Arc Length of a Parametric Curve

Part 1 · Hook

Why this matters

A cyclist's app shows a map of the route and the distance ridden. The map is a curve; the distance is its length. If the app logs how fast the bike moves east and how fast it moves north, it can add up the distance without ever knowing an equation for the route. That is arc length for a parametric curve: build each tiny step with the Pythagorean theorem, then integrate.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The length of $y = f(x)$ from $x = a$ to $x = b$ is

$\int_a^b \sqrt{1 + (f'(x))^2}\,dx$
$\int_a^b \sqrt{1 + f(x)^2}\,dx$
$\int_a^b (1 + f'(x))\,dx$
$\int_a^b |f(x)|\,dx$

Show the answer

Each piece is the hypotenuse of a triangle with legs $dx$ and $f'(x)\,dx$ .

Correct: $\int_a^b \sqrt{1 + (f'(x))^2}\,dx$ :
$\int_a^b \sqrt{1 + f(x)^2}\,dx$ :
$\int_a^b (1 + f'(x))\,dx$ :
$\int_a^b |f(x)|\,dx$ :

2. For $x = t^2$ , $y = 3t$ , the derivatives with respect to $t$ are

$x' = 2t$ , $y' = 3$
$x' = 2t$ , $y' = 3t$
$x' = t^2$ , $y' = 3$
$x' = 2$ , $y' = 0$

Show the answer

Differentiate each coordinate with respect to $t$ .

Correct: $x' = 2t$ , $y' = 3$ :
$x' = 2t$ , $y' = 3t$ :
$x' = t^2$ , $y' = 3$ :
$x' = 2$ , $y' = 0$ :

Part 4 · See it

See it first

A curved path is replaced by five dashed straight segments joining points on it. On one segment a right triangle is drawn with a horizontal leg delta x, a vertical leg delta y and the segment delta s as its hypotenuse. Beside it: delta s is about the square root of delta x squared plus delta y squared; delta x is about x prime of t times delta t and delta y about y prime of t times delta t, so delta s is about the square root of x prime squared plus y prime squared, times delta t. Adding the pieces and shrinking delta t gives L, the integral of the square root of x prime of t squared plus y prime of t squared, dt. — Each short piece of the curve is close to a straight segment, the hypotenuse of a triangle with legs $x'(t)\,\Delta t$ and $y'(t)\,\Delta t$ . Adding the pieces and letting $\Delta t$ shrink gives the arc length integral. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Over a short time $\Delta t$ the point moves $\Delta x \approx x'(t)\,\Delta t$ across and $\Delta y \approx y'(t)\,\Delta t$ upthe piece of curve is close to a segment of length $\sqrt{(x'(t))^2 + (y'(t))^2}\,\Delta t$
Add the pieces from $t = a$ to $t = b$ and let $\Delta t \to 0$ $L$ $= \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$
The integrand is the speed of the pointthe integral is the distance traveled, which is the curve's length when the curve is traced once
Most of these integrals have no simple antiderivativeon the exam they are calculator questions: write the setup, then three decimals

Part 6 · Key ideas

Key ideas

$L$ $= \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$ : square each derivative, add, take the root, integrate in $t$ .
The limits are values of $t$ , not of $x$ . Check that $t$ from $a$ to $b$ traces the curve once; a path traced twice gives twice the length.
For $y = f(x)$ , take $x = t$ : the formula becomes the 8.13 formula $\int_a^b \sqrt{1 + (f'(x))^2}\,dx$ .
Exact answers come from integrands that simplify, often by $\cos^2 t + \sin^2 t = 1$ or a perfect square under the root.

Part 7 · Misconception

A common mistake

The wrong idea: The length of the segment $x = 2 + 5t$ , $y = 1 - 12t$ for $0 \le t \le 1$ is $\int_0^1 (5 + 12)\,dt$ $= 17$ .

What actually happens: The point moves 5 across and 12 down per unit of $t$ at the same time, along the diagonal: its speed is $\sqrt{5^2 + 12^2} = 13$ . The length is $\int_0^1 13\,dt$ $= 13$ , which matches the distance from $(2, 1)$ to $(7, -11)$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

One arch traced by a point on a rolling wheel

A wheel of radius 1 rolls along the

x

-axis. A point on its rim traces the path

x = t - \sin t

,

y = 1 - \cos t

. One arch is traced for

0 \le t \le 2\pi

.

One arch of the path x = t minus sin t, y = 1 minus cos t for t from 0 to 2 pi: it starts at the origin, rises to a highest point of height 2 above x = pi, and comes back down to (2 pi, 0), with arrows showing it is traced from left to right. A dashed wheel of radius 1 is drawn at t = 2 with the traced point on its rim marked. — LevlPrep original diagram.

1. Which integral gives the length of one arch?

$\int_0^{2\pi} \sqrt{(t - \sin t)^2 + (1 - \cos t)^2}\,dt$
$\int_0^{2\pi} \sqrt{(1 - \cos t)^2 + (\sin t)^2}\,dt$
$\int_0^{2\pi} \left((1 - \cos t) + \sin t\right)\,dt$
$\int_0^{2\pi} \sqrt{1 + \left(\frac{\sin t}{1 - \cos t}\right)^2}\,dt$

Show the answer

Length is $\int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$ with $x'(t) = 1 - \cos t$ and $y'(t) = \sin t$ , over the $t$ -interval that traces the arch once.

$\int_0^{2\pi} \sqrt{(t - \sin t)^2 + (1 - \cos t)^2}\,dt$ : This squares the coordinates $x$ and $y$ ; the formula squares their derivatives.
Correct: $\int_0^{2\pi} \sqrt{(1 - \cos t)^2 + (\sin t)^2}\,dt$ : Right: the square root of the sum of the squared derivatives.
$\int_0^{2\pi} \left((1 - \cos t) + \sin t\right)\,dt$ : Without squares and a square root this is not a length; the legs of the small triangle combine by the Pythagorean theorem.
$\int_0^{2\pi} \sqrt{1 + \left(\frac{\sin t}{1 - \cos t}\right)^2}\,dt$ : $\sqrt{1 + \left(\frac{dy}{dx}\right)^2}$ goes with $dx$ , not $dt$ ; with $dt$ it needs the factor $\frac{dx}{dt}$ , which this leaves out.

2. At the top of the arch, $t = \pi$ . Find $\sqrt{(x'(\pi))^2 + (y'(\pi))^2}$ , the speed of the rim point there.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$x'(\pi) = 1 - \cos\pi = 2$ and $y'(\pi) = \sin\pi = 0$ , so the speed is $\sqrt{2^2 + 0^2} = 2$ : at the top, the rim point moves twice as fast as the wheel's center.

Answer: 2

3. Using $(1 - \cos t)^2$ $+ \sin^2 t$ $= 2 - 2\cos t$ $= 4\sin^2\left(\frac{t}{2}\right)$ , find the exact length of one arch.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

On $0 \le t \le 2\pi$ , $\sin\left(\frac{t}{2}\right) \ge 0$ , so the integrand is $\sqrt{4\sin^2\left(\frac{t}{2}\right)}$ $= 2\sin\left(\frac{t}{2}\right)$ . Then $\int_0^{2\pi} 2\sin\left(\frac{t}{2}\right)\,dt$ $= -4\cos\left(\frac{t}{2}\right)\Big|_0^{2\pi}$ $= 4$ $+ 4$ $= 8$ .

Answer: 8

4. The wheel keeps rolling, and the rim point traces the path for $0 \le t \le 4\pi$ . How long is the path it traces?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Each arch has length 8, and $t$ from 0 to $4\pi$ traces two of them: $\int_0^{4\pi} 2\left|\sin\left(\frac{t}{2}\right)\right|\,dt$ $= 16$ .

Answer: 16

5. Find the length of the curve $x = t^2$ , $y = \frac{2}{3}t^3$ for $0 \le t \le \sqrt{3}$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$x' = 2t$ and $y' = 2t^2$ , so the integrand is $\sqrt{4t^2 + 4t^4}$ $= 2t\sqrt{1 + t^2}$ for $t \ge 0$ . With $u = 1 + t^2$ : $\int_1^4 u^{1/2}\,du$ $= \frac{2}{3}u^{3/2}\Big|_1^4$ $= \frac{2}{3}(8 - 1)$ $= \frac{14}{3}$ .

Answer: $\frac{14}{3}$ ≈ 4.667

6. For $x = \cos(2t)$ , $y = \sin(2t)$ , $0 \le t \le 2\pi$ , the integral $\int_0^{2\pi} \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$ equals $4\pi$ . Which statement is correct?

The circle has circumference $4\pi$ , so its radius is 2
The point goes around the unit circle twice, so the integral is twice the circumference $2\pi$
The integral should be $2\pi$ ; a value of $4\pi$ means an arithmetic slip
The integral measures the area inside the circle, $4\pi$

Show the answer

$x^2 + y^2 = 1$ : the unit circle, of circumference $2\pi$ . The speed is $\sqrt{4\sin^2(2t) + 4\cos^2(2t)}$ $= 2$ , and $2t$ runs from 0 to $4\pi$ : two full turns. The integral is the distance the point travels, $4\pi$ , which counts the circle twice.

The circle has circumference $4\pi$ , so its radius is 2: $x^2$ $+ y^2$ $= \cos^2(2t)$ $+ \sin^2(2t)$ $= 1$ , so the radius is 1; the integral counts the circle twice.
Correct: The point goes around the unit circle twice, so the integral is twice the circumference $2\pi$ : Right: the formula gives the length of the curve only when it is traced once.
The integral should be $2\pi$ ; a value of $4\pi$ means an arithmetic slip: The integral is right; it measures the distance traveled, which includes both turns.
The integral measures the area inside the circle, $4\pi$ : This integral adds up small lengths, not areas: the area of the unit circle is $\pi$ .

7. The curve $y = x^2$ , $0 \le x \le 1$ , can be written as $x = t$ , $y = t^2$ , $0 \le t \le 1$ . Which integral gives its length?

$\int_0^1 \sqrt{1 + 2t}\,dt$
$\int_0^1 \sqrt{t^2 + t^4}\,dt$
$\int_0^1 \sqrt{1 + 4t^2}\,dt$
$\int_0^1 (1 + 2t)\,dt$

Show the answer

$x' = 1$ and $y' = 2t$ , so the integrand is $\sqrt{1^2 + (2t)^2}$ $= \sqrt{1 + 4t^2}$ . It is the same integral as the 8.13 formula $\int_0^1 \sqrt{1 + (2x)^2}\,dx$ .

$\int_0^1 \sqrt{1 + 2t}\,dt$ : $y' = 2t$ must be squared: $(2t)^2 = 4t^2$ .
$\int_0^1 \sqrt{t^2 + t^4}\,dt$ : This squares $x$ and $y$ ; the formula squares their derivatives.
Correct: $\int_0^1 \sqrt{1 + 4t^2}\,dt$ : Right: $\sqrt{(x')^2 + (y')^2}$ with $x' = 1$ and $y' = 2t$ .
$\int_0^1 (1 + 2t)\,dt$ : Lengths combine by the Pythagorean theorem: square, add, take the root.

Part 9 · Summary

Summary

The length of a parametric curve traced once for

a \le t \le b

is

\int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt

, the integral of the speed of the moving point. The limits are values of

t

; a curve traced more than once is counted more than once. With

x = t

the formula reduces to the

y = f(x)

formula, and on calculator questions you write the integral and give three decimal places.

Part 10 · Up next

What comes next

Part 11 · Connections