Multivariable Calculus Quiz: Fundamental Theorem For Line Integrals
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Fundamental Theorem For Line IntegralsQuestion 1 of 14

The line integral of a conservative vector field F⃗=∇f\vec{F} = \nabla f along a path CC from point A=(1,2,0)A=(1,2,0) to point BB is equal to 20. If the potential function is given by f(x,y,z)=x2y−yz2+5f(x,y,z) = x^2 y - y z^2 + 5, what is the value of f(B)f(B)?

27
22
13
-13
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Fundamental Theorem For Line Integrals

Practice Fundamental Theorem For Line Integrals in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Fundamental Theorem For Line Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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Question 1

The line integral of a conservative vector field F⃗=∇f\vec{F} = \nabla f along a path CC from point A=(1,2,0)A=(1,2,0) to point BB is equal to 20. If the potential function is given by f(x,y,z)=x2y−yz2+5f(x,y,z) = x^2 y - y z^2 + 5, what is the value of f(B)f(B)?

  1. 27 (correct answer)
  2. 22
  3. 13
  4. -13
Explanation: According to the Fundamental Theorem for Line Integrals, for a conservative vector field F⃗=∇f\vec{F} = \nabla f, the line integral along a curve CC from point AA to point BB is given by ∫CF⃗⋅dr⃗=f(B)−f(A)\int_C \vec{F} \cdot d\vec{r} = f(B) - f(A). We are given that the value of the integral is 20, so f(B)−f(A)=20f(B) - f(A) = 20. We need to find the value of the potential function ff at the starting point A=(1,2,0)A=(1,2,0). f(A)=f(1,2,0)=(1)2(2)−(2)(0)2+5=2−0+5=7f(A) = f(1,2,0) = (1)^2(2) - (2)(0)^2 + 5 = 2 - 0 + 5 = 7. Now we can solve for f(B)f(B): f(B)−7=20f(B) - 7 = 20. f(B)=20+7=27f(B) = 20 + 7 = 27.

Question 2

Let F⃗(x,y)=⟨−yx2+y2,xx2+y2⟩\vec{F}(x, y) = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle. Evaluate the line integral ∮CF⃗⋅dr⃗\oint_C \vec{F} \cdot d\vec{r}, where CC is the circle x2+y2=9x^2+y^2=9 traversed counter-clockwise.

  1. 00
  2. 2π2\pi (correct answer)
  3. −2π-2\pi
  4. 18π18\pi
Explanation: Let P=−yx2+y2P = \frac{-y}{x^2+y^2} and Q=xx2+y2Q = \frac{x}{x^2+y^2}. We can check the condition for a conservative field: ∂Q∂x=(x2+y2)(1)−x(2x)(x2+y2)2=y2−x2(x2+y2)2\frac{\partial Q}{\partial x} = \frac{(x^2+y^2)(1) - x(2x)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2} ∂P∂y=(x2+y2)(−1)−(−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)(-1) - (-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} Since ∂Q∂x=∂P∂y\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}, the field satisfies the component condition for being conservative. However, the Fundamental Theorem for Line Integrals over a closed path only applies if the vector field is conservative on a region containing the path. The vector field F⃗\vec{F} is undefined at the origin (0,0)(0,0), which is inside the circle CC. Since the domain R2∖{(0,0)}\mathbb{R}^2 \setminus \{(0,0)\} is not simply connected, we cannot conclude that F⃗\vec{F} is conservative or that the integral is zero. We must evaluate the integral by parameterization. Let CC be parameterized by r⃗(t)=⟨3cos⁡t,3sin⁡t⟩\vec{r}(t) = \langle 3\cos t, 3\sin t \rangle for t∈[0,2π]t \in [0, 2\pi]. Then dr⃗=r⃗′(t)dt=⟨−3sin⁡t,3cos⁡t⟩dtd\vec{r} = \vec{r}'(t) dt = \langle -3\sin t, 3\cos t \rangle dt. On the path CC, x2+y2=9x^2+y^2 = 9. So, F⃗(r⃗(t))=⟨−3sin⁡t9,3cos⁡t9⟩=⟨−13sin⁡t,13cos⁡t⟩\vec{F}(\vec{r}(t)) = \left\langle \frac{-3\sin t}{9}, \frac{3\cos t}{9} \right\rangle = \left\langle -\frac{1}{3}\sin t, \frac{1}{3}\cos t \right\rangle. The integral is: ∫02πF⃗(r⃗(t))⋅r⃗′(t) dt=∫02π⟨−13sin⁡t,13cos⁡t⟩⋅⟨−3sin⁡t,3cos⁡t⟩ dt\int_0^{2\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt = \int_0^{2\pi} \left\langle -\frac{1}{3}\sin t, \frac{1}{3}\cos t \right\rangle \cdot \langle -3\sin t, 3\cos t \rangle \, dt =∫02π(sin⁡2t+cos⁡2t) dt=∫02π1 dt=2π= \int_0^{2\pi} (\sin^2 t + \cos^2 t) \, dt = \int_0^{2\pi} 1 \, dt = 2\pi.

Question 3

Let f(x,y,z)=xcos⁡(y)+zf(x,y,z) = x \cos(y) + z and let F⃗=∇f\vec{F} = \nabla f. Evaluate the line integral ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r}, where CC is the helix r⃗(t)=⟨cos⁡(t),sin⁡(t),t⟩\vec{r}(t) = \langle \cos(t), \sin(t), t \rangle from t=0t=0 to t=π/2t=\pi/2.

  1. 00
  2. 1−π21 - \frac{\pi}{2}
  3. π2\frac{\pi}{2}
  4. π2−1\frac{\pi}{2} - 1 (correct answer)
Explanation: When you encounter a line integral of a gradient vector field, immediately think about the Fundamental Theorem for Line Integrals. Since F⃗=∇f\vec{F} = \nabla f, this integral equals simply f(endpoint)−f(starting point)f(\text{endpoint}) - f(\text{starting point}), regardless of the path taken. First, find the endpoints of your helix. At t=0t = 0: r⃗(0)=⟨1,0,0⟩\vec{r}(0) = \langle 1, 0, 0 \rangle. At t=π/2t = \pi/2: r⃗(π/2)=⟨0,1,π/2⟩\vec{r}(\pi/2) = \langle 0, 1, \pi/2 \rangle. Now evaluate the scalar function f(x,y,z)=xcos⁡(y)+zf(x,y,z) = x\cos(y) + z at these points:
  • f(1,0,0)=1⋅cos⁡(0)+0=1⋅1+0=1f(1,0,0) = 1 \cdot \cos(0) + 0 = 1 \cdot 1 + 0 = 1
  • f(0,1,π/2)=0⋅cos⁡(1)+π/2=0+π/2=π/2f(0,1,\pi/2) = 0 \cdot \cos(1) + \pi/2 = 0 + \pi/2 = \pi/2
Therefore: ∫CF⃗⋅dr⃗=f(0,1,π/2)−f(1,0,0)=π/2−1\int_C \vec{F} \cdot d\vec{r} = f(0,1,\pi/2) - f(1,0,0) = \pi/2 - 1 Answer A (00) suggests incorrectly that this is a closed curve or that the function values are equal. Answer B (1−π/21 - \pi/2) reverses the subtraction order—a common error when applying the fundamental theorem. Answer C (π/2\pi/2) forgets to subtract the initial value, using only the final endpoint. Key strategy: When you see F⃗=∇f\vec{F} = \nabla f in a line integral, skip the parametric integration entirely. Just evaluate the scalar function at the endpoints and subtract. This saves enormous time and eliminates computational errors from differentiating the parametric curve.

Question 4

Evaluate ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r} where F⃗(x,y,z)=⟨2xy,x2+z2,2yz⟩\vec{F}(x,y,z) = \langle 2xy, x^2 + z^2, 2yz \rangle and CC is the curve of intersection of the parabolic cylinder y=x2y=x^2 and the plane z=xz=x from the point (0,0,0)(0,0,0) to the point (2,4,2)(2,4,2).

  1. 16
  2. 24
  3. 32 (correct answer)
  4. 40
Explanation: First, we check if the vector field F⃗\vec{F} is conservative by computing its curl. ∇×F⃗=∣i^j^k^∂∂x∂∂y∂∂z2xyx2+z22yz∣=i^(2z−2z)−j^(0−0)+k^(2x−2x)=0⃗\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2xy & x^2+z^2 & 2yz \end{vmatrix} = \hat{i}(2z-2z) - \hat{j}(0-0) + \hat{k}(2x-2x) = \vec{0}. Since the curl is zero and the domain is R3\mathbb{R}^3, the field is conservative. We can find a potential function f(x,y,z)f(x,y,z). f=∫2xy dx=x2y+g(y,z)f = \int 2xy \, dx = x^2y + g(y,z). ∂f∂y=x2+∂g∂y=x2+z2  ⟹  ∂g∂y=z2  ⟹  g(y,z)=yz2+h(z)\frac{\partial f}{\partial y} = x^2 + \frac{\partial g}{\partial y} = x^2+z^2 \implies \frac{\partial g}{\partial y} = z^2 \implies g(y,z) = yz^2 + h(z). So f=x2y+yz2+h(z)f = x^2y + yz^2 + h(z). ∂f∂z=2yz+h′(z)=2yz  ⟹  h′(z)=0\frac{\partial f}{\partial z} = 2yz + h'(z) = 2yz \implies h'(z)=0. Thus, a potential function is f(x,y,z)=x2y+yz2f(x,y,z) = x^2y + yz^2. By the Fundamental Theorem for Line Integrals, the integral's value depends only on the endpoints. The complex path description is extraneous information. Start point: A=(0,0,0)A=(0,0,0). End point: B=(2,4,2)B=(2,4,2). Value = f(B)−f(A)=f(2,4,2)−f(0,0,0)f(B) - f(A) = f(2,4,2) - f(0,0,0). f(2,4,2)=(22)(4)+(4)(22)=(4)(4)+(4)(4)=16+16=32f(2,4,2) = (2^2)(4) + (4)(2^2) = (4)(4) + (4)(4) = 16 + 16 = 32. f(0,0,0)=0f(0,0,0) = 0. The value of the integral is 32−0=3232-0=32.

Question 5

Let F⃗1=⟨y,−x⟩\vec{F}_1 = \langle y, -x \rangle and F⃗2=⟨x,y⟩\vec{F}_2 = \langle x, y \rangle. Let C1C_1 be the line segment from (1,0)(1,0) to (−1,0)(-1,0) and C2C_2 be the upper semicircle from (1,0)(1,0) to (−1,0)(-1,0). Define the integrals I1=∫C1F⃗1⋅dr⃗I_1 = \int_{C_1} \vec{F}_1 \cdot d\vec{r}, I2=∫C2F⃗1⋅dr⃗I_2 = \int_{C_2} \vec{F}_1 \cdot d\vec{r}, J1=∫C1F⃗2⋅dr⃗J_1 = \int_{C_1} \vec{F}_2 \cdot d\vec{r}, and J2=∫C2F⃗2⋅dr⃗J_2 = \int_{C_2} \vec{F}_2 \cdot d\vec{r}. Which statement is correct?

  1. I1=I2I_1 = I_2 and J1=J2J_1 = J_2
  2. I1=I2I_1 = I_2 and J1≠J2J_1 \neq J_2
  3. I1≠I2I_1 \neq I_2 and J1=J2J_1 = J_2 (correct answer)
  4. I1≠I2I_1 \neq I_2 and J1≠J2J_1 \neq J_2
Explanation: When you encounter line integrals of vector fields along different paths, you should immediately think about conservative vector fields and path independence. A vector field is conservative if its line integral depends only on the endpoints, not the path taken between them. To determine if a 2D vector field F⃗=⟨P,Q⟩\vec{F} = \langle P, Q \rangle is conservative, check if ∂P∂y=∂Q∂x\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}. For F⃗1=⟨y,−x⟩\vec{F}_1 = \langle y, -x \rangle: ∂∂y(y)=1\frac{\partial}{\partial y}(y) = 1 and ∂∂x(−x)=−1\frac{\partial}{\partial x}(-x) = -1. Since 1≠−11 \neq -1, F⃗1\vec{F}_1 is not conservative, so I1≠I2I_1 \neq I_2. For F⃗2=⟨x,y⟩\vec{F}_2 = \langle x, y \rangle: ∂∂y(x)=0\frac{\partial}{\partial y}(x) = 0 and ∂∂x(y)=0\frac{\partial}{\partial x}(y) = 0. Since 0=00 = 0, F⃗2\vec{F}_2 is conservative, so J1=J2J_1 = J_2. Now examining the choices: Option A claims both pairs are equal, but we showed I1≠I2I_1 \neq I_2. Option B claims I1=I2I_1 = I_2, which is false. Option C correctly states I1≠I2I_1 \neq I_2 and J1=J2J_1 = J_2, matching our analysis. Option D claims both pairs are unequal, but J1=J2J_1 = J_2 since F⃗2\vec{F}_2 is conservative. Study tip: Always test for conservative vector fields first using the partial derivative condition. Conservative fields give path-independent integrals, while non-conservative fields generally give different values along different paths. This saves time compared to computing the integrals directly.

Question 6

Let F⃗\vec{F} be a continuously differentiable vector field on R3\mathbb{R}^3. The line integral ∫C1F⃗⋅dr⃗\int_{C_1} \vec{F} \cdot d\vec{r} along a specific path C1C_1 from point AA to point BB is found to be 5. To guarantee that the line integral along any smooth path from AA to BB is also 5, which additional condition must be met?

  1. The divergence of F⃗\vec{F} is zero everywhere.
  2. The curl of F⃗\vec{F} is zero everywhere. (correct answer)
  3. The magnitude of F⃗\vec{F} is constant.
  4. The path must be a straight line segment.
Explanation: The Fundamental Theorem for Line Integrals states that the line integral of a conservative vector field depends only on the endpoints of the path, not the path itself. For a vector field defined on a simply connected domain like R3\mathbb{R}^3, the condition for it to be conservative is that its curl is zero. If ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0} everywhere, then F⃗\vec{F} is conservative. This implies that the line integral ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r} is path-independent. Therefore, if we know the integral is 5 for one path C1C_1 from AA to BB, it will be 5 for any other path from AA to BB. Choice A: Zero divergence is related to incompressible fields and the Divergence Theorem, not path independence of line integrals. Choice C: Constant magnitude does not imply the field is conservative. Choice D: Restricting the path does not provide a condition on the vector field that ensures path independence for all possible paths.

Question 7

Let F⃗(r⃗)=r⃗∣r⃗∣3\vec{F}(\vec{r}) = \frac{\vec{r}}{|\vec{r}|^3} be a vector field in R3\mathbb{R}^3, where r⃗=⟨x,y,z⟩\vec{r} = \langle x,y,z \rangle. Evaluate the line integral of F⃗\vec{F} along a smooth curve CC from point A=(1,2,2)A=(1,2,2) to point B=(3,0,4)B=(3,0,4) that does not pass through the origin.

  1. −2-2
  2. −2/15-2/15
  3. 22
  4. 2/152/15 (correct answer)
Explanation: When you encounter a line integral with a vector field that resembles an inverse square law (like F⃗(r⃗)=r⃗∣r⃗∣3\vec{F}(\vec{r}) = \frac{\vec{r}}{|\vec{r}|^3}), always check if the field is conservative. This particular field is the gradient of a scalar potential function ϕ(r⃗)=−1∣r⃗∣\phi(\vec{r}) = -\frac{1}{|\vec{r}|}, making it conservative everywhere except at the origin. Since the curve doesn't pass through the origin, we can use the Fundamental Theorem for Line Integrals. For a conservative field, the line integral depends only on the endpoints: ∫CF⃗⋅dr⃗=ϕ(B)−ϕ(A)\int_C \vec{F} \cdot d\vec{r} = \phi(B) - \phi(A). Let's calculate the potential at each point. At A=(1,2,2)A = (1,2,2): ∣r⃗A∣=12+22+22=3|\vec{r}_A| = \sqrt{1^2 + 2^2 + 2^2} = 3, so ϕ(A)=−13\phi(A) = -\frac{1}{3}. At B=(3,0,4)B = (3,0,4): ∣r⃗B∣=32+02+42=5|\vec{r}_B| = \sqrt{3^2 + 0^2 + 4^2} = 5, so ϕ(B)=−15\phi(B) = -\frac{1}{5}. Therefore: ∫CF⃗⋅dr⃗=−15−(−13)=13−15=5−315=215\int_C \vec{F} \cdot d\vec{r} = -\frac{1}{5} - \left(-\frac{1}{3}\right) = \frac{1}{3} - \frac{1}{5} = \frac{5-3}{15} = \frac{2}{15} A) -2 likely comes from incorrectly using ϕ=1∣r⃗∣\phi = \frac{1}{|\vec{r}|} (wrong sign) and making arithmetic errors. B) -2/15 uses the correct calculation but with the wrong sign for the potential. C) 2 suggests using the wrong potential function entirely, perhaps confusing this with a different type of field. Strategy tip: Always verify if a vector field is conservative before computing line integrals the hard way. Inverse square fields are almost always conservative away from singularities.

Question 8

Consider the vector field F⃗=⟨3x2+yz,3y2+xz,3z2+xy⟩\vec{F} = \langle 3x^2 + yz, 3y^2 + xz, 3z^2 + xy \rangle. Let CC be the curve parameterized by r⃗(t)=⟨cos⁡(πt),sin⁡(πt),cos⁡(2πt)⟩\vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), \cos(2\pi t) \rangle for t∈[0,2]t \in [0, 2]. Which of the following is true about the line integral ∫CF⃗⋅dr⃗\int_C \vec{F} \cdot d\vec{r}?

  1. The integral is 0 because CC is a closed curve.
  2. The integral is 0 because F⃗\vec{F} is conservative and CC is a closed curve. (correct answer)
  3. The integral is non-zero and must be computed by direct parameterization.
  4. The integral's value depends on the orientation of the surface bounded by CC.
Explanation: First, we determine if the vector field F⃗\vec{F} is conservative by checking if its curl is zero. Let F⃗=⟨P,Q,R⟩\vec{F} = \langle P, Q, R \rangle. ∂R∂y=x\frac{\partial R}{\partial y} = x, ∂Q∂z=x\frac{\partial Q}{\partial z} = x. So they are equal. ∂P∂z=y\frac{\partial P}{\partial z} = y, ∂R∂x=y\frac{\partial R}{\partial x} = y. So they are equal. ∂Q∂x=z\frac{\partial Q}{\partial x} = z, ∂P∂y=z\frac{\partial P}{\partial y} = z. So they are equal. Since ∇×F⃗=0⃗\nabla \times \vec{F} = \vec{0} and the domain is R3\mathbb{R}^3, F⃗\vec{F} is a conservative vector field. Next, we determine if the curve CC is a closed loop by checking if its start and end points are the same. Start point: r⃗(0)=⟨cos⁡(0),sin⁡(0),cos⁡(0)⟩=⟨1,0,1⟩\vec{r}(0) = \langle \cos(0), \sin(0), \cos(0) \rangle = \langle 1, 0, 1 \rangle. End point: r⃗(2)=⟨cos⁡(2π),sin⁡(2π),cos⁡(4π)⟩=⟨1,0,1⟩\vec{r}(2) = \langle \cos(2\pi), \sin(2\pi), \cos(4\pi) \rangle = \langle 1, 0, 1 \rangle. Since the start and end points are identical, the curve CC is a closed loop. The Fundamental Theorem for Line Integrals states that the line integral of a conservative vector field over a closed loop is zero. Therefore, ∫CF⃗⋅dr⃗=0\int_C \vec{F} \cdot d\vec{r} = 0. Choice B provides the correct value and the complete reasoning. Choice A is incomplete because a line integral over a closed loop is not always zero; the field must also be conservative.

Question 9

Let F(x,y)=(2xy+ex,x2+cos⁡y)\mathbf{F}(x,y) = (2xy + e^x, x^2 + \cos y) and let CC be any piecewise smooth curve from (0,0)(0,0) to (1,π)(1,\pi). If f(x,y)f(x,y) is a potential function for F\mathbf{F}, which of the following correctly represents ∫CF⋅dr\int_C \mathbf{F} \cdot d\mathbf{r}?

  1. f(1,π)−f(0,0)f(1,\pi) - f(0,0) where f(x,y)=x2y+ex+sin⁡y+Cf(x,y) = x^2y + e^x + \sin y + C (correct answer)
  2. f(1,π)−f(0,0)f(1,\pi) - f(0,0) where f(x,y)=x2y+ex−sin⁡y+Cf(x,y) = x^2y + e^x - \sin y + C
  3. f(0,0)−f(1,π)f(0,0) - f(1,\pi) where f(x,y)=x2y+ex+sin⁡y+Cf(x,y) = x^2y + e^x + \sin y + C
  4. The integral cannot be evaluated using a potential function since F\mathbf{F} is not conservative
Explanation: First, we verify that F\mathbf{F} is conservative by checking that ∂F1∂y=∂F2∂x\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}. We have ∂∂y(2xy+ex)=2x\frac{\partial}{\partial y}(2xy + e^x) = 2x and ∂∂x(x2+cos⁡y)=2x\frac{\partial}{\partial x}(x^2 + \cos y) = 2x. Since these are equal, F\mathbf{F} is conservative. To find the potential function, we integrate: ∂f∂x=2xy+ex\frac{\partial f}{\partial x} = 2xy + e^x gives f(x,y)=x2y+ex+g(y)f(x,y) = x^2y + e^x + g(y). Then ∂f∂y=x2+g′(y)=x2+cos⁡y\frac{\partial f}{\partial y} = x^2 + g'(y) = x^2 + \cos y, so g′(y)=cos⁡yg'(y) = \cos y and g(y)=sin⁡y+Cg(y) = \sin y + C. Thus f(x,y)=x2y+ex+sin⁡y+Cf(x,y) = x^2y + e^x + \sin y + C. By the Fundamental Theorem for Line Integrals, ∫CF⋅dr=f(1,π)−f(0,0)\int_C \mathbf{F} \cdot d\mathbf{r} = f(1,\pi) - f(0,0). Choice B has the wrong sign for sin⁡y\sin y, Choice C reverses the order of evaluation, and Choice D incorrectly claims the field is not conservative.

Question 10

Let F(x,y)=(P(x,y),Q(x,y))\mathbf{F}(x,y) = (P(x,y), Q(x,y)) be a conservative vector field in a simply connected region DD, and let ff be a potential function for F\mathbf{F}. If ∫CF⋅dr=15\int_C \mathbf{F} \cdot d\mathbf{r} = 15 where CC is a curve from point A(1,2)A(1,2) to point B(3,4)B(3,4), what is ∫C′F⋅dr\int_{C'} \mathbf{F} \cdot d\mathbf{r} where C′C' is any curve from B(3,4)B(3,4) to A(1,2)A(1,2)?

  1. −15-15, because line integrals of conservative fields depend only on endpoints (correct answer)
  2. 1515, because the fundamental theorem applies regardless of direction
  3. 00, because C′C' combined with CC forms a closed loop
  4. Cannot be determined without knowing the specific potential function ff
Explanation: Since F\mathbf{F} is conservative with potential function ff, the Fundamental Theorem for Line Integrals states that ∫CF⋅dr=f(B)−f(A)=f(3,4)−f(1,2)=15\int_C \mathbf{F} \cdot d\mathbf{r} = f(B) - f(A) = f(3,4) - f(1,2) = 15. For the curve C′C' from BB to AA, we have ∫C′F⋅dr=f(A)−f(B)=f(1,2)−f(3,4)=−(f(3,4)−f(1,2))=−15\int_{C'} \mathbf{F} \cdot d\mathbf{r} = f(A) - f(B) = f(1,2) - f(3,4) = -(f(3,4) - f(1,2)) = -15. Choice B incorrectly ignores the direction change. Choice C confuses the result with the fact that the integral around a closed loop is zero (which would be ∫CF⋅dr+∫C′F⋅dr=0\int_C \mathbf{F} \cdot d\mathbf{r} + \int_{C'} \mathbf{F} \cdot d\mathbf{r} = 0). Choice D incorrectly suggests we need the explicit form of ff, when the fundamental theorem gives us the relationship directly.

Question 11

Consider two potential functions f1(x,y)=x2y+xy2+5f_1(x,y) = x^2y + xy^2 + 5 and f2(x,y)=x2y+xy2−3f_2(x,y) = x^2y + xy^2 - 3. Let F1=∇f1\mathbf{F}_1 = \nabla f_1 and F2=∇f2\mathbf{F}_2 = \nabla f_2. If ∫C1F1⋅dr=7\int_{C_1} \mathbf{F}_1 \cdot d\mathbf{r} = 7 where C1C_1 goes from AA to BB, and ∫C2F2⋅dr=k\int_{C_2} \mathbf{F}_2 \cdot d\mathbf{r} = k where C2C_2 goes from BB to AA, what is the value of kk?

  1. k=1k = 1 since the constant terms affect the final integral value
  2. k=−15k = -15 since we must account for both the constant difference and direction
  3. k=−7k = -7 since the fields differ only by a constant and direction is reversed (correct answer)
  4. k=7k = 7 since gradients of functions differing by constants are identical
Explanation: This question tests your understanding of conservative vector fields and the Fundamental Theorem for Line Integrals. When you see gradient fields and line integrals between specific points, immediately think about how the integral depends only on the endpoints, not the path. First, let's analyze what we know. Both F1\mathbf{F}_1 and F2\mathbf{F}_2 are gradient fields (conservative), so line integrals depend only on the potential function values at the endpoints. For any conservative field F=∇f\mathbf{F} = \nabla f, we have ∫CF⋅dr=f(endpoint)−f(starting point)\int_C \mathbf{F} \cdot d\mathbf{r} = f(\text{endpoint}) - f(\text{starting point}). Since f1f_1 and f2f_2 differ only by a constant (f1=f2+8f_1 = f_2 + 8), their gradients are identical: ∇f1=∇f2\nabla f_1 = \nabla f_2. This means F1=F2\mathbf{F}_1 = \mathbf{F}_2. For the first integral: ∫C1F1⋅dr=f1(B)−f1(A)=7\int_{C_1} \mathbf{F}_1 \cdot d\mathbf{r} = f_1(B) - f_1(A) = 7 For the second integral, C2C_2 goes from BB to AA (opposite direction), so: ∫C2F2⋅dr=f2(A)−f2(B)\int_{C_2} \mathbf{F}_2 \cdot d\mathbf{r} = f_2(A) - f_2(B) Since F1=F2\mathbf{F}_1 = \mathbf{F}_2 and the direction is reversed, k=−7k = -7. Answer A incorrectly assumes constants affect gradients. Answer B wrongly incorporates the constant difference into the calculation. Answer D misses that reversing direction changes the sign of the integral. Key takeaway: When dealing with conservative fields, remember that adding constants to potential functions doesn't change their gradients, and reversing the path direction flips the integral's sign.

Question 12

A particle moves through a conservative force field F\mathbf{F} with potential energy function U(x,y,z)=−f(x,y,z)U(x,y,z) = -f(x,y,z) where ff is the potential function for F\mathbf{F}. The particle moves from point P(1,0,2)P(1,0,2) where U=10U = 10 to point Q(3,1,0)Q(3,1,0) where U=4U = 4. If the particle's kinetic energy at PP is KP=8K_P = 8, what is its kinetic energy KQK_Q at point QQ?

  1. KQ=2K_Q = 2 by conservation of mechanical energy since KP+UP=KQ+UQK_P + U_P = K_Q + U_Q
  2. KQ=14K_Q = 14 by conservation of mechanical energy since KP+UP=KQ+UQK_P + U_P = K_Q + U_Q (correct answer)
  3. KQ=14K_Q = 14 by the work-energy theorem since work done equals UP−UQU_P - U_Q
  4. KQ=2K_Q = 2 by the work-energy theorem since work done equals UQ−UPU_Q - U_P
Explanation: When you encounter problems involving particles moving through conservative force fields, the key principle is conservation of mechanical energy. In a conservative field, the total mechanical energy (kinetic + potential) remains constant throughout the particle's motion. For this particle, we can apply conservation of energy: KP+UP=KQ+UQK_P + U_P = K_Q + U_Q. Substituting the given values: 8+10=KQ+48 + 10 = K_Q + 4, which gives us 18=KQ+418 = K_Q + 4, so KQ=14K_Q = 14. Looking at the answer choices: Choice A incorrectly calculates KQ=2K_Q = 2, which would require 8+10=2+4=68 + 10 = 2 + 4 = 6, violating energy conservation. Choice C reaches the correct numerical answer of KQ=14K_Q = 14 but justifies it incorrectly using the work-energy theorem. While the work-energy theorem is valid (work done by conservative forces equals the negative change in potential energy), the statement "work done equals UP−UQU_P - U_Q" is backwards—work done should equal −(UQ−UP)=−(4−10)=6-(U_Q - U_P) = -(4-10) = 6. Choice D makes both errors: wrong calculation and incorrect work-energy application. Choice B correctly applies conservation of mechanical energy with the right calculation, giving KQ=14K_Q = 14. Study tip: For conservative force problems, always start with energy conservation: Kinitial+Uinitial=Kfinal+UfinalK_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}}. Remember that when potential energy decreases, kinetic energy must increase by the same amount to maintain total energy.

Question 13

A vector field F\mathbf{F} is defined on R3\mathbb{R}^3 such that ∇×F=0\nabla \times \mathbf{F} = \mathbf{0} and ∇⋅F=x+y+z\nabla \cdot \mathbf{F} = x + y + z. If ff is a potential function for F\mathbf{F} and ∫CF⋅dr=12\int_C \mathbf{F} \cdot d\mathbf{r} = 12 where CC goes from (1,1,1)(1,1,1) to (3,2,4)(3,2,4), which statement about the potential function is necessarily true?

  1. The given conditions are contradictory since ∇×F=0\nabla \times \mathbf{F} = \mathbf{0} implies ∇⋅F=0\nabla \cdot \mathbf{F} = 0
  2. f(3,2,4)−f(1,1,1)=12f(3,2,4) - f(1,1,1) = 12 but ∇2f=0\nabla^2 f = 0 since curl is zero
  3. f(1,1,1)−f(3,2,4)=12f(1,1,1) - f(3,2,4) = 12 and ∇2f=x+y+z\nabla^2 f = x + y + z
  4. f(3,2,4)−f(1,1,1)=12f(3,2,4) - f(1,1,1) = 12 and ∇2f=x+y+z\nabla^2 f = x + y + z (correct answer)
Explanation: When you encounter vector field problems involving curl, divergence, and potential functions, you need to understand how these concepts relate to each other and what conditions must hold for conservative fields. Since ∇×F=0\nabla \times \mathbf{F} = \mathbf{0}, the vector field F\mathbf{F} is conservative, which means it has a potential function ff where F=∇f\mathbf{F} = \nabla f. For conservative fields, the fundamental theorem for line integrals tells us that ∫CF⋅dr=f(endpoint)−f(startpoint)\int_C \mathbf{F} \cdot d\mathbf{r} = f(\text{endpoint}) - f(\text{startpoint}). Therefore, ∫CF⋅dr=f(3,2,4)−f(1,1,1)=12\int_C \mathbf{F} \cdot d\mathbf{r} = f(3,2,4) - f(1,1,1) = 12. Additionally, since F=∇f\mathbf{F} = \nabla f, we have ∇⋅F=∇⋅(∇f)=∇2f\nabla \cdot \mathbf{F} = \nabla \cdot (\nabla f) = \nabla^2 f. Given that ∇⋅F=x+y+z\nabla \cdot \mathbf{F} = x + y + z, we conclude ∇2f=x+y+z\nabla^2 f = x + y + z. Choice A is wrong because curl and divergence are independent properties—a zero curl doesn't require zero divergence. Choice B incorrectly states that ∇2f=0\nabla^2 f = 0, confusing the condition for harmonic functions with the general relationship between divergence and the Laplacian of the potential. Choice C has the correct Laplacian relationship but reverses the line integral formula, giving f(1,1,1)−f(3,2,4)=12f(1,1,1) - f(3,2,4) = 12 instead of the correct f(3,2,4)−f(1,1,1)=12f(3,2,4) - f(1,1,1) = 12. Remember: for conservative vector fields, always use ∇⋅F=∇2f\nabla \cdot \mathbf{F} = \nabla^2 f and the fundamental theorem ∫CF⋅dr=f(end)−f(start)\int_C \mathbf{F} \cdot d\mathbf{r} = f(\text{end}) - f(\text{start}).

Question 14

Calculate the work done by the force field F⃗(x,y,z)=⟨yz+2x,xz,xy⟩\vec{F}(x,y,z) = \langle yz+2x, xz, xy \rangle on a particle that moves along a path from (2,1,1)(2,1,1) to (1,2,3)(1,2,3).

  1. 1 (correct answer)
  2. -1
  3. 4
  4. -3
Explanation: The work done is given by the line integral W=∫CF⃗⋅dr⃗W = \int_C \vec{F} \cdot d\vec{r}. We first check if the vector field F⃗\vec{F} is conservative by calculating its curl. ∇×F⃗=∣i^j^k^∂∂x∂∂y∂∂zyz+2xxzxy∣=i^(∂∂y(xy)−∂∂z(xz))−j^(∂∂x(xy)−∂∂z(yz+2x))+k^(∂∂x(xz)−∂∂y(yz+2x))=i^(x−x)−j^(y−y)+k^(z−z)=0⃗\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ yz+2x & xz & xy \end{vmatrix} = \hat{i}(\frac{\partial}{\partial y}(xy) - \frac{\partial}{\partial z}(xz)) - \hat{j}(\frac{\partial}{\partial x}(xy) - \frac{\partial}{\partial z}(yz+2x)) + \hat{k}(\frac{\partial}{\partial x}(xz) - \frac{\partial}{\partial y}(yz+2x)) = \hat{i}(x-x) - \hat{j}(y-y) + \hat{k}(z-z) = \vec{0}. Since the curl is zero and the domain is R3\mathbb{R}^3, the field is conservative. We can find a potential function ff such that ∇f=F⃗\nabla f = \vec{F}. f(x,y,z)=∫(yz+2x) dx=xyz+x2+g(y,z)f(x,y,z) = \int (yz+2x) \, dx = xyz + x^2 + g(y,z). ∂f∂y=xz+∂g∂y=xz  ⟹  ∂g∂y=0\frac{\partial f}{\partial y} = xz + \frac{\partial g}{\partial y} = xz \implies \frac{\partial g}{\partial y} = 0, so gg is a function of zz alone. ∂f∂z=xy+g′(z)=xy  ⟹  g′(z)=0\frac{\partial f}{\partial z} = xy + g'(z) = xy \implies g'(z) = 0, so gg is a constant. We can choose g=0g=0. The potential function is f(x,y,z)=xyz+x2f(x,y,z) = xyz + x^2. By the Fundamental Theorem for Line Integrals, the work is f(end point)−f(start point)f(\text{end point}) - f(\text{start point}). Start point A=(2,1,1)A = (2,1,1), end point B=(1,2,3)B = (1,2,3). f(B)=f(1,2,3)=(1)(2)(3)+(1)2=6+1=7f(B) = f(1,2,3) = (1)(2)(3) + (1)^2 = 6+1=7. f(A)=f(2,1,1)=(2)(1)(1)+(2)2=2+4=6f(A) = f(2,1,1) = (2)(1)(1) + (2)^2 = 2+4=6. Work = f(B)−f(A)=7−6=1f(B) - f(A) = 7 - 6 = 1.