Finding Arc Lengths of Curves Given by Parametric Equations

AP Calculus BC · Parametric, Polar & Vector-Valued Functions

Arc Lengths of Parametric Curves

Pythagoras on the two coordinate rates — the integrand is simply the speed.

key idea$L=\int_{t_1}^{t_2}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\;dt$

Picture the region

For a parametrised curve the arc element is symmetric in the two coordinates: move $\dfrac{dx}{dt}\,dt$ across and $\dfrac{dy}{dt}\,dt$ up, and Pythagoras gives the step length. There is no $1$ under the root here — that $1$ only appears when the parameter is $x$.

One arch of the cycloid $x=t-\sin t$, $y=1-\cos t$. The arrows mark the direction of travel and the amber vector is the velocity at $t=2$. Its length is the speed, and integrating that speed from $t=0$ to $t=2\pi$ gives the length of the arch.

Setting up the integral

Building a parametric arc-length integral

Differentiate both coordinate functions with respect to $t$.
Square each and add. Look for a trigonometric identity — exam problems are almost always built so that this sum collapses.
Take the square root. The result is the speed, and it is never negative.
Set the limits as the two parameter values, not as $x$-values. This is the most common slip in the whole topic.
Integrate. If the integrand does not simplify, set it up correctly and evaluate numerically.

four things that go wrong

No $1$ appears under the root — that term belongs to the $y=f(x)$ form only.
The integrand $\sqrt{\left(x'\right)^{2}+\left(y'\right)^{2}}$ is the speed, so this integral is total distance travelled.
The limits are values of $t$. Converting them to $x$-values is a set-up error, not a shortcut.
If the curve retraces itself, the integral counts the retraced part again — distance travelled, not length of the picture.

Curve given as	Arc length integrand	Variable
$x=f(t)$, $y=g(t)$	$\sqrt{\left(f'\right)^{2}+\left(g'\right)^{2}}$	$dt$
$y=f(x)$	$\sqrt{1+\left(f'\right)^{2}}$	$dx$
$x=g(y)$	$\sqrt{1+\left(g'\right)^{2}}$	$dy$
$\vec{r}(t)=\langle f,g\rangle$	$\left\lVert\vec{r}\,'(t)\right\rVert$	$dt$

Rows 2 and 3 are the special case where one coordinate is the parameter: then one of the two derivatives is $1$, and squaring it produces the familiar $1$ under the root. There is really only one formula here.

Worked examples

EX 1

A trigonometric collapse

the cycloid arch

Find the length of one arch of $x=t-\sin t$, $y=1-\cos t$, for $0\le t\le2\pi$.

Differentiate

$\dfrac{dx}{dt}=1-\cos t$ and $\dfrac{dy}{dt}=\sin t$.

Square and add

$\left(1-\cos t\right)^{2}+\sin^{2}t=1-2\cos t+\cos^{2}t+\sin^{2}t=2-2\cos t$.

Use a half-angle identity

$2-2\cos t=4\sin^{2}\dfrac{t}{2}$, so the integrand is $2\left\lvert\sin\dfrac{t}{2}\right\rvert$.

Drop the absolute value

On $[0,2\pi]$, $\dfrac{t}{2}\in[0,\pi]$ where the sine is non-negative. So the integrand is $2\sin\dfrac{t}{2}$.

Integrate

$\displaystyle\int_{0}^{2\pi}2\sin\dfrac{t}{2}\,dt=\left[-4\cos\dfrac{t}{2}\right]_{0}^{2\pi}=4+4=8$.

EX 2

A circle traced once

check against $2\pi r$

Find the length of $x=3\cos t$, $y=3\sin t$ for $0\le t\le2\pi$.

Differentiate

$\dfrac{dx}{dt}=-3\sin t$ and $\dfrac{dy}{dt}=3\cos t$.

Square and add

$9\sin^{2}t+9\cos^{2}t=9$.

Take the root

The speed is the constant $3$ — the point moves at uniform rate.

Integrate

$\displaystyle\int_{0}^{2\pi}3\,dt=6\pi$.

Check

A circle of radius $3$ has circumference $2\pi(3)=6\pi$. ✓

EX 3

Retracing counts twice

distance, not picture

For the same circle but $0\le t\le4\pi$, what does the integral give, and what does it mean?

Note what changed

Only the upper limit; the speed is still the constant $3$.

Integrate

$\displaystyle\int_{0}^{4\pi}3\,dt=12\pi$.

Interpret

The point goes twice around, so it travels $12\pi$ units.

Distinguish the two questions

The curve drawn is still a circle of circumference $6\pi$; the distance travelled is $12\pi$.

State the rule

This integral always measures distance travelled. If a question asks for the length of the curve, restrict $t$ to one traversal first.

EX 4

When it does not simplify

set up and evaluate numerically

Set up the length of $x=t^{2}$, $y=t^{3}$ for $0\le t\le2$.

Differentiate

$\dfrac{dx}{dt}=2t$ and $\dfrac{dy}{dt}=3t^{2}$.

Square and add

$4t^{2}+9t^{4}$.

Factor what you can

$=t^{2}\left(4+9t^{2}\right)$, so the integrand is $\lvert t\rvert\sqrt{4+9t^{2}}$, which is $t\sqrt{4+9t^{2}}$ for $t\ge0$.

Recognise a substitution

$u=4+9t^{2}$ gives $du=18t\,dt$, so $L=\dfrac{1}{18}\displaystyle\int_{4}^{40}\sqrt{u}\,du$.

Evaluate

$\dfrac{1}{27}\left[u^{3/2}\right]_{4}^{40}=\dfrac{40\sqrt{40}-8}{27}\approx9.07$. Factoring out $t^{2}$ was what made this elementary.

Watch out

No stray $1$ under the root

$\sqrt{\left(x'\right)^{2}+\left(y'\right)^{2}}$. Adding a $1$ imports the Cartesian formula into a setting where it does not belong.

The limits are parameter values

Integrate from $t_1$ to $t_2$. Substituting $x$-endpoints into a $dt$ integral produces a number with no meaning.

Take the positive root

Speed cannot be negative. When a factor like $\lvert t\rvert$ or $\left\lvert\sin\tfrac{t}{2}\right\rvert$ appears, check its sign on the interval before dropping the bars.

Retracing is counted

The integral measures distance travelled. A parameter interval that covers the curve twice returns twice the length — usually the intended answer, but read the question.

Course map. Topic 9.3 · LO CHA-6.B — determine the length of a curve in the plane defined by parametric functions, using a definite integral. BC-only.