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The distance along a curve as the integral of its speed, why that number does not depend on how the curve was parametrised, the arc length function $s(t)$, and the unit-speed parametrisation it defines.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to set up and evaluate the arc length integral for a curve in space, explain why the answer is a property of the curve rather than of the parametrisation, build the arc length function from a given speed, and say what a unit-speed parametrisation is and why it is worth having even when it cannot be written down.
Arc length in the plane from Calculus II, where a curve's length was an integral of a square root, and velocity and speed from the last lesson. Here the same integral is written once for space, and then read backwards: if length can be computed from a parameter, a curve can be given a parameter that is its length.
The arc length of a curve is the distance along it, measured by following the curve rather than by cutting across. The arc length function $s(t)$ is the distance covered from a chosen starting time up to $t$. A reparametrisation replaces the parameter by another one without changing the set of points; a curve is parametrised by arc length, or unit speed, when its speed is one at every instant, so the parameter and the distance travelled are the same number. A parametrisation is smooth on an interval when $\mathbf{r}'$ is continuous and never zero there.
For a curve $\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$ traced once as $t$ runs from $a$ to $b$,
$$L = \int_{a}^{b} |\mathbf{r}'(t)|\,dt = \int_{a}^{b} \sqrt{f'(t)^{2} + g'(t)^{2} + h'(t)^{2}}\;dt.$$
Why. Over a short interval $\Delta t$ the curve is nearly straight and the particle nearly keeps its velocity, so it covers about $|\mathbf{r}'(t)|\,\Delta t$. Adding those up and shrinking $\Delta t$ is exactly the integral. Distance is speed times time, one instant at a time — the whole formula is that sentence.
The length does not depend on the parametrisation. If $t = \varphi(s)$ is a smooth increasing change of parameter, the chain rule multiplies the speed by $\varphi'(s)$ and the substitution replaces $dt$ by $\varphi'(s)\,ds$. The two factors cancel, and the integral is unchanged. So arc length belongs to the curve, not to the schedule — unlike velocity, speed and acceleration, which belong to the schedule.
The arc length function. Fix a starting time $a$ and define
$$s(t) = \int_{a}^{t} |\mathbf{r}'(u)|\,du, \qquad\text{so}\qquad \frac{ds}{dt} = |\mathbf{r}'(t)|$$
by the fundamental theorem. $s$ is increasing wherever the speed is positive, so it can be inverted: $t$ can be written in terms of $s$, and substituting gives a parametrisation $\mathbf{r}(t(s))$ whose speed is
$$\left|\frac{d\mathbf{r}}{ds}\right| = \left|\frac{d\mathbf{r}}{dt}\right|\Big/\frac{ds}{dt} = 1.$$
That is the unit speed or arc length parametrisation. It exists for every smooth curve, is almost never possible to write down in closed form, and is used constantly in theory for exactly that reason: it lets a definition be stated without a speed cluttering it up.
Another way: picture
Imagine laying a piece of string along the curve and then straightening it against a ruler. The number you read is the arc length, and it plainly does not depend on how quickly anybody ran along the curve while laying the string down. Parametrising by arc length is marking the string in metres before you lay it, so that every point of the curve carries its own distance from the start.
Another way: steps
To find the length of a curve:
The square root in the arc length formula almost never simplifies. Three families of curves are the exceptions, and they are the ones that appear in every exercise set:
Everything else is integrated numerically. The ellipse is the famous case: its arc length integral has no elementary antiderivative at all, and the functions invented to express it are named after it. Knowing this is not defeatism; it is what tells you that a messy root in a real problem is the normal situation rather than a sign that you differentiated wrongly.
This is also why so much of the theory in the next lesson is stated with respect to $s$ rather than $t$. The arc length parametrisation is a device for writing definitions cleanly, and it earns its keep even though nobody ever computes it.
Integrating the position instead of the speed. $\int \mathbf{r}(t)\,dt$ is a vector, and it is not a length of anything.
Taking the length after integrating. $\left|\int \mathbf{r}'\,dt\right|$ is the straight-line distance between the end points, which is the short cut across, not the distance along. The order of the operations is the whole difference between a chord and an arc.
Letting the curve be traced more than once. $\langle \cos t, \sin t, 0 \rangle$ on $0 \le t \le 4\pi$ gives $4\pi$, which is twice round the unit circle. The formula counts distance travelled, so it is right; it is the question that was wrong.
Assuming a unit-speed parametrisation can be written down. It exists; it is usually not expressible in elementary functions. Use it to state a definition, not to compute with.
$\left|\int_{a}^{b} \mathbf{r}'(t)\,dt\right|$ and $\int_{a}^{b} |\mathbf{r}'(t)|\,dt$ differ by nothing more than where the length bars are, and they are two different quantities. The first integrates the velocity vectors — which partly cancel whenever the curve doubles back — and reports the straight-line displacement from start to finish. The second adds up distances, which never cancel, and reports the distance travelled.
For a closed curve the first is zero and the second is the perimeter, which is as large a gap as two formulas this similar can have. The rule is simple: take the length of the velocity before integrating, never after. A curve's length and a particle's displacement are different questions, and only one of them can be answered by looking at the two end points.
Let $\mathbf{r}(t) = \langle 3\cos t,\, 3\sin t,\, 4t \rangle$. Then $\mathbf{r}'(t) = \langle -3\sin t,\, 3\cos t,\, 4 \rangle$.
Differentiate componentwise.
The square of the speed is $9\sin^{2}t + 9\cos^{2}t + 16 = 9 + 16 = 25$, so the speed is $5$, constant.
The Pythagorean identity removes the variable.
One full turn is $0 \le t \le 2\pi$, so the length is $5 \times 2\pi = 10\pi$. The helix climbs $8\pi$ while going $6\pi$ around, and $\sqrt{(6\pi)^{2} + (8\pi)^{2}} = 10\pi$ — the helix is a right-angled triangle rolled around a cylinder.
A constant speed makes the integral a multiplication.
Let $\mathbf{r}(t) = \langle 1 + 2t,\; 3t,\; 6t \rangle$, a straight line. Then $\mathbf{r}'(t) = \langle 2, 3, 6 \rangle$, constant.
A line has constant velocity.
The speed is $\sqrt{4 + 9 + 36} = 7$, so $s(t) = \int_{0}^{t} 7\,du = 7t$.
The arc length function is the integral of the speed.
Inverting, $t = s/7$, and $\mathbf{r}(s/7) = \langle 1 + 2s/7,\; 3s/7,\; 6s/7 \rangle$ has derivative $\langle 2/7, 3/7, 6/7 \rangle$ of length one. Here the unit-speed parametrisation can actually be written down, which is why the line is the example everybody meets first.
Inverting the arc length function gives unit speed.
A curve is traced with $|\mathbf{r}'(t)| = 2t$. The length from $t = 0$ to $t = 3$ is $\displaystyle\int_{0}^{3} 2t\,dt$.
Length is the integral of speed.
The antiderivative is $t^{2}$, so the value is $9 - 0 = 9$.
The arc length function is $s(t) = t^{2}$, so $t = \sqrt{s}$ and the unit-speed parameter runs over $0 \le s \le 9$ while the original ran over $0 \le t \le 3$. The two parameters measure different things and agree about the answer, which is the point of the whole lesson.
Put the steps for finding the length of a curve $\mathbf{r}(t)$ from $t = 0$ to $t = 8$ in order.
Number the steps in order (write the number in the box):
For the helix $\mathbf{r}(t) = \langle 6\cos t,\; 6\sin t,\; 8t \rangle$, fill in the three quantities below.
| Value | |
|---|---|
| The square of the speed | |
| The speed | |
| The length from the start to the given time |
A curve is traced with speed $|\mathbf{r}'(t)| = 6t$ for $t \ge 0$. How long is the piece traced from $t = 0$ to $t = 3$?
Answer:
The curve $\mathbf{r}(t)$ on $0 \le t \le 8$ is retraced by $\mathbf{q}(s) = \mathbf{r}(8s)$ on $0 \le s \le 1$. Why do the two give the same length?
Put the steps for finding the length of a curve $\mathbf{r}(t)$ from $t = 0$ to $t = 2$ in order.
Number the steps in order (write the number in the box):
A rover is programmed so that its speed along its track is exactly one metre per second at every instant. Its clock reads $9$ seconds at one marker and $11$ seconds at the next. How far apart are the markers, measured along the track?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For the helix $\mathbf{r}(t) = \langle 6\cos t,\; 6\sin t,\; 8t \rangle$, fill in the three quantities below.
| Value | |
|---|---|
| The square of the speed | |
| The speed | |
| The length from the start to the given time |
You can compute the length of a curve as the integral of its speed, and you can say why the distance along a curve is not the distance between its end points. Next: how sharply the curve bends, measured per unit of that same length.
10. Your turn: the length of the curve with speed equal to twice the time, step 3