Curvature Invariance Under Regular Reparametrization

Curvature Belongs to the Curve, Not Its Parameter

A space curve can be described using different parameters. One parameter may make a point move quickly along the curve, while another makes it move slowly; a third may trace the same path in the opposite direction. These changes affect the derivatives used in curvature calculations, but they do not change how sharply the curve bends at a given point.

This tutorial proves that invariance precisely. The key condition is that the change of parameter is regular: its derivative must not vanish. The proof will also show why the result holds when the new parameter reverses orientation.

Curvature and Regular Reparametrization

Let \(\mathbf r(t)\) be a twice-differentiable curve in \(\mathbb R^3\). It is regular at \(t\) if \(\mathbf r'(t)\ne\mathbf 0\). Its curvature at \(t\) is defined by the magnitude of the cross product of its first two derivatives, divided by the cube of its speed.

Definition: At a point where \(\mathbf r'(t)\ne\mathbf 0\), the curvature of \(\mathbf r\) is

\kappa_{\mathbf r}(t) = \frac{\lVert\mathbf r'(t)\times\mathbf r''(t)\rVert} {\lVert\mathbf r'(t)\rVert^3}.

The norm makes curvature nonnegative.

Suppose a second parameter \(u\) is related to \(t\) by \(t=\phi(u)\), and define the reparametrized curve by \(\mathbf R(u)=\mathbf r(\phi(u))\). We use a \(C^2\) function \(\phi\) on an interval and require \(\phi'(u)\ne0\) there. This condition ensures that \(\mathbf R\) remains regular wherever \(\mathbf r\) is regular. On an interval, the continuous derivative \(\phi'\) cannot change sign without passing through zero. Thus \(\phi\) either preserves orientation throughout the interval or reverses it throughout.

Definition: A regular reparametrization is a change \(t=\phi(u)\) with \(\phi\) twice differentiable and \(\phi'(u)\ne0\) on the parameter interval. It describes the same curve locally, with a possibly different rate or direction of traversal.

The Invariance Formula

Differentiate \(\mathbf R(u)=\mathbf r(\phi(u))\) using the chain rule. The second derivative contains two terms, but only one contributes to the cross product: the other is parallel to the first derivative.

Differentiate once.
\(\mathbf R'(u)=\mathbf r'(\phi(u))\phi'(u)\).

Differentiate again.
\(\mathbf R''(u)=\mathbf r''(\phi(u))(\phi'(u))^2+\mathbf r'(\phi(u))\phi''(u)\).

Form the cross product.
The term involving \(\mathbf r'\times\mathbf r'\) is zero, leaving a factor of \((\phi')^3\).

Compare the speeds.
The speed of \(\mathbf R\) is \(\lVert\mathbf r'(\phi(u))\rVert|\phi'(u)|\).

Proof of Curvature Invariance

We now calculate each part of the curvature formula. In the equations below, the derivatives of \(\mathbf r\) are evaluated at \(\phi(u)\). Bilinearity of the cross product gives

\begin{aligned} \mathbf R'(u)\times\mathbf R''(u) &= \bigl(\mathbf r'\phi'\bigr) \times \bigl(\mathbf r''(\phi')^2+\mathbf r'\phi''\bigr)\\ &= (\phi')^3(\mathbf r'\times\mathbf r'') +\phi'\phi''(\mathbf r'\times\mathbf r')\\ &= (\phi')^3(\mathbf r'\times\mathbf r''). \end{aligned}

The last equality follows because the cross product of a vector with itself is zero. Taking magnitudes therefore gives a factor of \(|\phi'|^3\), including when \(\phi'\) is negative. Meanwhile, the first derivative formula gives the speed:

\lVert\mathbf R'(u)\rVert = \lVert\mathbf r'(\phi(u))\rVert|\phi'(u)|.

Since \(\phi'(u)\ne0\), the reparametrized curve is regular, and its curvature is defined. Substituting these expressions into the curvature formula, then cancelling the positive factor \(|\phi'(u)|^3\), proves the result.

\begin{aligned} \kappa_{\mathbf R}(u) &= \frac{\lVert\mathbf R'(u)\times\mathbf R''(u)\rVert} {\lVert\mathbf R'(u)\rVert^3}\\ &= \frac{|\phi'(u)|^3 \lVert\mathbf r'(\phi(u))\times\mathbf r''(\phi(u))\rVert} {\lVert\mathbf r'(\phi(u))\rVert^3|\phi'(u)|^3}\\ &= \kappa_{\mathbf r}(\phi(u)). \end{aligned}

Theorem: If \(\mathbf r\) is a twice-differentiable regular space curve and \(t=\phi(u)\) is a \(C^2\) change of parameter with \(\phi'(u)\ne0\), then

\kappa_{\mathbf R}(u)=\kappa_{\mathbf r}(\phi(u)), \qquad \mathbf R(u)=\mathbf r(\phi(u)).

The equality compares curvature at corresponding points. It holds whether \(\phi\) preserves or reverses orientation.

The proof explains why the speed and the cross product change by exactly matching factors. A change of parameter scales the speed by \(|\phi'|\), so the cube of speed scales by \(|\phi'|^3\). The cross-product magnitude scales by that same amount. The extra term involving \(\phi''\) changes the acceleration, but points in the tangent direction and disappears from the cross product.

Worked Examples

Worked Example: A Nonlinear Change of Parameter

Consider \(\mathbf r(t)=(t,t^2,0)\) for \(t>0\), and reparametrize using \(t=u^2\) for \(u>0\). First calculate the curvature in the original parameter:

\mathbf r'(t)=(1,2t,0),\qquad \mathbf r''(t)=(0,2,0),\qquad \mathbf r'(t)\times\mathbf r''(t)=(0,0,2).

Thus \(\lVert\mathbf r'(t)\rVert=\sqrt{1+4t^2}\), and \(\kappa_{\mathbf r}(t)=2/(1+4t^2)^{3/2}\). At \(u=1\), the corresponding parameter is \(t=1\), so the original curvature is \(2/(5\sqrt5)\).

Now \(\mathbf R(u)=(u^2,u^4,0)\). Its derivatives at \(u=1\) are

\mathbf R'(1)=(2,4,0),\qquad \mathbf R''(1)=(2,12,0),\qquad \mathbf R'(1)\times\mathbf R''(1)=(0,0,16).

The speed is \(\sqrt{2^2+4^2}=\sqrt{20}\), so its cube is \(20\sqrt{20}\). Therefore \(\kappa_{\mathbf R}(1)=16/(20\sqrt{20})=2/(5\sqrt5)\), matching the curvature at \(t=1\). Substitution into the original formula gives the same value, verifying the calculation.

Worked Example: Reversing the Direction on a Helix

Let \(\mathbf r(t)=(\cos t,\sin t,t)\), and set \(t=3-2u\). Since \(dt/du=-2\), this change reverses orientation. For the original curve,

\begin{aligned} \mathbf r'(t)&=(-\sin t,\cos t,1),\\ \mathbf r''(t)&=(-\cos t,-\sin t,0),\\ \mathbf r'(t)\times\mathbf r''(t)&=(\sin t,-\cos t,1). \end{aligned}

The cross-product magnitude is \(\sqrt{\sin^2t+\cos^2t+1}=\sqrt2\), and the speed is also \(\sqrt2\). Hence \(\kappa_{\mathbf r}(t)=\sqrt2/(\sqrt2)^3=1/2\).

For \(\mathbf R(u)=\mathbf r(3-2u)\), the derivative formulas give \(\mathbf R'=-2\mathbf r'\) and \(\mathbf R''=4\mathbf r''\), with the derivatives on the right evaluated at \(3-2u\). Thus \(\lVert\mathbf R'\times\mathbf R''\rVert=8\sqrt2\), while \(\lVert\mathbf R'\rVert=2\sqrt2\) and \(\lVert\mathbf R'\rVert^3=16\sqrt2\). The reparametrized curvature is \(8\sqrt2/(16\sqrt2)=1/2\), as required. In particular, at \(u=1\), both descriptions refer to \(t=1\).

Worked Example: A Nonlinear Change That Reverses Orientation

Take the unit circle \(\mathbf r(t)=(\cos t,\sin t,0)\), and define \(t=1-u-u^2\) near \(u=0\). Here \(\phi'(0)=-1\), so the change reverses orientation near that point, while remaining regular on a sufficiently small interval around \(0\). At \(t=1\),

\mathbf r'(1)=(-\sin1,\cos1,0),\qquad \mathbf r''(1)=(-\cos1,-\sin1,0).

The cross product is \((0,0,1)\), and the speed is \(1\); therefore the original curvature is \(1\). At \(u=0\), \(\phi(0)=1\), \(\phi'(0)=-1\), and \(\phi''(0)=-2\). Consequently, \(\mathbf R'(0)=-\mathbf r'(1)\) and \(\mathbf R''(0)=\mathbf r''(1)-2\mathbf r'(1)\). Their cross product is \(-\mathbf r'(1)\times\mathbf r''(1)=(0,0,-1)\), with magnitude \(1\); the speed is also \(1\). Thus \(\kappa_{\mathbf R}(0)=1\), agreeing with the curvature of the original circle at \(t=1\).

Why the Absolute Value Matters

A common mistake is to carry the signed factor \((\phi')^3\) from the cross-product calculation into the curvature itself. For the reversed helix in the previous example, \(\phi'=-2\). The signed cross-product vector is multiplied by \(-8\), but curvature uses its magnitude, so its magnitude is multiplied by \(8\), not \(-8\). Curvature cannot be negative.

\frac{(-2)^3\lVert\mathbf r'\times\mathbf r''\rVert} {\bigl(\lVert\mathbf r'\rVert\,|-2|\bigr)^3} \quad\text{is not the curvature formula: its numerator must be a magnitude.}

The correct substitution is

\frac{|{-2}|^3\lVert\mathbf r'\times\mathbf r''\rVert} {\bigl(\lVert\mathbf r'\rVert\,|{-2}|\bigr)^3} = \frac{8\lVert\mathbf r'\times\mathbf r''\rVert} {8\lVert\mathbf r'\rVert^3} = \kappa_{\mathbf r}.

Another important condition is \(\phi'\ne0\). If the derivative vanishes, the reparametrized curve may fail to be regular at that point, and the curvature formula using its parameter derivatives may not be defined there. For example, \(t=u^2\) has \(\phi'(0)=0\). The invariance theorem applies to the example with \(u>0\), not at \(u=0\). A zero derivative is not an orientation reversal; it is a failure of the regular-change-of-parameter condition.

Geometric Meaning

Curvature measures local bending rather than speed or direction of travel. Reparametrization can change both speed and the sign of the velocity, but neither changes the curve's bending at a corresponding point. The cancellation in the proof makes this geometric fact exact: the parameter-dependent factors appear equally in the numerator and denominator of the curvature formula.

This distinction is useful whenever a curve is rewritten for convenience. One may choose a parameter that simplifies the equations or follows the curve at a different rate, without changing the curvature values along the trace. The only comparison needed is between corresponding points: a parameter value \(u\) on the new description represents the point with old parameter \(t=\phi(u)\).

Check Your Understanding

Use the definitions and proof above to answer these questions.

What condition on \(\phi'\) ensures that a change of parameter is regular?
Why does the term involving \(\phi''\) disappear from the cross product?
What factor multiplies the cross-product magnitude under a regular change of parameter?
Why does curvature remain nonnegative when the change reverses orientation?
Why does the theorem not directly apply at a point where \(\phi'=0\)?