Multivariable-Calculus · Unit 19 · Video 4 · Interactive Practice

Curl in the Plane: Measuring How a Field Rotates

IKey Formulas

FormulaNameWhat it says
curl⁡F⃗=Nx−My\operatorname{curl}\vec F = N_x - M_yCurl in the planeF⃗=∇f⇒curl⁡F⃗=0\vec F = \nabla f \Rightarrow \operatorname{curl}\vec F = 0
⟨a,b⟩\langle a, b\rangle, ⟨x,y⟩\langle x, y\rangle: curl =0= 0Constant and radial flowsTranslation and expansion do not rotate
⟨−ωy, ωx⟩\langle -\omega y,\ \omega x\rangle: curl =2ω= 2\omegaRigid rotationCurl is twice the angular velocity
torquemoment of inertia=angular acceleration\dfrac{\text{torque}}{\text{moment of inertia}} = \text{angular acceleration}Force field readingCurl of a force field measures this, up to the factor 22

Key Insight: a small floating object placed at a point spins with angular velocity 12curl⁡F⃗\tfrac12\operatorname{curl}\vec F there; the sign gives the direction (positive = counterclockwise).

IIA Floating Wheel Spins at Half the Curl

Only the rotation part of a flow turns a small floating wheel, and its angular velocity is 12(Nx−My)\tfrac12(N_x - M_y) at that point.

IIICurl Times Area Matches the Work Around a Loop

For the rotation field, curl 22 times the disk's area πρ2\pi\rho^2 equals the work around the circle — Green's theorem, next lecture.

IVQuiz Questions

Problem 1 · Compute a Curl

Given: F⃗=⟨3y, 5x⟩\vec F = \langle 3y,\ 5x\rangle — find curl⁡F⃗\operatorname{curl}\vec F.

✅ Correct! Nx−My=5−3=2N_x - M_y = 5 - 3 = 2.
❌ It is a difference, Nx−MyN_x - M_y, not a sum.
❌ Check the order. The curl is Nx−MyN_x - M_y, not My−NxM_y - N_x.
❌ Wrong partials. MxM_x and NyN_y are both 00, but the curl uses NxN_x and MyM_y.
❌ Not quite. Compute NxN_x and MyM_y.
Show solution

M=3yM = 3y, N=5xN = 5x, so Nx=5N_x = 5 and My=3M_y = 3:

curl⁡F⃗=Nx−My=5−3=2.\operatorname{curl}\vec F = N_x - M_y = 5 - 3 = 2.

Since the curl is not zero, F⃗\vec F is not a gradient field.

Problem 2 · Curl Versus Angular Velocity

Given: the velocity field v⃗=⟨−3y, 3x⟩\vec v = \langle -3y,\ 3x\rangle of a fluid. Find the angular velocity of a small floating object.

✅ Correct! The curl is 3−(−3)=63 - (-3) = 6, and the angular velocity is half of it: 33, counterclockwise.
❌ That is the curl. The angular velocity is half the curl.
❌ Check the sign. A positive curl means counterclockwise rotation.
❌ Not quite. Compute the curl, then use ω=12curl⁡\omega = \tfrac12\operatorname{curl}.
Show solution

Nx=3N_x = 3 and My=−3M_y = -3, so curl⁡v⃗=3−(−3)=6\operatorname{curl}\vec v = 3 - (-3) = 6. For a velocity field the curl is twice the angular velocity:

ω=12⋅6=3.\omega = \tfrac12 \cdot 6 = 3.

Directly: ⟨−3y,3x⟩\langle -3y, 3x\rangle is the rotation field with ω=3\omega = 3; a point at distance rr moves at speed 3r3r.

Problem 3 · Where Does It Turn Clockwise?

Given: the velocity field v⃗=⟨x2−y2, 2xy⟩\vec v = \langle x^2 - y^2,\ 2xy\rangle. Where does a small floating object spin clockwise?

✅ Correct! curl⁡v⃗=2y−(−2y)=4y\operatorname{curl}\vec v = 2y - (-2y) = 4y, negative exactly when y<0y < 0.
❌ Check the sign convention. Positive curl is counterclockwise; clockwise needs 4y<04y < 0.
❌ Check which variable. Nx=∂x(2xy)=2yN_x = \partial_x(2xy) = 2y and My=−2yM_y = -2y, so the curl depends on yy.
❌ Recompute MyM_y. ∂y(x2−y2)=−2y\partial_y(x^2 - y^2) = -2y, not 2y2y.
❌ Not quite. Compute Nx−MyN_x - M_y and find where it is negative.
Show solution
Nx=∂∂x(2xy)=2y,My=∂∂y(x2−y2)=−2y,curl⁡v⃗=4y.N_x = \frac{\partial}{\partial x}(2xy) = 2y, \qquad M_y = \frac{\partial}{\partial y}(x^2 - y^2) = -2y, \qquad \operatorname{curl}\vec v = 4y.

The object spins at 12⋅4y=2y\tfrac12\cdot 4y = 2y: counterclockwise where y>0y > 0, clockwise where y<0y < 0, and not at all on the xx-axis. The curl is a function of position, not a constant.

Problem 4 · A Force Field

Given: the force field F⃗=⟨y2, 2xy+x⟩\vec F = \langle y^2,\ 2xy + x\rangle, defined on the whole plane. Which statement is correct?

✅ Correct! Nx−My=(2y+1)−2y=1N_x - M_y = (2y + 1) - 2y = 1, a nonzero constant.
❌ Missed a term. ∂x(2xy+x)=2y+1\partial_x(2xy + x) = 2y + 1, not 2y2y.
❌ It is a difference. The 2y2y terms cancel in Nx−MyN_x - M_y.
❌ Check the order. The curl is Nx−MyN_x - M_y.
❌ Not quite. Compute NxN_x and MyM_y carefully.
Show solution
Nx=∂∂x(2xy+x)=2y+1,My=∂∂y(y2)=2y,curl⁡F⃗=1.N_x = \frac{\partial}{\partial x}(2xy + x) = 2y + 1, \qquad M_y = \frac{\partial}{\partial y}(y^2) = 2y, \qquad \operatorname{curl}\vec F = 1.

A nonzero curl means F⃗\vec F is not a gradient field, so it is not conservative. As a force field, its curl measures torque per unit moment of inertia; here it is the same constant at every point, counterclockwise.

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