=r\cos\theta$. Worked example, $r=a(1+\cos\theta)$, $0<\theta<\pi$:
• $y=a(1+\cos\theta)\sin\theta$ gives $\frac{dy}{d\theta}=a(2\cos\theta-1)(\cos\theta+1)=0$, so $\theta=\frac{\pi}{3}$ and $r=\frac{3a}{2}$.
• $x=a(\cos\theta+\cos^2\theta)$ gives $\frac{dx}{d\theta}=-a\sin\theta(1+2\cos\theta)=0$, so $\theta=\frac{2\pi}{3}$, $r=\frac a2$ and the tangent is $x=-\frac a4$. Always say which range of $\theta$ you are using, and reject solutions outside it, such as the pole where $r=0$.