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Pythagoras on a short step gives $ds = \sqrt{1 + (y')^2}\,dx$, and the mean value theorem is what turns the rise into a slope.
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 build the arc length integrand in the right order — differentiate, square, add one, take the root — derive the formula from a chord, Pythagoras and the mean value theorem, say what the $1$ is and why the length is never less than the horizontal span, integrate in the other variable when the curve is naturally $x = g(y)$, recognise that most arc length integrals have no elementary antiderivative and that this is not an error, and check any answer against the straight-line distance between the endpoints.
You believe that the straight-line distance between two points is $\sqrt{(\Delta x)^2 + (\Delta y)^2}$, and you believe that a curve can be approximated by a chain of short straight pieces.
Put those together and arc length is already defined: the length of a curve is the limit of the lengths of inscribed polygons as the pieces shrink. Everything in this lesson is that sentence turned into an integral, and the mean value theorem is the tool that does the turning.
A chord joins two points of the curve by a straight segment. An inscribed polygon is a chain of chords through points of the curve; its length is always at most the curve's, since a straight line is the shortest route between its ends.
A curve is rectifiable when those polygon lengths have a finite supremum — that supremum being, by definition, its length. Not every continuous curve is: a graph can be continuous, bounded, and of infinite length.
The element of arc length is $ds = \sqrt{(dx)^2 + (dy)^2}$, and the whole subject is different ways of factoring that expression.
Theorem. If $f'$ is continuous on $[a, b]$, the graph of $f$ has length $$L = \int_a^b \sqrt{1 + f'(x)^2}\,dx.$$
Where it comes from. Cut $[a,b]$ into pieces. One chord has length $\sqrt{(\Delta x)^2 + (\Delta y)^2}$. The mean value theorem gives $\Delta y = f'(c)\Delta x$ for some $c$ in the piece, so the chord is $\sqrt{1 + f'(c)^2}\,\Delta x$. Adding them is a Riemann sum, and its limit is the integral.
The hypothesis is used twice. Differentiability is needed for the mean value theorem; continuity of $f'$ is needed for the Riemann sum to converge to the integral. Without them the length may still exist and may not be given by this formula.
Reading the formula. The integrand is at least $1$, always, so $L \ge b - a$: a curve is never shorter than its horizontal span. It equals it only for a horizontal line.
The other way round. If the curve is naturally $x = g(y)$, use $$L = \int_c^d\sqrt{1 + g'(y)^2}\,dy,$$ which is the same $ds$ with $dy$ factored out instead. Choose whichever makes the derivative manageable.
The integrals are usually impossible. $\sqrt{1 + f'^2}$ is a square root of a general function and almost never has an elementary antiderivative. The arc length of a parabola needs a logarithm and a root; the arc length of an ellipse defines a new class of functions. This is a formula that is usually evaluated numerically, and that is not a failure of the method.
Another way: picture
A staircase drawn along the curve: each tread $dx$ across, each riser $dy$ up, and the curve running diagonally through the corners. The length of the curve is the sum of the diagonals, not of the treads and not of the risers. Summing the treads gives $b - a$; summing the risers gives the total rise; the diagonals give something larger than either, and that is the length.
Another way: steps
An arc length integral is elementary only when $1 + f'^2$ happens to be a perfect square. That is a strong coincidence, and the textbook examples are exactly the functions engineered to produce it.
$y = \tfrac23 x^{3/2}$ has $f' = x^{1/2}$, so $1 + f'^2 = 1 + x$ — a root that comes out. $y = \tfrac{x^3}{6} + \tfrac{1}{2x}$ has $f' = \tfrac{x^2}{2} - \tfrac{1}{2x^2}$, whose square is $\tfrac{x^4}{4} - \tfrac12 + \tfrac{1}{4x^4}$; adding one turns the $-\tfrac12$ into $+\tfrac12$ and makes it $\left(\tfrac{x^2}{2} + \tfrac{1}{2x^2}\right)^2$. Every such example is built by starting from the answer.
Meanwhile the parabola $y = x^2$ gives $\int\sqrt{1 + 4x^2}\,dx$, which needs a trigonometric substitution and produces a logarithm; the sine curve gives an integral with no elementary form at all; and the ellipse gives the elliptic integrals, a family of functions defined by these very integrals because nothing simpler describes them.
So the honest summary is that arc length is easy to set up and usually impossible to evaluate in closed form. Setting it up correctly is the skill; a numerical value is a separate and routine matter.
Adding the one before squaring. $\sqrt{(1 + y')^2}$ is $|1 + y'|$, which is not a length of anything.
Using $f$ instead of $f'$. The height of a curve has nothing to do with its length; shift the whole graph up and the length is unchanged, while $\int\sqrt{1 + f^2}$ changes.
Forgetting the root. $\int(1 + f'^2)dx$ has no geometric meaning.
Integrating in the wrong variable. $\sqrt{1 + g'(y)^2}$ must be integrated $dy$, over the $y$-range.
Expecting a clean answer. Most arc lengths are not elementary. An integral you cannot do is usually correct.
Ignoring a corner. $y = |x|$ on $[-1,1]$ has length $2\sqrt2$, but $f'$ does not exist at $0$; split the integral at the corner rather than quoting the theorem across it.
Checking nothing. The answer must exceed the straight-line distance between the endpoints. That takes one line and catches most slips.
Every previous application in this unit produced an integral that came out: areas of polynomial regions, volumes of cones and spheres, all elementary. Arc length does not, and the first time a learner meets $\int_0^1\sqrt{1 + 4x^2}\,dx$ the natural conclusion is that the setup went wrong.
It did not. The square root of a general expression almost never has an elementary antiderivative, and $1 + f'^2$ is a general expression. The textbook examples that work are reverse-engineered — chosen because $1 + f'^2$ is a perfect square — and they are a small and artificial family. The length of a parabola needs a logarithm; the length of a sine wave has no closed form; the length of an ellipse defines a family of functions named after the problem.
This matters because it changes what "doing" an arc length problem means. The work is setting up the integral correctly and knowing what the answer should roughly be; evaluating it is often a matter for a numerical method, which is a routine and reliable thing to do.
And it gives a useful discipline: check the size of the answer against the chord. The length must exceed the straight-line distance between the endpoints and exceed the horizontal span. Two comparisons, neither requiring the integral to be evaluated, and they catch almost every error in the setup.
$y = \tfrac23 x^{3/2}$ on $[0, 3]$: $y' = x^{1/2}$, so $(y')^2 = x$ and $1 + (y')^2 = 1 + x$.
Differentiate, square, add one.
$L = \displaystyle\int_0^3\sqrt{1 + x}\,dx = \left[\tfrac23(1+x)^{3/2}\right]_0^3 = \tfrac23(8 - 1) = \dfrac{14}{3}$.
The root is of a linear expression, so substitution finishes it.
Check: the endpoints are $(0,0)$ and $(3, 2\sqrt3)$, a straight-line distance of $\sqrt{9 + 12} = \sqrt{21} \approx 4.58$, and $\tfrac{14}{3} \approx 4.67$ is a little more. A curve slightly longer than its chord is exactly what the picture shows.
$y = x^2$ on $[0, 1]$: $y' = 2x$, so $L = \displaystyle\int_0^1\sqrt{1 + 4x^2}\,dx$.
Set up exactly as before.
A constant plus a square under a root: $2x = \tan\theta$, and the integral becomes $\tfrac12\int\sec^3\theta\,d\theta$ — a reduction-formula integral from lesson 2.
Every technique of unit 1 is needed for one arc length.
The result is $\tfrac{\sqrt5}{2} + \tfrac14\ln(2 + \sqrt5) \approx 1.479$, against a chord of $\sqrt2 \approx 1.414$. A logarithm has appeared in the length of a parabola, which is a fair warning about what these integrals are like.
$y' = \dfrac{x^{2}}{2} - \dfrac{1}{2x^{2}}$, so $(y')^{2} = \dfrac{x^{4}}{4} - \dfrac12 + \dfrac{1}{4x^{4}}$.
Differentiate and square carefully; the cross term is the interesting one.
Adding one turns $-\tfrac12$ into $+\tfrac12$, giving $\left(\dfrac{x^{2}}{2} + \dfrac{1}{2x^{2}}\right)^{2}$ — a perfect square, which is why this function was chosen.
The whole design of the example is in that sign flip.
$L = \displaystyle\int_1^2\left(\dfrac{x^{2}}{2} + \dfrac{1}{2x^{2}}\right)dx = \left[\dfrac{x^{3}}{6} - \dfrac{1}{2x}\right]_1^2 = \dfrac{17}{12}$.
Find the length of $y = \dfrac{6}{8}x$ from $x = 0$ to $x = 32$.
Answer:
Put the steps of building the arc length integrand for $y = 2x^{2}$ into the order you do them.
Number the steps in order (write the number in the box):
In $\displaystyle\int\sqrt{1 + (y')^{2}}\,dx$, what is the $1$?
Build the derivation of $L = \displaystyle\int_a^b\sqrt{1 + f'(x)^{2}}\,dx$ from a polygon of $4$ chords.
This task has no paper form; do it on a device.
For the line $y = \dfrac{7}{24}x$, build the arc length integrand one column at a time.
| Value | |
|---|---|
| $y'$ | 7/24 |
| $(y')^{2}$ | |
| $1 + (y')^{2}$ | |
| $\sqrt{1 + (y')^{2}}$ |
A cable runs in a straight line from the top of a mast down to a ground anchor $45$ metres away horizontally and $24$ metres below. How long is the cable?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Find the length of $y = \dfrac{8}{15}x$ from $x = 0$ to $x = 15$.
Answer:
You can set up the length of any curve and say why the setup is the skill and the evaluation often is not. Say in your own words why the height of a curve plays no part in its length. Next: the same $ds$, swept round an axis, giving a surface.
10. Your turn: the length of $y = \dfrac{x^{3}}{6} + \dfrac{1}{2x}$ from $x = 1$ to $x = 2$, step 3