Multivariable-Calculus · Unit 19 · Video 3 · Interactive Practice

Finding a Potential Function by Partial Antiderivatives

IKey Formulas

Step	For $\vec F = \langle 4x^2 + 8xy,\ 3y^2 + 4x^2\rangle$	Why
Integrate $f_x = M$ in $x$	$f = \tfrac43 x^3 + 4x^2 y + g(y)$	The "constant" may depend on $y$
Differentiate in $y$ , match $f_y = N$	$4x^2 + g'(y) = 3y^2 + 4x^2 \Rightarrow g'(y) = 3y^2$	$g'$ must contain no $x$
Integrate $g'$	$g = y^3 + c$ , so $f = \tfrac43 x^3 + 4x^2 y + y^3 + c$	Now $c$ is a true constant

Key Insight: integrate one equation, then differentiate and match the other — never integrate both and merge. If an $x$ survives in $g'(y)$ , the field is not a gradient field (or there is a slip).

IIThe Missing Piece Is a Function of y Alone

The gradient of the partial answer misses $\vec F$ by exactly $\langle 0,\ g'(y)\rangle$ , a gap that depends on $y$ only.

Integrate first in Step 1

Step 1 — Integrate $f_x = 4x^2 + 8xy$ in $x$

f = \tfrac43 x^3 + 4x^2 y + g(y)

Step 2 — Differentiate in $y$ and match $f_y = N$

f_y = 4x^2 + g'(y) = 3y^2 + 4x^2

g'(y) = 3y^2 \quad\text{(no } x\text{)}

Step 3 — Integrate $g'$

g = y^3 + c, \qquad f = \tfrac43 x^3 + 4x^2 y + y^3 + c

Step 1 — Integrate $f_y = 3y^2 + 4x^2$ in $y$

f = y^3 + 4x^2 y + h(x)

Step 2 — Differentiate in $x$ and match $f_x = M$

f_x = 8xy + h'(x) = 4x^2 + 8xy

h'(x) = 4x^2 \quad\text{(no } y\text{)}

Step 3 — Integrate $h'$

h = \tfrac43 x^3 + c, \qquad f = y^3 + 4x^2 y + \tfrac43 x^3 + c

IIIThe Method Checks Itself

With $8xy$ replaced by $Axy$ , the leftover $g'(y) = 3y^2 + \big(4 - \tfrac{A}{2}\big)x^2$ is free of $x$ only when $A = 8$ .

A 6

💡 A legitimate $g'(y)$ is the same all along each horizontal line, so every row of arrows must be uniform.

IVQuiz Questions

Problem 1 · The Right Integration Constant

Given: $\vec F = \langle 2xy + 3,\ x^2 + 4y\rangle$ . After integrating $f_x = 2xy + 3$ with respect to $x$ , which expression is correct?

✅ Correct! Anything with zero

x

-derivative can be added, so the constant is a function

g(y)

❌ Too narrow. A term such as

2y^2

also has zero partial in

x

; the constant must be allowed to depend on

y

❌ Wrong variable. A function of

x

has a nonzero

x

-derivative in general; the unknown depends on

y

❌ Not quite. Integrate in

x

, treating

y

as constant.

Show solution

Treat $y$ as a constant: $\int (2xy + 3)\,dx = x^2 y + 3x$ , plus anything whose $x$ -derivative is zero, i.e. a function of $y$ :

f = x^2 y + 3x + g(y).

Problem 2 · Match the Other Equation

Continue Problem 1: using $f_y = N = x^2 + 4y$ , find $g'(y)$ .

✅ Correct!

f_y = x^2 + g'(y)

must equal

x^2 + 4y

, so

g' = 4y

— no

x

, as it must be. Then

f = x^2 y + 3x + 2y^2 + c

❌ Subtract what is already there.

f_y

x^2 y + 3x

x^2

, which cancels against

N

x^2

❌ An $x$ survived — that signals an algebra slip here, since the field does pass the test.

❌ That is $g$ , not $g'$ .

❌ Not quite. Differentiate

x^2 y + 3x + g(y)

y

and match

N

Show solution

f_y = \frac{\partial}{\partial y}\big(x^2 y + 3x + g(y)\big) = x^2 + g'(y).

Matching $f_y = x^2 + 4y$ : the $x^2$ terms cancel, leaving $g'(y) = 4y$ , so $g = 2y^2 + c$ and $f = x^2 y + 3x + 2y^2 + c$ .

Problem 3 · A Trigonometric Potential

Given: $\vec F = \langle y\cos(xy) + 2x,\ x\cos(xy) + 3y^2\rangle$ , a gradient field. Find a potential $f$ .

✅ Correct!

f_x = y\cos(xy) + 2x

and

f_y = x\cos(xy) + 3y^2

❌ Counted $\sin(xy)$ twice. That is the risk of integrating both equations and merging.

❌ Check the antiderivative.

\partial_x \cos(xy) = -y\sin(xy)

; you need

\sin(xy)

❌ Missing $g(y)$ . Matching

f_y

leaves

g'(y) = 3y^2

❌ Not quite. Integrate in

x

, then match

f_y

Show solution

Integrate $f_x$ in $x$ : $f = \sin(xy) + x^2 + g(y)$ .

Differentiate in $y$ : $f_y = x\cos(xy) + g'(y)$ . Matching $x\cos(xy) + 3y^2$ gives $g'(y) = 3y^2$ , so $g = y^3 + c$ :

f = \sin(xy) + x^2 + y^3 + c.

Problem 4 · When an x Survives

Given: $\vec F = \langle 3x^2 y + a\,y^2,\ x^3 + 4xy\rangle$ . Integrating $f_x$ in $x$ gives $f = x^3 y + a\,x y^2 + g(y)$ . Which value of $a$ leaves $g'(y)$ free of $x$ , and what is $g'(y)$ then?

✅ Correct!

g'(y) = (4 - 2a)xy

is free of

x

only for

a = 2

, and then

g' = 0

; the test agrees:

M_y = 3x^2 + 2ay

N_x = 3x^2 + 4y

❌ Check the derivative.

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

, so matching

4xy

needs

2a = 4

❌ That $g'$ still contains $x$ . With

a = 2

the

4xy

terms cancel.

❌ Not quite. Differentiate

f

y

and match

x^3 + 4xy

Show solution

f_y = x^3 + 2a\,xy + g'(y) = x^3 + 4xy \;\Rightarrow\; g'(y) = (4 - 2a)\,xy.

A function of $y$ alone cannot contain $x$ , so $4 - 2a = 0$ : $a = 2$ , and then $g'(y) = 0$ , $g = c$ , $f = x^3 y + 2xy^2 + c$ .

The $x$ -derivative of the leftover, $(4 - 2a)y$ , is exactly $N_x - M_y = (3x^2 + 4y) - (3x^2 + 2ay)$ .

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