CP-7 Polar coordinates — printable coverage pack

CP-7.1 · Understand and use polar coordinates and be able to convert between polar and Cartesian coordinates.

Polar coordinates $(r,\theta)$ locate a point a directed distance $r$ from the pole at angle $\theta$ measured anticlockwise from the initial line.
Convert to Cartesian coordinates using $x=r\cos\theta$ and $y=r\sin\theta$ ; conversely, $r^2=x^2+y^2$ and $\tan\theta=y/x$ with the quadrant checked.
To convert a polar curve, multiply by $r$ when useful and replace $r\cos\theta$ by $x$ , $r\sin\theta$ by $y$ and $r^2$ by $x^2+y^2$ .
Polar coordinates are not unique: $(r,\theta)$ and $(-r,\theta+\pi)$ represent the same point. A common error is to use $\arctan(y/x)$ without correcting its quadrant.

1. Convert the polar coordinates $(4,\pi/6)$ to Cartesian coordinates.[2 marks]
Answer
- $(x,y)=(2\sqrt3,2)$
Method: $x=4\cos(\pi/6)=4(\sqrt3/2)=2\sqrt3$ and $y=4\sin(\pi/6)=4(1/2)=2$ .

1. Express the Cartesian point $(-3,3\sqrt3)$ in polar form, taking $r>0$ and $0\leq\theta<2\pi$ .[3 marks]
Answer
- $(r,\theta)=(6,2\pi/3)$
Method: $r=\sqrt{(-3)^2+(3\sqrt3)^2}=\sqrt{9+27}=6$ . The point lies in quadrant II and $\tan\theta=(3\sqrt3)/(-3)=-\sqrt3$ , so $\theta=2\pi/3$ .

Build a polar sketch from symmetry, zeros of $r$ , maxima and minima of $r$ , and values on the initial line; plot negative $r$ in the opposite direction.
If replacing $\theta$ by $-\theta$ leaves the equation unchanged, the curve is symmetric about the initial line; related tests identify symmetry about $\theta=\pi/2$ or the pole.
Curves such as $r=a\cos\theta$ and $r=a\sin\theta$ are circles, while $r=a\sin(n\theta)$ and $r=a\cos(n\theta)$ generate rose curves whose petals follow the extreme values of $r$ .
Tangents parallel to the initial line occur where $\frac{d}{d\theta}(r\sin\theta)=0$ , and tangents perpendicular to it where $\frac{d}{d\theta}(r\cos\theta)=0$ . A common error is to discard negative values of $r$ , which can remove an entire loop or petal.

1. Sketch the polar curve $r=3$ , labelling its key geometric features.[2 marks]
Answer
- A circle centred at the pole with radius $3$ .
- It passes through $(3,0)$ at $\theta=0$ and $(-3,0)$ at $\theta=\pi$ .
Method: Every point has constant distance $3$ from the pole while $\theta$ ranges through a complete turn. The locus is therefore the circle $x^2+y^2=9$ , centred at the pole with radius $3$ ; it meets the positive initial line at $(3,0)$ and its negative continuation at $(-3,0)$ .

The area swept by a polar curve from $\theta=\alpha$ to $\theta=\beta$ is $A=\dfrac12\int_\alpha^\beta r^2\,d\theta$ .
Find the angular limits from intersections, zeros of $r$ or symmetry before integrating; a complete loop may occupy only part of a full turn.
For the area between two polar curves over the same angles, integrate $\frac12(r_{\text{outer}}^2-r_{\text{inner}}^2)$ .
Sketch and shade the intended region first. Common errors are omitting the factor $1/2$ , using $r$ instead of $r^2$ , or counting a symmetric loop twice.

1. Find the exact area of the sector enclosed by $r=2$ and the rays $\theta=0$ and $\theta=\pi/3$ .[3 marks]
Answer
- Area $=\dfrac{2\pi}{3}$
Method: $A=\frac12\int_0^{\pi/3}2^2\,d\theta=2[\theta]_0^{\pi/3}=2\pi/3$ .

1. Determine the exact area enclosed by the loop $r=3\sin\theta$ .[5 marks]
Answer
- Area $=\dfrac{9\pi}{4}$
Method: The loop is traced once for $0\leq\theta\leq\pi$ . Hence $A=\frac12\int_0^\pi9\sin^2\theta\,d\theta$ . Since $\int_0^\pi\sin^2\theta\,d\theta=\pi/2$ , the area is $9\pi/4$ .

1. The curves $r=2$ and $r=4\cos\theta$ enclose a region that lies inside $r=4\cos\theta$ but outside $r=2$ . Find its exact area.[7 marks]
Answer
- Area $=\dfrac{4\pi}{3}+2\sqrt3$
Method: The curves meet when $2=4\cos\theta$ , so $\theta=\pm\pi/3$ for the required region. Thus $A=\frac12\int_{-\pi/3}^{\pi/3}(16\cos^2\theta-4)\,d\theta$ . Using $\cos^2\theta=(1+\cos2\theta)/2$ , the integral simplifies to $\frac12(16\pi/3+4\sqrt3-8\pi/3)=4\pi/3+2\sqrt3$ .