- #1
latentcorpse
- 1,444
- 0
Show that the parabolic coordinates [itex](u,v,\phi)[/itex] defined by
[itex]x=uv \cos{\phi} , y=uv \sin{\phi} , z=\frac{1}{2}(u^2-v^2)[/itex]
now I am a bit uneasy here because to do this i first need to find the basis vector right?
so if i try and rearrange for u say and then normalise to 1 that will give me [itex]\vec{e_u}[/itex]
[itex]u^2v^2=x^2+y^2[/itex] and [itex]u^2-2z=v^2[/itex]
[itex]u^2(u^2-2z)=x^2+y^2 \Rightarrow u^4-2u^2z=x^2+y^2[/itex] - i.e. my problem is I am finding it impossible to rearrange for u...
[itex]x=uv \cos{\phi} , y=uv \sin{\phi} , z=\frac{1}{2}(u^2-v^2)[/itex]
now I am a bit uneasy here because to do this i first need to find the basis vector right?
so if i try and rearrange for u say and then normalise to 1 that will give me [itex]\vec{e_u}[/itex]
[itex]u^2v^2=x^2+y^2[/itex] and [itex]u^2-2z=v^2[/itex]
[itex]u^2(u^2-2z)=x^2+y^2 \Rightarrow u^4-2u^2z=x^2+y^2[/itex] - i.e. my problem is I am finding it impossible to rearrange for u...