# William Daly

EMAT6680

Write Up #11

Summer 03

Straight Line in Polar Coordinates

Consider the graph of the equation in polar coordinates:

For many explorations in polar coordinates, we come to expect a symmetry or periodicity about the origin.  So why does this graph have the appearance of a straight line y=x+1 (in Cartesian coordinates):

To begin this study, a natural starting point may be an understanding of each of the denominator terms, sinθ and cosθ.  These are graphed over the range 0 < θ < π.

Plot these again from 0 to π/4 to get a sense of what each function is doing as θ.  But it is clear that at θ=0, sinθ=0 and cosθ=1.   Also θ=0, sinθ=0 and cosθ= π/2 sinθ=1 and cosθ=0.  This explains the Cartesian points (-1,0) and (0,1).

At this point is the development, we can observe a few points.  Let the value of the denominator be d=sin θ – cos θ.  It may be useful to graphs d(θ) to visualize the problem.

# From looking at this graph of d(θ), it is clear that when θ=0, r=1/d=-1.   Also, as θ approaches π/4 from the left, r → -∞;  as θ approaches π/4 from the right, r → ∞.This occurs again at 5π/4.Note the θ =π/4 and θ =5π/4 corresponds to the line y=x+1, so that r→±∞ at these values of θ explains the linear behavior near θ =π/4 and θ =5π/4.

Now that the behavior at the points(-1,0) and (0,1) as well as near θ =π/4 and θ =5π/4 shown to be consistent with the line y=x+1, what about the remaining points near the origin.  There does not appear to be anything intuitive about the graph of d(θ) nor the graphs of r=sinθ and r=cosθ that would insist on the points in the neighborhood of (-1,0) and (0,1) to lie on the line y=x+1.

To prove that this should be the case, it appears that one ought to look toward the relationship between the Cartesian and polar coordinate systems themselves.  Recall that:

x=r cosθ and y=r sinθ

If these are substituted into the original function, we obtain:

or     .

After canceling the r from both sides of the equation (note that r≠0), we are left with:

y-x=1   or   y=x+1.