Ok here's the problem:

F: R^2 -> R^2

(t is theta)

F(r,t) = (r*cos(t), r*sin(t))

What is F^-1(0,0)?

I know the answer is {(0,0) | -inf < t < inf }

but I'm not exactly sure how to get to that point.

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- Apr 21st 2008, 05:51 PMmerzbowInverse of a Multiple Variable Function
Ok here's the problem:

F: R^2 -> R^2

(t is theta)

F(r,t) = (r*cos(t), r*sin(t))

What is F^-1(0,0)?

I know the answer is {(0,0) | -inf < t < inf }

but I'm not exactly sure how to get to that point. - Apr 21st 2008, 06:12 PMTheEmptySet
lets start with this

$\displaystyle F(r,\theta)=(r\cos(\theta),r\sin(\theta))$

consider $\displaystyle F(r,\theta)=(0,0)$

so we would need to know when

$\displaystyle r\cos(\theta)=0 \mbox{ and } r\sin(\theta)=0$

since cosine and sine are never zero at the same value of theta r must be zero. If that is the case then the equation

$\displaystyle 0\cos(\theta)=0 \mbox{ and } 0\sin(\theta)=0$

is true for all values of theta so the inverse immage of (0,0) is

$\displaystyle [{(r,\theta)|r=0, -\infty < \theta < \infty}]$ - Apr 21st 2008, 06:18 PMmerzbow
Ah, thanks man. I knew it was something easy, I just couldn't figure it out.