Vector Calculus Questions and Answers

Answer: B

Moving along the surface leaves \( \phi \) unchanged, so the directional derivative in every tangential direction vanishes and the gradient is perpendicular to all of them. That is what makes the gradient the natural way to find a normal, and hence a tangent plane, to a surface given implicitly

Answer: A

The gradient \( 2\mathbf{i} + 4\mathbf{j} + 4\mathbf{k} \) is normal there, and dividing by two gives the simpler normal \( (1, 2, 2) \). The plane through the point with that normal is

\[ 1(x-1) + 2(y-2) + 2(z-2) \;=\; 0 \quad \Longrightarrow \quad x + 2y + 2z \;=\; 9. \]

Substituting the point confirms \( 1 + 4 + 4 = 9 \)

Answer: D

Writing out any component gives a difference of mixed second derivatives, such as \( \phi_{zy} - \phi_{yz} \), which vanishes whenever those derivatives are continuous. So

\[ \nabla \times \nabla\phi \;=\; \mathbf{0}. \]

This is why every gradient field is irrotational, and conversely why an irrotational field can be written as a gradient

Answer: C

Expanding gives six mixed third derivatives that cancel in pairs, so

\[ \nabla \cdot (\nabla \times \mathbf{A}) \;=\; 0. \]

Every curl field is therefore solenoidal, which is why a solenoidal field can be represented by a vector potential and why magnetic flux density, having zero divergence, is written as the curl of one

Answer: B

The operator differentiates both factors, and the term acting on the scalar has to be contracted with the vector to stay a scalar:

\[ \nabla \cdot (\phi\mathbf{A}) \;=\; \phi\,(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot \nabla\phi. \]

The sign is positive, as in an ordinary product rule; the minus sign belongs to no standard identity of this kind

Answer: A

Solenoidal means divergence-free, so the flux out of every closed surface is zero and the field lines close on themselves rather than beginning or ending. A curl-free field is irrotational instead, and that is the one expressible as a gradient. A field may be either, both, or neither

Answer: D

Irrotational means \( \nabla \times \mathbf{F} = \mathbf{0} \), and since the curl of any gradient vanishes identically, a scalar potential is the natural representation. A vector potential belongs to the solenoidal case, where the divergence rather than the curl is zero

Answer: C

Only the component of the force along the motion does work, so each element contributes \( \mathbf{F} \cdot d\mathbf{r} \) and

\[ W \;=\; \int_{C} \mathbf{F} \cdot d\mathbf{r}. \]

Integrating the magnitude alone would count a sideways force as though it were pushing along the path, and a cross product would give a vector rather than an amount of work

Answer: B

On the path \( y = x \), so with \( x = t \) and \( y = t \) for \( t \) from zero to one,

\[ \int_{C}\left(y\,dx + x\,dy\right) \;=\; \int_{0}^{1}\left(t + t\right)dt \;=\; \left[t^{2}\right]_{0}^{1} \;=\; 1. \]

The field is the gradient of \( xy \), so the value is just \( xy \) evaluated between the ends, \( 1 - 0 \), whatever path is taken

Answer: A

Path independence, a vanishing circulation round every closed loop, the existence of a scalar potential, and an identically zero curl on a simply connected region are four equivalent statements. A vanishing divergence is a different condition altogether, describing a solenoidal rather than a conservative field