Deriving the Cross-Product Curvature Formula

Why a Cross Product Gives Curvature

A space curve can bend even when its parameter does not measure distance at a constant rate. Curvature must describe the curve’s turning independently of how quickly the parameter moves along it. The unit tangent provides the geometric quantity of interest: it records direction without recording speed.

The earlier material on curvature defines curvature as the magnitude of the change in the unit tangent per unit arc length. This tutorial derives a useful formula for evaluating that quantity directly from the first two derivatives of a parametrization. The derivation will also explain why the speed appears to the third power in the denominator.

Central formula: If $\mathbf r(t)$ is a twice-differentiable regular space curve, then its curvature is $$ \kappa(t)=\frac{|\mathbf r'(t)\times\mathbf r''(t)|}{|\mathbf r'(t)|^3}. $$ Regularity means $\mathbf r'(t)\ne\mathbf 0$ at the point in question. The formula is not defined at a parameter value where the curve is not regular.

Definitions and Conditions

Let $\mathbf r(t)$ be a vector-valued parametrization of a curve in three-dimensional space, and write its speed as $v(t)=|\mathbf r'(t)|$. The curve is regular at $t$ if $v(t)>0$. For the derivation below, assume $\mathbf r$ is twice differentiable on an interval and regular there.

The unit tangent vector is the velocity vector divided by its magnitude:

\mathbf T(t)=\frac{\mathbf r'(t)}{|\mathbf r'(t)|}.

Let $s$ denote arc length along the curve. Since $ds/dt=|\mathbf r'(t)|=v(t)>0$, the chain rule gives $d/ds=(1/v)d/dt$. Curvature is the magnitude of the unit tangent’s derivative with respect to arc length:

\kappa=\left|\frac{d\mathbf T}{ds}\right| =\frac{|\mathbf T'(t)|}{|\mathbf r'(t)|}.

This definition measures turning per unit distance traveled. It is not simply $|\mathbf T'(t)|$, because that derivative measures change with respect to the parameter $t$, whose rate of passage along the curve may vary.

Deriving the Cross-Product Formula

Set $v=|\mathbf r'|$, so that $\mathbf r'=v\mathbf T$. Differentiate this identity with respect to $t$. The product rule gives $\mathbf r''=v'\mathbf T+v\mathbf T'$. Now take the cross product with $\mathbf r'$. The term involving $v'$ vanishes because a vector crossed with a scalar multiple of itself is zero.

Write velocity using the unit tangent.
Since the speed $v$ is positive, $\mathbf r'=v\mathbf T$.

Differentiate.
The product rule gives $\mathbf r''=v'\mathbf T+v\mathbf T'$.

Take the cross product.
Using $\mathbf T\times\mathbf T=\mathbf 0$, obtain $\mathbf r'\times\mathbf r''=v^2(\mathbf T\times\mathbf T')$.

Take magnitudes and use the definition.
Because $\mathbf T$ is a unit vector, $\mathbf T'$ is perpendicular to $\mathbf T$, so $|\mathbf T\times\mathbf T'|=|\mathbf T'|$. Divide by $v^3$.

For completeness, the perpendicularity in the last step follows by differentiating $\mathbf T\cdot\mathbf T=1$. This gives $2\mathbf T\cdot\mathbf T'=0$, hence $\mathbf T\cdot\mathbf T'=0$. The magnitude of a cross product is the product of the magnitudes times the sine of the angle between the vectors. Here that angle is a right angle, so $|\mathbf T\times\mathbf T'|=|\mathbf T||\mathbf T'|=|\mathbf T'|$.

\begin{aligned} \mathbf r'\times\mathbf r'' &=(v\mathbf T)\times(v'\mathbf T+v\mathbf T')\\ &=vv'(\mathbf T\times\mathbf T)+v^2(\mathbf T\times\mathbf T')\\ &=v^2(\mathbf T\times\mathbf T'),\\[4pt] |\mathbf r'\times\mathbf r''| &=v^2|\mathbf T'|,\\[4pt] \kappa &=\frac{|\mathbf T'|}{v} =\frac{|\mathbf r'\times\mathbf r''|}{v^3} =\frac{|\mathbf r'\times\mathbf r''|}{|\mathbf r'|^3}. \end{aligned}

This proves the formula under the stated differentiability and regularity conditions. The cross product discards the part of acceleration parallel to velocity: that part changes speed but does not turn the tangent. The remaining, perpendicular component is what contributes to curvature.

Worked Examples

Worked Example: Curvature of a Helix

Find the curvature of $\mathbf r(t)=(3\cos t,3\sin t,4t)$. This curve is regular wherever it is defined.

Differentiate each component:

\mathbf r'(t)=(-3\sin t,3\cos t,4), \qquad \mathbf r''(t)=(-3\cos t,-3\sin t,0).

The speed is $ |\mathbf r'(t)|=\sqrt{9\sin^2t+9\cos^2t+16}=\sqrt{25}=5$. The cross product is

\begin{aligned} \mathbf r'\times\mathbf r'' &=\bigl(3\cos t\cdot0-4(-3\sin t),\\ &\qquad 4(-3\cos t)-(-3\sin t)\cdot0,\\ &\qquad (-3\sin t)(-3\sin t)-(3\cos t)(-3\cos t)\bigr)\\ &=(12\sin t,-12\cos t,9). \end{aligned}

Its magnitude is $\sqrt{144\sin^2t+144\cos^2t+81}=\sqrt{225}=15$. Therefore

\kappa(t)=\frac{15}{5^3}=\frac{15}{125}=\frac{3}{25}.

Check using the definition: $\mathbf T=(-3\sin t/5,3\cos t/5,4/5)$, so $\mathbf T'=(-3\cos t/5,-3\sin t/5,0)$ has magnitude $3/5$. Thus $|\mathbf T'|/|\mathbf r'|=(3/5)/5=3/25$, agreeing with the formula.

Worked Example: Curvature of a Twisted Cubic at the Origin

For $\mathbf r(t)=(t,t^2,t^3)$, find the curvature at $t=0$. Its first derivative is $(1,2t,3t^2)$, whose first component is always $1$, so the curve is regular.

Differentiate again, then compute the cross product:

\begin{aligned} \mathbf r'(t)&=(1,2t,3t^2),& \mathbf r''(t)&=(0,2,6t),\\ \mathbf r'\times\mathbf r'' &=(12t^2-6t^2,\,-6t,\,2)\\ &=(6t^2,-6t,2). \end{aligned}

At $t=0$, the speed is $|\mathbf r'(0)|=|(1,0,0)|=1$, and the cross-product magnitude is $|(0,0,2)|=2$. Hence

\kappa(0)=\frac{2}{1^3}=2.

For a check, $\mathbf T(t)=(1,2t,3t^2)/\sqrt{1+4t^2+9t^4}$. Differentiating at zero gives $\mathbf T'(0)=(0,2,0)$: the denominator equals $1$ there and has derivative $0$ there. Since the speed at zero is $1$, the definition gives $|\mathbf T'(0)|/|\mathbf r'(0)|=2$, as required.

Worked Example: Curvature at a Point on a Spatial Curve

Consider $\mathbf r(t)=(\cos t,\sin t,t^2)$, and calculate the curvature at $t=0$. First find

\mathbf r'(t)=(-\sin t,\cos t,2t), \qquad \mathbf r''(t)=(-\cos t,-\sin t,2).

The velocity has a second component $\cos t$, so it is nonzero near zero. At zero, $\mathbf r'(0)=(0,1,0)$, and the speed is $1$. At that point, $\mathbf r''(0)=(-1,0,2)$, giving

\mathbf r'(0)\times\mathbf r''(0) =(0,1,0)\times(-1,0,2) =(2,0,1).

The magnitude is $\sqrt{2^2+0^2+1^2}=\sqrt{5}$, so $\kappa(0)=\sqrt{5}/1^3=\sqrt{5}$.

To check, the unit tangent is $\mathbf T(t)=(-\sin t,\cos t,2t)/\sqrt{1+4t^2}$. Differentiating at zero gives $\mathbf T'(0)=(-1,0,2)$, with magnitude $\sqrt{5}$. The speed at zero is $1$, so the unit-tangent definition also gives curvature $\sqrt{5}$.

Common Mistake: Using the Wrong Power of Speed

Another tempting but invalid step is to apply the formula where $\mathbf r'=\mathbf 0$. The denominator then vanishes, and the unit tangent itself is undefined there. A parametrization may fail to be regular even if the underlying geometric trace has a meaningful description; this formula applies only at parameter values where the velocity is nonzero.

Practical check: Before using the formula, verify $\mathbf r'(t)\ne\mathbf 0$. Then compute $\mathbf r'$, $\mathbf r''$, their cross product and its magnitude, and the speed cubed. Keep the cross-product magnitude nonnegative and use the same parameter value in every quantity.

Why the Formula Is Useful

The expression uses only the first and second parameter derivatives, so it can be more direct than explicitly differentiating the normalized tangent. It is especially useful when the cross product and speed simplify, as they do for the helix. The derivation also clarifies the geometry: acceleration in the direction of motion affects the speed, while the component that changes the direction of motion produces turning.

The formula does not remove the need to check the hypotheses or the arithmetic. A zero cross product at a regular point gives zero curvature, because velocity and acceleration are parallel there and the tangent has no instantaneous change in direction. A nonzero cross product produces positive curvature. In either case, the denominator is the cube of the speed, not the square.

Check Your Understanding

Use the definitions and derivation above to answer these questions.

What regularity condition is required to use the cross-product curvature formula at a parameter value?
Why does the term $v'\mathbf T$ disappear when deriving $\mathbf r'\times\mathbf r''$?
For $\mathbf r(t)=(t,2t,t)$, compute the curvature wherever the curve is regular.
For $\mathbf r(t)=(2\cos t,2\sin t,3t)$, find the speed and the denominator in the cross-product formula.
Explain why a cross product of zero gives zero curvature at a regular point.