Equation of motion from given 2D Potential

[Adoniram]

Homework Statement

A particle of mass m moves in two dimensions under the following potential energy
function:
V($\vec{r}$) = ½ k (x² + 4y²)

Find the resulting motion, given the initial condition at t=0:
x = a, y = 0, x' = 0, y' = v_o

Homework Equations

F = ma = -dV/dr

The Attempt at a Solution

This will obviously involve a 2nd order diff eq, and there are enough initial conditions to solve for the unknown constants. If the potential were given with the r variable instead of x and y, it would be simpler. As such, I'm not sure how to take dV/dr when V is V(x,y) not V(r)...

If I use x = r Cosθ, y = r Sinθ, I can put it as:
V($\vec{r}$) = (3/2) k r² Sin²θ

But now that I have θ in the formula, is it ok to take dV/dr as such and set it in F = -dV/dr?

 

[Noctisdark]

You are doing it the wrong way, there are vector involved and $\frac{d}{d \vec r} = \nabla$ and everyone already says it: force is the gradient of the potential, can you work it out now ?
[Edit: In case I wasn't clear, $\nabla = \lt \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \gt$ ]

 

[Adoniram]

*facepalm* Thank you!

$\vec{F}$ = -k( x $\hat{x}$ + 4y $\hat{y}$)

Then set to m$\vec{a}$ = m($\ddot{x}$ $\hat{x}$ + $\ddot{y}$ $\hat{y}$), and compare associated vector components... yes?

 

[Adoniram]

Just tried that, for anyone who wants to check:

X(t) = a $\cos$($\sqrt{k/m} t$)

Y(t) =$\frac{v_{o}}{\sqrt{4k/m}}\sin(\sqrt{4k/m} t)$