Math Problem Statement

Consider the vector field
F =⟨x 3 y 4 ,x 4 y 3 ⟩ Is this vector field Conservative? If so: Find a function f so that
F =∇f f(x,y)=+k Use your answer to evaluate ∫ C ​

F ⋅d r along the curve C: r (t)=3cos(t) i +3sin(t) j ​ ,0≤t≤ 4 π ​

Solution

To determine if the vector field F=x3y4,x4y3\mathbf{F} = \langle x^3y^4, x^4y^3 \rangle is conservative and evaluate the given line integral, we follow these steps:


Step 1: Check if the field is conservative

A vector field F=P,Q\mathbf{F} = \langle P, Q \rangle is conservative if there exists a scalar potential function f(x,y)f(x, y) such that f=F\nabla f = \mathbf{F}. This means: fx=Pandfy=Q.\frac{\partial f}{\partial x} = P \quad \text{and} \quad \frac{\partial f}{\partial y} = Q.

The field is conservative if Py=Qx\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}. For F=x3y4,x4y3\mathbf{F} = \langle x^3y^4, x^4y^3 \rangle:

  • P=x3y4P = x^3y^4, so Py=4x3y3\frac{\partial P}{\partial y} = 4x^3y^3,
  • Q=x4y3Q = x^4y^3, so Qx=4x3y3\frac{\partial Q}{\partial x} = 4x^3y^3.

Since Py=Qx\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, the field is conservative.


Step 2: Find the potential function f(x,y)f(x, y)

We find f(x,y)f(x, y) such that f=F\nabla f = \mathbf{F}:

  1. From fx=P=x3y4\frac{\partial f}{\partial x} = P = x^3y^4, integrate with respect to xx: f(x,y)=x3y4dx=x44y4+g(y),f(x, y) = \int x^3y^4 \, dx = \frac{x^4}{4}y^4 + g(y), where g(y)g(y) is an arbitrary function of yy.

  2. From fy=Q=x4y3\frac{\partial f}{\partial y} = Q = x^4y^3, differentiate f(x,y)f(x, y) with respect to yy: fy=x44(4y3)+g(y)=x4y3+g(y).\frac{\partial f}{\partial y} = \frac{x^4}{4}(4y^3) + g'(y) = x^4y^3 + g'(y). Equating this to Q=x4y3Q = x^4y^3, we find g(y)=0g'(y) = 0, so g(y)g(y) is constant.

Thus, the potential function is: f(x,y)=x4y44+k,f(x, y) = \frac{x^4y^4}{4} + k, where kk is an arbitrary constant.


Step 3: Evaluate the line integral

Since F\mathbf{F} is conservative, the line integral depends only on the potential difference between the endpoints of the curve CC: CFdr=f(x2,y2)f(x1,y1).\int_C \mathbf{F} \cdot d\mathbf{r} = f(x_2, y_2) - f(x_1, y_1).

The curve is r(t)=3cos(t)i+3sin(t)j\mathbf{r}(t) = 3\cos(t)\mathbf{i} + 3\sin(t)\mathbf{j}, with 0t4π0 \leq t \leq 4\pi. Substituting:

  • Start point (t=0t = 0): (x1,y1)=(3,0)(x_1, y_1) = (3, 0),
  • End point (t=4πt = 4\pi): (x2,y2)=(3,0)(x_2, y_2) = (3, 0).

At both points, f(x,y)=(3)4(0)44+k=kf(x, y) = \frac{(3)^4(0)^4}{4} + k = k. Thus: f(x2,y2)f(x1,y1)=kk=0.f(x_2, y_2) - f(x_1, y_1) = k - k = 0.

The line integral evaluates to: CFdr=0.\int_C \mathbf{F} \cdot d\mathbf{r} = 0.


Final Answer:

  1. The vector field is conservative.
  2. The potential function is f(x,y)=x4y44+kf(x, y) = \frac{x^4y^4}{4} + k.
  3. The line integral CFdr\int_C \mathbf{F} \cdot d\mathbf{r} evaluates to 0.

Would you like further clarification or additional examples?

Related Questions:

  1. What is the general process to check if a vector field is conservative?
  2. How do you handle vector fields defined in three dimensions for conservativeness?
  3. Can conservative vector fields have closed-path integrals that are nonzero?
  4. How would the solution change if the curve CC was not closed?
  5. What is the physical significance of conservative vector fields?

Tip:

For closed curves in conservative vector fields, the line integral is always zero because the potential difference depends only on the endpoints!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Vector Fields
Conservative Fields
Line Integrals
Potential Functions

Formulas

Conservative field condition: ∂P/∂y = ∂Q/∂x
Gradient of a potential function: ∇f = F
Line integral in a conservative field: ∫C F ⋅ dr = f(x2, y2) - f(x1, y1)

Theorems

Fundamental Theorem of Line Integrals
Condition for Conservativeness of Vector Fields

Suitable Grade Level

Grades 11-12 and College