# Contour Map

• Mar 7th 2009, 01:10 PM
ch2kb0x
Contour Map
I am fairly familiar with contour maps in the form f(x,y).

But there is this problem:
f(r,theta) = 1+sin(theta)-r with z = 1,3/4,1/2,1/4,0.

I was thinking about transforming that into rectangular coordinates and somehow solve it that way, but even so, I do not know how to convert that.

i TRIED to convert it and got y = root [(z-sin(theta)-1)^2-x^2]. If that is correct. how would I even map that?

• Mar 7th 2009, 03:47 PM
Mentia
Do you have any polar graph paper? That would make it easy to graph. Just set f(r,theta) equal to each of your z's and then solve for r in terms of theta. So for instance:

z = 1 = 1+ sin(T) - r ----> r = sin(T)

So start at theta=0 on your polar graph paper and then work your way counterclockwise calculating r along the way. Then do this for each z.

Alternatively, if you must have cartesian coordinates, remember:

r = (x^2+y^2)^(1/2)
sin(T) = y/r = y*(x^2+y^2)^(-1/2)

then f(r,theta) -> f(x,y) = 1 + y*(x^2+y^2)^(-1/2) - (x^2+y^2)^(1/2) = $1-\frac{x^2+(y-1) y}{\sqrt{x^2+y^2}}$

Good luck solving this for z... better stick with polar coordinates. Most calculators can be set into polar coordinates too.
• Mar 7th 2009, 05:01 PM
ch2kb0x
another question
thank you, i drew the contour map and i guess it's correct.

however, another question came up and it was, based on the contours in the previous problem, sketch the graph of the function f(r,theta) = 1+sin(theta)-r
(same function) but for 0< f(r,theta) <1

Isn't this the same thing as the previous problem? because from 0 to 1 is basically all the z values.