# Advanced Calculus -- Finding the local inverses of f

• Nov 9th 2011, 01:01 PM
MissMousey
Advanced Calculus -- Finding the local inverses of f
Problem: For each of the following transformations (u,v) = f(x, y): (i) compute the Jacobian det Df, (ii) draw a sketch of the images of some of the lines x = constant and y = constant in the uv-plane, and (iii) find the formulas for the local inverses of f when they exist.
(a) u = (e^x)(cosy), v = (e^x)(siny)
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I've completed (i) and (ii) but I'm stuck on (iii). I know I have to solve for x and y but don't know how to start. Hints?
• Nov 9th 2011, 01:30 PM
TKHunny
Re: Advanced Calculus -- Finding the local inverses of f
I'm tempted to divide by e^x (which never is zero), and then square.
• Nov 9th 2011, 01:44 PM
MissMousey
Re: Advanced Calculus -- Finding the local inverses of f
Quote:

Originally Posted by TKHunny
I'm tempted to divide by e^x (which never is zero), and then square.

Hmmmm....well, I'm supposed to get x = x(u,v) and y = y(u,v) so if do as such, I get:
u = (e^x)(cosy)
u/(e^x) = cosy
I take the arccos and get
\$\displaystyle arccos (u/e^x) = y\$

If I solve v for x and try to substitute, it just starts looking troublesome.
• Nov 9th 2011, 01:55 PM
TheEmptySet
Re: Advanced Calculus -- Finding the local inverses of f
Quote:

Originally Posted by MissMousey
Problem: For each of the following transformations (u,v) = f(x, y): (i) compute the Jacobian det Df, (ii) draw a sketch of the images of some of the lines x = constant and y = constant in the uv-plane, and (iii) find the formulas for the local inverses of f when they exist.
(a) u = (e^x)(cosy), v = (e^x)(siny)
----------------------------------------------------------------------------

I've completed (i) and (ii) but I'm stuck on (iii). I know I have to solve for x and y but don't know how to start. Hints?

You may find this about log polar coordinates interesting

Log-polar coordinates - Wikipedia, the free encyclopedia
• Nov 14th 2011, 07:45 AM
MissMousey
Re: Advanced Calculus -- Finding the local inverses of f
Ok, I got it. Thank you, everyone.