A potential energy function for a two-dimensional force is of the form U = 3x3y – 7x. Find the force that acts at the point (x, y).
The x force is -dU/dx (partial derivative)
The y force is -dU/dy (partial derivative)
You should place a ^ before exponents, as in
U = 3x^3* y -7x
I assume you know how to take partial derivatives. Treat the other variables as constants.
9x^2y-7,3x^3
Since F = -(dU/dx)i -(dU/dx)j
-(dU/dx) = - (9x^2y - 7)i
-(dU/dy) = - (3x^3)
F = -(dU/dx)i -(dU/dy)j = (-9x^2y+7)i - 3x^3j
To find the force that acts at a particular point (x, y) given a potential energy function, we need to calculate the negative gradient of the potential energy function. The negative gradient provides us with the force vector at any given point.
The force vector is defined as F = -∇U, where ∇U represents the gradient of the potential energy function U.
In this case, the potential energy function is U = 3x^3y - 7x. To find the force vector at the point (x, y), we need to calculate the partial derivatives of U with respect to x and y. Let's calculate them one by one.
Partial derivative with respect to x:
∂U/∂x = ∂(3x^3y - 7x)/∂x
= 9x^2y - 7
Partial derivative with respect to y:
∂U/∂y = ∂(3x^3y - 7x)/∂y
= 3x^3
Now that we have the partial derivatives, we can assemble the force vector. The force vector is a two-dimensional vector that consists of the partial derivatives of U with respect to x and y.
F = (-∂U/∂x) i + (-∂U/∂y) j
Substituting the derived partial derivatives, we get:
F = (-9x^2y + 7) i + (-3x^3) j
So, the force vector that acts at the point (x, y) is given by (-9x^2y + 7) i + (-3x^3) j.