Wize AP Calculus (BC) Textbook > Parametric & Polar Curves

Polar Coordinates

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Polar Coordinates
Every point on the Cartesian plane can be represented by a distance and an angle denoted by (r,θ)\left(r,\theta\right)(r,θ):
rrr is the distance between the point and the origin
θ\thetaθ 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=rcos⁡θx=r\cos\thetax=rcosθ
y=rsin⁡θy=r\sin\thetay=rsinθ

Convert Cartesian to Polar Coordinates
r=x2+y2r=\sqrt{x^2+y^2}r=x2+y2​
θ=arctan⁡(yx)\theta=\arctan\left(\frac{y}{x}\right)θ=arctan(xy​)

Polar Curves
We can graph curves of the form r(θ)=...θ...r\left(\theta\right)=...\theta...r(θ)=...θ... when the interval for θ\thetaθ is given.
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Example: Converting Coordinates
a.) Plot the polar points (2,π3)\left(2,\frac{\pi}{3}\right)(2,3π​), (−2, 4π3)\left(-2,\ \frac{4\pi}{3}\right)(−2, 34π​), and (2, 7π3)\left(2,\ \frac{7\pi}{3}\right)(2, 37π​)on the Cartesian plane. Then convert them to Cartesian coordinates.

In Cartesian coordinates, we have (x,y)=(rcos⁡(θ),rsin⁡(θ))(x,y)=(r\cos(\theta),r\sin(\theta))(x,y)=(rcos(θ),rsin(θ))..
(2,π3)  →  (2cos⁡(π/3),2sin⁡(π/3))=(212,232)=(1,3).\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}).(2,3π​)  →  (2cos(π/3),2sin(π/3))=(221​,223​​)=(1,3​).
(−2,4π3)  →  (−2cos⁡(4π3),−2sin⁡(4π3))=(1,3)\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)(−2,34π​)  →  (−2cos(34π​),−2sin(34π​))=(1,3​)
(2, 7π3)  →  (2cos⁡(7π3), 2sin⁡(7π3))=(1,3)\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)(2, 37π​)  →  (2cos(37π​), 2sin(37π​))=(1,3​)
Therefore, the cartesian coordinates are all (1,3)\left(1,\sqrt{3}\right)(1,3​).  

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

In Polar Coordinates r=x2+y2=(−5)2+122=13r=\sqrt{x^2+y^2}=\sqrt{\left(-5\right)^2+12^2}=13r=x2+y2​=(−5)2+122​=13 and θ=arctan⁡(yx)=arctan⁡(−125)\theta=\arctan\left(\frac{y}{x}\right)=\arctan\left(-\frac{12}{5}\right)θ=arctan(xy​)=arctan(−512​).
Therefore, the polar coordinates are (r, θ)=(13, arctan⁡(−125))\left(r,\ \theta\right)=\left(13,\ \arctan\left(-\frac{12}{5}\right)\right)(r, θ)=(13, arctan(−512​)).

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Graphing Polar Curves
We can graph the polar curve r=f(θ)r=f\left(\theta\right)r=f(θ) in the Cartesian Plane.

Strategy for Graphing
Set up a chart with nice values of θ\thetaθ (like 0, π4, 2π4, 3π4, ...,8π40,\ \frac{\pi}{4},\ \frac{2\pi}{4},\ \frac{3\pi}{4},\ ...,\frac{8\pi}{4}0, 4π​, 42π​, 43π​, ...,48π​), then calculate the corresponding rrr values.

Example
Plot the polar curve r=2cos⁡θr=2\cos\thetar=2cosθ

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Some Common Polar Curves

Practice Question
Plot the polar curves r=1+sin⁡θr=1+\sin\thetar=1+sinθ and r=1+sin⁡2θr=1+\sin2\thetar=1+sin2θ.

r=1+sin⁡θr=1+\sin\thetar=1+sinθ

r=1+sin⁡2θr=1+\sin2\thetar=1+sin2θ

r=1+sin⁡θr=1+\sin\thetar=1+sinθ

r=1+sin⁡2θr=1+\sin2\thetar=1+sin2θ

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

I don't know

Practice Question
Convert the equation of the curve x2+y2=3x2+y2−4xx^2 + y^2 = 3\sqrt{x^2 + y^2}- 4xx2+y2=3x2+y2​−4x to polar coordinates. 
  

r=3−4sin⁡θr=3-4\sin\thetar=3−4sinθ

r=3+4sin⁡θr=3+4\sin\thetar=3+4sinθ

r=3−4cos⁡θr=3-4\cos\thetar=3−4cosθ

r=3+4cos⁡θr=3+4\cos\thetar=3+4cosθ

r=3θ−4sin⁡θr=3\theta-4\sin\thetar=3θ−4sinθ

I don't know

Practice Question
Match the following graphs with its corresponding polar curve equation.

a.) r=sin⁡3θr=\sin3\thetar=sin3θ
b.) r=2sin⁡2θr=2\sin^2\thetar=2sin2θ
c.) r=12θr=\frac{1}{2}\thetar=21​θ
d.) r=1−2cos⁡θr=1-2\cos\thetar=1−2cosθ

A.

r=12θr=\frac{1}{2}\thetar=21​θ

B.

r=sin⁡3θr=\sin3\thetar=sin3θ

C.

r=1−2cos⁡θr=1-2\cos\thetar=1−2cosθ

D.

r=2sin⁡2θr=2\sin^2\thetar=2sin2θ

i.

ii.

iii.

iv.

I don't know