Differentiating a polar function

kgal
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Homework Statement


let z=f(x,y) be a differentiable function. If we change to polar coordinates, we make the substitution x=rcos(θ), y=rsin(θ), x^2+y^2=r^2 and tan(θ) = y/x.
a. Find expressions ∂z/∂r and ∂z/∂θ involving ∂z/∂x and ∂z/∂y.
b. Show that (∂z/∂x)^2 + (∂z/∂y)^2 = (∂z/∂r)^2 + (1/r^2)(∂z/∂θ)^2.


The Attempt at a Solution



a. i understand that f(x,y) in polar is f(r,θ) but don't understand how to calculate the partial derivatives of ∂z/∂x and ∂z/∂y because there is not know function for z...
 
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Do you know the chain rule for functions of two variables?
 
in the specific case of this problem they come out like this:
∂z/∂r = (∂z/∂x)(∂x/∂r) + (∂z/∂y)(∂y/∂r)
∂z/∂θ = (∂z/∂x)(∂x/∂θ ) + (∂z/∂y)(∂y/∂θ)

right?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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