Multivariable-Calculus · Unit 17 · Video 1 · Interactive Practice

Vector Fields: Sketching Constant, Radial and Rotating Fields

IKey Formulas

Field	Name	What it does
$\vec F = M(x,y)\,\hat\imath + N(x,y)\,\hat\jmath$	Vector field	One vector at every point, drawn with its tail at that point
$\vec F = 2\hat\imath + \hat\jmath$	Constant field	The same arrow $\langle 2, 1\rangle$ everywhere
$\vec F = x\,\hat\imath + y\,\hat\jmath$	Radial field	Points away from the origin, $\|\vec F\| = \sqrt{x^2+y^2} = r$
$\vec F = -y\,\hat\imath + x\,\hat\jmath$	Rotation field	Position vector turned $90\degree$ counterclockwise, $\|\vec F\| = r$

Key Insight: $\langle x, y\rangle \cdot \langle -y, x\rangle = -xy + yx = 0$ , so the rotation field is perpendicular to the radius everywhere and its arrows circle the origin; $\langle y, -x\rangle$ is the same rotation run clockwise.

IIOne Vector at Every Point

At each point $P$ the field gives one vector $\langle M(P), N(P)\rangle$ , drawn with its tail at $P$ , not at the origin.

Field

IIIUniform Rotation

In the field $-y\,\hat\imath + x\,\hat\jmath$ a particle at distance $r$ moves at speed $r$ , so every particle turns one radian per unit time.

Field time t 0.00

IVQuiz Questions

Problem 1 · Evaluate the Field at a Point

Given: $\vec F = -y\,\hat\imath + x\,\hat\jmath$ — find the vector the field attaches to the point $(2, -1)$ .

✅ Correct!

M = -y = 1

and

N = x = 2

; the arrow

\langle 1, 2\rangle

is drawn with its tail at

(2, -1)

❌ That is the position vector. The rotation field turns

\langle x, y\rangle

90\degree

: its components are

-y

and

x

❌ Wrong direction of rotation.

\langle -1, -2\rangle

\langle y, -x\rangle

, the clockwise field.

❌ Check the sign of $M$ . With

y = -1

M = -y = -(-1) = 1

Show solution

Substitute $x = 2$ , $y = -1$ into each component:

M = -y = -(-1) = 1, \qquad N = x = 2, \qquad \vec F(2, -1) = \langle 1, 2\rangle.

Check: $\langle 2, -1\rangle \cdot \langle 1, 2\rangle = 2 - 2 = 0$ , so the arrow is perpendicular to the radius, as it must be for the rotation field.

Problem 2 · Describe the Sketch

Given: $\vec F = y\,\hat\imath$ — which description matches its sketch?

✅ Correct!

N = 0

makes every arrow horizontal, and

M = y

sets its signed length.

❌ That is the sketch of $x\,\hat\imath$ . Here the length depends on

y

, so it is zero on the line

y = 0

, the

x

-axis.

❌ Not quite. The vector is

y\,\hat\imath = \langle y, 0\rangle

: the

y

sets the length, but

\hat\imath

points horizontally.

❌ Check the sign below the axis. For

y < 0

\langle y, 0\rangle

has a negative first component, so it points left.

Show solution

$\vec F = \langle y, 0\rangle$ . The second component is $0$ , so every arrow is horizontal. On the $x$ -axis $y = 0$ , giving the zero vector. At $y = 1$ the arrow is $\hat\imath$ , at $y = 2$ it is $2\hat\imath$ ; at $y = -1$ it is $-\hat\imath$ , pointing left. Moving left or right changes nothing, because $x$ does not appear.

Problem 3 · Speed and Period in the Rotation Field

Given: fluid moves with velocity field $\vec F = -y\,\hat\imath + x\,\hat\jmath$ and a particle is at $(3, 4)$ — find its speed and the time it takes to go once around the origin.

What is the particle's speed?

How long does one full turn take?

✅ Correct! Speed

r = 5

on a circle of length

10\pi

: one turn takes

10\pi/5 = 2\pi

, the same for every radius.

❌ Check the speed.

|\langle -4, 3\rangle| = \sqrt{16 + 9}

— the length of the vector, not the sum or the squared length of its components.

❌ Check the period. Time is distance over speed: the circle through

(3,4)

has length

2\pi \cdot 5

, covered at speed

5

Show solution

At $(3, 4)$ the field is $\langle -4, 3\rangle$ , so the speed is

|\vec F| = \sqrt{(-4)^2 + 3^2} = 5 = r.

The particle stays on the circle of radius $r = 5$ (the velocity is always perpendicular to the radius), whose length is $2\pi r = 10\pi$ . At speed $5$ ,

\text{time} = \frac{2\pi r}{r} = \frac{10\pi}{5} = 2\pi.

Problem 4 · A Faster Rotation

Given: the velocity field $\vec F = -2y\,\hat\imath + 2x\,\hat\jmath$ — find how long a particle takes to go once around the origin.

✅ Correct! The speed is

2r

, so the angle grows at

2

radians per unit time and one turn takes

2\pi/2 = \pi

❌ Not quite.

2\pi

is the period of

\langle -y, x\rangle

; doubling every arrow doubles the speed.

❌ Wrong direction. A faster particle needs less time: divide the circumference by the speed

2r

, don't multiply.

❌ Too fast. Speed

2r

halves the period of

2\pi

; it does not quarter it.

Show solution

$\vec F = 2\langle -y, x\rangle$ is perpendicular to the radius, so particles move on circles about the origin, with speed

|\vec F| = 2\sqrt{x^2 + y^2} = 2r.

From $s = r\theta$ , $\dfrac{ds}{dt} = r\dfrac{d\theta}{dt} = 2r$ , so $\dfrac{d\theta}{dt} = 2$ for every radius. One turn is $\theta = 2\pi$ , reached at $t = \pi$ .

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