📝 Vector valued functions of two variables (14 MCQs)

📖 Explanation: The divergence is computed as $\nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} = 2xy + 2yx = 4xy$. At (1,2), this equals 8. Option B is correct. Common errors include computing the gradient instead of divergence, or evaluating partial derivatives incorrectly as $x^2 + y^2$ leading to 5, which is not listed. This tests the definition and application of divergence.

📖 Explanation: The scalar curl in 2D is $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$. Here, $\frac{\partial Q}{\partial x} = e^{x}\sin y$ and $\frac{\partial P}{\partial y} = -e^{x}\sin y$. Their difference is $e^{x}\sin y - (-e^{x}\sin y) = 2e^{x}\sin y$. Wait, that gives option D. Let's recalc: $Q = e^x \sin y$, $P = e^x \cos y$. $\partial Q/\partial x = e^x \sin y$, $\partial P/\partial y = -e^x \sin y$. Difference = $2e^x \sin y$. So correct is D. But I wrote C as 0 erroneously; let me correct: The correct answer should be D. However, the question asks for true statement; if curl is zero, it's conservative. Here it's not zero. So correct option is D. I'll adjust the correct answer.

📖 Explanation: For $\mathbf{F} = (-y, x)$, $P = -y$, $Q = x$. Curl = $\partial Q/\partial x - \partial P/\partial y = 1 - (-1) = 2$. So the flow has non-zero curl, meaning it is rotational. The student's error is thinking that the visual circular motion implies irrotationality. In reality, curl measures local rotation, not global path shape. Option D correctly identifies the computed value. This tests the distinction between global and local rotation.

📖 Explanation: The Jacobian matrix is $\begin{pmatrix} 2u & -2v \\ 2v & 2u \end{pmatrix}$. Its determinant is $(2u)(2u) - (-2v)(2v) = 4u^2 + 4v^2 = 4(u^2+v^2)$. This is positive for all non-zero $u,v$, indicating orientation preservation. This transformation is essentially the complex square function. Option A is correct. Common mistakes include forgetting the cross terms or sign errors.

📖 Explanation: The surface area element is $|\mathbf{r}_u \times \mathbf{r}_v|$. Here $\mathbf{r}_u = (1,0,2u)$, $\mathbf{r}_v = (0,1,2v)$. Cross product = $(-2u, -2v, 1)$, magnitude = $\sqrt{4u^2+4v^2+1}$. So option A is correct. Option B misses the '1' from the vertical component. Option C has wrong coefficients. This requires multi-step reasoning: partial derivatives, cross product, norm, then setting up the double integral.

📖 Explanation: Divergence of $\mathbf{F} = 2+2 = 4$. By divergence theorem, flux = $\int \int \nabla \cdot \mathbf{F} \, dA = 4 \times \text{area} = 4\pi$. Direct computation with outward normal also yields $4\pi$. The second student likely forgot the Jacobian or used wrong normal. So option A is correct. This tests understanding of the divergence theorem and common computational errors in flux integrals.

📖 Explanation: A field is conservative if its curl is zero everywhere. For $(y,x)$, $P=y, Q=x$, curl = $\partial Q/\partial x - \partial P/\partial y = 1 - 1 = 0$. Also, it has a potential $f(x,y)=xy$. Option B has curl 2, not conservative. Option C has curl 0? $\partial Q/\partial x = 0, \partial P/\partial y = 0$, so curl 0, but wait, it is conservative too? Actually $(x^2, y^2)$ has potential $x^3/3 + y^3/3$, so it is conservative. But the question asks for 'which is conservative' and A and C both are. But we need one correct. Let's check: Option D has curl $2e^x \sin y$, not zero. So A and C are conservative. To make it single correct, I'll modify the question in the CSV to have only one correct. Let's change option C to $(x^2, -y^2)$ which has curl 0? $\partial Q/\partial x = 0, \partial P/\partial y = 0$ still 0. Actually any gradient field is conservative. So I'll change to $(x^2 y, x y^2)$ which has curl $y^2 - x^2$ not zero. So I'll adjust. In the actual CSV, I'll ensure only A is correct. Let's set C = $(x^2 y, x y^2)$, D = $(e^x \cos y, -e^x \sin y)$ which has curl? $\partial Q/\partial x = -e^x \sin y, \partial P/\partial y = -e^x \sin y$, difference = 0, so that would be conservative too. I need to design carefully. I'll set options: A) (y, x) - conservative, B) (-y, x) - not, C) (x^2, -y^2) - conservative (potential $x^3/3 - y^3/3$), so that's two. So I'll change C to (x^2 y, x y^2) and D to (e^x cos y, e^x sin y) as before. Then only A is conservative? Check D: curl = $e^x sin y - (-e^x sin y) = 2e^x sin y$ not zero. So A is correct. C has curl $y^2 - x^2$ not zero. So A is unique. So correct answer is A.

📖 Explanation: The Jacobian determinant is $\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -2$. Absolute value is 2. So area scales by 2 everywhere. This is constant because the transformation is linear. Option A is correct. Students often compute the determinant incorrectly as 0 or forget absolute value. This tests the geometric meaning of the Jacobian as a local area scaling factor.

📖 Explanation: Divergence = $\partial P/\partial x + \partial Q/\partial y = 2x - 2y$. Curl = $\partial Q/\partial x - \partial P/\partial y = 2x - 2y$. So they are equal everywhere. But the question asks 'at which point' and options are points. Since they are equal for all points, any point works. But we need a single correct. Let's check: At (1,0): div=2, curl=2. At (0,1): div=-2, curl=-2. At (1,1): div=0, curl=0. At (0,0): div=0, curl=0. So all satisfy. The question is flawed. I'll change it to ask where divergence is twice the curl. Then solve: $2x - 2y = 2(2x - 2y)$ implies $2x-2y = 0$ so $x=y$. Among options, (1,1) is the only one with x=y. So I'll set correct as C. This requires solving an equation involving both divergence and curl.

📖 Explanation: Compute $\mathbf{r}_u = (\cos v, \sin v, 0)$, $\mathbf{r}_v = (-u \sin v, u \cos v, 1)$. Cross product = $(\sin v \cdot 1 - 0, 0 - \cos v \cdot 1, \cos v \cdot u \cos v - \sin v \cdot (-u \sin v))$ = $(\sin v, -\cos v, u)$. Magnitude = $\sqrt{\sin^2 v + \cos^2 v + u^2} = \sqrt{1+u^2}$. At u=1, v=0, magnitude = $\sqrt{2}$. So option A. This tests surface parameterization and cross product computation.

📖 Explanation: Check $\partial f/\partial x = 2xy$ matches P, and $\partial f/\partial y = x^2 + 2y$ matches Q. So the student's expression is correct up to an additive constant. Option B correctly notes that any constant can be added. Option A is technically correct but incomplete; the presence of '+C' is important. The field is conservative (curl = $2x - 2x = 0$). So the best answer is B, as it highlights the arbitrary constant. This tests the concept of potential functions and the constant of integration.

📖 Explanation: Radially outward means vectors point away from origin. Magnitude increasing with distance suggests $\sqrt{x^2+y^2}$ times unit radial vector, which is $(x,y)$. Option A is exactly that. Option B is unit radial (magnitude constant =1). Option C is rotational. Option D is constant. So graph description matches A. This tests the ability to translate visual vector field behavior into algebraic form.

📖 Explanation: The Jacobian determinant is $4(u^2+v^2)$ as computed earlier. Area = $\int_0^1 \int_0^1 4(u^2+v^2) \, du\,dv = 4[\int_0^1 u^2 du \cdot 1 + 1 \cdot \int_0^1 v^2 dv] = 4(1/3+1/3)=8/3$. Option A is correct. This requires computing the Jacobian, setting up the integral, and evaluating. Option B misses factor 2, C is wrong norm, D is wrong cross term. This is multi-step and Olympiad-style due to exact integration and transformation.

📖 Explanation: Parameterize $r(t) = (t,t), t:0\to1$. Then $\mathbf{F}(r(t)) = (t, t)$, r'(t) = (1,1). Dot product = $t+t = 2t$. Integral = $\int_0^1 2t dt = 1$. Since the field is conservative with potential $xy$, the integral equals $f(1,1)-f(0,0)=1-0=1$. Option A is correct. Common error: using wrong parameterization or forgetting dot product. This tests both direct computation and use of potential function.