# The potential energy function of a particle moving in one-dimension is

1. Nov 30, 2011

### pittuniv

1. The problem statement, all variables and given/known data

U = k(x^2 + y^2) What is the force exerted on the particle?

2. Relevant equations

F = -(¶U/¶x ihat +¶U/¶y jhat +¶U/¶z khat) <--couldnt get the del symbol right
determining force from potential energy

3. The attempt at a solution

F = -¶/¶x[k(x^2+y^2)]ihat - ¶/¶y[k(x^2 +y^2)]jhat
=-[2kx + Y^2]ihat - [kx^2 + 2ky]jhat

Im new to calculus, and im pretty sure that im not doing the derivative of this right...any help would be greatly appreciated.

2. Nov 30, 2011

### Staff: Mentor

When you're doing partial derivatives of a function of several variables (x,y,z,...) with respect to a given variable, all the other independent variables are treated as constants. So,

$-\frac{\partial U}{\partial x} = -2 k x ~~~~~~~~~~-\frac{\partial U}{\partial y} = -2 k y$

3. Nov 30, 2011

### Andrew Mason

The partial derivative with respect to one variable is taken with the other variable held constant. So $\partial U/\partial x = 2kx\hat x$ and $\partial U/\partial y = 2ky\hat y$.

AM