Visual walkthrough — Line integrals — scalar and vector, work done

We build the whole thing from zero. If a word like "dot product" shows up, we draw it first.

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Step 1 — What a curve is, and why one number $t$ locates you on it

We call that one number $t$. As $t$ ticks from a start value $a$ to an end value $b$, a moving dot traces the curve. The dot's position is written

$\mathbf{r}(t)=(x(t),y(t)).$

• $\mathbf{r}$ (bold) — the position vector: an arrow from the origin to where the dot currently sits.
• $x(t),y(t)$ — the dot's coordinates as functions of the clock $t$.
• $t\in [a,b]$ — the clock runs from $a$ (start) to $b$ (finish).

WHY parametrise? Because once every point on the wiggly curve has an address $t$, any sum "over the curve" becomes an ordinary integral in the single variable $t$ — something we already know how to do.

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Step 2 — Velocity ${\mathbf{r}}^{\prime }(t)$: the arrow that points along the curve

• $\Delta$ ("delta") — plain shorthand for "a small change in". $\Delta \mathbf{r}$ is a vector: an arrow with both length and direction.
• $\Delta t$ — the tiny slice of clock-time for this step.

WHY divide by $\Delta t$? To get a rate that doesn't shrink to nothing as the step shrinks. Shrinking $\Delta t\to 0$:

${\mathbf{r}}^{\prime }(t)={\lim }_{\Delta t\to 0}\frac{\Delta \mathbf{r}}{\Delta t}=(x^{\prime }(t),y^{\prime }(t)).$

• ${\mathbf{r}}^{\prime }(t)$ — the velocity vector. Its direction is tangent to the curve (points where the dot is heading); its length $|{\mathbf{r}}^{\prime }(t)|$ is the speed (how fast arc-length is being eaten up).
• $x^{\prime }(t),y^{\prime }(t)$ — the derivatives of each coordinate: how fast $x$ and $y$ separately change.

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Step 3 — Two ways to measure one tiny step: length $ds$ vs displacement $d\mathbf{r}$

This is the fork in the road that produces two different integrals.

Length of the tiny step. Take the length of the displacement arrow:

$\Delta s=|\Delta \mathbf{r}|\approx |{\mathbf{r}}^{\prime }(t)|\Delta t.$

Shrinking gives the arc-length element $ds=|{\mathbf{r}}^{\prime }(t)|dt,|{\mathbf{r}}^{\prime }(t)|=\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}.$

• $ds$ — a scalar, always $\ge 0$ (it's a length; see Arc length).
• $|{\mathbf{r}}^{\prime }(t)|$ — the speed, from the square-root of squared component-rates. Squares kill signs, so direction is thrown away here — remember this.

Displacement of the tiny step. Keep the whole arrow: $d\mathbf{r}={\mathbf{r}}^{\prime }(t)dt.$

• $d\mathbf{r}$ — a vector. It remembers direction. Reverse your travel and ${\mathbf{r}}^{\prime }$ flips sign, so $d\mathbf{r}$ flips sign too.

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Step 4 — Scalar line integral: add up (value)$\times$(tiny length)

WHY the $\Delta s_i$ factor? Because mass = density $\times$ length, and $\Delta s_i$ is the length of the piece. Let the pieces shrink; the sum becomes an integral. Substitute the Step-3 fact $ds=|{\mathbf{r}}^{\prime }(t)|dt$ so everything is in $t$:

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Step 5 — Dot product: pulling out only the "forward" part of a push

Before the work integral, we must build the tool it uses. That tool is the dot product.

The number that extracts "how much of $\mathbf{F}$ lies along $\mathbf{d}$" is the dot product (full details in Dot product):

$\mathbf{F}\cdot \mathbf{d}=F_x{d}_x+F_y{d}_y=|\mathbf{F}||\mathbf{d}|\cos \theta .$

• $\theta$ — the angle between the two arrows.
• $\cos \theta$ — the dial that runs from $+1$ (same direction, full help) through $0$ (perpendicular, no help) to $-1$ (opposite, full fight).

WHY this tool and not another? Multiplication alone would ignore direction; the dot product is the unique simple operation that keeps only the aligned component — exactly what "work" needs.

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Step 6 — Work: add up (forward-push)$\times$(tiny displacement)

Substitute $d\mathbf{r}={\mathbf{r}}^{\prime }(t)dt$ from Step 3 and add up:

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Step 7 — The bridge: why the vector integral is a tangential scalar integral

Then, using Step 3's two facts together: $d\mathbf{r}={\mathbf{r}}^{\prime }dt=\underset{\mathbf{T}}{\underbrace{\frac{{\mathbf{r}}^{\prime }}{|{\mathbf{r}}^{\prime }|}}}\underset{ds}{\underbrace{|{\mathbf{r}}^{\prime }|dt}}=\mathbf{T}ds.$

So the work integral is literally a scalar line integral whose density is the forward-component $\mathbf{F}\cdot \mathbf{T}$:

${\int }_C\mathbf{F}\cdot d\mathbf{r}={\int }_C(\mathbf{F}\cdot \mathbf{T})ds.$

• $\mathbf{F}\cdot \mathbf{T}$ — a plain number at each point: "how strongly the field pushes forward here."

This is why work "feels like mass of a wire" but with a signed density.

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Step 8 — Every case: reversing orientation, sideways fields, zero speed

We must show what happens in the corner cases, or you'll meet one unprepared.

Case A — reverse the curve (walk it backwards). The travel direction flips, so ${\mathbf{r}}^{\prime }\to -{\mathbf{r}}^{\prime }$ and hence $d\mathbf{r}\to -d\mathbf{r}$ and $\mathbf{T}\to -\mathbf{T}$.

• Work: $W\to -W$. Dragging against the field undoes work. Sign flips.
• Scalar: $ds=|{\mathbf{r}}^{\prime }|dt$ has that square-root, so $|-{\mathbf{r}}^{\prime }|=|{\mathbf{r}}^{\prime }|$ — no change. Mass of a wire is the same whichever end you start from.

Case B — force exactly perpendicular to motion. Then $\theta =90^{\circ }$, $\cos \theta =0$, so $\mathbf{F}\cdot d\mathbf{r}=0$ on every step: zero work, no matter how strong $\mathbf{F}$ is. (A magnetic-type push that only steers, never speeds up.)

Case C — degenerate: the dot stops, ${\mathbf{r}}^{\prime }(t)=0$. At an instant where velocity is zero, $ds=0$ and $d\mathbf{r}=0$: that instant contributes nothing to either integral. Both formulas stay valid; a momentary pause adds no mass and no work. (If the dot is stuck for a whole interval, reparametrise to skip it.)

Recall Why is

$\mathbf{F}=(y,x)$ path-independent? Because $\mathbf{F}=\nabla \varphi$ with $\varphi =xy$, so $W=\varphi (\text{end})-\varphi (\text{start})$ — endpoints only. Preview of Fundamental Theorem for Line Integrals and Gradient and conservative fields. Check $0\to 1$: $\varphi (1,1)-\varphi (0,0)=1-0=1.$ ✓

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The one-picture summary

One tiny step splits into two children: its length $ds$ (direction erased by the square-root) feeds the scalar/mass integral; its displacement $d\mathbf{r}$ (direction kept) is dotted with $\mathbf{F}$ to feed the work integral. That single fork explains every difference — including why reversing the curve leaves mass alone but flips the sign of work.

Recall Feynman retelling (cover and explain aloud)

An ant walks a bent wire. Cut the trip into steps so short each is basically straight. For mass: each step has a little length; multiply by how heavy the wire is there; add all the pieces. Length has no arrow, so walking the wire backwards gives the same total — mass doesn't care about direction. For work: there's a wind at every point. On each short step I only count the part of the wind pushing the ant forward — that's the dot product, the tool that keeps the aligned part and drops the sideways part. Add "forward-push × step-length" all along. Walk backwards and forward-push becomes backward-fight, so the total flips sign — work does care about direction. Both totals come from the same little step; they only differ in whether we keep the step's arrow ($d\mathbf{r}$) or just its length ($ds$).

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Connections

• Line integrals — scalar and vector, work done — the parent this walkthrough unfolds.
• Parametric curves — Step 1's address system $\mathbf{r}(t)$.
• Arc length — Step 3/4 with $f=1$.
• Dot product — the engine built in Step 5.
• Gradient and conservative fields · Fundamental Theorem for Line Integrals — Step 8's path-independence.
• Green's Theorem — closed-curve version of the work integral.