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

[pittuniv]

Homework Statement

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

Homework Equations

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

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 I am pretty sure that I am not doing the derivative of this right...any help would be greatly appreciated.

 

[gneill]

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,

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

 

[Andrew Mason]

Homework Statement

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

Homework Equations

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

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 I am pretty sure that I am not doing the derivative of this right...any help would be greatly appreciated.

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