9.5 Arc Length - Avidemia

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Length of a Curve

In this section, we want to find the length of the curve $y = f (x)$ from $x = a$ to $x = b$ . A piece of a curve that lies between two specific points is called an arc. Like the previous sections of this chapter, we construct a formula for the length of an arc by considering infinitesimals.

Let $s$ be the length of the arc from the fixed point $(a, f (a))$ to a variable point $(x, f (x))$ as shown in Figure 1. Assume $s$ increases by an infinitesimally small amount $d s$ , and $x$ and $y$ by $d x$ and $d y$ , respectively. Because $d s$ is so small, this part of the curve is virtually straight and by the Pythagorean theorem we can write $d s = \sqrt{(d x)^{2} + (d y)^{2}}$ or $d s = \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x .$ But $d y / d x = f^{'} (x) .$ Therefore, $d s = \sqrt{1 + [f^{'} (x)]^{2}} d x .$ As $d s$ sweeps along the curve from $(a, f (a))$ to $(b, f (b))$ , we add up all the infinitesimal lengths: $\begin{aligned} length of arc & = \int_{a}^{b} d s = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x \\ = \int_{a}^{b} \sqrt{1 + [f^{'} (x)]^{2}} d x . \end{aligned}$

If we want to compute the length of the curve $x = h (y)$ between $(h (c), c)$ and $(h (d), d)$ , then we can still use $d s = \sqrt{(d x)^{2} + (d y)^{2}} .$ Therefore, $\begin{aligned} d s & = \sqrt{{(\frac{d x}{d y})}^{2} + 1} d y \\ = \sqrt{[h^{'} (y)]^{2} + 1} d y \end{aligned}$ and $\begin{aligned} length of arc & = \int_{c}^{d} d s = \int_{c}^{d} \sqrt{{(\frac{d x}{d y})}^{2} + 1} d y \\ = \int_{c}^{d} \sqrt{[h^{'} (y)]^{2} + 1} d y \end{aligned}$

In the above discussion we assumed that the curve is rectifiable; that is, the curve has a finite arc length. It is possible to provide an example of a continuous function $y = f (x)$ ( $a \leq x \leq b$ ) whose graph is not rectifiable.¹ A sufficient condition for the curve $y = f (x)$ to be rectifiable is that $f^{'}$ be bounded. That is, if there is some number $K$ such that $| f^{'} (x) | < K$ for $x$ between $a$ and $b$ .

Example 1

Find the length of the curve

y^{2} = x^{3}

between the points

(0, 0)

and

(4, 8)

Figure 2

Solution 1

Solving $y^{2} = x^{3}$ for $y$ , we obtain $y = x^{\frac{3}{2}} (for y > 0) \Rightarrow \frac{d y}{d x} = \frac{3}{2} x^{1 / 2}$ The arc length formula yields $\begin{aligned} L = & \int_{0}^{4} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x \\ = & \int_{0}^{4} \sqrt{1 + \frac{9}{4} x} d x \end{aligned}$ Let $u = 1 + \frac{9}{4} x$ $d u = \frac{9}{4} d x$ $\Rightarrow d x = \frac{4}{9} d u$
$x = 0 \Leftrightarrow u = 1$ $x = 4 \Leftrightarrow u = 10$ Therefore, $L = \int_{1}^{10} \frac{4}{9} \sqrt{u} d u = \frac{4}{9} (\frac{2}{3}) u^{\frac{3}{2}} |_{u = 1}^{u = 10} = \frac{8}{27} (\sqrt{1000} - 1) \approx 9.07342 .$

Example 2

Find the length of the curve

y = \ln \sec x

[0, \frac{π}{4}]

Figure 3: Graph of

y = \ln \sec x

Solution 2

y = \ln \sec x

\sec x = u \Rightarrow y^{'} = u^{'} \frac{1}{u} = (\sec x \tan x) (\frac{1}{\sec x}) = \tan x

So the arc length formula

\begin{aligned} L & = \int_{0}^{\frac{π}{4}} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x \\ = \int_{0}^{\frac{π}{4}} \sqrt{1 + \tan^{2} x} d x \\ = \int_{0}^{\frac{π}{4}} \sqrt{\sec^{2} x} d x (1 + \tan^{2} x = \sec^{2} x) \\ = \int_{0}^{\frac{π}{4}} | \sec x | d x \\ = \int_{0}^{\frac{π}{4}} \sec x d x (\sec x > 0 for 0 \leq x \leq \frac{π}{4}) \\ = [\ln | \sec x + \tan x |]_{x = 0}^{x = \frac{π}{4}} \\ = \ln | \sec \frac{π}{4} + \tan \frac{π}{4} | - \ln | \sec 0 + \tan 0 | \\ = \ln | \frac{1}{\cos \frac{π}{4}} + \tan \frac{π}{4} | - \ln | \frac{1}{\cos 0} + \tan 0 | \\ = \ln | \frac{12}{\sqrt{2}} + 1 | - \ln | 1 + 0 | \\ = \ln (\frac{2}{\sqrt{2}} + 1) - 0 \\ = \ln (\sqrt{2} + 1) . \end{aligned}

Example 3

Find the length of the curve

y = a \cosh \frac{x}{a}

[0, a]

Solution 3

Recall that

\frac{d}{d x} \cosh x = \sinh x

. Therefore

\begin{aligned} y = a \cosh \underset{u}{\underset{⏟}{\frac{x}{a}}} \Rightarrow y^{'} & = a u^{'} \sinh u \\ = a \frac{1}{a} \sinh \frac{x}{a} \\ = \sinh \frac{x}{a} \end{aligned}

The arc length formula gives

\begin{aligned} L & = \int_{0}^{a} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x \\ = \int_{0}^{a} \sqrt{1 + \sinh^{2} \frac{x}{a}} d x \\ = \int_{0}^{a} \cosh \frac{x}{a} d x (\cosh^{2} x - \sinh^{2} x = 1 and \cosh x > 0) \\ = a {\sinh \frac{x}{a} |}_{x = 0}^{x = a} \\ = a (\sinh 1 - \sinh 0) \\ = a \sinh 1 \end{aligned}

[Recall that

\int \cosh x d x = \sinh x + C

and hence

\begin{aligned} \int \cosh \frac{x}{a} d x = & \int \cosh u \frac{d u}{u} (x = a u \Rightarrow d x = a d u) \\ = & \frac{1}{a} \sinh u + C \\ = & \frac{1}{a} \sinh \frac{x}{a} + C] \end{aligned}

Discontinuities in dy/dx

Even when $d y / d x$ is not defined at some points on the curve $y = f (x)$ , it is possible that $d x / d y$ is defined everywhere on this curve. An example of such a situation is when the curve $y = f (x)$ has a vertical tangent. In such cases, to compute the arc length of the curve $y = f (x)$ , we can express $x$ in terms of $y$ and write $d s = \sqrt{d x^{2} + d y^{2}}$ $\Rightarrow \frac{d s}{d y} = \sqrt{{(\frac{d x}{d y})}^{2} + 1} .$

Example 4

Find the length of the curve

y = x^{2 / 3}

from

x = 0

x = 8

Figure 4

Solution 4

y = x^{2 / 3} \Rightarrow y^{'} = \frac{2}{3} x^{- 1 / 3}

Because

y^{'}

is not defined at

x = 0

, we cannot find the length of the curve. However, if we express

x

in terms of

y

, we will find

x = y^{3 / 2} \Rightarrow \frac{d x}{d y} = \frac{3}{2} y^{1 / 2}

and

x = 0 \Leftrightarrow y = 0

x = 8 \Leftrightarrow y = 4

The length of the curve is

\begin{aligned} s & = \int_{0}^{4} \sqrt{{(\frac{d x}{d y})}^{2} + 1} d y \\ = \int_{0}^{4} \sqrt{\frac{9}{4} y + 1} d y \\ = \frac{1}{2} \int_{0}^{4} \sqrt{9 y + 4} d y \\ = \frac{1}{2} \int_{4}^{40} \sqrt{u} \frac{d u}{9} (u = 9 y + 4 then d u = 9 d y) \\ = \frac{1}{18} {[\frac{2}{3} u^{3 / 2}]}_{4}^{40} \\ = \frac{1}{27} (40^{3 / 2} - 8) \\ = \frac{1}{27} (8 \sqrt{10^{3}} - 8) \\ = \frac{8}{27} (\sqrt{1000} - 1) \approx 9.07342 . \end{aligned}

The Arc Length Function

The length of the curve $y = f (x)$ from $(a, f (a))$ to $(b, f (b))$ is $\int_{a}^{b} \sqrt{1 + [f^{'} (x)]^{2}} d x .$ Let $s (x)$ be a function that measures the arc length from $(a, f (a))$ to a variable point $(x, f (x))$ . To find the formula of $s (x)$ , we just need replace $b$ in the above formula with $x$ : $s (x) = \int_{a}^{x} \sqrt{1 + [f^{'} (t)]^{2}} d t .$ [Because we can denote the variable of integration by any arbitrary letter and because $x$ appears as the upper limit of integration, we have replaced the variable of integration by $t$ so that $x$ does not have two different meanings.] This function is called the arc length function.

By the Fundamental Theorem of Calculus : $\frac{d s}{d x} = \sqrt{1 + [f^{'} (x)]^{2}} = \sqrt{1 + {(\frac{d y}{d x})}^{2}}$ or $d s = \sqrt{d x^{2} + d y^{2}} .$
Note that $s (x)$ is an increasing function. To mathematically prove this property, we need to show $s^{'} (x) > 0$ : $s^{'} (x) = \sqrt{1 + [f^{'} (x)]^{2}} > 0.$

Example 5

Find the arc length function for the curve

y = \frac{1}{2} x^{2} - \frac{1}{4} \ln x

taking

(1, 0.5)

as the starting point. Then, using this function, find the arc length along this curve from

(1, 0.5)

(e, 0.5 e^{2} - 0.25)

Solution 5

f (x) = \frac{1}{2} x^{2} - \frac{1}{4} \ln x \Rightarrow f^{'} (x) = x - \frac{1}{4 x}

then

\begin{aligned} \sqrt{1 + [f^{'} (x)]^{2}} & = \sqrt{1 + (x - \frac{1}{4 x})^{2}} \\ = \sqrt{1 + x^{2} - \frac{1}{2} + \frac{1}{16 x^{2}}} \\ = \sqrt{(x + \frac{1}{4 x})^{2}} \\ = x + \frac{1}{4 x} (because x > 0) \end{aligned}

Therefore, the arc length function is

\begin{aligned} s (x) & = \int_{1}^{x} \sqrt{1 + [f^{'} (t)]^{2}} d t \\ = \int_{1}^{x} (t + \frac{1}{4 t}) d t \\ = {[\frac{1}{2} t^{2} + \frac{1}{4} \ln t]}_{1}^{x} \\ = \frac{1}{2} x^{2} + \frac{1}{4} \ln x - \frac{1}{2} - \frac{1}{4} \underset{= 0}{\underset{⏟}{\ln 1}} \\ = \frac{1}{2} x^{2} + \frac{1}{4} \ln x - \frac{1}{2} . \end{aligned}

The arc length along this curve from

(1, 1 / 2)

(e, 0.5 e^{2} - 0.25)

s (e) = \frac{1}{2} e^{2} + \frac{1}{4} \underset{= 1}{\underset{⏟}{\ln e}} - \frac{1}{2} = \frac{e^{2}}{2} - \frac{1}{4} \approx 3.44453 .

Let $f (x) = {\begin{cases} x \sin \frac{π}{x} & x \neq 0 \\ 0 & x = 0 \end{cases}$ The function $f$ is not rectifiable on $[0, 1]$ .↩︎