For the harmonic potential V(x,y) = x^2 + y^2
a) Find the total differential, dV.
For this I got dV = 2x.dx + 2y.dy
b) Given that -dV = F(x).dx + F(y).dy, where F(x) and F(y) 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.
I am really not sure about this. Should I integrate the whole equation and then differentiate with respect to x?
To find the differential equation describing the change of potential energy with position x, you need to consider the relationship between the force and the potential energy.
In this case, the force can be derived from the potential function by taking the negative gradient:
F(x) = -dV/dx = -2x
F(y) = -dV/dy = -2y
The negative sign is included because the force is in the opposite direction to the potential energy gradient.
Now, you can use the relationship -dV = F(x).dx + F(y).dy to write the differential equation. Since dV = -2x.dx - 2y.dy, you have:
-2x.dx - 2y.dy = F(x).dx + F(y).dy
Substituting the expressions for F(x) and F(y):
-2x.dx - 2y.dy = (-2x).dx + (-2y).dy
Now, rearrange the equation:
(-2x + 2x).dx + (-2y + 2y).dy = 0
Simplifying further:
0.dx + 0.dy = 0
This equation is identically true, which means that it holds for all values of x and y.
Therefore, the differential equation describing the change of potential energy with position x is 0, indicating that the potential energy does not change with position in this case.