Solve Elliptical Motion Homework: Find r', r'', θ', θ

[camorin]

Homework Statement

A particle of mass m is subjected to an isotropic, two dimensional, harmonic central force, F=-kr.
(r=(x,y)). At t=0 the particle is at r=A

and has velocity V y^hat.

Find r' r'' θ' θ'' in 2D spherical polar coordinates.

Homework Equations

F_x=-kx
F_y=-ky

The Attempt at a Solution

I have found the equations of motion, as well as the particular solutions using the initial conditions.
So far I have:
x(t)=-Acos(wt-π)
y(t)=V/wcos(wt-π/2)

From here I found r' to be Awsin(wt-π)-Vsin(wt-π/2)
and r'': Aw²cos(wt-π)-Vwcos(wt-π/2)
Im fairly certain everything up to this point is correct, but I have no idea what to do about theta.

I know in spherical polar coordinates θ=arctan(y/x) but I don't think I can just take the derivative in this form. I have tried setting x(t)=-Acos(wt) and y(t)=V/wsin(wt) by using the relation between sines and cosines offset by pi/2.This has brought me to
θ=arctan((V/wa)tan(wt)). Again, I hit a wall with finding the derivatives.

 

[Simon Bridge]

Is the primed notation a space or a time derivative?

I have no idea what to do about theta

Write down an expression for theta and take the appropriate derivative.

I know in spherical polar coordinates θ=arctan(y/x) but I don't think I can just take the derivative in this form.

Why not? Did you try?

I have tried setting x(t)=-Acos(wt) and y(t)=V/wsin(wt) by using the relation between sines and cosines offset by pi/2.This has brought me to
θ=arctan((V/wa)tan(wt)).

... what happens when you take the appropriate derivative of that then?

 

[camorin]

Im not sure of how to take the derivative of that arctan expression. Is there a way to simplify it? I've tried finding methods but I'm having trouble.

The primed notation is a time derivative, sorry.

 

[Simon Bridge]

Look it up.
... that's what everyone else does.

 

[paranormal]

To find θ' and θ'', you can use the chain rule for derivatives.
θ' = dθ/dt = dθ/dx * dx/dt + dθ/dy * dy/dt
= (1/(1+(y/x)^2)) * (-V/wsin(wt)) + (-x/y^2) * (V/wcos(wt))
= (-V/wsin(wt)) / (1+(y/x)^2) - (x/y^2) * (V/wcos(wt))
Similarly,
θ'' = d^2θ/dt^2 = d/dt (dθ/dt)
= d/dt ((-V/wsin(wt)) / (1+(y/x)^2) - (x/y^2) * (V/wcos(wt)))
= (V^2/w^2cos^2(wt)) / (1+(y/x)^2)^2 * (y/x) * (dy/dt) + (V/wsin(wt)) / (1+(y/x)^2) * (x/y^2) * (dy/dt)
+ (V/wsin(wt)) / (1+(y/x)^2) * (dx/dt) + (x/y^2) * (V/wcos(wt)) * (dx/dt)
= (V^2/w^2cos^2(wt)) / (1+(y/x)^2)^2 * (y/x) * (V/wcos(wt)) + (V/wsin(wt)) / (1+(y/x)^2) * (x/y^2) * (V/wcos(wt))
+ (V/wsin(wt)) / (1+(y/x)^2) * (-Awsin(wt)) + (x/y^2) * (V/wcos(wt)) * (Awsin(wt))
= (V^2/wcos^2(wt)) / (1+(y/x)^2)^3 * (y/x) * (V/w) + (V/wsin(wt)) / (1+(y/x)^2) * (-Awsin(wt)) + (x/y^2) * (V/wcos(wt)) * (Awsin(wt))
Now you can substitute the expressions for x(t) and y(t) to get a final answer for θ'' in terms of the