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Calculus IIIChapter 5: Line IntegralsSection 5-6

Conservative Vector Fields β€” Question 1

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Question 1

Determine whether 𝑭(x,y)=⟨2xy+3,x2+4y⟩\mathbf F(x,y)=\langle 2xy+3,\,x^2+4y\rangle is conservative on ℝ2\mathbb R^2. If it is, find a potential function ff.

Tasks

  1. Compare the relevant mixed partial derivatives.

  2. Construct ff and determine the one-variable correction term.

  3. Differentiate the result to verify the field.

Original worksheet page 1: question and worked solution for 5-6-001
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Question 1 – Solution

Strategy. Test Py=QxP_y=Q_x on the simply connected domain, then integrate one component and recover the missing function.

Step 1: Test the field With P=2xy+3P=2xy+3 and Q=x2+4yQ=x^2+4y, Py=2x,Qx=2x.P_y=2x, \qquad Q_x=2x. The components have continuous first partial derivatives on all of ℝ2\mathbb R^2, which is simply connected. Thus the equality guarantees that 𝑭\mathbf F is conservative.

See the diagram in the original worksheet below.

Step 2: Construct a potential Integrating fx=Pf_x=P with respect to xx gives f=x2y+3x+g(y).f=x^2y+3x+g(y). Then fy=x2+gβ€²(y)=Q=x2+4yf_y=x^2+g'(y)=Q=x^2+4y, so gβ€²(y)=4yg'(y)=4y and g(y)=2y2+Cg(y)=2y^2+C. f(x,y)=x2y+3x+2y2+C.\boxed{f(x,y)=x^2y+3x+2y^2+C}.

Verification Differentiation gives βˆ‡f=⟨2xy+3,x2+4y⟩=𝑭\nabla f=\langle 2xy+3,x^2+4y\rangle=\mathbf F. The arbitrary constant has zero gradient.

Original worksheet page 2: question and worked solution for 5-6-001

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