# GRAPHING POLAR EQUATIONS

Let P be the rectangular coordinate in the form (x, y), we should convert it into the form of (r, θ).

Then,

r2  =  x2 + y2

Let us see some example problems to understand how to graph the given polar equations.

Example 1 :

Graph each of the following polar equations. Convert to rectangular coordinates, if possible.

(a)  r  =  3   (b) θ  =  π/6    (c) r sinθ  =  2    (d)  r  =  sec θ

Solution :

(a)  r  =  3

r is the directed distance from origin. (b) θ  =  π/6

Angle formed in the positive along x axis. (c) r sinθ  =  2

Since r sinθ  =  x

By replacing r sinθ by x, we get

x  =  2

It is a straight line which is at 2 units distance from the origin. (d)  r  =  sec θ

r/sec θ  =  1

r cos θ  =  1

x  =  1 Example 2 :

Consider the equation r(θ)  =  2 sinθ  Solution :

(a)

 If θ  =  0r(θ)  =  2 sinθ r(0)  =  2 sin0r(0)  =  0 If θ  =  π/6r(π/6) = 2 sinπ/6 r(π/6)  =  2(1/2)r(π/6)  =  1 If θ  =  π/3r(π/3) = 2 sinπ/3 r(π/3)  = 1.732 If θ  =  π/2r(π/2) = 2 sinπ/2 r(π/2)  = 2 If θ  =  2π/3r(2π/3) = 2 sin 2π/3 r(π/3)  = 1.732 If θ  =  5π/6r(5π/6) = 2 sin 5π/6 r(5π/6)  = 1If θ  =  πr(π)  =  2sin π r(π)  =  0 (b)  Sketch the equation again but this time in terms of y vs x. (c)  If the domain is changed as [0, 2π], the graph will be same.

Example 3 :

Graph each polar equation then convert it into rectangular form.

(a)  1  =  r sin θ    (b)  r  =  -3/cos θ

Solution :

r  =  1/sin θ

If θ  =  30, r  =  2

If θ  =  45, r  =  √2

If θ  =  60, r  =  2/√3 Converting the given into rectangular form :

Polar Equation  :

1  =  r sinθ

r sinθ  =  x

Rectangular form :

y  =  1

(b)  r  =  -3/cos θ

Solution :

r  =  -3/cos θ

If θ  =  0, r  =  -3

If θ  =  30, r  =  -2√3

If θ  =  45, r  =  -3√2

If θ  =  60, r  =  -6 Converting the given into rectangular form :

Polar Equation  :

r  =  -3/cosθ

r cos  =  -3

Rectangular form :

x  =  -3 Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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