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Computing Curvature from Parameter Derivatives

Learn to calculate curvature from a curve’s parameter derivatives by finding its unit tangent, differentiating it, and dividing by the curve’s speed.

Intermediate 10 min read

What You'll Learn

  • State the smoothness and regularity conditions needed for the parameter-derivative curvature formula.
  • Find the unit tangent by normalizing the velocity vector.
  • Differentiate a normalized tangent, including when its components depend on the parameter.
  • Calculate curvature by dividing the tangent derivative’s magnitude by speed.
  • Check curvature calculations and identify the error in omitting the speed factor.

Introduction

Curvature measures how quickly a curve changes direction as distance along the curve increases. For a parametrized curve, the parameter does not necessarily measure distance: the curve may move faster in some parts of the parameter interval than in others. A curvature calculation must account for both the changing direction and the curve’s speed.

The method here starts with the unit tangent vector, differentiates it with respect to the parameter, and then divides by the speed. The key is to normalize the velocity before differentiating; differentiating the velocity alone does not, in general, give the change in direction.

Conditions and the Curvature Formula

Definition: Let \(\mathbf r(t)\) be a twice continuously differentiable curve on an open interval, and suppose it is regular there, meaning \(\mathbf r'(t)\ne\mathbf 0\) at every parameter value under consideration. Its speed is \(v(t)=\|\mathbf r'(t)\|\), and its unit tangent is \(\mathbf T(t)=\mathbf r'(t)/\|\mathbf r'(t)\|\). The curvature is \(\displaystyle \kappa(t)=\frac{\|\mathbf T'(t)\|}{\|\mathbf r'(t)\|}\).

The regularity condition makes the speed positive, so the unit tangent is defined. The twice-continuous differentiability condition ensures that the tangent can be differentiated. More generally, the formula applies at a particular point if \(\mathbf r\) is regular there and \(\mathbf T'(t)\) exists there. Differentiability and regularity alone are not enough to guarantee that this derivative exists.

For a unit-speed parametrization, the speed is \(1\), so curvature is simply \(\|\mathbf T'(t)\|\). For a general parameter, the denominator is essential: it adjusts the rate of change of the tangent to the distance actually traveled.

Why the Formula Divides by Speed

Let \(s\) denote arc length measured along the curve. For a regular curve, the arc-length function satisfies \(ds/dt=\|\mathbf r'(t)\|\). Curvature is the magnitude of the change in the unit tangent per unit arc length, \(\|d\mathbf T/ds\|\). Applying the chain rule gives the parameter formula.

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

This also explains why speed belongs in the denominator. If the tangent changes by a fixed amount over a parameter interval, that change represents less turning per unit distance when the curve travels faster.

Differentiating the Normalized Tangent

In many problems, the main work is differentiating \(\mathbf T(t)=\mathbf r'(t)/\|\mathbf r'(t)\|\). The denominator depends on \(t\), so it must be differentiated as well as the numerator. A useful expression follows directly from the product and chain rules.

Formula: If \(\mathbf r\) is twice differentiable and regular near \(t\), then, writing \(v=\|\mathbf r'(t)\|\), \(\displaystyle \mathbf T'(t)=\frac{\mathbf r''(t)}{v}-\frac{\mathbf r'(t)\cdot\mathbf r''(t)}{v^3}\mathbf r'(t)\).

To verify it, first differentiate the speed. Since \(v=(\mathbf r'\cdot\mathbf r')^{1/2}\), the chain rule gives \(v'=(\mathbf r'\cdot\mathbf r'')/v\). Then differentiate \(\mathbf T=\mathbf r'/v\):

$$ \mathbf T' = \frac{\mathbf r''}{v} - \frac{v'}{v^2}\mathbf r' = \frac{\mathbf r''}{v} - \frac{\mathbf r'\cdot\mathbf r''}{v^3}\mathbf r'. $$

This identity is one way to carry out the calculation, but it is not necessary to memorize it. Directly finding \(\mathbf T\) and differentiating its components is often clearer, especially when the normalized components simplify.

Worked Examples

Worked Example: Curvature of a Circle

Find the curvature of \(\mathbf r(t)=\langle 3\cos t,3\sin t\rangle\).

First differentiate the position vector to get the velocity, then find its magnitude:

$$ \mathbf r'(t)=\langle-3\sin t,3\cos t\rangle, \qquad \|\mathbf r'(t)\| = \sqrt{9\sin^2t+9\cos^2t} = 3. $$

The speed is positive, so the parametrization is regular. Normalize the velocity:

$$ \mathbf T(t) = \frac{\langle-3\sin t,3\cos t\rangle}{3} = \langle-\sin t,\cos t\rangle. $$

Differentiate the unit tangent and calculate its magnitude:

$$ \mathbf T'(t)=\langle-\cos t,-\sin t\rangle, \qquad \|\mathbf T'(t)\| = \sqrt{\cos^2t+\sin^2t} = 1. $$

Therefore,

$$ \kappa(t) = \frac{\|\mathbf T'(t)\|}{\|\mathbf r'(t)\|} = \frac{1}{3}. $$

As a check, the computed tangent has unit length because \(\sin^2t+\cos^2t=1\); its derivative has magnitude \(1\), and the speed is \(3\). Substituting those values into the curvature formula gives \(1/3\) at every parameter value.

Worked Example: Curvature of a Helix

Find the curvature of the space curve \(\mathbf r(t)=\langle 2\cos t,2\sin t,t\rangle\).

Differentiate and calculate the speed:

$$ \mathbf r'(t)=\langle-2\sin t,2\cos t,1\rangle, \qquad \|\mathbf r'(t)\| = \sqrt{4\sin^2t+4\cos^2t+1} = \sqrt{5}. $$

Since the speed is never zero, the curve is regular. Divide each velocity component by \(\sqrt{5}\) to obtain the unit tangent:

$$ \mathbf T(t) = \frac{1}{\sqrt{5}}\langle-2\sin t,2\cos t,1\rangle. $$

Differentiate this normalized vector. The constant third component has derivative zero:

$$ \mathbf T'(t) = \frac{1}{\sqrt{5}}\langle-2\cos t,-2\sin t,0\rangle, \qquad \|\mathbf T'(t)\| = \frac{1}{\sqrt{5}}\sqrt{4\cos^2t+4\sin^2t} = \frac{2}{\sqrt{5}}. $$

Thus,

$$ \kappa(t) = \frac{2/\sqrt{5}}{\sqrt{5}} = \frac{2}{5}. $$

For a check, the displayed tangent has squared magnitude \((4\sin^2t+4\cos^2t+1)/5=1\), as required of a unit tangent. Its derivative has magnitude \(2/\sqrt{5}\), and dividing by the speed \(\sqrt{5}\) gives \(2/5\) for every \(t\).

Worked Example: Differentiating a Tangent with a Variable Denominator

Find the curvature of \(\mathbf r(t)=\langle t,t^2\rangle\) at \(t=1\). This example requires differentiating a unit tangent whose components share a parameter-dependent denominator.

The velocity and speed are

$$ \mathbf r'(t)=\langle1,2t\rangle, \qquad \|\mathbf r'(t)\|=\sqrt{1+4t^2}. $$

The speed is positive for every real \(t\), so the curve is regular. Normalize the velocity:

$$ \mathbf T(t) = \left\langle \frac{1}{\sqrt{1+4t^2}}, \frac{2t}{\sqrt{1+4t^2}} \right\rangle. $$

Differentiate the first component using the chain rule. For the second component, use the product rule:

$$ \frac{d}{dt}(1+4t^2)^{-1/2} = -4t(1+4t^2)^{-3/2}, $$ $$ \frac{d}{dt}\left(2t(1+4t^2)^{-1/2}\right) = 2(1+4t^2)^{-1/2} - 8t^2(1+4t^2)^{-3/2} = \frac{2}{(1+4t^2)^{3/2}}. $$

Therefore,

$$ \mathbf T'(t) = \left\langle \frac{-4t}{(1+4t^2)^{3/2}}, \frac{2}{(1+4t^2)^{3/2}} \right\rangle. $$

Its magnitude simplifies as follows:

$$ \|\mathbf T'(t)\| = \frac{\sqrt{16t^2+4}}{(1+4t^2)^{3/2}} = \frac{2\sqrt{1+4t^2}}{(1+4t^2)^{3/2}} = \frac{2}{1+4t^2}. $$

At \(t=1\), the speed is \(\sqrt{5}\) and the tangent derivative has magnitude \(2/5\). Hence,

$$ \kappa(1) = \frac{2/5}{\sqrt{5}} = \frac{2}{5\sqrt{5}}. $$

To check, \(\mathbf T(1)=\langle1,2\rangle/\sqrt{5}\) has magnitude \(1\), and substituting \(t=1\) into the differentiated tangent gives \(\mathbf T'(1)=\langle-4,2\rangle/(5\sqrt{5})\), whose magnitude is \(\sqrt{20}/(5\sqrt{5})=2/5\). Dividing by the speed \(\sqrt{5}\) confirms the stated curvature.

A Common Error: Leaving Out the Speed

In the parabola example, \(\|\mathbf T'(1)\|=2/5\). It would be incorrect to report this as the curvature merely because it is the magnitude of the tangent’s parameter derivative. The parameter \(t\) does not measure distance along this curve, so that magnitude is not the turning per unit distance.

$$ \text{Incorrect: }\quad \kappa(1)=\|\mathbf T'(1)\|=\frac{2}{5}. \qquad \text{Correct: }\quad \kappa(1)=\frac{\|\mathbf T'(1)\|}{\|\mathbf r'(1)\|} =\frac{2/5}{\sqrt{5}} =\frac{2}{5\sqrt{5}}. $$

The correction is to calculate the speed from \(\mathbf r'(t)\) and divide by it. Omitting that step is valid only for a unit-speed parametrization, where the speed is \(1\).

Using the Method Reliably

For a regular parametrization, curvature is nonnegative because it is a ratio of magnitudes and the speed is positive. If \(\mathbf T'(t)=\mathbf 0\), the formula gives zero curvature at that point: the unit tangent has no instantaneous change with respect to the parameter there. The formula does not apply where \(\mathbf r'(t)=\mathbf 0\), because the unit tangent and the denominator are then undefined.

A disciplined calculation keeps the following quantities distinct: the velocity \(\mathbf r'(t)\), the speed \(\|\mathbf r'(t)\|\), the unit tangent \(\mathbf T(t)\), and the magnitude \(\|\mathbf T'(t)\|\). In particular, the tangent must be normalized before it is differentiated. Otherwise, changes in the magnitude of the velocity may be mistaken for changes in direction.

When checking work, verify that the calculated tangent has magnitude \(1\), differentiate the normalized expression carefully, and only then form the curvature ratio. These checks are especially useful when the speed contains a variable expression under a square root.

Check Your Understanding

For each question, identify the relevant speed and distinguish it from the magnitude of the unit tangent’s parameter derivative.

  1. What condition on \(\mathbf r'(t)\) ensures that the unit tangent is defined?
  2. Why does \(\kappa(t)=\|\mathbf T'(t)\|\) hold for a unit-speed curve?
  3. For \(\mathbf r(t)=\langle 3\cos t,3\sin t\rangle\), what are the speed and curvature?
  4. For \(\mathbf r(t)=\langle t,t^2\rangle\), what is the speed at \(t=1\), and why must it be included in the curvature calculation?
  5. Why do twice differentiability and regularity provide appropriate conditions for applying the formula on an interval?

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