SME
Calculus III (Multivariable) · Problem type
Work in a gradient field from its potential
Use this for “F = ∇φ with φ = 2x² − xy + y². Find the work done by F along any path from (1, 1) to (2, 3).”
Diagram
F = ⟨2x + 3y, 3x⟩
Work along the path W = ∫F·dr = 7
This is an example. Type your own values to start a new problem.
φ(x,
y)
=
a
x²
+
b
xy
+
c
y²
W
=
φ(
x₂
,
y₂
)
−
φ(
x₁
,
y₁
)
=
φ(B)
−
φ(A)
=
Formulas
φ(A)
=
a
×
x₁
2
+
b
×
x₁
×
y₁
+
c
×
y₁
2
0
=
1
×
0
2
+
3
×
0
×
0
+
0
×
0
2
φ(B)
=
a
×
x₂
2
+
b
×
x₂
×
y₂
+
c
×
y₂
2
7
=
1
×
1
2
+
3
×
1
×
2
+
0
×
2
2
W
=
φ(B)
−
φ(A)
7 = 7 − 0
Assumptions
•
F = ∇φ = ⟨2ax + by, bx + 2cy⟩ for the potential φ = ax² + bxy + cy².
•
Pᵧ = b = Qₓ everywhere in the plane, so F is conservative: the work depends only on where the path starts and ends.
•
The fundamental theorem for line integrals: W = ∫ F · dr = φ(B) − φ(A) on any path from A to B, and 0 round any closed loop.
Step-by-step
WE KNOW
a = 1, b = 3, c = 0, x₁ = 0, y₁ = 0, x₂ = 1, y₂ = 2
FIND
potential at A (φ(A)), potential at B (φ(B)), work from A to B (W)
Step 1 · Find potential at A (φ(A))
Use φ(A) = a × x₁² + b × x₁ × y₁ + c × y₁²
Put the start A into the potential φ.
φ(A)
=
1
×
0
2
+
3
×
0
×
0
+
0
×
0
2
φ(A) = 0 + 0 + 0
φ(A) = 0
Step 2 · Find potential at B (φ(B))
Use φ(B) = a × x₂² + b × x₂ × y₂ + c × y₂²
Put the end B into the potential φ.
φ(B)
=
1
×
1
2
+
3
×
1
×
2
+
0
×
2
2
φ(B) = 1 + 6
φ(B) = 7
Step 3 · Find work from A to B (W)
Use W = φ(B) − φ(A)
The fundamental theorem for line integrals: the potential at the end minus the potential at the start. No path is needed.
W
=
7
−
0
W
=
7
Check
Put the numbers back in. Both sides should match.
0
=
1
×
0
2
+
3
×
0
×
0
+
0
×
0
2
✓
7
=
1
×
1
2
+
3
×
1
×
2
+
0
×
2
2
✓
7 = 7 − 0 ✓
Topic check
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