For the harmonic potential V(x,y) = x^2 + y^2
a)Find the total differential, dV.
b) Show that dV is exact
c) Given that -dV = Fx.dx + Fy.dy where Fx and Fy is the force in the x and y direction, respectively, write a differential equation describing the change of potential energy, V(x), with position x.
a) To find the total differential, dV, we need to compute its partial derivatives with respect to x and y.
The potential function V(x, y) = x^2 + y^2, so taking partial derivatives, we have:
∂V/∂x = 2x
∂V/∂y = 2y
Therefore, the total differential, dV, is given by:
dV = (∂V/∂x)dx + (∂V/∂y)dy
= (2x)dx + (2y)dy
b) To show that dV is exact, we need to verify if its mixed partial derivatives are equal.
Taking the second partial derivatives of dV with respect to x and y, we have:
∂^2V/∂x∂y = ∂^2V/∂y∂x = 0
Since the mixed partial derivatives are equal to zero, dV is exact.
c) Given -dV = Fx·dx + Fy·dy, where Fx and Fy are the forces in the x and y directions respectively, we can equate the total differential to -dV and substitute the values of dV from part a:
(2x)dx + (2y)dy = Fx·dx + Fy·dy
To describe the change of potential energy, V(x), with position x, we need to eliminate y from the equation. We can rearrange the equation as follows:
(2x - Fx)dx = (Fy - 2y)dy
Now we can equate the coefficients of dx and dy to obtain the differential equation describing the change of potential energy with position x:
2x - Fx = Fy - 2y