Reparametrization Invariance

Why the Parameter Should Not Determine the Work

A parametrization describes how a point moves along a curve. One parametrization might move quickly through one part of the curve and slowly through another; a different one might move at a different rate. If both describe the same route in the same direction, the work done by a force field should not depend on that choice of clock.

The vector line integral captures this distinction. It accounts for the force along the curve and for the infinitesimal displacement at each point. Under a valid change of parameter, the displacement changes in just the way needed to offset the altered rate of traversal. This tutorial establishes that fact precisely and verifies it by calculation.

Vector Line Integrals and Reparametrization

Let $\mathbf{r}:[a,b]\to\mathbb{R}^n$ be a continuously differentiable parametrization of a curve, and let $\mathbf{F}$ be a continuous vector field defined along that curve. The vector line integral, also called the work integral, is defined by

\int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt.

Here $\mathbf{r}'(t)$ is the velocity vector, and the dot product with the force gives the force component in the direction of motion, multiplied by the rate of displacement. The integral adds these contributions along the path.

Definition: Let $\mathbf{r}:[a,b]\to\mathbb{R}^n$ be a continuously differentiable curve. A smooth increasing change of parameter is a continuously differentiable, strictly increasing function $\phi:[c,d]\to[a,b]$ such that $\phi(c)=a$ and $\phi(d)=b$. The reparametrized curve is $\mathbf{q}(u)=\mathbf{r}(\phi(u))$, for $u\in[c,d]$. It traces the same curve from the same initial endpoint to the same final endpoint.

Strict increase ensures that the parameter moves forward through the original curve rather than reversing its direction. The derivative $\phi'(u)$ is nonnegative, though it may equal zero at some points. A vanishing derivative at an isolated point does not invalidate the result.

Theorem: Suppose $\mathbf{r}:[a,b]\to\mathbb{R}^n$ and $\phi:[c,d]\to[a,b]$ satisfy the conditions above, and $\mathbf{F}$ is continuous along the image of $\mathbf{r}$. Then $\mathbf{q}=\mathbf{r}\circ\phi$ has the same vector line integral as $\mathbf{r}$: $$ \int_c^d \mathbf{F}(\mathbf{q}(u))\cdot\mathbf{q}'(u)\,du = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt. $$

The hypotheses matter. The new parameter must cover the same interval of the original parameter, and it must proceed in the increasing direction. If either endpoint condition fails, the new curve may cover only part of the path. If the direction is reversed, the orientation is not fixed.

Why the Invariance Holds

The chain rule explains the key cancellation. Since $\mathbf{q}(u)=\mathbf{r}(\phi(u))$, differentiation gives

\mathbf{q}'(u)=\mathbf{r}'(\phi(u))\phi'(u).

The factor $\phi'(u)$ records how quickly the original parameter changes as $u$ changes. It must be included when computing the new velocity. To prove the theorem, set

h(t)=\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t).

Continuity of $\mathbf{F}$ along the curve and continuous differentiability of $\mathbf{r}$ ensure that $h$ is continuous. The Fundamental Theorem of Calculus gives an antiderivative $H(t)=\int_a^t h(s)\,ds$, with $H'(t)=h(t)$. The chain rule then yields a complete justification:

Write the reparametrized integrand.
Using $\mathbf{q}'(u)=\mathbf{r}'(\phi(u))\phi'(u)$, its dot product is $h(\phi(u))\phi'(u)$.

Recognize a derivative.
By the chain rule, $H'(\phi(u))\phi'(u)=h(\phi(u))\phi'(u)$. Thus the new integrand is the derivative of $H(\phi(u))$.

Evaluate at the endpoints.
The Fundamental Theorem of Calculus gives $\int_c^d h(\phi(u))\phi'(u)\,du=H(\phi(d))-H(\phi(c))=H(b)-H(a)=\int_a^b h(t)\,dt$.

Since $h$ is exactly the original work integrand, this proves that the two vector line integrals are equal. The parameter can change the speed and the time spent at each point, but the velocity factor in the integral adjusts accordingly.

Worked Examples with Two Parametrizations

Worked Example: A Parabolic Path with a Slower Start

Let $\mathbf{r}(t)=(t,t^2)$, for $0\le t\le1$, and let $\mathbf{F}(x,y)=(y,x)$. First calculate the integral using $t$. The velocity is $\mathbf{r}'(t)=(1,2t)$, while $\mathbf{F}(\mathbf{r}(t))=(t^2,t)$. Therefore,

\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =(t^2,t)\cdot(1,2t) =t^2+2t^2 =3t^2, \qquad \int_0^1 3t^2\,dt =\left[t^3\right]_0^1 =1.

Now use the increasing change $t=\phi(u)=u^2$, for $0\le u\le1$. The new parametrization is $\mathbf{q}(u)=(u^2,u^4)$, and $\mathbf{q}'(u)=(2u,4u^3)$. The force along it is $\mathbf{F}(\mathbf{q}(u))=(u^4,u^2)$, so

\mathbf{F}(\mathbf{q}(u))\cdot\mathbf{q}'(u) =(u^4,u^2)\cdot(2u,4u^3) =2u^5+4u^5 =6u^5, \qquad \int_0^1 6u^5\,du =\left[u^6\right]_0^1 =1.

The endpoint check is $\mathbf{q}(0)=(0,0)=\mathbf{r}(0)$ and $\mathbf{q}(1)=(1,1)=\mathbf{r}(1)$. The change is increasing because $\phi'(u)=2u\ge0$ and $\phi$ is strictly increasing on $[0,1]$. Both parametrizations give the same work, $1$.

Worked Example: A Line Segment with a Cubic Change of Parameter

Consider $\mathbf{r}(t)=(t,2t)$, $0\le t\le1$, with force field $\mathbf{F}(x,y)=(x+y,x-y)$. The velocity is $(1,2)$, and the force along the curve is $(3t,-t)$. Thus

\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =(3t,-t)\cdot(1,2) =3t-2t =t, \qquad \int_0^1t\,dt=\left[\frac{t^2}{2}\right]_0^1=\frac12.

Reparametrize with $t=\phi(u)=u^3$, for $0\le u\le1$. Then $\mathbf{q}(u)=(u^3,2u^3)$, $\mathbf{q}'(u)=(3u^2,6u^2)$, and $\mathbf{F}(\mathbf{q}(u))=(3u^3,-u^3)$. Direct calculation gives

\mathbf{F}(\mathbf{q}(u))\cdot\mathbf{q}'(u) =(3u^3,-u^3)\cdot(3u^2,6u^2) =9u^5-6u^5 =3u^5, \qquad \int_0^1 3u^5\,du =\left[\frac{u^6}{2}\right]_0^1 =\frac12.

Here $\phi'(u)=3u^2\ge0$, and the endpoints remain $(0,0)$ and $(1,2)$. The result agrees with the original calculation.

Worked Example: A Quarter-Circle Traversed with a Different Clock

Let $\mathbf{r}(t)=(\cos t,\sin t)$, for $0\le t\le\pi/2$, and take $\mathbf{F}(x,y)=(-y,x)$. Along the curve, $\mathbf{F}(\mathbf{r}(t))=(-\sin t,\cos t)$, while $\mathbf{r}'(t)=(-\sin t,\cos t)$. Hence

\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) =\sin^2t+\cos^2t =1, \qquad \int_0^{\pi/2}1\,dt=\frac{\pi}{2}.

Set $t=\phi(u)=\frac{\pi}{2}u^2$, for $0\le u\le1$. Then $\phi'(u)=\pi u$, and $\mathbf{q}(u)=(\cos(\frac{\pi}{2}u^2),\sin(\frac{\pi}{2}u^2))$. The chain rule gives $\mathbf{q}'(u)=\mathbf{r}'(\phi(u))\pi u$. Since $\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)=1$ for every $t$,

\mathbf{F}(\mathbf{q}(u))\cdot\mathbf{q}'(u) =\pi u, \qquad \int_0^1\pi u\,du =\left[\frac{\pi u^2}{2}\right]_0^1 =\frac{\pi}{2}.

The endpoints are $(1,0)$ and $(0,1)$ for both parametrizations. The increasing change of parameter alters the speed but not the oriented path or the integral.

A Common Mistake: Dropping the Speed Factor

In the quarter-circle example, it is tempting to substitute the new parameter into the force and the old velocity, but forget the factor $\phi'(u)$. That gives $\mathbf{F}(\mathbf{q}(u))\cdot\mathbf{r}'(\phi(u))=1$, and the incorrect calculation $\int_0^1 1\,du=1$. It does not give the work integral for $\mathbf{q}$, because the velocity of $\mathbf{q}$ is not just $\mathbf{r}'(\phi(u))$.

The correction is to differentiate the composition before forming the dot product:

\mathbf{q}'(u)=\mathbf{r}'(\phi(u))\phi'(u) =\mathbf{r}'(\phi(u))\pi u.

Including this factor gives $\int_0^1\pi u\,du=\pi/2$, as required. The chain rule is not an optional adjustment; it is how the changed parameter's rate of motion enters the integral.

Why Reparametrization Invariance Matters

The theorem allows a calculation to use whichever smooth increasing parameter makes the algebra most convenient. A curve may be described with a linear, quadratic, or other increasing change of parameter, and the work remains the same provided the same oriented path is covered. This is useful because a convenient parameter can simplify the curve, the force evaluation, or the resulting integral.

It also clarifies what the integral measures. It does not measure how quickly a path is traversed. Its integrand combines the force with the actual displacement, including the velocity factor associated with the chosen parameter. Different parameter speeds redistribute the contributions over the parameter interval without changing their total.

Key takeaway: If $\mathbf{q}(u)=\mathbf{r}(\phi(u))$ for a smooth increasing change of parameter with matching endpoints, then $\mathbf{q}'(u)=\mathbf{r}'(\phi(u))\phi'(u)$. Including this chain-rule factor makes the vector line integral invariant.

Check Your Understanding

Use the definitions and result above to answer these questions.

State the endpoint and direction conditions on $\phi:[c,d]\to[a,b]$ for it to be a change of parameter covered by the theorem.
If $\mathbf{q}(u)=\mathbf{r}(\phi(u))$, what is the expression for $\mathbf{q}'(u)$, and what does the factor $\phi'(u)$ represent in the calculation?
For $\mathbf{r}(t)=(t,t^2)$, $0\le t\le1$, and $\phi(u)=u^2$, write down $\mathbf{q}(u)$ and $\mathbf{q}'(u)$.
In the quarter-circle example, what incorrect value results from omitting $\phi'(u)$, and what is the correct integral?
Why does the proof require the new parameter to reach both original endpoints?

What You'll Learn