Multivariable-Calculus · Unit 19 · Video 1 · Interactive Practice

The Gradient Field Test: My = Nx, and Where It Breaks

IKey Formulas

Formula	Name	What it says
$\vec F = \nabla f \;\Rightarrow\; M_y = N_x$	The test	From $M = f_x$ , $N = f_y$ and $f_{xy} = f_{yx}$
$M_y = N_x$ everywhere, $\vec F$ defined on the whole plane $\;\Rightarrow\; \vec F = \nabla f$	The converse	Needs $\vec F$ continuously differentiable at every point
$\langle -y,\, x\rangle$ : $M_y = -1 \ne 1 = N_x$	Rotation field	Fails the test: not a gradient field
$\vec G = \dfrac{\langle -y,\, x\rangle}{x^2+y^2}$ , $\displaystyle\oint_{x^2+y^2=1} \vec G \cdot d\vec r = 2\pi$	The counterexample	Passes the test off the origin, yet is not a gradient field

Key Insight: $M_y = N_x$ must hold identically, as expressions in $x$ and $y$ ; it is never an equation to solve for $x$ .

IIAgreement Everywhere, Not at One Point

For $\langle 4x^2 + Axy,\ 3y^2 + 4x^2\rangle$ , $M_y = Ax$ and $N_x = 8x$ ; only $A = 8$ makes every loop's work zero.

A 4

💡 Challenge: with $A \ne 8$ , find where to put the loop so its work is zero, and explain why that does not make the field conservative.

IIIOne Missing Point Breaks the Converse

$\vec G$ satisfies $M_y = N_x$ wherever it is defined, but a loop around the origin picks up work $2\pi$ .

loop radius 1

IVQuiz Questions

Problem 1 · Apply the Test

Which of these fields, all defined on the whole plane, is a gradient field?

✅ Correct!

M_y = 1 = N_x

and the field is defined everywhere; indeed

\langle y, x\rangle = \nabla(xy)

❌ Not quite. That is the rotation field:

M_y = -1

but

N_x = 1

❌ Not quite. Here

M_y = 1

and

N_x = -1

, so the test fails.

❌ Not quite. The test pairs

M

with

y

and

N

with

x

M_y = x

and

N_x = y

, which differ.

❌ Not quite. Compute

M_y

and

N_x

for each field.

Show solution

For each field compute $M_y$ and $N_x$ :

$\langle -y, x\rangle$ : $M_y = -1$ , $N_x = 1$ — fails.
$\langle y, x\rangle$ : $M_y = 1$ , $N_x = 1$ — passes.
$\langle y, -x\rangle$ : $M_y = 1$ , $N_x = -1$ — fails.
$\langle xy, xy\rangle$ : $M_y = x$ , $N_x = y$ — not equal as expressions, fails.

Only $\langle y, x\rangle$ passes, and since it is defined on the whole plane the converse makes it a gradient field: $\nabla(xy) = \langle y, x\rangle$ .

Problem 2 · Find the Constant

Given: $\vec F = (a\,x y^2 + 2x)\,\hat\imath + (6x^2 y - 1)\,\hat\jmath$ — find every value of $a$ that makes $\vec F$ a gradient field.

✅ Correct!

M_y = 2axy

and

N_x = 12xy

agree for all

x, y

exactly when

a = 6

❌ Check the derivative.

\partial_y(a x y^2) = 2axy

, not

axy

❌ That is solving for $x$ . The test asks for

M_y = N_x

at every point, as expressions.

❌ Not quite. Compute

M_y

and

N_x

and compare coefficients of

xy

Show solution

M_y = \frac{\partial}{\partial y}\big(a x y^2 + 2x\big) = 2a\,xy, \qquad N_x = \frac{\partial}{\partial x}\big(6x^2 y - 1\big) = 12\,xy.

Both are multiples of $xy$ ; they agree for every $(x, y)$ only when the coefficients agree: $2a = 12$ , so $a = 6$ . The components are polynomials, so $\vec F$ is defined on the whole plane and the converse applies.

Equality at $x = 0$ (or $y = 0$ ) holds for every $a$ , but that is not the question.

Problem 3 · Passing the Test Is Not Enough

Given: $\vec G = \dfrac{-y\,\hat\imath + x\,\hat\jmath}{x^2+y^2}$ , defined for $(x, y) \ne (0, 0)$ , with $\displaystyle\oint_C \vec G\cdot d\vec r = 2\pi$ around the unit circle. Which statement is correct?

✅ Correct! The converse needs

\vec G

defined everywhere; the missing origin breaks it, and the nonzero loop settles the question.

❌ Not quite. The converse requires the field to be defined at every point of the plane;

\vec G

is not defined at the origin.

❌ Recompute. Both partials equal

\dfrac{y^2 - x^2}{(x^2+y^2)^2}

❌ Not quite. The test alone cannot decide, but the work

2\pi

around a closed curve can.

❌ Not quite. Compute

M_y

and

N_x

, then use the work around the circle.

Show solution

With $M = \dfrac{-y}{x^2+y^2}$ and $N = \dfrac{x}{x^2+y^2}$ :

M_y = \frac{-(x^2+y^2) + 2y^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}, \qquad N_x = \frac{(x^2+y^2) - 2x^2}{(x^2+y^2)^2} = \frac{y^2-x^2}{(x^2+y^2)^2}.

So $\vec G$ passes the test wherever it is defined. But a gradient field is conservative, and its work around every closed curve would be $0$ . The work around the unit circle is $2\pi$ , so $\vec G$ is not a gradient field. The converse does not apply because $\vec G$ is undefined at the origin.

Problem 4 · Two Unknown Constants

Given: $\vec F = (a\,x^2 y + y^3)\,\hat\imath + (x^3 + b\,x y^2)\,\hat\jmath$ — find the values of $a$ and $b$ that make $\vec F$ a gradient field.

✅ Correct!

M_y = a x^2 + 3y^2

and

N_x = 3x^2 + b y^2

; matching the

x^2

and

y^2

coefficients gives

a = 3

b = 3

❌ Wrong pairing. Comparing

M_x = 2axy

with

N_y = 2bxy

is not the test; differentiate

M

y

and

N

x

❌ Check the $y^2$ terms.

\partial_y(y^3) = 3y^2

must match

\partial_x(b x y^2) = b y^2

❌ Not quite. Compute

M_y

and

N_x

and match coefficients term by term.

Show solution

M_y = a x^2 + 3y^2, \qquad N_x = 3x^2 + b y^2.

For these to agree at every point, the coefficients of $x^2$ and of $y^2$ must agree: $a = 3$ and $b = 3$ . The field is polynomial, so it is defined everywhere and the converse applies. (Indeed $\vec F = \nabla(x^3 y + x y^3)$ .)

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