Fundamental Theorem for Line Integrals - Multivariable Calculus

1. Let $\mathbf{F}(x,y) = (P(x,y), Q(x,y))$ be a conservative vector field in a simply connected region $D$, and let $f$ be a potential function for $\mathbf{F}$. If $\int_C \mathbf{F} \cdot d\mathbf{r} = 15$ where $C$ is a curve from point $A(1,2)$ to point $B(3,4)$, what is $\int_{C'} \mathbf{F} \cdot d\mathbf{r}$ where $C'$ is any curve from $B(3,4)$ to $A(1,2)$?
   Answer: $-15$, because line integrals of conservative fields depend only on endpoints Since $\mathbf{F}$ is conservative with potential function $f$, the Fundamental Theorem for Line Integrals states that $\int_C \mathbf{F} \cdot d\mathbf{r} = f(B) - f(A) = f(3,4) - f(1,2) = 15$. For the curve $C'$ from $B$ to $A$, we have $\int_{C'} \mathbf{F} \cdot d\mathbf{r} = f(A) - f(B) = f(1,2) - f(3,4) = -(f(3,4) - f(1,2)) = -15$. 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 $\int_C \mathbf{F} \cdot d\mathbf{r} + \int_{C'} \mathbf{F} \cdot d\mathbf{r} = 0$). Choice D incorrectly suggests we need the explicit form of $f$, when the fundamental theorem gives us the relationship directly.
2. Consider the vector field $\vec{F} = \langle 3x^2 + yz, 3y^2 + xz, 3z^2 + xy \rangle$. Let $C$ be the curve parameterized by $\vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), \cos(2\pi t) \rangle$ for $t \in [0, 2]$. Which of the following is true about the line integral $\int_C \vec{F} \cdot d\vec{r}$?
   Answer: The integral is 0 because $\vec{F}$ is conservative and $C$ is a closed curve. First, we determine if the vector field $\vec{F}$ is conservative by checking if its curl is zero. Let $\vec{F} = \langle P, Q, R \rangle$. $\frac{\partial R}{\partial y} = x$, $\frac{\partial Q}{\partial z} = x$. So they are equal. $\frac{\partial P}{\partial z} = y$, $\frac{\partial R}{\partial x} = y$. So they are equal. $\frac{\partial Q}{\partial x} = z$, $\frac{\partial P}{\partial y} = z$. So they are equal. Since $\nabla \times \vec{F} = \vec{0}$ and the domain is $\mathbb{R}^3$, $\vec{F}$ is a conservative vector field. Next, we determine if the curve $C$ is a closed loop by checking if its start and end points are the same. Start point: $\vec{r}(0) = \langle \cos(0), \sin(0), \cos(0) \rangle = \langle 1, 0, 1 \rangle$. End point: $\vec{r}(2) = \langle \cos(2\pi), \sin(2\pi), \cos(4\pi) \rangle = \langle 1, 0, 1 \rangle$. Since the start and end points are identical, the curve $C$ 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, $\int_C \vec{F} \cdot d\vec{r} = 0$. 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.
3. Let $\vec{F}(x, y) = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right\rangle$. Evaluate the line integral $\oint_C \vec{F} \cdot d\vec{r}$, where $C$ is the circle $x^2+y^2=9$ traversed counter-clockwise.
   Answer: $2\pi$ Let $P = \frac{-y}{x^2+y^2}$ and $Q = \frac{x}{x^2+y^2}$. We can check the condition for a conservative field: $\frac{\partial Q}{\partial x} = \frac{(x^2+y^2)(1) - x(2x)}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}$ $\frac{\partial P}{\partial y} = \frac{(x^2+y^2)(-1) - (-y)(2y)}{(x^2+y^2)^2} = \frac{-x^2-y^2+2y^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}$ Since $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$, 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 $\vec{F}$ is undefined at the origin $(0,0)$, which is inside the circle $C$. Since the domain $\mathbb{R}^2 \setminus \{(0,0)\}$ is not simply connected, we cannot conclude that $\vec{F}$ is conservative or that the integral is zero. We must evaluate the integral by parameterization. Let $C$ be parameterized by $\vec{r}(t) = \langle 3\cos t, 3\sin t \rangle$ for $t \in [0, 2\pi]$. Then $d\vec{r} = \vec{r}'(t) dt = \langle -3\sin t, 3\cos t \rangle dt$. On the path $C$, $x^2+y^2 = 9$. So, $\vec{F}(\vec{r}(t)) = \left\langle \frac{-3\sin t}{9}, \frac{3\cos t}{9} \right\rangle = \left\langle -\frac{1}{3}\sin t, \frac{1}{3}\cos t \right\rangle$. The integral is: $\int_0^{2\pi} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt = \int_0^{2\pi} \left\langle -\frac{1}{3}\sin t, \frac{1}{3}\cos t \right\rangle \cdot \langle -3\sin t, 3\cos t \rangle \, dt$ $= \int_0^{2\pi} (\sin^2 t + \cos^2 t) \, dt = \int_0^{2\pi} 1 \, dt = 2\pi$.
4. The line integral of a conservative vector field $\vec{F} = \nabla f$ along a path $C$ from point $A=(1,2,0)$ to point $B$ is equal to 20. If the potential function is given by $f(x,y,z) = x^2 y - y z^2 + 5$, what is the value of $f(B)$?
   Answer: 27 According to the Fundamental Theorem for Line Integrals, for a conservative vector field $\vec{F} = \nabla f$, the line integral along a curve $C$ from point $A$ to point $B$ is given by $\int_C \vec{F} \cdot d\vec{r} = f(B) - f(A)$. We are given that the value of the integral is 20, so $f(B) - f(A) = 20$. We need to find the value of the potential function $f$ at the starting point $A=(1,2,0)$. $f(A) = f(1,2,0) = (1)^2(2) - (2)(0)^2 + 5 = 2 - 0 + 5 = 7$. Now we can solve for $f(B)$: $f(B) - 7 = 20$. $f(B) = 20 + 7 = 27$.
5. Consider two potential functions $f_1(x,y) = x^2y + xy^2 + 5$ and $f_2(x,y) = x^2y + xy^2 - 3$. Let $\mathbf{F}_1 = \nabla f_1$ and $\mathbf{F}_2 = \nabla f_2$. If $\int_{C_1} \mathbf{F}_1 \cdot d\mathbf{r} = 7$ where $C_1$ goes from $A$ to $B$, and $\int_{C_2} \mathbf{F}_2 \cdot d\mathbf{r} = k$ where $C_2$ goes from $B$ to $A$, what is the value of $k$?
   Answer: $k = -7$ since the fields differ only by a constant and direction is reversed 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 $\mathbf{F}_1$ and $\mathbf{F}_2$ are gradient fields (conservative), so line integrals depend only on the potential function values at the endpoints. For any conservative field $\mathbf{F} = \nabla f$, we have $\int_C \mathbf{F} \cdot d\mathbf{r} = f(\text{endpoint}) - f(\text{starting point})$. Since $f_1$ and $f_2$ differ only by a constant ($f_1 = f_2 + 8$), their gradients are identical: $\nabla f_1 = \nabla f_2$. This means $\mathbf{F}_1 = \mathbf{F}_2$. For the first integral: $\int_{C_1} \mathbf{F}_1 \cdot d\mathbf{r} = f_1(B) - f_1(A) = 7$ For the second integral, $C_2$ goes from $B$ to $A$ (opposite direction), so: $\int_{C_2} \mathbf{F}_2 \cdot d\mathbf{r} = f_2(A) - f_2(B)$ Since $\mathbf{F}_1 = \mathbf{F}_2$ and the direction is reversed, $k = -7$. 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.
6. Calculate the work done by the force field $\vec{F}(x,y,z) = \langle yz+2x, xz, xy \rangle$ on a particle that moves along a path from $(2,1,1)$ to $(1,2,3)$.
   Answer: 1 The work done is given by the line integral $W = \int_C \vec{F} \cdot d\vec{r}$. We first check if the vector field $\vec{F}$ is conservative by calculating its curl. $\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ yz+2x & xz & xy \end{vmatrix} = \hat{i}(\frac{\partial}{\partial y}(xy) - \frac{\partial}{\partial z}(xz)) - \hat{j}(\frac{\partial}{\partial x}(xy) - \frac{\partial}{\partial z}(yz+2x)) + \hat{k}(\frac{\partial}{\partial x}(xz) - \frac{\partial}{\partial y}(yz+2x)) = \hat{i}(x-x) - \hat{j}(y-y) + \hat{k}(z-z) = \vec{0}$. Since the curl is zero and the domain is $\mathbb{R}^3$, the field is conservative. We can find a potential function $f$ such that $\nabla f = \vec{F}$. $f(x,y,z) = \int (yz+2x) \, dx = xyz + x^2 + g(y,z)$. $\frac{\partial f}{\partial y} = xz + \frac{\partial g}{\partial y} = xz \implies \frac{\partial g}{\partial y} = 0$, so $g$ is a function of $z$ alone. $\frac{\partial f}{\partial z} = xy + g'(z) = xy \implies g'(z) = 0$, so $g$ is a constant. We can choose $g=0$. The potential function is $f(x,y,z) = xyz + x^2$. By the Fundamental Theorem for Line Integrals, the work is $f(\text{end point}) - f(\text{start point})$. Start point $A = (2,1,1)$, end point $B = (1,2,3)$. $f(B) = f(1,2,3) = (1)(2)(3) + (1)^2 = 6+1=7$. $f(A) = f(2,1,1) = (2)(1)(1) + (2)^2 = 2+4=6$. Work = $f(B) - f(A) = 7 - 6 = 1$.
7. Let $\vec{F}$ be a continuously differentiable vector field on $\mathbb{R}^3$. The line integral $\int_{C_1} \vec{F} \cdot d\vec{r}$ along a specific path $C_1$ from point $A$ to point $B$ is found to be 5. To guarantee that the line integral along any smooth path from $A$ to $B$ is also 5, which additional condition must be met?
   Answer: The curl of $\vec{F}$ is zero everywhere. 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 $\mathbb{R}^3$, the condition for it to be conservative is that its curl is zero. If $\nabla \times \vec{F} = \vec{0}$ everywhere, then $\vec{F}$ is conservative. This implies that the line integral $\int_C \vec{F} \cdot d\vec{r}$ is path-independent. Therefore, if we know the integral is 5 for one path $C_1$ from $A$ to $B$, it will be 5 for any other path from $A$ to $B$. 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.
8. Evaluate $\int_C \vec{F} \cdot d\vec{r}$ where $\vec{F}(x,y,z) = \langle 2xy, x^2 + z^2, 2yz \rangle$ and $C$ is the curve of intersection of the parabolic cylinder $y=x^2$ and the plane $z=x$ from the point $(0,0,0)$ to the point $(2,4,2)$.
   Answer: 32 First, we check if the vector field $\vec{F}$ is conservative by computing its curl. $\nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2xy & x^2+z^2 & 2yz \end{vmatrix} = \hat{i}(2z-2z) - \hat{j}(0-0) + \hat{k}(2x-2x) = \vec{0}$. Since the curl is zero and the domain is $\mathbb{R}^3$, the field is conservative. We can find a potential function $f(x,y,z)$. $f = \int 2xy \, dx = x^2y + g(y,z)$. $\frac{\partial f}{\partial y} = x^2 + \frac{\partial g}{\partial y} = x^2+z^2 \implies \frac{\partial g}{\partial y} = z^2 \implies g(y,z) = yz^2 + h(z)$. So $f = x^2y + yz^2 + h(z)$. $\frac{\partial f}{\partial z} = 2yz + h'(z) = 2yz \implies h'(z)=0$. Thus, a potential function is $f(x,y,z) = x^2y + yz^2$. 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: $A=(0,0,0)$. End point: $B=(2,4,2)$. Value = $f(B) - f(A) = f(2,4,2) - f(0,0,0)$. $f(2,4,2) = (2^2)(4) + (4)(2^2) = (4)(4) + (4)(4) = 16 + 16 = 32$. $f(0,0,0) = 0$. The value of the integral is $32-0=32$.
9. Let $\vec{F}(\vec{r}) = \frac{\vec{r}}{|\vec{r}|^3}$ be a vector field in $\mathbb{R}^3$, where $\vec{r} = \langle x,y,z \rangle$. Evaluate the line integral of $\vec{F}$ along a smooth curve $C$ from point $A=(1,2,2)$ to point $B=(3,0,4)$ that does not pass through the origin.
   Answer: $2/15$ When you encounter a line integral with a vector field that resembles an inverse square law (like $\vec{F}(\vec{r}) = \frac{\vec{r}}{|\vec{r}|^3}$), always check if the field is conservative. This particular field is the gradient of a scalar potential function $\phi(\vec{r}) = -\frac{1}{|\vec{r}|}$, 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: $\int_C \vec{F} \cdot d\vec{r} = \phi(B) - \phi(A)$. Let's calculate the potential at each point. At $A = (1,2,2)$: $|\vec{r}_A| = \sqrt{1^2 + 2^2 + 2^2} = 3$, so $\phi(A) = -\frac{1}{3}$. At $B = (3,0,4)$: $|\vec{r}_B| = \sqrt{3^2 + 0^2 + 4^2} = 5$, so $\phi(B) = -\frac{1}{5}$. Therefore: $\int_C \vec{F} \cdot d\vec{r} = -\frac{1}{5} - \left(-\frac{1}{3}\right) = \frac{1}{3} - \frac{1}{5} = \frac{5-3}{15} = \frac{2}{15}$ A) -2 likely comes from incorrectly using $\phi = \frac{1}{|\vec{r}|}$ (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.
10. Let $\mathbf{F}(x,y) = (2xy + e^x, x^2 + \cos y)$ and let $C$ be any piecewise smooth curve from $(0,0)$ to $(1,\pi)$. If $f(x,y)$ is a potential function for $\mathbf{F}$, which of the following correctly represents $\int_C \mathbf{F} \cdot d\mathbf{r}$?
    Answer: $f(1,\pi) - f(0,0)$ where $f(x,y) = x^2y + e^x + \sin y + C$ First, we verify that $\mathbf{F}$ is conservative by checking that $\frac{\partial F_1}{\partial y} = \frac{\partial F_2}{\partial x}$. We have $\frac{\partial}{\partial y}(2xy + e^x) = 2x$ and $\frac{\partial}{\partial x}(x^2 + \cos y) = 2x$. Since these are equal, $\mathbf{F}$ is conservative. To find the potential function, we integrate: $\frac{\partial f}{\partial x} = 2xy + e^x$ gives $f(x,y) = x^2y + e^x + g(y)$. Then $\frac{\partial f}{\partial y} = x^2 + g'(y) = x^2 + \cos y$, so $g'(y) = \cos y$ and $g(y) = \sin y + C$. Thus $f(x,y) = x^2y + e^x + \sin y + C$. By the Fundamental Theorem for Line Integrals, $\int_C \mathbf{F} \cdot d\mathbf{r} = f(1,\pi) - f(0,0)$. Choice B has the wrong sign for $\sin y$, Choice C reverses the order of evaluation, and Choice D incorrectly claims the field is not conservative.
11. Let $\vec{F}_1 = \langle y, -x \rangle$ and $\vec{F}_2 = \langle x, y \rangle$. Let $C_1$ be the line segment from $(1,0)$ to $(-1,0)$ and $C_2$ be the upper semicircle from $(1,0)$ to $(-1,0)$. Define the integrals $I_1 = \int_{C_1} \vec{F}_1 \cdot d\vec{r}$, $I_2 = \int_{C_2} \vec{F}_1 \cdot d\vec{r}$, $J_1 = \int_{C_1} \vec{F}_2 \cdot d\vec{r}$, and $J_2 = \int_{C_2} \vec{F}_2 \cdot d\vec{r}$. Which statement is correct?
    Answer: $I_1 \neq I_2$ and $J_1 = J_2$ 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 $\vec{F} = \langle P, Q \rangle$ is conservative, check if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. For $\vec{F}_1 = \langle y, -x \rangle$: $\frac{\partial}{\partial y}(y) = 1$ and $\frac{\partial}{\partial x}(-x) = -1$. Since $1 \neq -1$, $\vec{F}_1$ is not conservative, so $I_1 \neq I_2$. For $\vec{F}_2 = \langle x, y \rangle$: $\frac{\partial}{\partial y}(x) = 0$ and $\frac{\partial}{\partial x}(y) = 0$. Since $0 = 0$, $\vec{F}_2$ is conservative, so $J_1 = J_2$. Now examining the choices: Option A claims both pairs are equal, but we showed $I_1 \neq I_2$. Option B claims $I_1 = I_2$, which is false. Option C correctly states $I_1 \neq I_2$ and $J_1 = J_2$, matching our analysis. Option D claims both pairs are unequal, but $J_1 = J_2$ since $\vec{F}_2$ 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.
12. Let $f(x,y,z) = x \cos(y) + z$ and let $\vec{F} = \nabla f$. Evaluate the line integral $\int_C \vec{F} \cdot d\vec{r}$, where $C$ is the helix $\vec{r}(t) = \langle \cos(t), \sin(t), t \rangle$ from $t=0$ to $t=\pi/2$.
    Answer: $\frac{\pi}{2} - 1$ When you encounter a line integral of a gradient vector field, immediately think about the Fundamental Theorem for Line Integrals. Since $\vec{F} = \nabla f$, this integral equals simply $f(\text{endpoint}) - f(\text{starting point})$, regardless of the path taken. First, find the endpoints of your helix. At $t = 0$: $\vec{r}(0) = \langle 1, 0, 0 \rangle$. At $t = \pi/2$: $\vec{r}(\pi/2) = \langle 0, 1, \pi/2 \rangle$. Now evaluate the scalar function $f(x,y,z) = x\cos(y) + z$ at these points: - $f(1,0,0) = 1 \cdot \cos(0) + 0 = 1 \cdot 1 + 0 = 1$ - $f(0,1,\pi/2) = 0 \cdot \cos(1) + \pi/2 = 0 + \pi/2 = \pi/2$ Therefore: $\int_C \vec{F} \cdot d\vec{r} = f(0,1,\pi/2) - f(1,0,0) = \pi/2 - 1$ Answer A ($0$) suggests incorrectly that this is a closed curve or that the function values are equal. Answer B ($1 - \pi/2$) reverses the subtraction order—a common error when applying the fundamental theorem. Answer C ($\pi/2$) forgets to subtract the initial value, using only the final endpoint. Key strategy: When you see $\vec{F} = \nabla f$ 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.
13. A vector field $\mathbf{F}$ is defined on $\mathbb{R}^3$ such that $\nabla \times \mathbf{F} = \mathbf{0}$ and $\nabla \cdot \mathbf{F} = x + y + z$. If $f$ is a potential function for $\mathbf{F}$ and $\int_C \mathbf{F} \cdot d\mathbf{r} = 12$ where $C$ goes from $(1,1,1)$ to $(3,2,4)$, which statement about the potential function is necessarily true?
    Answer: $f(3,2,4) - f(1,1,1) = 12$ and $\nabla^2 f = x + y + z$ 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 $\nabla \times \mathbf{F} = \mathbf{0}$, the vector field $\mathbf{F}$ is conservative, which means it has a potential function $f$ where $\mathbf{F} = \nabla f$. For conservative fields, the fundamental theorem for line integrals tells us that $\int_C \mathbf{F} \cdot d\mathbf{r} = f(\text{endpoint}) - f(\text{startpoint})$. Therefore, $\int_C \mathbf{F} \cdot d\mathbf{r} = f(3,2,4) - f(1,1,1) = 12$. Additionally, since $\mathbf{F} = \nabla f$, we have $\nabla \cdot \mathbf{F} = \nabla \cdot (\nabla f) = \nabla^2 f$. Given that $\nabla \cdot \mathbf{F} = x + y + z$, we conclude $\nabla^2 f = x + y + z$. Choice A is wrong because curl and divergence are independent properties—a zero curl doesn't require zero divergence. Choice B incorrectly states that $\nabla^2 f = 0$, 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 $f(1,1,1) - f(3,2,4) = 12$ instead of the correct $f(3,2,4) - f(1,1,1) = 12$. Remember: for conservative vector fields, always use $\nabla \cdot \mathbf{F} = \nabla^2 f$ and the fundamental theorem $\int_C \mathbf{F} \cdot d\mathbf{r} = f(\text{end}) - f(\text{start})$.
14. A particle moves through a conservative force field $\mathbf{F}$ with potential energy function $U(x,y,z) = -f(x,y,z)$ where $f$ is the potential function for $\mathbf{F}$. The particle moves from point $P(1,0,2)$ where $U = 10$ to point $Q(3,1,0)$ where $U = 4$. If the particle's kinetic energy at $P$ is $K_P = 8$, what is its kinetic energy $K_Q$ at point $Q$?
    Answer: $K_Q = 14$ by conservation of mechanical energy since $K_P + U_P = K_Q + U_Q$ 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: $K_P + U_P = K_Q + U_Q$. Substituting the given values: $8 + 10 = K_Q + 4$, which gives us $18 = K_Q + 4$, so $K_Q = 14$. Looking at the answer choices: Choice A incorrectly calculates $K_Q = 2$, which would require $8 + 10 = 2 + 4 = 6$, violating energy conservation. Choice C reaches the correct numerical answer of $K_Q = 14$ 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 $U_P - U_Q$" is backwards—work done should equal $-(U_Q - U_P) = -(4-10) = 6$. 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 $K_Q = 14$. Study tip: For conservative force problems, always start with energy conservation: $K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}}$. Remember that when potential energy decreases, kinetic energy must increase by the same amount to maintain total energy.

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