- #1

AmagicalFishy

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Consider a particle of mass

**m**subject to an isotropic two-dimensional harmonic central force F= −k[itex]\vec{r}[/itex], where

**k**is a positive constant. At t=0, we throw the particle from position [itex]\vec{r}_0[/itex] = A[itex]\hat{x}[/itex] with velocity [itex]\vec{v}_0[/itex] = V[itex]\hat{y}[/itex]. Show that the trajectory of the particle is, in general, an ellipse.

So my plan is just to solve the equation for Newton's 2nd law of motion, get a 2nd order differential equation, etc. The confusion comes in when I ask myself "When do I need to make use of the chain rule?"

It seems easy enough to me to just take the equation [itex]k\vec{r} + m\ddot{\vec{r}} = 0[/itex] at face value but (and this may seem like a silly question)...

... isn't [itex]\vec{r} = r\hat{r}[/itex]?

So [itex]\dot{\vec{r}} = \dot{r}\hat{r} + r\dot{\hat{r}}[/itex]?

... and then I use the chain rule

*again*to get the second derivative: [itex]\hat{r}\ddot{r} + \dot{r}\dot{\hat{r}} + \dot{r}\dot{\hat{r}} + \ddot{\hat{r}}r[/itex]

I'm sure I'm over-complicating things, but this is the type of confusion I

**always**end up wasting tons of my time on, and I have a real hard time finding an answer in textbooks or Wikipedia.

The only way I can consolidate the two methods is by thinking: The force isn't dependent on θ, so the 1st and 2nd derivatives of θ are zero. Since the 1st and 2nd derivatives of [itex]\hat{r}[/itex] depend on the derivatives on theta, those are also zero—and that ugly combination above simplifies to [itex]\ddot{r}\hat{r}[/itex].

Is this correct? It sounds fine to me, but questions like this (can I do this? Or should I approach it, mathematically, like

*this*? etc.) take up so much of my time that I end up spending multiple hours on problems I realistically should spend only a couple of minutes on.