Multivariable Calculus Quiz: Conservative Fields And Potentials
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Conservative Fields And PotentialsQuestion 1 of 8

Consider the vector field F(x,y)=(2xy+y2)i+(x2+2xy+3)j\mathbf{F}(x,y) = (2xy + y^2)\mathbf{i} + (x^2 + 2xy + 3)\mathbf{j}. Which statement about this field is correct?

The field is conservative with potential function f(x,y)=x2y+xy2+3y+Cf(x,y) = x^2y + xy^2 + 3y + C
The field is conservative with potential function f(x,y)=x2y+xy2+3x+Cf(x,y) = x^2y + xy^2 + 3x + C
The field is not conservative because ∂P∂y≠∂Q∂x\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}
The field is not conservative because it contains a constant term in the j\mathbf{j} component
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Conservative Fields And Potentials

Practice Conservative Fields And Potentials in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Conservative Fields And Potentials, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Consider the vector field F(x,y)=(2xy+y2)i+(x2+2xy+3)j\mathbf{F}(x,y) = (2xy + y^2)\mathbf{i} + (x^2 + 2xy + 3)\mathbf{j}. Which statement about this field is correct?

  1. The field is conservative with potential function f(x,y)=x2y+xy2+3y+Cf(x,y) = x^2y + xy^2 + 3y + C (correct answer)
  2. The field is conservative with potential function f(x,y)=x2y+xy2+3x+Cf(x,y) = x^2y + xy^2 + 3x + C
  3. The field is not conservative because ∂P∂y≠∂Q∂x\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}
  4. The field is not conservative because it contains a constant term in the j\mathbf{j} component
Explanation: First check if the field is conservative by testing if ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}. Here P=2xy+y2P = 2xy + y^2 and Q=x2+2xy+3Q = x^2 + 2xy + 3. We have ∂P∂y=2x+2y\frac{\partial P}{\partial y} = 2x + 2y and ∂Q∂x=2x+2y\frac{\partial Q}{\partial x} = 2x + 2y. Since these are equal, the field is conservative. To find the potential function, we integrate: ∂f∂x=2xy+y2\frac{\partial f}{\partial x} = 2xy + y^2 gives f=x2y+xy2+g(y)f = x^2y + xy^2 + g(y). Then ∂f∂y=x2+2xy+g′(y)=x2+2xy+3\frac{\partial f}{\partial y} = x^2 + 2xy + g'(y) = x^2 + 2xy + 3, so g′(y)=3g'(y) = 3 and g(y)=3y+Cg(y) = 3y + C. Choice B has the wrong integration for the constant term (should be 3y3y not 3x3x). Choice C incorrectly claims the partial derivatives are unequal. Choice D incorrectly assumes constant terms prevent conservativity.

Question 2

Suppose F\mathbf{F} is a vector field such that ∫CF⋅dr=5\int_C \mathbf{F} \cdot d\mathbf{r} = 5 where CC is the path from (0,0)(0,0) to (2,1)(2,1) along the curve y=x24y = \frac{x^2}{4}, and ∫DF⋅dr=3\int_D \mathbf{F} \cdot d\mathbf{r} = 3 where DD is the straight line from (0,0)(0,0) to (2,1)(2,1). What can you conclude about F\mathbf{F}?

  1. F\mathbf{F} is definitely conservative because both integrals have the same endpoints
  2. F\mathbf{F} is definitely not conservative because the line integrals have different values (correct answer)
  3. F\mathbf{F} might be conservative; more information about the domain is needed
  4. F\mathbf{F} is not conservative because one path is curved and the other is straight
Explanation: If a vector field F\mathbf{F} is conservative, then line integrals depend only on the endpoints, not on the path taken. Since we have two different paths CC and DD with the same endpoints (0,0)(0,0) and (2,1)(2,1), but ∫CF⋅dr=5≠3=∫DF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r} = 5 \neq 3 = \int_D \mathbf{F} \cdot d\mathbf{r}, the field cannot be conservative (assuming both paths lie in a region where F\mathbf{F} is defined and has continuous partial derivatives). Choice A is wrong because having the same endpoints is necessary but not sufficient for conservativity. Choice C is wrong because the different values of the line integrals already provide definitive information. Choice D gives an irrelevant reason - the shape of the path doesn't matter, only whether the integrals are path-independent.

Question 3

A vector field F\mathbf{F} has the property that ∇×F=0\nabla \times \mathbf{F} = \mathbf{0} everywhere in its domain, which is R3\mathbb{R}^3 minus the zz-axis. Which statement is most accurate?

  1. F\mathbf{F} is conservative throughout its domain since the curl is zero
  2. More information about the behavior of F\mathbf{F} near the zz-axis is needed
  3. F\mathbf{F} cannot be conservative because its domain is not simply connected
  4. F\mathbf{F} is conservative on any ball that doesn't intersect the zz-axis (correct answer)
Explanation: When you encounter vector field problems involving curl and conservativity, you need to carefully consider both the mathematical conditions and the topology of the domain. The key insight is that zero curl is necessary but not always sufficient for a conservative vector field. Since ∇×F=0\nabla \times \mathbf{F} = \mathbf{0} everywhere in the domain, we know the curl condition is satisfied. However, conservativity also depends on the connectedness properties of the domain. The domain here is R3\mathbb{R}^3 minus the zz-axis, which is not simply connected—you can draw loops around the missing zz-axis that cannot be continuously shrunk to a point while staying in the domain. Answer D is correct because any ball that doesn't intersect the zz-axis is simply connected (it's topologically equivalent to a solid sphere with no holes). On such regions, zero curl does guarantee conservativity by the fundamental theorem for line integrals. Answer A is wrong because it ignores the topology issue—zero curl alone doesn't guarantee conservativity in non-simply connected domains. Answer B is incorrect because we already have sufficient information; the behavior near the zz-axis doesn't change the topological constraints. Answer C overstates the case—while the field cannot be guaranteed conservative on the entire domain, it can still be conservative on appropriately chosen subregions. Remember: Zero curl plus simple connectedness gives you conservativity. When domains have "holes" or missing pieces, focus on finding simply connected subregions where the standard theorems apply.

Question 4

Let f(x,y,z)=xy2z+x3−yz2f(x,y,z) = xy^2z + x^3 - yz^2 and consider F=∇f\mathbf{F} = \nabla f. If CC is any piecewise smooth curve from point AA to point BB, then ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r} equals:

  1. f(B)−f(A)f(B) - f(A) only if CC lies in a simply connected region
  2. f(A)−f(B)f(A) - f(B) regardless of the path CC
  3. f(B)−f(A)f(B) - f(A) regardless of the path CC (correct answer)
  4. Zero, since F\mathbf{F} is the gradient of a scalar function
Explanation: When you encounter a vector field that's the gradient of a scalar function, you're dealing with a conservative vector field. This is a fundamental concept in multivariable calculus that connects gradient fields to path independence. Since F=∇f\mathbf{F} = \nabla f, we can apply the Fundamental Theorem for Line Integrals. This theorem states that for any conservative vector field F=∇f\mathbf{F} = \nabla f, the line integral ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r} depends only on the endpoints, not the path. Specifically, ∫CF⋅dr=f(B)−f(A)\int_C \mathbf{F} \cdot d\mathbf{r} = f(B) - f(A) where AA and BB are the initial and final points of curve CC. This makes answer C correct: the integral equals f(B)−f(A)f(B) - f(A) regardless of the path CC. Answer A is incorrect because it suggests the result only holds in simply connected regions. While conservative fields are indeed path-independent in simply connected domains, gradient fields like ∇f\nabla f are conservative everywhere they're defined (which is all of R3\mathbb{R}^3 for this polynomial function). Answer B has the wrong sign order. The Fundamental Theorem gives us f(final)−f(initial)f(\text{final}) - f(\text{initial}), not the reverse. Answer D reflects a common misconception. Conservative vector fields don't make line integrals zero; they make them path-independent. The integral only equals zero when the curve is closed (starts and ends at the same point). Key takeaway: Whenever you see F=∇f\mathbf{F} = \nabla f, immediately think "path independence" and apply the Fundamental Theorem: ∫CF⋅dr=f(end)−f(start)\int_C \mathbf{F} \cdot d\mathbf{r} = f(\text{end}) - f(\text{start}).

Question 5

Consider a vector field F(x,y)=P(x,y)i+Q(x,y)j\mathbf{F}(x,y) = P(x,y)\mathbf{i} + Q(x,y)\mathbf{j} defined on a simply connected domain DD. Given that ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} throughout DD, and that F(1,0)=3i+2j\mathbf{F}(1,0) = 3\mathbf{i} + 2\mathbf{j}, what additional information is needed to uniquely determine a potential function ff such that ∇f=F\nabla f = \mathbf{F}?

  1. The gradient of F\mathbf{F} must be specified throughout DD
  2. The values of F\mathbf{F} at least three non-collinear points are needed
  3. The value of ff at one point in DD is sufficient to determine ff uniquely (correct answer)
  4. No additional information is needed; ff is already uniquely determined
Explanation: When you encounter a conservative vector field problem, you're dealing with the fundamental relationship between vector fields and their potential functions. The given condition ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} tells you that F\mathbf{F} is conservative on the simply connected domain DD, which guarantees that a potential function ff exists such that ∇f=F\nabla f = \mathbf{F}. The key insight is understanding what "uniquely determine" means for potential functions. Since ∇f=F\nabla f = \mathbf{F}, we know that ∂f∂x=P(x,y)\frac{\partial f}{\partial x} = P(x,y) and ∂f∂y=Q(x,y)\frac{\partial f}{\partial y} = Q(x,y). This system of partial differential equations has infinitely many solutions that differ only by a constant. To pin down one specific function ff, you need exactly one additional piece of information: the value of ff at any single point in DD. This is analogous to how knowing a derivative doesn't uniquely determine an antiderivative until you specify an initial condition. Option A is incorrect because the gradient of F\mathbf{F} (which would be the Hessian of ff) provides information about second derivatives, not the constant of integration. Option B misunderstands the problem—knowing F\mathbf{F} at multiple points doesn't help since we already know F\mathbf{F} throughout DD. Option D ignores that potential functions are determined up to an additive constant. Remember: conservative vector fields always have potential functions determined up to a constant. One function value fixes that constant completely.

Question 6

Let F(x,y,z)=(yz+2x)i+(xz+y)j+(xy+z2)k\mathbf{F}(x,y,z) = (yz + 2x)\mathbf{i} + (xz + y)\mathbf{j} + (xy + z^2)\mathbf{k}. Which of the following statements is true?

  1. F\mathbf{F} is conservative and a potential function is f(x,y,z)=xyz+x2+y22+z33f(x,y,z) = xyz + x^2 + \frac{y^2}{2} + \frac{z^3}{3} (correct answer)
  2. F\mathbf{F} is conservative and a potential function is f(x,y,z)=xyz+x2+y22+z2f(x,y,z) = xyz + x^2 + \frac{y^2}{2} + z^2
  3. F\mathbf{F} is not conservative because ∇×F≠0\nabla \times \mathbf{F} \neq \mathbf{0}
  4. F\mathbf{F} is not conservative because it is defined in three dimensions
Explanation: To check if F\mathbf{F} is conservative, we compute ∇×F\nabla \times \mathbf{F}. With P=yz+2xP = yz + 2x, Q=xz+yQ = xz + y, R=xy+z2R = xy + z^2, we have: ∂R∂y−∂Q∂z=x−x=0\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = x - x = 0, ∂P∂z−∂R∂x=y−y=0\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = y - y = 0, ∂Q∂x−∂P∂y=z−z=0\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = z - z = 0. Since ∇×F=0\nabla \times \mathbf{F} = \mathbf{0}, the field is conservative. To find the potential function: ∂f∂x=yz+2x\frac{\partial f}{\partial x} = yz + 2x gives f=xyz+x2+g(y,z)f = xyz + x^2 + g(y,z). Then ∂f∂y=xz+∂g∂y=xz+y\frac{\partial f}{\partial y} = xz + \frac{\partial g}{\partial y} = xz + y, so ∂g∂y=y\frac{\partial g}{\partial y} = y and g=y22+h(z)g = \frac{y^2}{2} + h(z). Finally, ∂f∂z=xy+h′(z)=xy+z2\frac{\partial f}{\partial z} = xy + h'(z) = xy + z^2, so h′(z)=z2h'(z) = z^2 and h(z)=z33h(z) = \frac{z^3}{3}. Choice B has the wrong integration of z2z^2. Choice C incorrectly claims the curl is nonzero. Choice D incorrectly assumes three-dimensional fields cannot be conservative.

Question 7

Consider the vector field G(x,y)=−yi+xjx2+y2\mathbf{G}(x,y) = \frac{-y\mathbf{i} + x\mathbf{j}}{x^2 + y^2} defined on R2∖{(0,0)}\mathbb{R}^2 \setminus \{(0,0)\}. Which statement best describes this field?

  1. G\mathbf{G} is conservative on its entire domain and has potential function f(x,y)=arctan⁡(y/x)f(x,y) = \arctan(y/x)
  2. G\mathbf{G} satisfies ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} but is not conservative due to topological constraints (correct answer)
  3. G\mathbf{G} is not conservative because ∂P∂y≠∂Q∂x\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x} at most points
  4. G\mathbf{G} is conservative on any simply connected subset of its domain
Explanation: Let P=−yx2+y2P = \frac{-y}{x^2 + y^2} and Q=xx2+y2Q = \frac{x}{x^2 + y^2}. Computing partial derivatives: ∂P∂y=−(x2+y2)−(−y)(2y)(x2+y2)2=−x2−y2+2y2(x2+y2)2=y2−x2(x2+y2)2\frac{\partial P}{\partial y} = \frac{-(x^2 + y^2) - (-y)(2y)}{(x^2 + y^2)^2} = \frac{-x^2 - y^2 + 2y^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}. Wait, let me recalculate: ∂P∂y=∂∂y(−yx2+y2)=−(x2+y2)−(−y)(2y)(x2+y2)2=−x2−y2+2y2(x2+y2)2=y2−x2(x2+y2)2\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}\left(\frac{-y}{x^2 + y^2}\right) = \frac{-(x^2 + y^2) - (-y)(2y)}{(x^2 + y^2)^2} = \frac{-x^2 - y^2 + 2y^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}. Similarly, ∂Q∂x=∂∂x(xx2+y2)=(x2+y2)−x(2x)(x2+y2)2=x2+y2−2x2(x2+y2)2=y2−x2(x2+y2)2\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x}{x^2 + y^2}\right) = \frac{(x^2 + y^2) - x(2x)}{(x^2 + y^2)^2} = \frac{x^2 + y^2 - 2x^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}. So ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, but the domain R2∖{(0,0)}\mathbb{R}^2 \setminus \{(0,0)\} is not simply connected (it has a 'hole' at the origin). The line integral ∮CG⋅dr\oint_C \mathbf{G} \cdot d\mathbf{r} around a circle centered at the origin equals 2π≠02\pi \neq 0, so G\mathbf{G} is not conservative on its full domain. Choice A is wrong because the field isn't conservative everywhere. Choice C is wrong because the partial derivatives are equal. Choice D is correct about simply connected subsets, but B is the better answer as it captures the key insight about topological constraints.

Question 8

Suppose F(x,y)=(ax+by)i+(cx+dy)j\mathbf{F}(x,y) = (ax + by)\mathbf{i} + (cx + dy)\mathbf{j} where aa, bb, cc, dd are constants. For what relationship between these constants is F\mathbf{F} conservative?

  1. b=cb = c and a=da = d, ensuring the field has the form of a gradient
  2. ad−bc=0ad - bc = 0, ensuring the determinant condition is satisfied
  3. a+d=0a + d = 0 and b−c=0b - c = 0, making the divergence zero
  4. b=cb = c regardless of the values of aa and dd (correct answer)
Explanation: When you encounter a vector field and need to determine if it's conservative, you're testing whether the field can be written as the gradient of some scalar function. This is a fundamental concept in multivariable calculus with important applications in physics and engineering. For a two-dimensional vector field F(x,y)=P(x,y)i+Q(x,y)j\mathbf{F}(x,y) = P(x,y)\mathbf{i} + Q(x,y)\mathbf{j} to be conservative, it must satisfy the condition ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}. This comes from the fact that mixed partial derivatives are equal for well-behaved functions. In your given field F(x,y)=(ax+by)i+(cx+dy)j\mathbf{F}(x,y) = (ax + by)\mathbf{i} + (cx + dy)\mathbf{j}, we have P(x,y)=ax+byP(x,y) = ax + by and Q(x,y)=cx+dyQ(x,y) = cx + dy. Taking the required partial derivatives: ∂P∂y=b\frac{\partial P}{\partial y} = b and ∂Q∂x=c\frac{\partial Q}{\partial x} = c. For the field to be conservative, we need b=cb = c, regardless of what aa and dd are. Answer A incorrectly suggests you need both b=cb = c and a=da = d. While b=cb = c is necessary, the condition a=da = d is not required for conservativity. Answer B confuses this with determinant conditions from other contexts. Answer C incorrectly relates this to divergence conditions (∇⋅F=0\nabla \cdot \mathbf{F} = 0), which determines if a field is incompressible, not conservative. Remember: for conservative vector fields, focus on the curl being zero. In 2D, this simplifies to checking that the mixed partial derivatives of the components are equal.