Conversion from Polar to Cartesian equations

[Helena54321]

I just did a quiz in a lecture and walked out crying. There was one question (which probably seems very easy to most :/ ) were you had to convert polar equations to cartesian ones. We also had to draw the cartesian graphs (2D).

a) rcos(th)
b)r=2asin(th)
c)r^2sin2(th)=2k
d)rsin(th+(pi/4))=a2^(1/2)

th=theta.

In our lecture notes for this course we have derivations for conversion from cylindrical-cartesian (3D) and spherical-cylindrical. (-=either way)

But when it comes to 2D cartesian and polar I'm like ?. I have no clue what to do. I know a) is a straight line where x=a and b) is a circle but only because my friend told me. I understand a) but none of the rest.

I feel quite lost. How do I approach polar and cartesian egtn conversions?? Thank you!

Helena. First year undergrad at The Chinese uni of Hong Kong x x x nd feeling quite lost! :(

 

[N-Gin]

Generally, polar coordinates can be converted to Cartesian using these equations:

[tex]x=r\cos\theta[/tex]

[tex]y=r\sin\theta[/tex]

[tex]r^2=x^2+y^2[/tex]

So, for the part b) you have

[tex]r=2a\sin\theta[/tex]

[tex]r^2=2ay[/tex]

[tex]x^2+y^2-2ay=0[/tex]

[tex]x^2+(y^2-2ay+a^2)=a^2[/tex]

[tex]x^2+(y-a)^2=a^2[/tex]

Now, the last equation is implicit equation of a circle with center in [itex](0,a)[/itex] and radius [itex]a[/itex].

Try to do similar manipulations with other parts of the question and see what you can get.

 

[fluidistic]

Instead of just memorizing the formula conversions (though it is so common in physics that you'll eventually know them by heart), the most important thing is to know how to derive them.

See http://en.wikipedia.org/wiki/Polar_...rting_between_polar_and_Cartesian_coordinates and take a look specially in the graph in the right side of the page. This is what you have to do in order to reach the formula and this is precisely what they asked you to do in your test. It's very important to understand the graph, which isn't more complicated than applying Pytharogas' theorem.