# Help with coordinate transformation problem

1. Sep 18, 2012

### xiphius75

1. The problem statement, all variables and given/known data

For elliptical cylindrical coordinates:

x = a * cosh (u) * cos (v)
y = a * sinh (u) * sin (v)
z = z

Derive the relations analogous to those of Equations (168b-e) for circular cylindrical coordinates. In particular, verify that

h_u = h_v = a * sqrt(cosh^2 (u) - cos^2 (v))
h_z = 1

u_1 = {[sinh (u) * cos (v)](i) + [cosh (u) * sin (v)](j)} / sqrt(cosh^2 (u) - cos^2 (v))
u_2 = {[cosh (u) * sin (v)](-i) + [sinh (u) * cos (v)](j)} / sqrt(cosh^2 (u) - cos^2 (v))

2. Relevant equations

From the book:
h_u = magnitude( dr/du_1)
and similarly for h_v and h_z

U_u = h_u * u_u

3. The attempt at a solution

Ok, so I think I am either getting confused between u's or am missing some vital trig identity.
From the equations given for the x,y and z coordinates, I get that the new relevant variables are u, v and z. So the position vector for the coordinate system can be written as:

r = (a * cosh (u) * cos (v)) + (a * sinh (u) * sin (v))[j] + z[k]

So, from here I can get:

u_u = (dr/du_u) = (a * sinh (u) * cos (v)) + (a * cosh (u) * sin (v))[j] + 0[k]
u_v = (dr/du_v) = (-a * cosh (u) * sin (v)) + (a * sinh (u) * cos (v))[j] + 0[k]
u_z = (dr/du_z) = 0 + 0[j] + 1[k]

from here, it follows from the formulas in the book that:

h_u = magnitude(dr/du_u) = sqrt [(a * sinh (u) * cos (v))^2 + (a * cosh (u) * sin (v))^2 + 0^2]

=a * sqrt[sinh^2 (u) cos^2 (v) +cosh^2 (u) sin^2 (v)]

This is as far as I can simplify it, and I do not know how they are getting their answer of a * sqrt(cosh^2 (u) - cos^2 (v)), unless there is some identity that I am unaware of or I screwed something up somewhere along the line. Any ideas or insight from someone who has done coordinate transformations before?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 18, 2012

### MathematicalPhysicist

You might find the next identities helpful:
sin^2=1-cos^2

sinh^2 = cosh^2 -1

3. Sep 18, 2012

### xiphius75

That helped greatly! Thanks! I was aware of the sin - cos identity, but had never been taught the hyperbolic identities. With those I easily get the answer they give.