Wize University Calculus 2 Textbook > Parametric & Polar Curves

Polar Coordinates

Polar Coordinates

Every point on the Cartesian plane can be represented by a distance and an angle denoted by $\left(r,\theta\right)$ :

$r$ is the distance between the point and the origin
$\theta$ is the angle measured counter-clockwise from the positive x-axis to the line joining the point and the origin

Convert Polar to Cartesian Coordinates

$x=r\cos\theta$
$y=r\sin\theta$

Convert Cartesian to Polar Coordinates

$r=\sqrt{x^2+y^2}$
$\theta=\arctan\left(\frac{y}{x}\right)$

Polar Curves

We can graph curves of the form $r\left(\theta\right)=...\theta...$ when the interval for $\theta$ is given.

Example: Converting Coordinates

a.) Plot the polar points $\left(2,\frac{\pi}{3}\right)$ , $\left(-2,\ \frac{4\pi}{3}\right)$ , and $\left(2,\ \frac{7\pi}{3}\right)$ on the Cartesian plane. Then convert them to Cartesian coordinates.

In Cartesian coordinates, we have $(x,y)=(r\cos(\theta),r\sin(\theta))$ ..

$\left(2,\frac{\pi}{3}\right)\ \ \to\ \ (2\cos(\pi/3),2\sin(\pi/3))=(2\frac{1}{2},2\frac{\sqrt{3}}{2})=(1,\sqrt{3}).$
$\left(-2,\frac{4\pi}{3}\right)\ \ \to\ \ \left(-2\cos\left(\frac{4\pi}{3}\right),-2\sin\left(\frac{4\pi}{3}\right)\right)=\left(1,\sqrt{3}\right)$
$\left(2,\ \frac{7\pi}{3}\right)\ \ \to\ \ \left(2\cos\left(\frac{7\pi}{3}\right),\ 2\sin\left(\frac{7\pi}{3}\right)\right)=\left(1,\sqrt{3}\right)$

Therefore, the cartesian coordinates are all $\left(1,\sqrt{3}\right)$ .

b.) Convert the Cartesian point $\left(-5,\ 12\right)$ to Polar Coordinates.

In Polar Coordinates $r=\sqrt{x^2+y^2}=\sqrt{\left(-5\right)^2+12^2}=13$ and $\theta=\arctan\left(\frac{y}{x}\right)=\arctan\left(-\frac{12}{5}\right)$ .

Therefore, the polar coordinates are $\left(r,\ \theta\right)=\left(13,\ \arctan\left(-\frac{12}{5}\right)\right)$ .

Graphing Polar Curves

We can graph the polar curve $r=f\left(\theta\right)$ in the Cartesian Plane.

Strategy for Graphing

Set up a chart with nice values of $\theta$ (like $0,\ \frac{\pi}{4},\ \frac{2\pi}{4},\ \frac{3\pi}{4},\ ...,\frac{8\pi}{4}$ ), then calculate the corresponding $r$ values.

Example

Plot the polar curve $r=2\cos\theta$

Image illustrating Graphing Polar Curves

Some Common Polar Curves

Practice Question

Plot the polar curves $r=1+\sin\theta$ and $r=1+\sin2\theta$ .

$r=1+\sin\theta$

Illustration for practice question: Practice Question

$r=1+\sin2\theta$

$r=1+\sin\theta$

$r=1+\sin2\theta$

One or more of the graphs in the above options are incorrect

I don't know

Practice Question

Convert the equation of the curve $x^2 + y^2 = 3\sqrt{x^2 + y^2}- 4x$ to polar coordinates.

$r=3-4\sin\theta$

$r=3+4\sin\theta$

$r=3-4\cos\theta$

$r=3+4\cos\theta$

$r=3\theta-4\sin\theta$

I don't know

Practice Question

Match the following graphs with its corresponding polar curve equation.

a.) $r=\sin3\theta$

b.) $r=2\sin^2\theta$

c.) $r=\frac{1}{2}\theta$

d.) $r=1-2\cos\theta$

$r=1-2\cos\theta$

$r=\frac{1}{2}\theta$

$r=2\sin^2\theta$

$r=\sin3\theta$

ii.

iv.

iii.

I don't know

Extra Practice

The point $(1, \sqrt{3})$ in rectangular coordinates has a polar coordinate representation as well. Give a polar coordinate representation of this point where $2 \pi \leq \theta < 4\pi$ and $r < 0$ .

Practice Question

Convert the equation of the curve $x^2 + y^2 = 3\sqrt{x^2 + y^2}- 4x$ to polar coordinates.

Practice Question

Plot the polar curves $r=1+\sin\theta$ and $r=1+\sin2\theta$ .

Practice Question

Convert the equation of the curve $x^2 + y^2 = 3\sqrt{x^2 + y^2}- 4x$ to polar coordinates.

Practice Question

Convert the equation of the curve $x^2 + y^2 = 3\sqrt{x^2 + y^2}- 4x$ to polar coordinates.

Polar Coordinates

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