Multivariable Calculus Quiz: Vector Fields And Field Lines

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Vector Fields And Field LinesQuestion 1 of 10

The vector field $\mathbf{F}(x,y) = \langle P(x,y), Q(x,y) \rangle$ has field lines that form a family of curves $y = x^2 + c$ . If the field has no sources or sinks (zero divergence everywhere), which relationship must hold between $P$ and $Q$ ?

P(x,y) = k

and

Q(x,y) = 2kx

for some nonzero constant

k

P(x,y) = 2kx

and

Q(x,y) = k

for some nonzero constant

k

P(x,y) = k(y-x^2)

and

Q(x,y) = 2kx(y-x^2)

for constant

k

P(x,y) = 2kx(x^2-y)

and

Q(x,y) = k(x^2-y)

for constant

k

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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Vector Fields And Field Lines

Practice Vector Fields And Field Lines in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Vector Fields And Field Lines, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$P(x,y) = k$ and $Q(x,y) = 2kx$ for some nonzero constant $k$ (correct answer)
$P(x,y) = 2kx$ and $Q(x,y) = k$ for some nonzero constant $k$
$P(x,y) = k(y-x^2)$ and $Q(x,y) = 2kx(y-x^2)$ for constant $k$
$P(x,y) = 2kx(x^2-y)$ and $Q(x,y) = k(x^2-y)$ for constant $k$

Explanation: Field lines y = x² + c have slope dy/dx = 2x along each curve. The field must be tangent to these lines, so Q/P = 2x, giving Q = 2xP. Zero divergence requires ∂P/∂x + ∂Q/∂y = 0. If P and Q depend only on position relative to the field lines, the simplest solution is P = k (constant) and Q = 2kx. Check: ∂P/∂x = 0, ∂Q/∂y = 0, so divergence = 0 + 0 = 0 ✓. Also Q/P = 2kx/k = 2x ✓. Choice B reverses P and Q. Choices C and D introduce unnecessary dependence on (y-x²) that doesn't match the simple parabolic field line family.

Question 2

The field lines of a 2D vector field $\vec{F}(x,y) = \langle P(x,y), Q(x,y) \rangle$ are curves that satisfy the differential equation $dy/dx = Q(x,y)/P(x,y)$ . Which of the following vector fields has field lines that are parabolas of the form $y = kx^2$ for some constant $k$ ?

$\vec{F}(x,y) = \langle x, 2y \rangle$ (correct answer)
$\vec{F}(x,y) = \langle 2x, y \rangle$
$\vec{F}(x,y) = \langle 1, 2x \rangle$
$\vec{F}(x,y) = \langle y, x \rangle$

Explanation: For a curve

y = kx^2

, the slope is

dy/dx = 2kx

. We need to find a vector field

\vec{F} = \langle P, Q \rangle

such that

Q/P = 2kx

. Since

k = y/x^2

for points on the parabola, we can substitute this into the equation for the slope:

dy/dx = 2(y/x^2)x = 2y/x

. We must find the vector field where

Q/P = 2y/x

. For choice A,

\vec{F}(x,y) = \langle x, 2y \rangle

, we have

P=x

and

Q=2y

. The ratio is

Q/P = 2y/x

. This matches the required slope, so its field lines are the parabolas

y=kx^2

Question 3

Let $\vec{R}(x,y) = \langle x, y \rangle$ be the standard radial vector field. Consider a modified field $\vec{F}(x,y) = \frac{1}{x^2+y^2}\vec{R}(x,y)$ for $(x,y) \neq (0,0)$ . Which statement accurately describes the field $\vec{F}$ ?

The field lines are unchanged from $\vec{R}$ , but the vectors point inward toward the origin.
The field lines are unchanged from $\vec{R}$ , but vector magnitudes decrease as distance from the origin increases. (correct answer)
The field lines become circles centered at the origin, and vector magnitudes are constant on these circles.
The field lines are unchanged from $\vec{R}$ , but vector magnitudes increase as distance from the origin increases.

Explanation: The vector field is

\vec{F}(x,y) = \frac{1}{x^2+y^2}\langle x, y \rangle

. Since

\frac{1}{x^2+y^2}

is a positive scalar, the direction of

\vec{F}

at any point

(x,y)

is the same as the direction of

\vec{R}(x,y)

, which is radially outward. Thus, the field lines, which show the direction of the field, are unchanged (they are rays from the origin). The magnitude of

\vec{F}

|\vec{F}| = \left| \frac{1}{x^2+y^2} \right| |\vec{R}| = \frac{1}{x^2+y^2} \sqrt{x^2+y^2} = \frac{\sqrt{x^2+y^2}}{x^2+y^2} = \frac{1}{\sqrt{x^2+y^2}}

. If

r = \sqrt{x^2+y^2}

is the distance from the origin, the magnitude is

1/r

. As

r

increases, the magnitude

1/r

decreases.

Question 4

A vector field $\vec{F}(x,y)$ in the plane is radially symmetric, meaning the vector at any point $(x,y)$ points directly toward or away from the origin and its magnitude depends only on the distance $r = \sqrt{x^2+y^2}$ . Which of the following formulas could represent such a field?

$\vec{F}(x,y) = \langle x^2, y^2 \rangle$
$\vec{F}(x,y) = \langle y, -x \rangle$
$\vec{F}(x,y) = \langle -y, x \rangle$
$\vec{F}(x,y) = \frac{1}{x^2+y^2+1}\langle x, y \rangle$ (correct answer)

Explanation: When you encounter radially symmetric vector fields, you need to check two key properties: the vectors must point directly toward or away from the origin, and the magnitude should depend only on the distance from the origin. For a vector field to be radially symmetric, it must have the form

\vec{F}(x,y) = f(r)\langle x, y \rangle

where

r = \sqrt{x^2+y^2}

and

f(r)

is some function of distance only. The unit vector

\langle x, y \rangle/r

points radially outward from the origin, so multiplying by any function of

r

preserves this radial direction. Option D fits this pattern perfectly:

\vec{F}(x,y) = \frac{1}{x^2+y^2+1}\langle x, y \rangle

. Here,

f(r) = \frac{1}{r^2+1}

, which depends only on distance, and the direction is purely radial. Option A gives

\langle x^2, y^2 \rangle

, which doesn't point radially. For example, at point

(1,2)

, this gives

\langle 1, 4 \rangle

, but the radial direction should be

\langle 1, 2 \rangle

. Options B and C represent rotational fields that are perpendicular to the radial direction. At any point

(x,y)

, option B gives

\langle y, -x \rangle

and option C gives

\langle -y, x \rangle

, both of which are orthogonal to the position vector

\langle x, y \rangle

. Study tip: For radially symmetric fields, always look for the pattern

f(r)\langle x, y \rangle

where the function multiplying the position vector depends only on distance

r

Question 5

Two vector fields $\mathbf{F}_1$ and $\mathbf{F}_2$ have identical field line patterns (same curves), but $|\mathbf{F}_2| = 3|\mathbf{F}_1|$ at every point. A particle released from rest will follow these field lines under the influence of each field as a force. How do the particle trajectories compare?

The trajectories follow identical paths but the particle under $\mathbf{F}_2$ travels exactly 3 times faster at each point (correct answer)
The trajectories follow identical paths but the particle under $\mathbf{F}_2$ reaches any given point in 1/3 the time
The trajectories follow identical paths and the particle under $\mathbf{F}_2$ travels $\sqrt{3}$ times faster at each point
The particles follow identical paths with identical speeds since field line geometry determines motion

Explanation: Since both fields have identical field line patterns, particles follow the same geometric paths. When the fields represent forces, the acceleration at each point is proportional to the field magnitude. If |F₂| = 3|F₁| everywhere, then the acceleration under F₂ is 3 times greater at each point. Starting from rest and following the same path with consistently 3 times the acceleration leads to 3 times the speed at each corresponding position along the trajectory. Choice B incorrectly relates to time rather than speed comparison. Choice C incorrectly uses √3. Choice D wrongly assumes field magnitude doesn't affect particle dynamics.

Question 6

The vector field $\mathbf{F}(x,y) = \langle f(x,y), g(x,y) \rangle$ has field lines that are concentric circles centered at the origin. If $\frac{\partial f}{\partial y} + \frac{\partial g}{\partial x} = 0$ everywhere, what can be concluded about the relationship between $f$ and $g$ ?

The functions satisfy $f(x,y) = -ky$ and $g(x,y) = kx$ for some nonzero constant $k$ (correct answer)
The functions satisfy $f(x,y) = kx$ and $g(x,y) = ky$ for some positive constant $k$
The functions are related by $f^2 + g^2 = k(x^2 + y^2)$ for some constant $k$
The functions satisfy $xf + yg = 0$ but their individual forms depend on the circle radius

Explanation: For field lines to be concentric circles, the vector field must be tangent to circles centered at the origin. This means F⊥⟨x,y⟩, so F·⟨x,y⟩ = xf + yg = 0. The condition ∂f/∂y + ∂g/∂x = 0 means the field has zero divergence. Combined with the tangency condition, this uniquely determines that f = -ky and g = kx for some constant k. Choice B gives a radial field (diverging from origin). Choice C gives a magnitude relationship but doesn't specify direction. Choice D correctly identifies the tangency condition but incorrectly suggests the form varies with radius.

Question 7

Consider the vector field in $\mathbb{R}^3$ given by $\vec{F}(x, y, z) = \langle -y, x, 1 \rangle$ . Which statement best describes the field lines of $\vec{F}$ ?

They are circles in planes parallel to the $xy$ -plane.
They are straight lines parallel to the $z$ -axis.
They are helices that spiral around the $z$ -axis. (correct answer)
They are straight lines radiating from the origin.

Explanation: The vector field can be understood by its components. The component in the

xy

-plane is

\langle -y, x \rangle

, which creates a rotational flow counter-clockwise around the

z

-axis. The component in the

z

-direction is

\langle 0, 0, 1 \rangle

, which is a constant upward flow. When combined, a particle moving along a field line will both rotate around the

z

-axis and move upward with a constant speed in the

z

-direction. This combined motion traces out a helix.

Question 8

Let $\vec{F}$ be a continuously differentiable vector field on $\mathbb{R}^2$ . Which of the following statements about the field lines of $\vec{F}$ is NOT always true?

Two distinct field lines cannot intersect at a point where $\vec{F}$ is non-zero.
The density of drawn field lines in a diagram indicates the magnitude of the vector field. (correct answer)
If $\vec{F}$ is a conservative field, its field lines cannot form closed loops.
A field line must be everywhere tangent to the vector field $\vec{F}$ .

Explanation: While it is a common convention in physics diagrams (e.g., for electric fields) to draw field lines closer together where the field is stronger, this is just a visualization aid, not a mathematical property. The mathematical definition of field lines only concerns their direction (they must be tangent to the field vectors). One can draw any number of field lines. The other statements are true. A) is a consequence of the uniqueness theorem for solutions of ordinary differential equations. D) is the definition of a field line. C) is true because the line integral of a conservative field over any closed path is zero, but the line integral along a closed field line must be positive, a contradiction.

Question 9

A vector field is given by $\vec{F}(x,y) = \langle 2y, -x \rangle$ . Which of the following parameterized curves $\vec{r}(t)$ corresponds to a field line of $\vec{F}$ ?

$\vec{r}(t) = \langle \cos(t), \sin(t) \rangle$
$\vec{r}(t) = \langle \cos(2t), \sin(2t) \rangle$
$\vec{r}(t) = \langle \sqrt{2}\cos(\sqrt{2}t), \sin(\sqrt{2}t) \rangle$ (correct answer)
$\vec{r}(t) = \langle e^{\sqrt{2}t}, e^{-\sqrt{2}t} \rangle$

Explanation: A curve

\vec{r}(t) = \langle x(t), y(t) \rangle

is a field line if its tangent vector,

\vec{r}'(t)

, is parallel to the vector field

\vec{F}(x(t), y(t))

at every point. This means

\vec{r}'(t) = k \vec{F}(\vec{r}(t))

for some scalar

k

. Let's test choice C:

\vec{r}(t) = \langle \sqrt{2}\cos(\sqrt{2}t), \sin(\sqrt{2}t) \rangle

. The tangent vector is

\vec{r}'(t) = \langle \frac{d}{dt}(\sqrt{2}\cos(\sqrt{2}t)), \frac{d}{dt}(\sin(\sqrt{2}t)) \rangle = \langle -2\sin(\sqrt{2}t), \sqrt{2}\cos(\sqrt{2}t) \rangle

. The vector field along the curve is

\vec{F}(\vec{r}(t)) = \langle 2y(t), -x(t) \rangle = \langle 2\sin(\sqrt{2}t), -\sqrt{2}\cos(\sqrt{2}t) \rangle

. Comparing

\vec{r}'(t)

and

\vec{F}(\vec{r}(t))

, we see that

\vec{r}'(t) = -1 \cdot \vec{F}(\vec{r}(t))

. Since the tangent vector is a scalar multiple of the field vector, this curve is a field line.

Question 10

Consider the vector field $\vec{F}(x,y) = \langle x - y^2, y - x^2 \rangle$ . How would the behavior of the flow near the origin $(0,0)$ be best described?

A source, where flow moves directly away from the origin. (correct answer)
A sink, where flow moves directly toward the origin.
A center, where flow circulates in closed loops around the origin.
A saddle point, with both incoming and outgoing flow lines.

Explanation: To analyze the behavior near an equilibrium point, we can compute the divergence and curl, or linearize the system. The origin is an equilibrium point since

\vec{F}(0,0) = \langle 0,0 \rangle

. The divergence is

\mathrm{div}(\vec{F}) = \frac{\partial}{\partial x}(x-y^2) + \frac{\partial}{\partial y}(y-x^2) = 1 + 1 = 2

. Since the divergence at the origin is positive, there is a net outflow from the origin, which characterizes a source. The curl is

\mathrm{curl}(\vec{F}) = \frac{\partial}{\partial x}(y-x^2) - \frac{\partial}{\partial y}(x-y^2) = -2x - (-2y) = 2y-2x

. At the origin, the curl is 0, indicating no local rotation. A source with no rotation means flow moves directly away from the origin.