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Distance travelled is speed accumulated: $\int \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$, the same formula written $\int \sqrt{1 + (dy/dx)^2}\,dx$ for a graph — and never the same thing as the displacement unless the path is straight.
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 can set up and evaluate an arc length integral for a parametric curve, assembling the speed $\sqrt{(dx/dt)^2 + (dy/dt)^2}$ before integrating it, and you can write the same integral for the graph of a function. You can say why the integral of a speed can never decrease, distinguish the distance travelled from the displacement, and check the formula against the circumference of a circle.
You can find the velocity of a moving point and its speed, and in calculus AB you integrated a speed on a line to get a distance. This lesson does exactly that in two dimensions. Nothing new is proved; the speed is simply assembled from two components first, and then integrated as before.
Arc length: the length of a curve, measured along it.
Distance travelled: the same number, read as a motion — the integral of the speed.
Displacement: the straight line from the start point to the end point, computed from those two points alone.
The arc length integral: $\displaystyle\int_a^b \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$.
Cut the path into pieces so short that each is nearly straight. A piece in which $x$ changes by $dx$ and $y$ by $dy$ has length $\sqrt{dx^2 + dy^2}$ by Pythagoras. Factor out $dt$:
$$\sqrt{dx^2 + dy^2} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\;dt$$
and adding up the pieces gives
$$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt.$$
The integrand is the speed, so the sentence is the familiar one: distance travelled is speed accumulated.
For the graph of a function the same formula applies with $x$ as its own parameter, so $dx/dx = 1$ and
$$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx.$$
It is one formula, not two.
Distance is not displacement. The integrand is a square root and so never negative, which means the integral can only grow: a path that doubles back adds length rather than cancelling it. The displacement uses only the endpoints. They agree exactly when the path is straight and never travelled backwards, and in every other case the distance is larger.
Most of these integrals cannot be done by hand, and that is expected. A square root of a sum of squares rarely has an elementary antiderivative. Setting the integral up correctly is the skill; evaluating it is often a calculator's job.
Another way: picture
Picture laying a piece of string along the curve and then pulling it straight to read its length against a ruler. The string follows every bend and every reversal, so it is always at least as long as the ruler held directly between the two ends — and longer as soon as the path is not straight. The string is the distance travelled; the ruler between the ends is the displacement.
Another way: steps
| Quantity | How it is found | Can it decrease? |
|---|---|---|
| Distance travelled | $\int \sqrt{x'^2 + y'^2}\,dt$ | no |
| Displacement | endpoints only, by Pythagoras | yes |
| Change in $x$ alone | $x(b) - x(a)$ | yes |
Only the first uses the whole path. A question that asks how much cable, how much fencing or how far something travelled wants the first; a question that asks how far it ended up from where it began wants the second.
Take $x = a\cos t$, $y = a\sin t$ over $0 \le t \le 2\pi$. The component derivatives are $-a\sin t$ and $a\cos t$; squaring and adding gives $a^2(\sin^2 t + \cos^2 t) = a^2$, so the speed is the constant $a$.
The length is then $a \times 2\pi = 2\pi a$, which is the circumference you have known since primary school. A formula that reproduces a fact you already know is a formula you have written down correctly, and this is the cheapest such check available for arc length.
It also shows what a constant speed buys: when the integrand does not depend on $t$, the integral collapses to speed times elapsed time and there is no antiderivative to find. That is why almost every arc length you can finish by hand has a constant speed or a perfect square hiding under the root.
Integrating before taking the square root. The root comes first: the speed is $\sqrt{x'^2 + y'^2}$, and it is the speed that is integrated. Integrating $x'^2 + y'^2$ and rooting afterwards gives a number with no meaning.
Reading the arc length as a displacement. The integral of a speed can never decrease. A path that goes out and comes back has a large length and zero displacement.
Adding the components instead of combining them by Pythagoras. $x' = 3$ and $y' = 4$ give a speed of $5$. The $7$ is the length of a path that goes along one direction and then the other, which is a different path.
$x = 3t$, $y = 4t$ for $0 \le t \le 2$: the components of velocity are $3$ and $4$.
Both constant.
Speed $= \sqrt{9 + 16} = 5$, so the integrand does not depend on $t$.
Pythagoras, not addition.
$L = 5 \times 2 = 10$ — and the displacement is also $10$, because this particular path happens to be straight.
$x = t^2 - 4t$, $y = 0$ for $0 \le t \le 4$: the motion is along a line, out to $x = -4$ at $t = 2$ and back to $x = 0$.
Speed $= |2t - 4|$.
Distance travelled $= \int_0^4 |2t - 4|\,dt = 4 + 4 = 8$.
The absolute value is the square root.
Displacement $= x(4) - x(0) = 0$. Same journey, two honest answers to two different questions.
Differentiate: the components of velocity are $6$ and $8$, both constant.
Nothing depends on $t$.
Square, add and root: $\sqrt{36 + 64} = \sqrt{100} = 10$, the speed.
Not $14$.
Integrate a constant over a range of length $5$: $L = 10 \times 5 = 50$.
Put the steps of finding the length of the path $x = 2t$, $y = t^2$ from $t = 0$ to $t = 1$ into the order you do them.
Number the steps in order (write the number in the box):
A particle moves with $x = 9t$ and $y = 12t$, from $t = 0$ to $t = 2$. How far does it travel?
Answer:
Match each expression to the quantity it measures.
| Distance travelled along a parametric path | Displacement from start to finish | Length of the graph of a function | Area swept in polar coordinates | |
|---|---|---|---|---|
| $\displaystyle\int_a^b \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$ | ||||
| $\sqrt{(x(b) - x(a))^2 + (y(b) - y(a))^2}$ | ||||
| $\displaystyle\int_a^b \sqrt{1 + (dy/dx)^2}\,dx$ | ||||
| $\tfrac12\displaystyle\int r^2\,d\theta$ |
A particle moves with $x = 3t$ and $y = 4t$ metres, with $t$ in seconds. Fill in its position and the distance it has travelled at each time.
| $x$ | $y$ | distance travelled | |
|---|---|---|---|
| after 1 second | |||
| after 2 seconds | |||
| after 3 seconds |
A particle travels along a curved path from one point to another. What does $\displaystyle\int \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$ over that journey give?
A particle moves with $x = 7\cos t$ and $y = 7\sin t$ from $t = 0$ to $t = 2\pi$. Give its speed, then the length of its path as a multiple of $\pi$.
speed = s, length = m times pi
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of finding the length of the path $x = 5t$, $y = t^2$ from $t = 0$ to $t = 1$ into the order you do them.
Number the steps in order (write the number in the box):
You can find a length along a curve rather than across it. Without looking: what is the integrand of an arc length integral, and when do the distance travelled and the displacement come out equal?
10. Your turn: the length of $x = 6t$, $y = 8t$ from $t = 0$ to $t = 5$, step 3