Arc Length of Parametric Curves

For a curve traced by

(x(t), y(t))

, a small change

\Delta t

in the parameter changes both

x

and

y

. Each chord along the curve has sides from both changes, as the chords for a graph did in Unit 8.

The arc length formula

Over a short interval of length

\Delta t

, the point travels about

\frac{dx}{dt}\Delta t

horizontally and

\frac{dy}{dt}\Delta t

vertically. The chord between the two points has length about

\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\;\Delta t.

Adding the chords gives a Riemann sum, and its limit is an integral.

With

x = t

and

y = f(t)

, the formula becomes the arc length of a graph from Unit 8. The integrand is the rate at which distance builds up along the curve.

Example 1

Find the length of the curve

x = \frac{t^5}{5} - t

y = \frac{2t^3}{3}

for

0 \le t \le 2

The derivatives are

\frac{dx}{dt} = t^4 - 1

and

\frac{dy}{dt} = 2t^2

. Then

\begin{aligned} &\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \\[4pt] &\quad = t^8 - 2t^4 + 1 + 4t^4 \\[4pt] &\quad = t^8 + 2t^4 + 1 = \left(t^4 + 1\right)^2. \end{aligned}

Since

t^4 + 1 > 0

, the square root is

t^4 + 1

. The point turns back in

x

t = 1

, but it rises the whole time, since

\frac{dy}{dt} > 0

for

t > 0

. So it never repeats a point, and the curve is traced once.

\begin{aligned} L &= \int_{0}^{2} \left(t^4 + 1\right)dt \\[4pt] &= \left(\frac{t^5}{5} + t\right)\Big|_{0}^{2} = \frac{32}{5} + 2 = \frac{42}{5}. \end{aligned}

The curve from (0, 0) at t = 0 to (22/5, 16/3) at t = 2. It travels left until t = 1, at (−4/5, 2/3), then right, rising the whole way.

Lengths with a calculator

A curve traced more than once

The integral adds up the distance the point travels. If the point goes over part of the curve twice, the integral counts that part twice.

Example 3

Let

x = \sin t

and

y = \sin^2 t

for

0 \le t \le 2\pi

. A calculator is allowed. (a) Describe the path of the point, including how many times it covers each part. (b) Find the value of the arc length integral over

[0, 2\pi]

. (c) Find the length of the curve itself, each point counted once.

(a) Since

y = x^2

, every point lies on the parabola, with

-1 \le x \le 1

. As

t

goes from 0 to

2\pi

x = \sin t

goes from 0 up to 1, back down to

-1

, and back to 0. The point sweeps the arc of the parabola from

(-1, 1)

(1, 1)

twice.

(b) With

\frac{dx}{dt} = \cos t

and

\frac{dy}{dt} = 2\sin t\cos t

\begin{aligned} &\int_{0}^{2\pi} \sqrt{\cos^2 t + 4\sin^2 t\cos^2 t}\,dt \\ &\qquad \approx 5.916. \end{aligned}

(c) On

\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]

the value of

x

decreases from 1 to

-1

, so the arc is traced exactly once there:

\begin{aligned} &\int_{\pi/2}^{3\pi/2} \sqrt{\cos^2 t + 4\sin^2 t\cos^2 t}\,dt \\ &\qquad \approx 2.958. \end{aligned}

The integral over

[0, 2\pi]

is twice the length of the curve. It's the distance the point travels, which is longer than the curve.

The arc of y = x² from (−1, 1) to (1, 1). For 0 ≤ t ≤ π/2 the point travels the highlighted part. Over the full period it sweeps the whole arc twice.