How to graph r=4/costheta-sintheta

Question

should i change it to rectangular form to graph it? is there any technique to graph directly from a polar equation?

Answer 1

You can convert to rectangular form:

 

x = rcosθ and y = rsinθ

 

If r = 4/(cosθ-sinθ), then r(cosθ-sinθ) = 4

 

                                   rcosθ - rsinθ = 4

 

                                        x - y = 4

 

                                            y = x - 4

 

The graph of y = x-4 is a line.  However, we have to be careful.  The denominator of the given equation is zero when cosθ = sinθ, which occurs when θ = π/4 or 5π/4.  So, those values of θ have to be excluded.

 

Therefore, the points (x,y) on the line that correspond to those values of θ need to be removed from the graph of y = x - 4.

 

These points are (4cos(π/4)), 4sin(π/4)) = (2√2, 2√2)

 

                 and (4cos(5π/4), 4sin(5π/4)) = (-2√2, -2√2)

 

So, the graph of the given equation is the line y = x - 4 with "holes" in the graph at (2√2, 2√2) and (-2√2, -2√2).  

Answer 2

How to graph r=4/costheta-sintheta

should i change it to rectangular form to graph it? is there any technique to graph directly from a polar equation?

 

Technique is unique: take theta, calculate R, and put the point (theta,R) on graph ( in polar coordinates).

That is it :))

 

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how do i get r in my abovementioned equation if i dont have the values of x & y? Can you do the technique on my eqtn so that i can visualize it? thanks so much!

 

Well: You simply don't care about x and y. You'll be working in (theta, r) coordinates (polar that is) instead of (x,y) decartian ones. So, if for a given theta you can surely find corresponding r from your formual, then first set of (theta, r) u have already.

just vary theta (say 30, 60, 90 , 120 ... degress) and find corresponding r's. and each pair of (theta,r) put on graph ...

message if something is not clear.

all the best:)