Multivariable-Calculus · Unit 17 · Video 2 · Interactive Practice

Work as a Line Integral: Computing ∫ F · dr Along a Path

IKey Formulas

Formula	Name	What it says
$W \approx \vec F \cdot \Delta\vec r$	Work of a small motion	Positive along the force, negative against it, zero when perpendicular
$\displaystyle W = \int_C \vec F \cdot d\vec r$	Line integral	The limit of $\sum_i \vec F \cdot \Delta\vec r_i$ as the pieces of $C$ shrink
$\displaystyle \int_C \vec F \cdot d\vec r = \int_{t_1}^{t_2} \vec F\big(x(t), y(t)\big) \cdot \frac{d\vec r}{dt}\,dt$	Computing it	Write the force and the velocity in terms of $t$

Key Insight: the path is part of the data. For $\vec F = \langle -y, x\rangle$ from $(0,0)$ to $(1,1)$ the work is $\tfrac13$ along $y = x^2$ but $-1$ up the $y$ -axis and across.

IISumming $\vec F \cdot \Delta\vec r_i$

Cut $C$ into $n$ pieces and add $\vec F \cdot \Delta\vec r_i$ ; as the pieces shrink the sum approaches the line integral.

Path pieces n 4

IIIEverything in Terms of $t$

At time $t$ the integrand is the force at the particle's position dotted with its velocity, an ordinary function of $t$ .

Path t 0.60

IVQuiz Questions

Problem 1 · Work Along a Parabola

Given: $\vec F = y\,\hat\imath + \hat\jmath$ and $C$ : $x = t$ , $y = t^2$ , $0 \le t \le 1$ — find $\displaystyle\int_C \vec F \cdot d\vec r$ .

✅ Correct!

\langle t^2, 1\rangle \cdot \langle 1, 2t\rangle = t^2 + 2t

, and

\int_0^1 (t^2 + 2t)\,dt = \tfrac13 + 1

❌ A term is missing. The

\hat\jmath

component of

\vec F

1

, and it multiplies

dy/dt = 2t

❌ Dot with the velocity, not the position. The integrand is

\vec F \cdot \dfrac{d\vec r}{dt}

, and

\dfrac{d\vec r}{dt} = \langle 1, 2t\rangle

❌ A term is missing. The

\hat\imath

component

y = t^2

multiplies

dx/dt = 1

Show solution

Along $C$ , $\vec F = \langle y, 1\rangle = \langle t^2, 1\rangle$ and $\dfrac{d\vec r}{dt} = \langle 1, 2t\rangle$ . So

\int_C \vec F \cdot d\vec r = \int_0^1 \big(t^2 \cdot 1 + 1 \cdot 2t\big)\,dt = \Big[\tfrac{t^3}{3} + t^2\Big]_0^1 = \tfrac13 + 1 = \tfrac43.

Problem 2 · Match the Components

Given: $\vec F = -y\,\hat\imath + x\,\hat\jmath$ and $C$ : $x = t$ , $y = 2$ , $0 \le t \le 3$ (the segment from $(0,2)$ to $(3,2)$ ) — find the work done by $\vec F$ along $C$ .

✅ Correct!

\langle -2, t\rangle \cdot \langle 1, 0\rangle = -2

: the force's component along the motion points backward the whole way.

❌ Check the sign.

M = -y = -2

, and the particle moves in the

+x

direction, so the force resists the motion.

❌ Components mismatched.

dx/dt = 1

pairs with

M = -y

, and

dy/dt = 0

pairs with

N = x

; you dotted

N

with

dx/dt

❌ Not perpendicular here.

\vec F = \langle -2, t\rangle

has a horizontal component

-2

along this horizontal path.

Show solution

Along $C$ , $\vec F = \langle -y, x\rangle = \langle -2, t\rangle$ and $\dfrac{d\vec r}{dt} = \langle 1, 0\rangle$ , so

W = \int_0^3 \big((-2)(1) + t \cdot 0\big)\,dt = \int_0^3 -2\,dt = -6.

Problem 3 · Same Endpoints, Two Paths

Given: $\vec F = y\,\hat\imath - x\,\hat\jmath$ — find the work from $(0,0)$ to $(1,1)$ along the segment $C_1$ : $x = t$ , $y = t$ , and along the parabola $C_2$ : $x = t$ , $y = t^2$ (both $0 \le t \le 1$ ).

Work along the segment $C_1$ ?

Work along the parabola $C_2$ ?

✅ Correct!

0

along the segment and

-\tfrac13

along the parabola: same force, same endpoints, different work.

❌ Check $C_1$ . Along

y = x

the force

\langle t, -t\rangle

is perpendicular to the velocity

\langle 1, 1\rangle

❌ Check $C_2$ .

\langle t^2, -t\rangle \cdot \langle 1, 2t\rangle = t^2 - 2t^2

; the path matters, so do not reuse the segment's answer.

Show solution

Segment: $\vec F = \langle t, -t\rangle$ , $\dfrac{d\vec r}{dt} = \langle 1, 1\rangle$ , so $\vec F \cdot \dfrac{d\vec r}{dt} = t - t = 0$ and the work is $0$ .

Parabola: $\vec F = \langle t^2, -t\rangle$ , $\dfrac{d\vec r}{dt} = \langle 1, 2t\rangle$ , so

W = \int_0^1 (t^2 - 2t^2)\,dt = \int_0^1 -t^2\,dt = -\tfrac13.

The two values differ, so this work depends on the path between the endpoints.

Problem 4 · Work and Kinetic Energy

Given: $\vec F$ is the only force acting on a particle of mass $2$ , and along the curve $C$ its speed rises from $1$ to $3$ — find $\displaystyle\int_C \vec F \cdot d\vec r$ .

✅ Correct! With

\vec F

the only force, the work equals the change in kinetic energy,

\tfrac12 \cdot 2 \cdot (9 - 1) = 8

❌ Not quite. Kinetic energy is

\tfrac12 m v^2

: square the speeds before subtracting.

❌ Missing the factor $\tfrac12$ .

\Delta\big(\tfrac12 m v^2\big) = \tfrac12 \cdot 2 \cdot (3^2 - 1^2)

❌ There is a shortcut here. When

\vec F

is the only force acting, its work equals the change in the particle's kinetic energy.

Show solution

If $\vec F$ is the only force acting, the work it does equals the change in kinetic energy:

\int_C \vec F \cdot d\vec r = \tfrac12 m v_2^2 - \tfrac12 m v_1^2 = \tfrac12 \cdot 2 \cdot 3^2 - \tfrac12 \cdot 2 \cdot 1^2 = 9 - 1 = 8.

In general the line integral needs $\vec F$ and $C$ ; this shortcut applies only because $\vec F$ alone drives the motion.

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Work as a Line Integral: Computing ∫ F · dr Along a Path

IKey Formulas

IISumming F⃗⋅Δr⃗i\vec F \cdot \Delta\vec r_iF⋅Δri​

IIIEverything in Terms of ttt

IVQuiz Questions

IISumming $\vec F \cdot \Delta\vec r_i$

IIIEverything in Terms of $t$