Conservative vector fields — potential functions: Visual walkthrough

This page is the picture-by-picture derivation of the boxed result on the parent note, Conservative vector fields — potential functions: $\int_{C} \nabla f \cdot d r = f (B) - f (A) .$

We will earn every piece of that formula.

Step 1 — What is a vector field? (an arrow at every point)

WHAT: We picture the plane as a field of grass, and at each blade we glue a little arrow.

WHY: Everything else — force, work, potential — is read off these arrows. We must see them first.

PICTURE: In the figure, the black arrows are $F$ . Look at one red arrow: its horizontal shadow is $P$ , its vertical shadow is $Q$ . That is all $P$ and $Q$ mean — the two shadows of one arrow.

Figure — Conservative vector fields — potential functions

Step 2 — The hidden hill: a scalar function $f$

WHAT: Instead of an arrow at each point, $f$ puts a height at each point.

WHY: A conservative field is one that came from such a hill. Before we can say "the arrows point downhill", we need the hill drawn.

PICTURE: The nested red loops are contour lines of $f$ — each loop is one fixed altitude. Where loops crowd together, the hill is steep; where they spread out, it is flat.

Step 3 — The gradient $\nabla f$ : the steepest-uphill arrow

WHAT: We build a new arrow at each point from the two directional slopes of the hill.

WHY use two partial slopes and not one? In one dimension a hill has a single slope. In two dimensions the steepness depends on which way you face, so we need two numbers — the rise going east and the rise going north — and stacking them into a vector gives the single steepest-uphill direction. This is exactly the tool Gradient and directional derivative provides.

PICTURE: At the marked point the red gradient arrow points straight uphill, perpendicular to the contour line it sits on. A conservative field $F = \nabla f$ means the black field arrows are these red gradient arrows.

Step 4 — Work along a path: the line integral

WHAT: At each instant we take the dot product of the field arrow with the velocity arrow, then add up all these instants.

WHY the dot product? The dot product $a \cdot b$ measures how much of $a$ points along $b$ . Work counts only the part of the force that pushes along your motion — a sideways force does no work. The dot product is the exact tool that keeps the along-part and discards the sideways-part. (More in Line integrals of vector fields.)

PICTURE: Along the curve, at each dot the red velocity arrow and the black field arrow meet at an angle; the dot product harvests only their aligned overlap.

Step 5 — The one magic move: the chain rule turns the integrand into a pure derivative

Now substitute $F = \nabla f$ into the work integral: $\int_{a}^{b} \nabla f (r (t)) \cdot r^{'} (t) d t .$

Look term by term at the integrand: $(f_{x}, f_{y}) \nabla f \cdot (x^{'} (t), y^{'} (t)) r^{'} = f_{x} x^{'} (t) + f_{y} y^{'} (t) .$

WHAT: We claim this sum is exactly the time-rate-of-change of the height $f$ as we ride along the curve: $\frac{d}{d t} f (x (t), y (t)) = \frac{\partial f}{\partial x} x^{'} (t) + \frac{\partial f}{\partial y} y^{'} (t) .$

WHY is this true? This is the multivariable chain rule. In a tiny time step $d t$ the dot moves right by $x^{'} d t$ and up by $y^{'} d t$ . Each bit of sideways motion changes the height by (slope in that direction) × (distance): the east part contributes $f_{x} x^{'} d t$ , the north part contributes $f_{y} y^{'} d t$ . Add them — that is the total height change $df$ . Divide by $d t$ and you have the formula. So the integrand is the derivative of altitude-along-the-curve.

PICTURE: Two small right-triangle steps stacked: a horizontal nudge $x^{'} d t$ climbing $f_{x} x^{'} d t$ , and a vertical nudge $y^{'} d t$ climbing $f_{y} y^{'} d t$ . Their heights add up to one climb $df$ (red).

Step 6 — Add up the height changes: telescoping to the endpoints

Because the integrand is a perfect derivative, the Fundamental Theorem of Calculus applies. It says: adding up all the tiny changes of a quantity gives its net change from start to finish.

$\int_{a}^{b} \frac{d}{d t} [f (r (t))] d t = height at end B f (r (b)) - height at start A f (r (a)) .$

WHAT: All the little climbs $df$ collapse ("telescope") into just final height minus starting height.

WHY: Every intermediate height is added once and subtracted once by the next step, so the whole middle cancels — only the two endpoints survive. This is why the wiggly path is irrelevant.

PICTURE: A staircase of small red rises; when you sum them, every internal step cancels and only the total vertical gap between $A$ and $B$ remains.

Step 7 — Two paths, same answer (path-independence made visible)

WHAT: Take two totally different curves $C_{1}$ and $C_{2}$ from the same $A$ to the same $B$ . Step 6 says both integrals equal $f (B) - f (A)$ , so they are equal to each other.

WHY it matters: This is the physical meaning of "conservative" — you cannot get a different amount of work by choosing a cleverer route.

PICTURE: A short straight red path and a long looping black path share endpoints $A, B$ . Both climb the same net height, so both do the same work.

Step 8 — The closed loop: net work is exactly zero (degenerate case $A = B$ )

WHAT: Now let the path return to where it started, so $B = A$ . Then $\oint_{C} \nabla f \cdot d r = f (A) - f (A) = 0.$

WHY this is the crucial edge case: A closed loop is the "free-energy" test. If you could gain energy by circling forever, the field would not be conservative. The formula forbids it: end height = start height ⟹ net work $= 0$ . This is Conservation of energy in physics in one line, and it links to Green's theorem and Curl of a vector field as the "no swirl inside" condition.

PICTURE: A closed red loop lying across the contour lines: it crosses each contour going up exactly as often as coming down, so the heights cancel to zero. Beside it, a loop around a hole where they do not cancel.

The one-picture summary

Everything above is a single sentence with a single picture: the field arrows are downhill arrows of a hill, so work = height climbed = end height − start height.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a hilly park where, at every spot, an arrow shows which way a ball rolls: straight downhill. That arrow-map is our field $F$ , and the hill's height-map is $f$ . Now walk any wiggly path from a bench to the swings. In each tiny step, the "push you get from the slope" times the distance you moved is exactly how much your altitude changed — that's the chain rule, and it's the one clever move. Since every step's work is just a little height-change, adding them all up gives only the total height change: swings height minus bench height. The zig-zags in the middle all cancel. So it doesn't matter which path you take, only where you start and stop. And if you walk a loop back to the bench, you end at the same height you began — total work is zero. You can never milk free energy by circling. The one warning: this needs a proper single hill with no holes underneath; if the ground has a puncture (like a whirlpool's center), the "height" can climb $2 π$ per lap and the trick breaks.

Recall Quick self-test

Why does path shape not matter for a conservative field? ::: Work equals $f (B) - f (A)$ , which uses only the endpoints. Which single theorem converts $\nabla f \cdot r^{'}$ into $\frac{d}{d t} f (r (t))$ ? ::: The multivariable chain rule. What does the dot product in the work integral throw away? ::: The part of the force perpendicular to the motion (only the aligned part does work). Why is a closed-loop integral zero? ::: Start and end heights are equal, so $f (A) - f (A) = 0$ . When can the loop integral fail to be zero despite $F = \nabla f$ locally? ::: When the domain has a hole (not simply connected), so no single-valued hill exists.