Hello,

I have a function f, defined parametrically. What exactly the function is can vary, but right now I'm trying to work on a quarter-circle of radius R, so

$\displaystyle x=R*cos(t), y= R*sin(t), {{-\pi} \over {2}} \leq t \leq 0$.

I need to find another function, g, which has the following properties:

1) At all times t, g(t) is at a fixed distance from f(t) (the distance is 1).

2) At all times t, the line between f(t) and g(t) is parallel to the tangent line to g(t).

I also have initial points for both functions (if f is a circle of radius R):

$\displaystyle f({{-\pi}\over{2}}) = (0,-R) $

$\displaystyle g({{-\pi}\over{2}}) = (-1,-R) $

*Mainly what I want to know is whether or not there is enough information here to solve this.*

What I have tried so far: I thought it would be a good idea to solve this with vectors, since g + (g's unit tangent) = f.

$\displaystyle Let f(t) = <R*cos(t), R*sin(t)>, \ and \ g = <x(t), y(t)> $

$\displaystyle Then \ {{g'(t)} \over {|g'(t)|}} + g(t) = f(t). $

$\displaystyle \dfrac {<x'(t), y'(t)>}{\sqrt{x'(t)^2 + y'(t)^2}} + <x(t), y(t)> = <R*cos(t), R*sin(t)>$

This gives two equations:

$\displaystyle \dfrac {x'(t)} {\sqrt{x'(t)^2 + y'(t)^2}} + x(t) = R*cos(t)$

$\displaystyle \dfrac {y'(t)} {\sqrt{x'(t)^2 + y'(t)^2}} + y(t) = R*sin(t)$

But after this I'm stuck. I know how to solve differential equations by separating variables, but not any other way. Again, I'm not interested in a solution to this, only whether or not it's possible with the given data and whether or not I'm on the right track with this vector thing.

Thanks!

On a completely unrelated note, could somebody tell me how to get line breaks in latex without having to wrap every line individually in math tags? It seems that neither \newline nor \\ work.