Polar and Cartesian coordinatesAQA A-Level Further Maths: Revision notes

A point can be located by its distance $r$ from the pole (the origin) and the angle $\theta$ measured anticlockwise from the initial line (the positive $x$-axis). This gives the polar coordinates $(r,\theta)$. Angles are in radians, and the principal range is usually either $0\le\theta<2\pi$ or $-\pi<\theta\le\pi$; unless told otherwise, take $r\ge0$. For example, $(4,\frac{5\pi}{6})$ lies $4$ units from the pole in the second quadrant. A point has many polar representations: $(r,\theta)$ and $(r,\theta+2\pi)$ are the same point.

From the right-angled triangle formed by the point and its projection on the $x$-axis, $x=r\cos\theta,\qquad y=r\sin\theta.$ Example: $\left(4,\frac{5\pi}{6}\right)$ gives $x=4\cos\frac{5\pi}{6}=-2\sqrt3$ and $y=4\sin\frac{5\pi}{6}=2$. Use exact values for $\theta=\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3}$ and their multiples, and take care with the signs of sine and cosine in each quadrant.

$r=\sqrt{x^2+y^2},\qquad \tan\theta=\frac yx.$ The equation $\tan\theta=\frac yx$ has two solutions in a full turn, so find the reference angle $\alpha=\tan^{-1}\left|\frac yx\right|$ and place it in the correct quadrant using the signs of $x$ and $y$:

• first quadrant: $\theta=\alpha$; second: $\pi-\alpha$; third: $\alpha-\pi$ (or $\pi+\alpha$); fourth: $-\alpha$ (or $2\pi-\alpha$).

Example: $\left(-\sqrt3,-1\right)$ has $r=2$ and $\alpha=\frac{\pi}{6}$. It is in the third quadrant, so $\theta=-\frac{5\pi}{6}$ (or $\frac{7\pi}{6}$). Points on an axis are best done by inspection, for instance $(0,-3)$ is $\left(3,-\frac{\pi}{2}\right)$.

Use $x=r\cos\theta$, $y=r\sin\theta$, $x^2+y^2=r^2$ and $\tan\theta=\frac yx$ to change a curve's equation from one system to the other.

• Cartesian to polar: $x^2+y^2=4x$ becomes $r^2=4r\cos\theta$, so $r=4\cos\theta$.
• Polar to Cartesian: multiply by $r$ (or $r^2$) until the equation is built from $r\cos\theta$, $r\sin\theta$ and $r^2$. For $r=6\cos\theta$: $r^2=6r\cos\theta$, so $x^2+y^2=6x$, a circle with centre $(3,0)$ and radius $3$. For $r^2=18\sin2\theta=36\sin\theta\cos\theta$: multiply by $r^2$, giving $r^4=36(r\sin\theta)(r\cos\theta)$ and $(x^2+y^2)^2=36xy$.
• A line $r(\cos\theta+\sin\theta)=6$ becomes $x+y=6$.

Find where the circle $r=6\cos\theta$ meets the line $r(\cos\theta+\sin\theta)=6$. Substitute: $6\cos\theta(\cos\theta+\sin\theta)=6$, so $\cos^2\theta+\sin\theta\cos\theta=1$. Since $\cos^2\theta=1-\sin^2\theta$, this gives $\sin\theta\cos\theta=\sin^2\theta$, so $\sin\theta(\cos\theta-\sin\theta)=0$. Then $\theta=0$ (with $r=6$) or $\tan\theta=1$, $\theta=\frac{\pi}{4}$ (with $r=3\sqrt2$). Check in Cartesian form: $(6,0)$ and $(3,3)$ both satisfy $x+y=6$ and $x^2+y^2=6x$.