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Arc Length and Curvature · Tutorial 200 of 999

Arc Length and Curvature Mixed Problem Workshop

Follow a complete arc-length and curvature analysis of one curve, with exact calculations and checks at each stage.

Intermediate 12 min read

What You'll Learn

  • Check whether a parametrized curve is regular on a specified interval
  • Find speed and compute arc length using an exact antiderivative
  • Calculate unit tangent and principal unit normal vectors
  • Compute curvature from parameter derivatives
  • Distinguish the derivative of the unit tangent from curvature
  • Verify vector lengths, orthogonality, and curvature calculations

A Complete Analysis of One Curve

A curve problem can ask for several quantities that are connected but not interchangeable. First we check that the parametrization is regular; then we use its speed to find arc length. The unit tangent describes the direction of motion, the principal unit normal describes the direction in which that tangent turns, and curvature measures the rate of that turning per unit distance.

In this workshop, we will analyze the same plane curve throughout:

$$ \mathbf{r}(t)=\langle t,t^2,0\rangle,\qquad 0\leq t\leq 1. $$

The interval matters: regularity and length are assessed on the portion of the curve specified by the parameter interval. We will also evaluate tangent, normal, and curvature at particular parameter values. Keeping the curve fixed makes it possible to see how each calculation supports the next.

Definitions and the Calculation Order

Definition: A differentiable parametrized curve \(\mathbf{r}(t)\) is regular on an interval if \(\mathbf{r}'(t)\neq\mathbf{0}\) at every parameter value in that interval. Its speed is \(\|\mathbf{r}'(t)\|\). The length from \(t=a\) to \(t=b\), where \(a\leq b\), is $$ L=\int_a^b \|\mathbf{r}'(t)\|\,dt. $$

For a regular curve, the unit tangent vector is \(\mathbf{T}(t)=\mathbf{r}'(t)/\|\mathbf{r}'(t)\|\). Where \(\mathbf{T}'(t)\neq\mathbf{0}\), the principal unit normal is \(\mathbf{N}(t)=\mathbf{T}'(t)/\|\mathbf{T}'(t)\|\). For a twice-differentiable regular curve, curvature is \(\kappa(t)=\|\mathbf{T}'(t)\|/\|\mathbf{r}'(t)\|\). This formula accounts for the fact that curvature measures turning per unit of arc length, not per unit of the parameter.

For our curve, the derivatives and speed are

$$ \mathbf{r}'(t)=\langle 1,2t,0\rangle,\qquad \mathbf{r}''(t)=\langle 0,2,0\rangle,\qquad \|\mathbf{r}'(t)\|=\sqrt{1+4t^2}. $$

Since \(1+4t^2\) is positive for every real \(t\), the velocity never vanishes. In particular, this curve is regular throughout \([0,1]\). That check justifies using the speed to calculate length and dividing by speed to define \(\mathbf{T}\).

Why Curvature Requires a Speed Factor

The parameter \(t\) need not measure distance along a curve. If \(s\) denotes arc length measured in the direction of increasing \(t\), then \(ds/dt=\|\mathbf{r}'(t)\|\) for a regular curve. Applying the chain rule to the unit tangent gives

$$ \frac{d\mathbf{T}}{dt} = \frac{d\mathbf{T}}{ds}\frac{ds}{dt} = \frac{d\mathbf{T}}{ds}\|\mathbf{r}'(t)\|. $$

Taking norms and rearranging yields the parameter formula for curvature:

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

This also explains the calculation order: find \(\mathbf{T}'(t)\), then divide its magnitude by the curve's speed. The vector \(\mathbf{T}'(t)\) points in the direction of the principal normal when it is nonzero, but its magnitude alone is generally not the curvature.

Worked Example: Regularity and Length on the Full Interval

Worked Example: Regularity and Length on the Full Interval

Find the speed of \(\mathbf{r}(t)=\langle t,t^2,0\rangle\), verify regularity on \([0,1]\), and calculate the length of this portion of the curve.

Differentiate each component:

$$ \mathbf{r}'(t)=\langle 1,2t,0\rangle. $$

Its magnitude is

$$ \|\mathbf{r}'(t)\| =\sqrt{1^2+(2t)^2+0^2} =\sqrt{1+4t^2}. $$

Because \(1+4t^2\geq 1>0\), the speed is positive for all \(t\), so \(\mathbf{r}'(t)\neq\mathbf{0}\) throughout \([0,1]\). The parametrization is regular there. Its length is therefore

$$ L=\int_0^1\sqrt{1+4t^2}\,dt. $$

An antiderivative is

$$ F(t)=\frac{t}{2}\sqrt{1+4t^2} +\frac{1}{4}\ln\!\left(2t+\sqrt{1+4t^2}\right). $$

To check it, differentiate the first term:

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

For the logarithmic term, with \(D=1+4t^2\),

$$ \frac{d}{dt}\left(\frac14\ln(2t+\sqrt{D})\right) =\frac14\frac{2+4t/\sqrt{D}}{2t+\sqrt{D}} =\frac{1}{2\sqrt{D}}. $$

Adding these derivatives gives \(\sqrt{D}\), as required. Evaluating at the endpoints gives

$$ L=F(1)-F(0) =\frac{\sqrt{5}}{2}+\frac14\ln(2+\sqrt{5}) -\frac14\ln(1) =\frac{\sqrt{5}}{2}+\frac14\ln(2+\sqrt{5}). $$

As a check, \(F(0)=0\), and differentiating \(F\) recovers the speed \(\sqrt{1+4t^2}\). The length is positive, as expected for a nonconstant curve over an interval of positive width.

Unit Tangent and Principal Unit Normal

For this curve, divide the velocity by its magnitude to obtain

$$ \mathbf{T}(t) =\frac{\langle 1,2t,0\rangle}{\sqrt{1+4t^2}} =\left\langle \frac{1}{\sqrt{1+4t^2}}, \frac{2t}{\sqrt{1+4t^2}}, 0 \right\rangle. $$

To find the principal unit normal, differentiate \(\mathbf{T}\). Writing \(D=1+4t^2\), the product rule gives

$$ \mathbf{T}'(t) =\left\langle -\frac{4t}{D^{3/2}}, \frac{2}{D^{3/2}}, 0 \right\rangle. $$

Its magnitude is

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

Because this is positive at every \(t\), the principal unit normal exists throughout the interval. Normalizing \(\mathbf{T}'\) gives

$$ \mathbf{N}(t) =\frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} =\left\langle-\frac{2t}{\sqrt{1+4t^2}}, \frac{1}{\sqrt{1+4t^2}}, 0\right\rangle. $$

Worked Example: Tangent and Normal at an Interior Point

Worked Example: Tangent and Normal at \(t=\frac12\)

Find and check the unit tangent and principal unit normal at \(t=\frac12\).

At this parameter value, \(1+4t^2=1+4(1/4)=2\). Substituting into the tangent formula gives

$$ \mathbf{T}\!\left(\frac12\right) =\left\langle\frac{1}{\sqrt2},\frac{1}{\sqrt2},0\right\rangle. $$

Substitution into the normal formula gives

$$ \mathbf{N}\!\left(\frac12\right) =\left\langle-\frac{1}{\sqrt2},\frac{1}{\sqrt2},0\right\rangle. $$

Check that both are unit vectors:

$$ \left\|\mathbf{T}\!\left(\frac12\right)\right\|^2 =\frac12+\frac12=1,\qquad \left\|\mathbf{N}\!\left(\frac12\right)\right\|^2 =\frac12+\frac12=1. $$

They are perpendicular because their dot product is

$$ \mathbf{T}\!\left(\frac12\right)\cdot\mathbf{N}\!\left(\frac12\right) =-\frac12+\frac12+0=0. $$

Thus both vectors have unit length, and the normal is orthogonal to the tangent, as required. The negative first component of the normal also agrees with the direction in which the tangent's first component is changing near this parameter value.

Curvature Along the Curve

We have already found \(\|\mathbf{T}'(t)\|=2/(1+4t^2)\) and \(\|\mathbf{r}'(t)\|=\sqrt{1+4t^2}\). The curve is regular, so the parameter curvature formula applies. Therefore

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

This curvature is positive throughout \([0,1]\), and it decreases as \(t\) increases: the denominator grows while the numerator stays fixed. It is not constant, even though the curve is smooth and regular everywhere on the interval.

Worked Example: Curvature at an Endpoint

Worked Example: Curvature at \(t=1\)

Find the unit tangent, principal unit normal, and curvature at \(t=1\), and verify the curvature calculation from the derivatives.

At \(t=1\), the speed is \(\sqrt{1+4}=\sqrt5\). Thus

$$ \mathbf{T}(1)=\left\langle\frac{1}{\sqrt5},\frac{2}{\sqrt5},0\right\rangle, \qquad \mathbf{N}(1)=\left\langle-\frac{2}{\sqrt5},\frac{1}{\sqrt5},0\right\rangle. $$

The derivative of the tangent at \(t=1\) is

$$ \mathbf{T}'(1) =\left\langle-\frac{4}{5\sqrt5},\frac{2}{5\sqrt5},0\right\rangle. $$

Its magnitude is

$$ \|\mathbf{T}'(1)\| =\sqrt{\frac{16}{125}+\frac{4}{125}} =\sqrt{\frac{20}{125}} =\frac{2}{5}. $$

Dividing by the speed gives the curvature:

$$ \kappa(1)=\frac{2/5}{\sqrt5} =\frac{2}{5\sqrt5}. $$

The general formula gives the same result:

$$ \kappa(1)=\frac{2}{(1+4)^{3/2}} =\frac{2}{5\sqrt5}. $$

Finally, the tangent has unit length because \((1+4)/5=1\); the normal also has unit length because \((4+1)/5=1\); and their dot product is \((-2+2)/5=0\). These checks confirm the endpoint vectors as well as the curvature value.

A Common Mistake: Forgetting to Divide by Speed

A frequent error is to calculate \(\|\mathbf{T}'(t)\|\) and report it as the curvature. At \(t=1\), that would give the incorrect result

$$ \kappa(1)\ \text{(incorrect)}=\|\mathbf{T}'(1)\|=\frac25. $$

The derivative \(\mathbf{T}'(t)\) measures turning per unit change in \(t\), while curvature measures turning per unit distance along the curve. Here the speed at \(t=1\) is \(\sqrt5\), not \(1\). The correction is to divide by that speed:

$$ \kappa(1)=\frac{\|\mathbf{T}'(1)\|}{\|\mathbf{r}'(1)\|} =\frac{2/5}{\sqrt5} =\frac{2}{5\sqrt5}. $$

The uncorrected calculation would only agree with curvature at a point where the speed equals \(1\). This is why checking the parameter speed is not a formality: it determines how parameter change relates to distance travelled.

Putting the Workflow Together

For a mixed arc-length and curvature problem, organize the work so that each formula is used only after its conditions have been checked.

1
Differentiate and check regularity.
Find \(\mathbf{r}'(t)\). Confirm that its magnitude is nonzero over the interval before using formulas that divide by speed.
2
Calculate speed and length.
Set the speed equal to \(\|\mathbf{r}'(t)\|\), then integrate it over the specified parameter interval.
3
Normalize to get the tangent.
Compute \(\mathbf{T}(t)=\mathbf{r}'(t)/\|\mathbf{r}'(t)\|\) and check that its magnitude is \(1\).
4
Differentiate the tangent.
Find \(\mathbf{T}'(t)\). If it is nonzero, normalize it to obtain the principal unit normal.
5
Compute curvature and verify.
Use \(\kappa(t)=\|\mathbf{T}'(t)\|/\|\mathbf{r}'(t)\|\), then compare with direct substitution into the resulting formula.

These steps form a connected analysis, not a list of unrelated calculations. Regularity permits division by speed; speed determines both arc length and the conversion from parameter rate to distance rate; the tangent and its derivative describe the curve's changing direction.

Check Your Understanding

Use \(\mathbf{r}(t)=\langle t,t^2,0\rangle\) on \(0\leq t\leq1\) to answer the following.

  1. What is the speed, and why does it show that the curve is regular on the interval?
  2. Write the definite integral for the length from \(t=0\) to \(t=\frac12\).
  3. What are \(\mathbf{T}(0)\) and \(\mathbf{N}(0)\)? Check that they are unit vectors and perpendicular.
  4. Find the curvature at \(t=0\) using the parameter formula.
  5. Explain why \(\|\mathbf{T}'(t)\|\) is not, in general, the curvature.

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