A potential energy function for a two-dimensional force is of the form U = 3 x^(4)y - 8 x. Find the force that acts at the point (x, y). (Use x and y as appropriate.)
I know its something about partial derivatives but I'm not sure how to go about it. And the answer has to be in vector components (ihat jhat)
Well, if a force is acting on something and energy is conserved and there is no change in kinetic energy then
Delta U = increase in potential energy from point a to point b = integral Fx dx + integral Fy dy where the integrals are from a to b.
That is the work we would do on the system. The force the system exerts is equal and opposite by Newton #3.
So I have a hunch that
Fx = -dU/dx
and
Fy = -dU/dy
and the vector F = Fx i + Fy j
so here
U = 3 x^(4)y - 8 x
-dU/dx = -12 x^3 y +8
-dU/dy = -3 x^4
and
F = (-12 x^3 y +8)i - 3 x^4 j
Yes! Thank You! I was very close but I messed up on my derivative somehow :)
To find the force that acts at the point (x, y), we need to take the negative gradient of the potential energy function U(x, y).
The negative gradient of a scalar function is a vector function, where each component of the vector is the negative partial derivative of the function with respect to its corresponding coordinate.
In this case, the potential energy function is given as U = 3x^4y - 8x. To find the force vector, we need to calculate the partial derivatives of U with respect to x and y.
Taking the partial derivative of U with respect to x:
∂U/∂x = (12x^3)y - 8
Taking the partial derivative of U with respect to y:
∂U/∂y = 3x^4
To obtain the force vector, we combine the partial derivatives as follows:
Force vector, F = - (∂U/∂x) î - (∂U/∂y) ĵ
= - [(12x^3)y - 8] î - (3x^4) ĵ
Therefore, the force that acts at the point (x, y) is given by F = - [(12x^3)y - 8] î - (3x^4) ĵ, where î and ĵ are the unit vectors in the x and y directions, respectively.
To find the force that acts at a given point (x, y), we need to find the gradient of the potential energy function U(x, y) at that point. The gradient is a vector that points in the direction of the steepest increase of the function and whose magnitude represents the rate of change.
In this case, the potential energy function U(x, y) is given as U = 3x^4y - 8x. To find the force as a vector, we need to calculate the partial derivatives of U with respect to x and y, and then express them as vector components.
1. Calculate the partial derivative of U with respect to x (∂U/∂x):
Taking the derivative of U with respect to x treats y as a constant, so the y-term becomes 0 when differentiating with respect to x. Thus, we have:
∂U/∂x = ∂(3x^4y - 8x)/∂x = 12x^3y - 8.
2. Calculate the partial derivative of U with respect to y (∂U/∂y):
Taking the derivative of U with respect to y treats x as a constant, so the x-term becomes 0 when differentiating with respect to y. Thus, we have:
∂U/∂y = ∂(3x^4y - 8x)/∂y = 3x^4.
3. Express the force as a vector:
The force vector acting at the point (x, y) is given by the negative gradient of the potential energy function, which is -∇U. The gradient is a vector that points in the direction of the steepest increase of the function, so the force vector at any point is exactly the opposite direction.
The force vector components can be expressed as:
F = - (∂U/∂x) î - (∂U/∂y) ĵ,
where î and ĵ are the unit vectors in the x and y directions, respectively.
Substituting the partial derivatives we calculated earlier, the force vector becomes:
F = - (12x^3y - 8) î - (3x^4) ĵ.
That's the expression for the force vector at the point (x, y).