Jtdylktuy

Let $(x, y)$ be Cartesian coordinates in the plane and let $(r, theta)$ be Polar coordinates. Express $frac{d^2z}{dtheta^2}$.

my attempt

$z = f(x, y)$

Note

$frac{du}{dtheta} = frac{dr}{dtheta}cos(theta) = -rsin(theta)$

$frac{dy}{dtheta} = rcos(theta)$

$frac{dz}{dtheta} = frac{df}{du}frac{du}{dtheta} + frac{df}{dy}frac{dy}{dtheta}$

$frac{dz}{dtheta} = -r frac{df}{du}sin(theta) + rfrac{df}{dy}cos(theta)$

$frac{1}{r}frac{d}{dtheta}(frac{dz}{dtheta}) = frac{d}{dtheta}(frac{df}{du})sin(theta) - cos(theta)frac{df}{du} + frac{d}{dtheta} (frac{df}{dy})cos(theta) - frac{df}{dy}sin(theta)$

$frac{1}{r}frac{d^2z}{dtheta^2} = -frac{d^2f}{dx^2}frac{dx}{dtheta}sin(theta) - frac{df}{du}cos(theta) + frac{d^2f}{dy^2}(frac{dy}{dtheta})cos(theta) - frac{df}{dy}sin(theta)$

thus,

$$frac{d^2z}{dtheta^2} = r^2sin^2(theta)frac{d^2f}{x^2} - frac{rdf}{dx}cos(theta) - r frac{df}{dy}sin(theta) + r^2 frac{d^2f}{dy^2}cos^2(theta)$$

Is this correct?

I think I have something. Was careful to apply the chain rule to terms like $frac{d}{dtheta}(frac{partial f}{partial x})$. This results in the second partial derivatives appearing in the below.

$$z=f(x,y)$$
$$dz/dtheta=frac{partial f}{partial x}frac{dx}{dtheta}+frac{partial f}{partial y}frac{dy}{dtheta}$$

$$frac{d^2z}{dtheta^2}=frac{d}{dtheta}(frac{partial f}{partial x})frac{dx}{dtheta}+frac{partial f}{partial x}frac{d^2x}{dtheta ^2}+frac{d}{dtheta}(frac{partial f}{partial y})frac{dy}{dtheta}+frac{partial f}{partial y}frac{d^2y}{dtheta^2}$$

$$frac{d^2z}{dtheta^2}=(frac{partial^2f}{partial x^2}frac{dx}{dtheta}+frac{partial ^2f}{partial xpartial y}frac{dy}{dtheta})(frac{dx}{dtheta})+frac{partial f}{partial x}frac{d^2x}{dtheta ^2}+frac{partial f}{partial y}frac{d^2y}{dtheta^2}+(frac{partial^2 f}{partial ypartial x}frac{dx}{dtheta}+frac{partial ^2f}{partial y^2}frac{dy}{dtheta})frac{dy}{dtheta}$$

$$frac{d^2z}{dtheta^2}=(frac{partial ^2f}{partial x^2})(frac{dx}{dtheta})^2+2frac{partial ^2 f}{partial x partial y}frac{dy}{dtheta}frac{dx}{dtheta}+frac{partial f}{partial x}frac{d^2x}{dtheta^2}+frac{partial f}{partial y}frac{d^2y}{dtheta ^2}+frac{partial ^2 f}{partial y^2}(frac{dy}{dtheta})^2$$

$x=rcos{theta}$

$y=rsin{theta}$

$dx/dtheta=-rsin{theta}$

$dy/dtheta= r cos{theta}$

$$frac{d^2z}{dtheta^2}=(frac{partial ^2f}{partial x^2})(r^2sin^2{theta})+2frac{partial ^2f}{partial x partial y}(-r^2sin{theta}cos{theta})+frac{partial f}{partial x}(-rcos{theta})+frac{partial f}{partial y}(-rsin{theta})+frac{partial ^2f}{partial y^2}(r^2cos^2{theta})$$

In my mind, the bulletproof way to calculate differential operators of transformations is the following:

1. Write down the equality of the transformation:

$$z(r,varphi)=f(rcos(varphi),rsin(varphi))$$

2. Now you can simply differentiate twice and need not to worry about forgetting anything because nothing is 'hidden'

$$frac{partial z}{partialvarphi}=f_x(rcos(varphi),rsin(varphi))(-rsin(varphi))+f_y(rcos(varphi),rsin(varphi))(rcos(varphi))$$
$$frac{partial^2 z}{partialvarphi^2}=f_{xx}(rcos(varphi),rsin(varphi))(-rsin(varphi))^2-2f_{xy}(rcos(varphi),rsin(varphi))r^2sin(varphi)cos(varphi)-f_x(rcos(varphi),rsin(varphi))(rcos(varphi))+f_{yy}(rcos(varphi),rsin(varphi))(rcos(varphi))^2-f_y(rcos(varphi),rsin(varphi))(rsin(varphi))$$

I thought of a vector approach.

$z=f(x,y)$, find $d^2z/dtheta^2$.

$frac{df}{dtheta}=nabla fcdot frac{dvec{s}}{dtheta}$

$frac{d^2f}{dtheta^2}=nabla(nabla fcdot frac{dvec{s}}{dtheta})cdot frac{dvec{s}}{dtheta}$

We have the identity: $nabla(vec{A}cdot vec{B})=(vec{A}cdot nabla)vec{B}+(vec{B}cdot nabla)vec{A}+vec{A}times(nabla times vec{B})+vec{B}times(nabla times vec{A})$

So: $frac{d^2f}{dtheta^2}=frac{dvec{s}}{dtheta}cdot [ (nabla fcdot nabla)frac{dvec{s}}{dtheta}+(frac{dvec{s}}{dtheta}cdot nabla)nabla f+nabla ftimes(nabla times frac{dvec{s}}{dtheta})+frac{dvec{s}}{dtheta}times(nabla times nabla f) ]$

This can be simplified some. By vector identity, $nabla times nabla f =0.$

So:

$frac{d^2f}{dtheta^2}=frac{dvec{s}}{dtheta}cdot [ (nabla fcdot nabla)frac{dvec{s}}{dtheta}+(frac{dvec{s}}{dtheta}cdot nabla)nabla f+nabla ftimes(nabla times frac{dvec{s}}{dtheta})]$

$frac{dvec{s}}{dtheta}=<-rsin{(theta)}, rcos{(theta),0}>=<-y,x,0>$

$nabla timesfrac{dvec{s}}{d theta}=<0,0,2>$