Conservative Fields & Potentials - Multivariable Calculus

1. Suppose $\mathbf{F}$ is a vector field such that $\int_C \mathbf{F} \cdot d\mathbf{r} = 5$ where $C$ is the path from $(0,0)$ to $(2,1)$ along the curve $y = \frac{x^2}{4}$, and $\int_D \mathbf{F} \cdot d\mathbf{r} = 3$ where $D$ is the straight line from $(0,0)$ to $(2,1)$. What can you conclude about $\mathbf{F}$?
   Answer: $\mathbf{F}$ is definitely not conservative because the line integrals have different values If a vector field $\mathbf{F}$ is conservative, then line integrals depend only on the endpoints, not on the path taken. Since we have two different paths $C$ and $D$ with the same endpoints $(0,0)$ and $(2,1)$, but $\int_C \mathbf{F} \cdot d\mathbf{r} = 5 \neq 3 = \int_D \mathbf{F} \cdot d\mathbf{r}$, the field cannot be conservative (assuming both paths lie in a region where $\mathbf{F}$ 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.
2. Consider a vector field $\mathbf{F}(x,y) = P(x,y)\mathbf{i} + Q(x,y)\mathbf{j}$ defined on a simply connected domain $D$. Given that $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ throughout $D$, and that $\mathbf{F}(1,0) = 3\mathbf{i} + 2\mathbf{j}$, what additional information is needed to uniquely determine a potential function $f$ such that $\nabla f = \mathbf{F}$?
   Answer: The value of $f$ at one point in $D$ is sufficient to determine $f$ uniquely 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 $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ tells you that $\mathbf{F}$ is conservative on the simply connected domain $D$, which guarantees that a potential function $f$ exists such that $\nabla f = \mathbf{F}$. The key insight is understanding what "uniquely determine" means for potential functions. Since $\nabla f = \mathbf{F}$, we know that $\frac{\partial f}{\partial x} = P(x,y)$ and $\frac{\partial f}{\partial y} = Q(x,y)$. This system of partial differential equations has infinitely many solutions that differ only by a constant. To pin down one specific function $f$, you need exactly one additional piece of information: the value of $f$ at any single point in $D$. 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 $\mathbf{F}$ (which would be the Hessian of $f$) provides information about second derivatives, not the constant of integration. Option B misunderstands the problem—knowing $\mathbf{F}$ at multiple points doesn't help since we already know $\mathbf{F}$ throughout $D$. 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.
3. Consider the vector field $\mathbf{F}(x,y) = (2xy + y^2)\mathbf{i} + (x^2 + 2xy + 3)\mathbf{j}$. Which statement about this field is correct?
   Answer: The field is conservative with potential function $f(x,y) = x^2y + xy^2 + 3y + C$ First check if the field is conservative by testing if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. Here $P = 2xy + y^2$ and $Q = x^2 + 2xy + 3$. We have $\frac{\partial P}{\partial y} = 2x + 2y$ and $\frac{\partial Q}{\partial x} = 2x + 2y$. Since these are equal, the field is conservative. To find the potential function, we integrate: $\frac{\partial f}{\partial x} = 2xy + y^2$ gives $f = x^2y + xy^2 + g(y)$. Then $\frac{\partial f}{\partial y} = x^2 + 2xy + g'(y) = x^2 + 2xy + 3$, so $g'(y) = 3$ and $g(y) = 3y + C$. Choice B has the wrong integration for the constant term (should be $3y$ not $3x$). Choice C incorrectly claims the partial derivatives are unequal. Choice D incorrectly assumes constant terms prevent conservativity.
4. Suppose $\mathbf{F}(x,y) = (ax + by)\mathbf{i} + (cx + dy)\mathbf{j}$ where $a$, $b$, $c$, $d$ are constants. For what relationship between these constants is $\mathbf{F}$ conservative?
   Answer: $b = c$ regardless of the values of $a$ and $d$ 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 $\mathbf{F}(x,y) = P(x,y)\mathbf{i} + Q(x,y)\mathbf{j}$ to be conservative, it must satisfy the condition $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. This comes from the fact that mixed partial derivatives are equal for well-behaved functions. In your given field $\mathbf{F}(x,y) = (ax + by)\mathbf{i} + (cx + dy)\mathbf{j}$, we have $P(x,y) = ax + by$ and $Q(x,y) = cx + dy$. Taking the required partial derivatives: $\frac{\partial P}{\partial y} = b$ and $\frac{\partial Q}{\partial x} = c$. For the field to be conservative, we need $b = c$, regardless of what $a$ and $d$ are. Answer A incorrectly suggests you need both $b = c$ and $a = d$. While $b = c$ is necessary, the condition $a = d$ is not required for conservativity. Answer B confuses this with determinant conditions from other contexts. Answer C incorrectly relates this to divergence conditions ($\nabla \cdot \mathbf{F} = 0$), 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.
5. Let $\mathbf{F}(x,y,z) = (yz + 2x)\mathbf{i} + (xz + y)\mathbf{j} + (xy + z^2)\mathbf{k}$. Which of the following statements is true?
   Answer: $\mathbf{F}$ is conservative and a potential function is $f(x,y,z) = xyz + x^2 + \frac{y^2}{2} + \frac{z^3}{3}$ To check if $\mathbf{F}$ is conservative, we compute $\nabla \times \mathbf{F}$. With $P = yz + 2x$, $Q = xz + y$, $R = xy + z^2$, we have: $\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = x - x = 0$, $\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = y - y = 0$, $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = z - z = 0$. Since $\nabla \times \mathbf{F} = \mathbf{0}$, the field is conservative. To find the potential function: $\frac{\partial f}{\partial x} = yz + 2x$ gives $f = xyz + x^2 + g(y,z)$. Then $\frac{\partial f}{\partial y} = xz + \frac{\partial g}{\partial y} = xz + y$, so $\frac{\partial g}{\partial y} = y$ and $g = \frac{y^2}{2} + h(z)$. Finally, $\frac{\partial f}{\partial z} = xy + h'(z) = xy + z^2$, so $h'(z) = z^2$ and $h(z) = \frac{z^3}{3}$. Choice B has the wrong integration of $z^2$. Choice C incorrectly claims the curl is nonzero. Choice D incorrectly assumes three-dimensional fields cannot be conservative.
6. Let $f(x,y,z) = xy^2z + x^3 - yz^2$ and consider $\mathbf{F} = \nabla f$. If $C$ is any piecewise smooth curve from point $A$ to point $B$, then $\int_C \mathbf{F} \cdot d\mathbf{r}$ equals:
   Answer: $f(B) - f(A)$ regardless of the path $C$ 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 $\mathbf{F} = \nabla f$, we can apply the Fundamental Theorem for Line Integrals. This theorem states that for any conservative vector field $\mathbf{F} = \nabla f$, the line integral $\int_C \mathbf{F} \cdot d\mathbf{r}$ depends only on the endpoints, not the path. Specifically, $\int_C \mathbf{F} \cdot d\mathbf{r} = f(B) - f(A)$ where $A$ and $B$ are the initial and final points of curve $C$. This makes answer C correct: the integral equals $f(B) - f(A)$ regardless of the path $C$. 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 $\nabla f$ are conservative everywhere they're defined (which is all of $\mathbb{R}^3$ for this polynomial function). Answer B has the wrong sign order. The Fundamental Theorem gives us $f(\text{final}) - f(\text{initial})$, 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 $\mathbf{F} = \nabla f$, immediately think "path independence" and apply the Fundamental Theorem: $\int_C \mathbf{F} \cdot d\mathbf{r} = f(\text{end}) - f(\text{start})$.
7. Consider the vector field $\mathbf{G}(x,y) = \frac{-y\mathbf{i} + x\mathbf{j}}{x^2 + y^2}$ defined on $\mathbb{R}^2 \setminus \{(0,0)\}$. Which statement best describes this field?
   Answer: $\mathbf{G}$ satisfies $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ but is not conservative due to topological constraints Let $P = \frac{-y}{x^2 + y^2}$ and $Q = \frac{x}{x^2 + y^2}$. Computing partial derivatives: $\frac{\partial P}{\partial y} = \frac{-(x^2 + y^2) - (-y)(2y)}{(x^2 + y^2)^2} = \frac{-x^2 - y^2 + 2y^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}$. Wait, let me recalculate: $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}\left(\frac{-y}{x^2 + y^2}\right) = \frac{-(x^2 + y^2) - (-y)(2y)}{(x^2 + y^2)^2} = \frac{-x^2 - y^2 + 2y^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}$. Similarly, $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x}{x^2 + y^2}\right) = \frac{(x^2 + y^2) - x(2x)}{(x^2 + y^2)^2} = \frac{x^2 + y^2 - 2x^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}$. So $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$, but the domain $\mathbb{R}^2 \setminus \{(0,0)\}$ is not simply connected (it has a 'hole' at the origin). The line integral $\oint_C \mathbf{G} \cdot d\mathbf{r}$ around a circle centered at the origin equals $2\pi \neq 0$, so $\mathbf{G}$ 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.
8. A vector field $\mathbf{F}$ has the property that $\nabla \times \mathbf{F} = \mathbf{0}$ everywhere in its domain, which is $\mathbb{R}^3$ minus the $z$-axis. Which statement is most accurate?
   Answer: $\mathbf{F}$ is conservative on any ball that doesn't intersect the $z$-axis 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 $\nabla \times \mathbf{F} = \mathbf{0}$ 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 $\mathbb{R}^3$ minus the $z$-axis, which is not simply connected—you can draw loops around the missing $z$-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 $z$-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 $z$-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.

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