Vector Calculus Questions and Answers

Answer: A

The difference \( \mathbf{r}(t + \Delta t) - \mathbf{r}(t) \) is a chord, and as \( \Delta t \to 0 \) its direction becomes the tangent's. Dividing by \( |d\mathbf{r}/dt| \) gives the unit tangent. The principal normal points towards the centre of curvature, at right angles to this, and the binormal is perpendicular to both

Answer: D

Differentiate and evaluate:

\[ \frac{d\mathbf{r}}{dt} \;=\; \mathbf{i} + 2t\mathbf{j} \;=\; \mathbf{i} + 2\mathbf{j} \quad (t = 1). \]

Its magnitude is \( \sqrt{1 + 4} = \sqrt{5} \), so the unit tangent is that vector divided by \( \sqrt{5} \). Omitting the division leaves a vector of length \( \sqrt{5} \) rather than one

Answer: C

Arc length is measured along the curve itself, so a unit change of \( s \) moves a unit distance and

\[ \left|\frac{d\mathbf{r}}{ds}\right| \;=\; 1. \]

This is what makes the arc-length parametrisation convenient: the derivative *is* the unit tangent, with no normalising required, and its own derivative then gives the curvature directly

Answer: B

Curvature measures the turning of the tangent, and torsion the turning of the binormal, which is the same as the tilting of the plane containing the tangent and principal normal. A plane curve has zero torsion whatever its curvature, since it never leaves its own plane

Answer: A

The rate of change along a direction is the component of the gradient in that direction, so

\[ \frac{d\phi}{ds} \;=\; \nabla\phi \cdot \mathbf{u} \;=\; |\nabla\phi|\cos\theta. \]

A cross product would give a vector rather than a rate, and the product of magnitudes discards the angle, which is exactly what decides how much of the gradient the direction picks up

Answer: D

Since \( \nabla\phi \cdot \mathbf{u} = |\nabla\phi|\cos\theta \), the value is largest when \( \theta = 0 \), and it equals \( |\nabla\phi| \) there. The opposite direction gives the steepest *decrease*, and any direction tangent to the level surface gives zero, the field not changing along a surface of constant value

Answer: C

The gradient is \( \nabla\phi = 2x\mathbf{i} + 2y\mathbf{j} + 2z\mathbf{k} \), which at the point is \( 2\mathbf{i} + 4\mathbf{j} + 4\mathbf{k} \). The greatest rate is its magnitude:

\[ |\nabla\phi| \;=\; \sqrt{2^{2} + 4^{2} + 4^{2}} \;=\; \sqrt{4 + 16 + 16} \;=\; \sqrt{36} \;=\; 6. \]

The value three is the distance of the point from the origin, not the rate

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