A potential function for F is: f(x, y) = (x^5y^4)/4 + (x^5y^5)/20 + c
We know that a potential function exists for F if F is conservative. To check if F is conservative, we can use the curl test. The curl of F is given by:
curl F = ∂Q/∂x - ∂P/∂y
where P = x^4y^5 and Q = x^5y^4. Calculating the partial derivatives, we get:
∂Q/∂x = 5x^4y^4 - 4x^3y^5
∂P/∂y = 4x^4y^5 - 5x^5y^4
Setting these two equal, we get:
5x^4y^4 - 4x^3y^5 = 4x^4y^5 - 5x^5y^4
This simplifies to:
x^5y^4 - x^3y^5 = 0
Factoring out x^3y^4, we get:
x^3y^4(x^2 - 1) = 0
This means that F is conservative if x^3y^4 = 0 or x^2 - 1 = 0. We can see that x^3y^4 = 0 for all points on the curve C, so F is conservative.
Now, we can find a potential function f for F. We know that the gradient of f is equal to F, so:
∇f = F
This means that:
∂f/∂x = x^4y^5
∂f/∂y = x^5y^4
Integrating the first equation with respect to x, we get:
f(x, y) = ∫ x^4y^5 dx + g(y)
where g(y) is an arbitrary function of y. Integrating by substitution, we get:
f(x, y) = (x^5y^5)/5 + g(y)
Integrating the second equation with respect to y, we get:
f(x, y) = (x^5y^4)/4 + h(x)
where h(x) is an arbitrary function of x. Setting these two expressions for f(x, y) equal to each other, we get:
(x^5y^5)/5 + g(y) = (x^5y^4)/4 + h(x)
Solving for g(y), we get:
g(y) = (x^5y^4)/20 + c
where c is a constant of integration. Therefore, a potential function for F is:
f(x, y) = (x^5y^4)/4 + (x^5y^5)/20 + c