Another algebra and trig question (maybe not though)

[FrogPad]

I have u(r,\theta)=r \cos \theta + r^{-1}\cos \theta and then am changing coordinates u(x,y)=|x|+\frac{|x|}{x^2+y^2}. I then have to compute the vector \vec v = (u_x,u_y).

The thing is, is the derivative of |x| is sign\,\,x. I don't think we should be dealing with this function. So I'm not sure how I should be dropping the absolute value to get to: u(x,y)=x+\frac{x}{x^2+y^2}.

I'm thinking it is from the conversion to polar to cartesian... but I'm not sure. Maybe it is within the definition of \tan^{-1}... but again I'm not sure. Well I'm sick of throwing random darts, so if anyone has any suggestions, I'm all ears :)
Please note that u(x,y) was my conversion from polar coordiantes to cartesian, so this could also be a problem, and might be incorrect.

Thankyou.

 

[Integral]

Consider the polar point (1, \pi) this should be the same as (-1,0) in Cartesian coordinates.

These points should yield the same result for your function. Do they?

Consider your rectangular function both with and without the abs, which gives the correct result?

 

[benorin]

Consider that u(r,\theta)=r \cos \theta + r^{-1}\cos \theta = r \cos \theta + \frac{r\cos \theta}{r^2} and you can get it from there, correct?

 

[FrogPad]

Integral:
Interesting. That makes sense.
u_{polar}(1,\pi) = \cos \pi + \cos \pi =-2

u_{cart}(-1,0) = |-1| + \frac{|-1|}{1}=2 \neq -2

So I should drop the abs functions for u(x,y) [/tex]. I&#039;m curious. How did you know to think of this as a counter example? <br /> <br /> Benorin:<br /> I wish I understood what you are conveying in the algebraic rewrite of the expression. I&#039;m sorry but I don&#039;t understand.

 

[benorin]

Use the transformation equations

Consider that u(r,\theta)=r \cos \theta + r^{-1}\cos \theta = r \cos \theta + \frac{r\cos \theta}{r^2} and you can get it from there, correct?

Recall that

x=r\cos \theta, \, y=r\sin \theta,\mbox{ and }x^2+y^2=r^2​

(even though we won't use the second equation) so that

u(r,\theta)=r \cos \theta + r^{-1}\cos \theta = r \cos \theta + \frac{r\cos \theta}{r^2} = x + \frac{x}{x^2+y^2}​

 

[FrogPad]

Oh snap!
I didn't see what you were showing before. That's cool. Thanks man :)

hehe, I don't know how I didn't see that!

 

[benorin]

Sure, glad to help: so post another.

 

[FrogPad]

lol, will do. I have another homework problem that is bugging me, but I need to get some sleep. It's been a fun st patricks day, and I'm not really in the proper mind to be working on it right now :)

 

[Integral]

Frog pad,
It is a very good practice on this type of problem to pick a easy to verify point. We needed some thing that would yield a negitive result, thus cos \pi.

Looks like every thing is happy now!