# Derivatives in polar coordinates

1. Aug 21, 2010

### JustAChemist

I appologise in the lack of distinction between curly d's and infinitesimals! All derivatives are partial and anything outside of brackets is an infinitesimal.

also, I sincerely apologise for any dodgy terminology, but I am for the most part self taught (regarding calculus) :/

(also, 0 is my poor attempt to represent theta :P)

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

More my obsessive compulsiveness when it's really something I'm just supposed to remember, but here goes...

Express (dz/dx) in polar coordinates for some arbitrary z = z(r, 0)

2. Relevant equations

the answer given by my book (Steiner - The Chemistry Maths Book):

(dz/dx) = (dz/dr)cos(0) - sin(theta)(1/r)(dz/d0)

3. The attempt at a solution

dz = (dz/dr)dr + (dz/d0)d0

(dz/dx) = (dz/dr)(dr/dx) + (dz/d0)(d0/dx)

essentially, my problem is in evaluating (dr/dx) and (d0/dx) (0 is theta)

ATTEMPT AT EVALUATING (dr/dx)

x = r cos(0) => r = x/cos(0)

=> (dr/dx) = 1/cos(0)

which is... wrong :/. I've also tried so much more silly sh*t like implicit differentiation, and inverting dx/dr, always arriving at the same answer ):

ATTEMPT AT EVAUATING (d0/dx)

x = r cos(0) => 0 = arccos(x/r)

and I have no idea how to start evaluating that...

Any help anyone could give would be greatly appreciated! (:

EDIT:

I am an idiot... after 3 days of ploughing through with this and 10 minutes typing the problem to physicsforums, I just worked out that r can be represented in terms of x and y as sqrt(x^2 + y^2) and theta as arctan(y/x)... the rest writes itself lol :P

I hate my life. :P

Last edited: Aug 21, 2010
2. Aug 21, 2010

### lanedance

good that you've got it now, and sounds like a much easier method, but just to point out the reason this fell over, is becauser theta is a function of x & y also,
so you must consider 0(x,y) when you take the derivative w.r.t. x