Potential Energy Function and Force

[kiwikahuna]

Homework Statement

A potential energy funtion for a two dimensional force is of the form
U = ax^3y + bx
where a =10.51 J/m^4 and b = -8 J/m.
Find the magnitude of the force that acts at the point (x,y) for x =93, y =68

Homework Equations

F(x) = dU/dx

The Attempt at a Solution

Given the PE function, I know that I have to take the derivative to find Force but I am confused as to how to do this with it being in 2-d. There are two variables, x and y. Help please.

 

[malawi_glenn]

if

U = ax^3y + bx

then you want to differentiate this with respect to x and then y, you just when you differentiate with respect to x treat ALL the others as constants. And the same way with the y.

$$\dfrac{dU}{dx} = 3ayx^{3y-1} + b$$

now you try

 

[m2003]

To find the magnitude of the force at the point (x,y), we need to take the gradient of the potential energy function. This involves taking the partial derivatives with respect to both x and y. So, we have:

F(x,y) = (∂U/∂x) i + (∂U/∂y) j

Where i and j are unit vectors in the x and y directions, respectively.

Taking the partial derivatives, we get:

(∂U/∂x) = 3ax^2y + b

(∂U/∂y) = ax^3

Substituting the given values of a and b, we get:

(∂U/∂x) = 3(10.51)(93)^2(68) + (-8) = 3.29 x 10^6 J/m

(∂U/∂y) = (10.51)(93)^3 = 9.38 x 10^6 J/m

Therefore, the magnitude of the force at the point (93,68) is:

|F(93,68)| = √[(3.29 x 10^6)^2 + (9.38 x 10^6)^2] = 1.00 x 10^7 J/m

This means that there is a force of 1.00 x 10^7 J/m acting at an angle of tan^-1(9.38/3.29) = 71.97 degrees from the positive x-axis at the point (93,68).