Rectangular and Polar Coordinates with variables

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 3K views
APolaris
Messages
8
Reaction score
0

Homework Statement



I'm trying to help a friend with these two questions, but given that I haven't studied this material in over a decade, it's one of the topics I cannot recall at all.

Convert the following from rectangular to polar coordinates:

(a) x2 + y2 = x

(b) y2 = 2x

Homework Equations



r2 = x2 + y2

Tan(T) = y/x, where (T) stands in for theta

The Attempt at a Solution



For the first one, I found r = x1/2, aka sqrt (x), but am completely stumped on what to do about finding theta. Arctan (sqrt(x - x2) / (x2 + y2))... yeah, I'm pretty sure that whole tangent thing is useless. Using the reverse procedure (polar to rectangular) gives [sqrt(x) cos (T)]2 + [sqrt(x) sin(T)]2 = sqrt(x) cos(T). Then x cos2(T) + x sin2(T) = sqrt(x) cos(T). Factoring out an x, the cos2 + sin2 identity just gives x = sqrt(x) cos(T), which is... I'm pretty sure it's what was already known.

For the second: No idea how the two relevant equations could even potentially be useful, but I did notice if you take the reverse procedure (going from polar to rectangular), then using x = r cos(T), y = r sin(T), then [r sin(T)]2 = 2 r cos(T). This eventually reduces to r = 2 cos(T) / sin2(T), or r = 2 cot(T)csc(T), or any number of other expressions that leaves me none the wiser about what the polar coordinates are supposed to resemble.

I suppose if you substitute y = sqrt(2x), that gives r = sqrt(x2 + 2x). Again, no idea how this is supposed to be useful.
 
Last edited:
Physics news on Phys.org
rock.freak667 said:
Just use the coordinate transformations you stated above.

x=rcosθ
y=rsinθ

you don't need to find the actual value of θ

I thought that was for converting polar coordinates to rectangular.

Anyway, does that mean my solution for the second one is complete, and that the answer should just be r = 2 cos(θ)/sin2(θ)?

For the second one, I tried redoing the manipulation without substituting sqrt(x) for r. Should the following be correct?

r2 cos2θ + r2 sin2θ = r cos θ

r2 = r cos θ

r = cos θ
 
Thank you very much. I'm certain my friend will appreciate this in the morning.

Funny how problems can be easier than they seem like this. The other day, I solved half of a Putnam question, then later failed to manage a basic geometry proof involving a circle. It's like sidestepping a pile of horse dung and falling down a manhole cover. There should really be an adjective describing people like me.
 
APolaris said:
Thank you very much. I'm certain my friend will appreciate this in the morning.

Funny how problems can be easier than they seem like this. The other day, I solved half of a Putnam question, then later failed to manage a basic geometry proof involving a circle. It's like sidestepping a pile of horse dung and falling down a manhole cover. There should really be an adjective describing people like me.

I think they are called over-thinkers :-p