Vector Calculus Questions and Answers

Answer: C

The vector \(\vec{c}\) is orthogonal to \(\vec{a} + \vec{b}\).

A cross product is perpendicular to both of its factors: \(\vec{c}\cdot\vec{a} = 0\) and \(\vec{c}\cdot\vec{b} = 0\). Adding these gives \(\vec{c}\cdot(\vec{a} + \vec{b}) = 0\), so \(\vec{c}\) is orthogonal to the sum, and indeed to every vector in the plane of \(\vec{a}\) and \(\vec{b}\).

\(\vec{c}\) is not zero, because a cross product vanishes only when the vectors are parallel (or one of them is zero), and these are not. The two 'alone' options are wrong because \(\vec{c}\) is orthogonal to both vectors, not to one of them only.

Answer: B

The field is curl-free but not divergence-free.

For \(\vec{F} = x\hat{\imath} + y\hat{\jmath} + z\hat{k}\):

\[ \begin{aligned} \nabla\cdot\vec{F} &= \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} \\ &\quad + \frac{\partial z}{\partial z} \\ &= 1 + 1 + 1 = 3 \end{aligned} \]

so it is not divergence-free; it is the field pointing straight outward from the origin, a uniform source. Each component of the curl, such as \(\partial F_{z}/\partial y - \partial F_{y}/\partial z = 0 - 0\), vanishes because each component depends only on its own coordinate, so \(\nabla\times\vec{F} = 0\). Indeed \(\vec{F}\) is the gradient of \((x^{2} + y^{2} + z^{2})/2\), and every gradient field is curl-free.

That rules out 'divergence-free and curl-free', 'divergence-free but not curl-free' and 'neither'.

Answer: A

The direction is that of the gradient, \(\nabla\phi\).

The rate of change of \(\phi\) in the direction of a unit vector \(\hat{n}\) is the directional derivative \(\nabla\phi \cdot \hat{n} = |\nabla\phi|\cos\alpha\), where \(\alpha\) is the angle between \(\hat{n}\) and the gradient. It is largest when \(\alpha = 0\), that is, along \(\nabla\phi\), and its largest value is \(|\nabla\phi|\).

\(\nabla \times (\phi\vec{r})\) equals \(\nabla\phi \times \vec{r}\), which is perpendicular to the gradient, so along it the rate of change is zero. \(\phi\vec{r}\) points along the position vector, which has nothing to do with how \(\phi\) varies. \((\nabla\phi \cdot d\vec{r})\vec{r}\) is also a multiple of \(\vec{r}\), scaled by the change in \(\phi\), and again points radially rather than along the gradient.

Answer: A

The integral is 0.

The field is conservative. Its curl (the z-component, for a plane field) is

\[ \begin{aligned} &\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \\ &= 2 - 2 = 0 \end{aligned} \]

Indeed \(\vec{F} = \nabla \phi\) with \(\phi = \tfrac{1}{2}x^{2} + 2xy + 2y^{2}\). By Stokes' (or Green's) theorem the circulation of a curl-free field round any closed contour is zero, whatever the shape or direction of the contour.

By Green's theorem the circulation equals the curl integrated over the enclosed area. The square has diagonals of length 2 and so an area of 2, which is what a curl of 1 would give; with a curl of zero, 2, 4 and 6 are all impossible. The integral along each single side is not zero (from O to P it is \(\phi(P) - \phi(O) = 4.5\)), but the four sides cancel.

Answer: B

The curl is the second option: \(\mathbf{I}(\frac{\partial h}{\partial y} - \frac{\partial g}{\partial z})\) \(- \mathbf{J}(\frac{\partial h}{\partial x} - \frac{\partial f}{\partial z})\) \(+ \mathbf{K}(\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y})\).

The curl is the determinant

\[ \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{I} & \mathbf{J} & \mathbf{K} \\ \partial_{x} & \partial_{y} & \partial_{z} \\ f & g & h \end{vmatrix} \]

Expanding along the first row, the I term is \(\partial_{y}h - \partial_{z}g\), the J term carries a minus sign in front of \(\partial_{x}h - \partial_{z}f\), and the K term is \(\partial_{x}g - \partial_{y}f\).

The third option has a plus sign before J, so its J component has the wrong sign. The first and fourth options reverse the order inside every bracket, which gives the negative of the curl (and the fourth also has the wrong J sign).

Answer: D

The work done is −15.

The resultant force and the displacement from A to B are

\[ \begin{aligned} \mathbf{F} &= (2 - 1)\mathbf{i} + (-5 + 2)\mathbf{j} + (6 - 1)\mathbf{k} \\ &= \mathbf{i} - 3\mathbf{j} + 5\mathbf{k} \\ \overrightarrow{AB} &= 2\mathbf{i} + 4\mathbf{j} - \mathbf{k} \end{aligned} \]

Work is their dot product:

\[ W = (1)(2) + (-3)(4) + (5)(-1) = 2 - 12 - 5 = -15 \]

The negative sign means the force opposes the displacement on balance. 2 and −5 are single terms of the sum, and 12 drops the signs.

Answer: D

The moment is \(-112\hat{i} - 16\hat{j} + 50\hat{k}\).

The moment of a force about a point A is \(\vec{r} \times \vec{F}\), where \(\vec{r}\) runs from A to any point on the force's line of action. The force passes through the origin, so \(\vec{r} = O - A = (-3, -4, -8)\).

\[ \begin{aligned} M_{x} &= (-4)(8) - (-8)(-10) = -112 \\ M_{y} &= (-8)(5) - (-3)(8) = -16 \\ M_{z} &= (-3)(-10) - (-4)(5) = 50 \end{aligned} \]

Using \(\vec{r}\) from the origin to A, (3, 4, 8), reverses every sign and gives \(112\hat{i} + 16\hat{j} - 50\hat{k}\), the moment about the origin of a force through A. The other two options rearrange the same three numbers into the wrong components.

Answer: A

The divergence of a curl is 0.

For a twice continuously differentiable vector field \(\mathbf{F} = (F_{1}, F_{2}, F_{3})\),

\[ \begin{aligned} \nabla \cdot (\nabla \times \mathbf{F}) &= \partial_{x}(\partial_{y}F_{3} - \partial_{z}F_{2}) \\ &\quad + \partial_{y}(\partial_{z}F_{1} - \partial_{x}F_{3}) \\ &\quad + \partial_{z}(\partial_{x}F_{2} - \partial_{y}F_{1}) \end{aligned} \]

The mixed partial derivatives cancel in pairs, since \(\partial_{x}\partial_{y}F_{3} = \partial_{y}\partial_{x}F_{3}\) and so on (equality of mixed partials, which is why the field must be twice continuously differentiable). The result is identically zero, so 1, 2 and infinity are all wrong. A curl field is always solenoidal.

Answer: D

The outward flux is \(4\pi abc\).

By the divergence theorem, the flux through the closed ellipsoid equals the integral of \(\nabla\cdot\vec{u}\) over the solid it bounds. Here

\[ \begin{aligned} \nabla\cdot\vec{u} &= 1 + 1 + 1 = 3 \\ \text{flux} &= 3\cdot\tfrac{4}{3}\pi abc = 4\pi abc \end{aligned} \]

using the volume \(\frac{4}{3}\pi abc\) of the ellipsoid. The condition \(a \gt b \gt c\) plays no part. \(\pi abc\), \(2\pi abc\) and \(3\pi abc\) come from using the wrong volume or dropping the factor 3 from the divergence; the volume alone would be \(\frac{4}{3}\pi abc\), which is not offered.

Answer: B

The work done is \(-\frac{\pi}{2}\).

Parametrise the upper semicircle by x = cos t, y = sin t, t from 0 to \(\pi\). On it \(x^{2} + y^{2} = 1\), so

\[ \begin{aligned} F \cdot dr &= (x+y)\,dx - (x^{2}+y^{2})\,dy \\ &= -(\cos t + \sin t)\sin t\,dt - \cos t\,dt \end{aligned} \]

\[ \begin{aligned} W &= \int_{0}^{\pi} (-\sin t\cos t - \sin^{2} t - \cos t)\,dt \\ &= 0 - \frac{\pi}{2} - 0 = -\frac{\pi}{2} \end{aligned} \]

The terms \(\sin t\cos t\) and \(\cos t\) integrate to 0 over [0, \(\pi\)], and \(\sin^{2} t\) integrates to \(\pi/2\). \(\pi/2\) has the sign reversed (the path traversed the other way), and \(\pm\pi\) would come from integrating \(\sin^{2} t\) as if it were 1.