117k views
4 votes
Show that each of the vector fields F=yi+x,j. G=y/x²+y i+-x/x²+y² j, and H = 5x/√x²+y² j are gradient vector fields on some domain (not necessarily the whole plane) by finding a potential function for each.

For F, a potential function is f(x, y) =
For G, a potential function is g(x, y) =
For Ĥ, a potential function is h(x, y) =

1 Answer

3 votes

Final answer:

To prove the vector fields F, G, and H are gradient fields, potential functions f(x, y) = x^2/2 + y^2/2, g(x, y) = arctan(y/x), and h(x, y) = 5arctan(y/x) are found, respectively, satisfying the relationship ∇f = F, ∇g = G, and ∇h = H within their specific domains.

Step-by-step explanation:

To show that the vector fields F, G, and H are gradient vector fields, we need to find the potential functions f(x,y), g(x,y), and h(x,y), respectively, that satisfy the property ∇f = F, ∇g = G, and ∇h = H. A gradient vector field is one where the vector field can be expressed as the gradient of some scalar potential function.

Finding Potential Functions

Vector Field F

Given F = yi + xj, to find a potential function f(x,y), we integrate the x-component with respect to x holding y constant and the y-component with respect to y holding x constant, and combine any functions of the other variable:

  • ∫ (y)dy = y2/2 + Cx(x)
  • ∫ (x)dx = x2/2 + Cy(y)

Therefore, the potential function f(x, y) = x2/2 + y2/2 satisfies ∇f = F.

Vector Field G

For G = y/x2 + y i - x/x2 + y2 j, we integrate similarly:

  • ∫ (y/x2 + y)dx = arctan(y/x) + Cy(y)
  • ∫ (-x/x2 + y2)dy = -arctan(x/y) + Cx(x)

If we choose the constant of integration appropriately, g(x, y) = arctan(y/x) is a potential function that works for G.

Vector Field H

H = 5x/√x2+y2 j requires integrating with respect to y:

  • ∫ (5x/√x2+y2)dy = 5arctan(y/x) + C(x)

The potential function h(x, y) = 5arctan(y/x) yields ∇h = H.

Note: For each potential function, the domain must exclude points where the denominators are zero or where the functions are not defined.

User Scott Kirkwood
by
8.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.