Arc Length

Since an arc on any curve is made up of a lot of little bits of straight lines joined end to end, if we could add all these little bits, we would get the length of the arc. But we have seen that to add a lot of little bits together is precisely what is called integration, so that it is likely that, since we know how to integrate, we can find also the length of an arc on any curve, provided that the equation of the curve is such that it lend itself to integration.

If $M N$ is an arc on any curve, the length $s$ of which is required (see the following figure), if we call “a little bit” of the are $d s$ , then we see at once that $(d s)^{2} = (d x)^{2} + (d y)^{2} .$ or either $d s = \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y or d s = \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x .$

Now the arc $M N$ is made up of the sum of all the little bits $d s$ between $M$ and $N$ , that is, between $x_{1}$ and $x_{2}$ , or between $y_{1}$ and $y_{2}$ , so that we get either

$s = \int_{x_{1}}^{x_{2}} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x or s = \int_{y_{1}}^{y_{2}} \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y .$

That is all!

The second integral is useful when there are several points of the curve corresponding to the given values of $x$ (as in the next figure). In this case the integral between $x_{1}$ and $x_{2}$ leaves a doubt as to the exact portion of the curve, the length of which is required. It may be $S T$ , instead of $M N$ , $P Q$ , or $S Q$ , by integrating between $y_{1}$ and $y_{2}$ the uncertainty is removed, and in this case one should use the second integral.

If instead of $x$ and $y$ coordinates,—or Cartesian coordinates, as they are named from the French mathematician Descartes, who invented them—we have $r$ and $θ$ coordinates (or polar coordinates); then, if $M N$ be a small arc of length $d s$ on any curve, the length $s$ of which is required (see the following figure), $O$ being the pole, then the distance $O N$ will generally differ from $O M$ by a small amount $d r$ . If the small angle $∠ M O N$ is called $d θ$ , then, the polar coordinates of the point $M$ being $θ$ and $r$ , those of $N$ are $(θ + d θ)$ and $(r + d r)$ . Let $M P$ be perpendicular to $O N$ , and let $O R = O M$ ; then $R N = d r$ , and this is very nearly the same as $P N$ , as long as $d θ$ is a very small angle. Also $R M = r d θ$ , and $R M$ is very nearly equal to $P M$ , and the arc $M N$ is very nearly equal to the chord $M N$ . In fact we can write $P N = d r, P M = r d θ$ , and arc $M N =$ chord $M N$ without appreciable error, so that we have:

$(d s)^{2} = (chord M N)^{2} = {\overset{―}{P N}}^{2} + {\overset{―}{P M}}^{2} = (d r)^{2} + r^{2} (d θ)^{2} .$

Dividing by $(d θ)^{2}$ we get ${(\frac{d s}{d θ})}^{2} = r^{2} + {(\frac{d r}{d θ})}^{2}$ ; hence $\frac{d s}{d θ} = \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} and d s = \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ;$ hence, since the length $s$ is made up of the sum of all the little bits $d s$ , between values of $θ = θ_{1}$ and $θ = θ_{2}$ we have $s = \int_{θ_{1}}^{θ_{2}} d s = \int_{θ_{1}}^{θ_{2}} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ .$

We can proceed at once to work out a few examples.

Example 22.1. The equation of a circle, the centre of which is at the origin—or intersection of the axis of $x$ with the axis of $y$ —is $x^{2} + y^{2} = r^{2}$ ; find the length of an arc of one quadrant.

Solution.

$y^{2} = r^{2} - x^{2}$ and $2 y d y = - 2 x d x,$ so that $\frac{d y}{d x} = - \frac{x}{y};$ hence $s = \int \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x = \int \sqrt{1 + \frac{x^{2}}{y^{2}}} d x$ and since $y^{2} = r^{2} - x^{2}$ , $s = \int \sqrt{1 + \frac{x^{2}}{r^{2} - x^{2}}} d x = \int \frac{r d x}{\sqrt{r^{2} - x^{2}}} .$

The length we want—one quadrant—extends from a point for which $x = 0$ to another point for which $x = r$ . We express this by writing

$s = \int_{x = 0}^{x = r} \frac{r d x}{\sqrt{r^{2} - x^{2}}},$ or, more simply, by writing $s = \int_{0}^{r} \frac{r d x}{\sqrt{r^{2} - x^{2}}},$ the 0 and $r$ to the right of the sign of integration merely meaning that the integration is only to be performed on a portion of the curve, namely that between $x = 0, x = r$ , as we have seen here.

Here is a fresh integral for you! Can you manage it?

In the chapter on the Derivatives of Trigonometric Functions, we have differentiated $y = \arcsin x$ (also denoted by $\sin^{- 1} x$ ) and found $\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}}$ . If you have tried all sorts of variations of the given examples (as you ought to have done!), you perhaps tried to differentiate something like $y = a \arcsin \frac{x}{a}$ , which gave $\frac{d y}{d x} = \frac{a}{\sqrt{a^{2} - x^{2}}} or d y = \frac{a d x}{\sqrt{a^{2} - x^{2}}},$ that is, just the same expression as the one we have to integrate here.

Hence $s = \int \frac{r d x}{\sqrt{r^{2} - x^{2}}} = r \arcsin \frac{x}{r} + C,$ $C$ being a constant.

As the integration is only to be made between $x = 0$ and $x = r$ , we write

$s = \int_{0}^{r} \frac{r d x}{\sqrt{r^{2} - x^{2}}} = {[r \arcsin \frac{x}{r} + C]}_{0}^{r};$ proceeding then as explained in Example [eg:Ch19-1] we get $s = r \arcsin \frac{r}{r} + C - r \arcsin \frac{0}{r} - C,$ or $s = r \times \frac{π}{2},$ since $\arcsin 1$ is $\frac{π}{2}$ and $\arcsin 0$ is zero, and the constant $C$ disappears, as has been shown.

The length of the quadrant is therefore $\frac{π r}{2}$ , and the length of the circumference, being four times this, is $4 \times \frac{π r}{2} = 2 π r$ .

Example 22.2. Find the length of the arc $A B$ between $x_{1} = 2$ and $x_{2} = 5$ , in the circumference $x^{2} + y^{2} = 6^{2}$ (see the following figure).

Solution. Here, proceeding as in previous example,

\begin{align} s & =\left[r \arcsin \left(\frac{x}{r}\right)+C\right]_{x_{1}}^{x_{2}}=\left[6 \arcsin \left(\frac{x}{6}\right)+C\right]_{2}^{5} \\ & =6\left[\arcsin \left(\frac{5}{6}\right)-\arcsin \left(\frac{2}{6}\right)\right]\\ &=6(0.985111-0.339837) \\ & =3.87164 \text { units of length (the arcs being expressed in radians). } \end{align}

It is always well to check results obtained by a new and yet unfamiliar method. This is easy, for $\cos (∠ A O C) = \frac{2}{6} = \frac{1}{3} and \cos (∠ B O C) = \frac{5}{6};$ hence $∠ A O C = \arccos \frac{1}{3} = 1.23096 radians, ∠ B O C = \arccos \frac{5}{6} = 0.585686 radians,$ and $∠ A O B = ∠ A O C - ∠ B O C = 0.645274 radians$

Therefore, the length of the arc $A B$ is $6 \times 0.645274 = 3.87164$ units of length.

[Recall that the length of a circular arc = arc angle (measured in radians) $\times$ circle radius.]

A slight discrepancy between the results obtained from the two methods may occur if the result of each calculation is not recorded with sufficient decimal places.

Example 22.3. Find the length of an arc of the curve

$y = \frac{a}{2} (e^{\frac{x}{a}} + e^{- \frac{x}{a}})$ between $x = 0$ and $x = a$ . (This curve is the catenary.)

Solution. \begin{align} y & =\frac{a}{2} e^{\frac{x}{a}}+\frac{a}{2} e^{-\frac{x}{a}}, \quad \frac{d y}{d x}=\frac{1}{2}\left(e^{\frac{x}{a}}-e^{-\frac{x}{a}}\right), \\ s & =\int \sqrt{1+\frac{1}{4}\left(e^{\frac{x}{a}}-e^{-\frac{x}{a}}\right)^{2}} d x \\ & =\frac{1}{2} \int \sqrt{4+e^{\frac{2 x}{a}}+e^{-\frac{2 x}{a}}-2 e^{\frac{x}{a}-\frac{x}{a}}} \, dx . \end{align} Now $e^{\frac{x}{a} - \frac{x}{a}} = e^{0} = 1,$ so that $s = \frac{1}{2} \int \sqrt{2 + e^{\frac{2 x}{a}} + e^{- \frac{2 x}{a}}} d x;$ we can replace 2 by $2 \times e^{0} = 2 \times e^{\frac{x}{a} - \frac{x}{a}}$ ; then \begin{align} s & =\frac{1}{2} \int \sqrt{e^{\frac{2 x}{a}}+2 e^{\frac{x}{a}-\frac{x}{a}}+e^{-\frac{2 x}{a}}} d x \\ & =\frac{1}{2} \int \sqrt{\left(e^{\frac{x}{a}}+e^{-\frac{x}{a}}\right)^{2}} d x\\ &=\frac{1}{2} \int\left(e^{\frac{x}{a}}+e^{-\frac{x}{a}}\right)\, d x \\ & =\frac{1}{2} \int e^{\frac{x}{a}} d x+\frac{1}{2} \int e^{-\frac{x}{a}} d x=\frac{a}{2}\left[e^{\frac{x}{a}}-e^{-\frac{x}{a}}\right] . \end{align} Here $s = \frac{a}{2} {[e^{\frac{x}{a}} - e^{- \frac{x}{a}}]}_{0}^{a} = \frac{a}{2} [e^{1} - e^{- 1} + 1 - 1]$ , and $s = \frac{a}{2} (e - \frac{1}{e}) .$

Example 22.4. A curve is such that the length of the tangent at any point $P$ (see the following figure) from $P$ to the intersection $T$ of the tangent with a fixed line $A B$ is a constant length $a$ . Find an expression for an arc of this curve,- which is called the tractrix,-and find the length, when $a = 3$ , between the ordinates $y = a$ and $y = 1$ .

Solution. We shall take the fixed line for the axis of $x$ . The point $D$ , with $D O = a$ , is a point on the curve, which must be tangent to $O D$ at $D$ . We take $O D$ as the axis of $y$ ; $A B$ and $O D$ are what are called axes of symmetry, that is the curve is symmetrical about them; $P T = a, P N = y, O N = x$ .

If we consider a small portion $d s$ of the curve, at $P$ , then $\sin θ = \frac{d y}{d s} = - \frac{y}{a}$ (minus because the curve slopes downwards to the right, see here).

Hence $\frac{d s}{d y} = - \frac{a}{y}, d s = - a \frac{d y}{y}, and s = - a \int \frac{d y}{y},$ that is $s = - a \ln y + C (\ln | y | = \ln y since y > 0)$ When $x = 0, s = 0, y = a$ , so that $0 = - a \ln a + C,$ and $C = a \ln a$ .

It follows that $s = a \ln a - a \ln y = a \ln \frac{a}{y}$ When $a = 3$ , $s$ between $y = a$ and $y = 1$ is therefore \begin{align} s&=3\left[\ln \frac{3}{y}\right]_{1}^{3}=3\left(\ln 1-\ln 3\right) =3 \times(0-1.0986) \\ & =-3.296 \text { or } 3.296, \end{align} as the sign $-$ refers merely to the direction in which the length was measured, from $D$ to $P$ , or from $P$ to $D$ .

Note that this result has been obtained without a knowledge of the equation of the curve. This is sometimes possible. In order to get the length of an arc between two points given by their abscissae (i.e. their $y$ -value), however, it is necessary to know the equation of the curve: this is easily obtained as follows: $\frac{d y}{d x} = - \tan θ = - \frac{y}{\sqrt{a^{2} - y^{2}}}, since P T = a;$ hence $d x = - \frac{\sqrt{a^{2} - y^{2}} d y}{y}$

The integration will give us a relation between $x$ and $y$ , which is the equation of the curve

$x = - \int \frac{\sqrt{a^{2} - y^{2}} d y}{y} .$

To integrate let $u = \sqrt{a^{2} - y^{2}}$ or $u^{2} = a^{2} - y^{2}$ . Then $2 u d u = - 2 y d y, or d y = - \frac{u}{y} d u$ and the integral becomes \begin{align} -\int \frac{\sqrt{a^{2}-y^{2}}\, dy}{y}&=-\int \frac{1}{u y}\left(-\frac{u}{y} du\right)\\ &=\int \frac{u^2}{a^2-u^2}du=-\int\frac{u^2}{u^2-a^2}du\\ &=-\int \frac{(u^2-a^2)+a^2}{u^2-a^2}du\\ &=-\int \left[1+\frac{a^2}{u^2-a^2}\right]du \end{align} Since $\frac{1}{u^{2} - a^{2}} = \frac{1}{(u - a) (u + a)} = \frac{1}{2 a} [\frac{1}{u - a} - \frac{1}{u + a}]$ we can rewrite the integral as \begin{align} \int \left[-1+\frac{a}{2(u+a)}-\frac{a}{2(u-a)}\right]du&=u+\frac{a}{2}\ln|u+a|-\frac{a}{2}\ln|u-a|+C\\ &=-u+\frac{a}{2}\ln\left|\frac{u+a}{u-a}\right|+C\\ &=-u+\frac{a}{2}\ln\left|\frac{(u+a)(u+a)}{(u-a)(u+a)}\right|+C\\ &=-u+\frac{a}{2}\ln\left|\frac{(u+a)^2}{u^2-a^2}\right|+C\\ &=-u+a\ln\left|\frac{u+a}{\sqrt{u^2-a^2}}\right|+C\\ &=-\sqrt{a^2-y^2}+a\ln\frac{\sqrt{a^2-y^2}+a}{y}+C. && (y>0) \end{align} The equation of the tractrix is therefore $x = a \ln \frac{a + \sqrt{a^{2} - y^{2}}}{y} - \sqrt{a^{2} - y^{2}} .$

If $a = 3$ , as before, and if the length of the arc from $x = 0$ to $x = 1$ is required, it is not an easy matter to calculate the value of $y$ corresponding to any given numerical value of $x$ . It is, however, easy to find graphically an approximation as near the correct value as we desire, when we are given the value of $a$ as follows:

Plot the graph, giving suitable values to $y$ , say 3, $2, 1.5, 1$ . From this graph, find what values of $y$ correspond to the two given values of $x$ determining the arc, the length of which is needed, as accurately is the scale of the graph allows. For $x = 0, y = 3$ of course; suppose that for $x = 1$ you find $y = 1.72$ on the graph. This is only approximate. Now plot again, on as large a scale as possible, taking only three values of $y$ , $1.6$ , $1.7$ , $1.8$ . On this second graph, which is nearly, but not quite a straight line, you will be probably able to read any value of $y$ correct to three places of decimals, and this is sufficient for our purpose. We find from the graph that $y = 1.723$ corresponds to $x = 1$ . Then

\begin{align} s&=3\left[\ln \frac{3}{y}\right]_{x=0}^{x=1} =3\left[\ln \frac{3}{y}\right]_{3}^{1 \cdot 723} \\ & =3\left(\ln 1.741-0\right)=1.66 . \end{align}

If we wanted a more accurate value of $y$ we could plot a third graph, taking for values of $y$ 1.722, 1.723, $1.724, \dots;$ this would give us, correct to five places of decimals, the value of $y$ corresponding to $x = 1$ , and so on, till the required accuracy is reached.

Example 22.5. Find the length of an arc of the logarithmic spiral $r = e^{θ}$ between $θ = 0$ and $θ = 1$ radian (the following figure).

Solution. Do you remember differentiating $y = e^{x} ?$ It is an easy one to remember, for it remains always the same whatever is done to it: $\frac{d y}{d x} = e^{x}$ (see page ).

Here, since $r = e^{θ}, \frac{d r}{d θ} = e^{θ} = r$ .

If we reverse the process and integrate $\int e^{θ} d θ$ we get back to $r + C$ , the constant $C$ being always introduced by such a process, as we have seen in Chapter 17.

It follows that

\begin{align} s & =\int \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}} d \theta=\int \sqrt{r^{2}+r^{2}} d \theta \\ & =\sqrt{2} \int r\, d \theta=\sqrt{2} \int e^{\theta}\, d \theta=\sqrt{2}\left(e^{\theta}+C\right) . \end{align}

Integrating between the two given values $θ = 0$ and $θ = 1$ , we get

\begin{align} s & =\int_{0}^{1} \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}} d \theta=\left[\sqrt{2}\left(e^{\theta}+C\right)\right]_{0}^{1} \\ & =\sqrt{2}\, e^{1}-\sqrt{2}\, e^{0}=\sqrt{2}(e-1) \\ & =1.41 \times 1.713=2.42 \text { units of length, } \end{align} since $r = e^{0} = 1$ when $θ = 0$ .

Example 22.6. Find the length of an arc of the logarithmic spiral $r = e^{θ}$ between $θ = 0$ and $θ = θ_{1}$ .

Solution. As we have just seen,

$s = \sqrt{2} \int_{0}^{θ_{1}} e^{θ} d θ = \sqrt{2} [e^{θ_{1}} - e^{0}] = \sqrt{2} (e^{θ_{1}} - 1) .$

Example 22.7. As a last example let us work fully a case leading to a typical integration which will be found useful for several of the exercises found at the end of this chapter. Let us find the expression for the length of an arc of the curve $y = \frac{a}{2} x^{2} + 3$ .

Solution. $\frac{d y}{d x} = a x, s = \int \sqrt{1 + a^{2} x^{2}} d x .$

Integrate this by parts: let $u = \sqrt{1 + a^{2} x^{2}} and d x = d v;$ then $x = v and d u = \frac{a^{2} x d x}{\sqrt{1 + a^{2} x^{2}}}$ by the method of differentiation explained in Chapter 9.

Since $\int u d v = u v - \int v d u$ (see integration by parts), we have \int \sqrt{1+a^{2} x^{2}}\ d x=x \sqrt{1+a^{2} x^{2}}-a^{2} \int \frac{x^{2}\ d x}{\sqrt{1+a^{2} x^{2}}}.\tag{1}

Also, we can write $\int \sqrt{1 + a^{2} x^{2}} d x = \int \frac{(1 + a^{2} x^{2}) d x}{\sqrt{1 + a^{2} x^{2}}};$ hence \int \sqrt{1+a^{2} x^{2}}\ d x=\int \frac{d x}{\sqrt{1+a^{2} x^{2}}}+a^2 \int \frac{x^{2}\ dx}{\sqrt{1+a^{2} x^{2}}}.\tag{2}

Adding (1) and (2) we get

2 \int \sqrt{1+a^{2} x^{2}} d x=x \sqrt{1+a^{2} x^{2}}+\int \frac{d x}{\sqrt{1+a^{2} x^{2}}}.\tag{3}

Remains to integrate $\int \frac{d x}{\sqrt{1 + a^{2} x^{2}}}$ ; for this purpose let $\sqrt{1 + a^{2} x^{2}} = v - a x;$ ¹ then

$1 + a^{2} x^{2} = v^{2} - 2 a v x + a^{2} x^{2} or 1 = v^{2} - 2 a v x .$ Differentiating this, to get rid of the constant, we get, $0 = 2 v d v - 2 a v d x - 2 a x d v or a v d x = v d v - a x d v;$ that is $d x = \frac{(v - a x) d v}{a v};$ replacing in $\int \frac{d x}{\sqrt{1 + a^{2} x^{2}}}$ we obtain $\int \frac{(v - a x) d v}{a v \sqrt{1 + a^{2} x^{2}}} = \frac{1}{a} \int \frac{(v - a x) d v}{v (v - a x)} = \frac{1}{a} \int \frac{d v}{v} = \frac{1}{a} \ln | v |;$ hence $\int \frac{d x}{\sqrt{1 + a^{2} x^{2}}} = \frac{1}{a} \ln (a x + \sqrt{1 + a^{2} x^{2}}) .$

Replacing in (3) and dividing by 2 we get, finally, \begin{align} s =\int \sqrt{1+a^{2} x^{2}} d x =\frac{x}{2} \sqrt{1+a^{2} x^{2}}+\frac{1}{2 a} \ln\left(a x+\sqrt{1+a^{2} x^{2}}\right), \end{align} which can easily be calculated between any given limits.

You ought now to be able to attempt with success the following exercises. You will find it interesting as well as instructive to plot the curves and verify your results by measurement where possible.

The integration is usually of the kind shown in Example 20.5, Example 20.6, or Example 22.7.

Exercises

Exercise 22.1. Find the length of the line $y = 3 x + 2$ between the two points for which $x = 1$ and $x = 4$ .

Answer

$s = 3 \sqrt{10} \approx 9.487$ .

Solution

\begin{align} y & =3 x+2 \Rightarrow \frac{d y}{d x}=3 \\ s & =\int_{1}^{4} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \\ & =\int_{1}^{4} \sqrt{1+3^{2}} d x \\ & =\sqrt{10} \int_{1}^{4} d x \\ & =\sqrt{10}[x]_{1}^{4}=\sqrt{10}(4-1) \\ & =3 \sqrt{10} . \end{align}

Checking the answer:

When $x = 1, y = 5$

When $x = 4, y = 14$

\begin{align} \text { Arc length } & =\sqrt{(14-5)^{2}+(4-1)^{2}}=\sqrt{9^{2}+3^{2}} \\ & =\sqrt{90}=\sqrt{9} \sqrt{10}=3 \sqrt{10} . \end{align}

Exercise 22.2. Find the length of the line $y = a x + b$ between the two points for which $x = a^{2}$ and $x = - 1$ .

Answer

$s = {(1 + a^{2})}^{\frac{3}{2}}$

Solution

$y = a x + b \Rightarrow \frac{d y}{d x} = a$

\begin{align} s & =\int_{-1}^{a^{2}} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \\ & =\int_{-1}^{a^{2}} \sqrt{1+a^{2}} d x \\ & =\sqrt{1+a^{2}}\big[x\big]_{x=-1}^{x=a^{2}} \\ & =\sqrt{1+a^{2}}\left(a^{2}+1\right) \\ & =\left(1+a^{2}\right)^{\frac{3}{2}} \end{align}

Exercise 22.3. Find the length of the curve $y = \frac{2}{3} x^{\frac{3}{2}}$ between the two points for which $x = 0$ and $x = 1$ .

Answer

$s \approx 1.22$ .

Solution

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

\begin{align} s & =\int_{0}^{1} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \\ & =\int_{0}^{1} \sqrt{1+x} d x \end{align}

Let $u = 1 + x$ , then $d u = d x$

$\int \sqrt{1 + x} d x = \int u^{\frac{1}{2}} d u = \frac{2}{3} u^{\frac{3}{2}} + c$

Therefore \begin{align} s&=\left[\frac{2}{3}(1+x)^{\frac{3}{2}}\right]_{0}^{1}\\ &=\frac{2}{3}\left(2^{\frac{2}{3}}-1\right)\\ &\approx 1.21895 \end{align}

Exercise 22.4. Find the length of the curve $y = x^{2}$ between the two points for which $x = 0$ and $x = 2$ .

Answer

\begin{align} \displaystyle s&=\int_{0}^{2} \sqrt{1+4 x^{2}} d x\\ &=\left[\frac{x}{2} \sqrt{1+4 x^{2}}+\frac{1}{4} \ln \left(2 x+\sqrt{1+4 x^2}\right)\right]_{0}^{2}\\ &\approx 4.65 \end{align}.

Solution

$y = x^{2} \Rightarrow \frac{d y}{d x} = 2 x$

\begin{align} s&=\int_{0}^{2} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\, dx\\ &=\int_0^2 \sqrt{1+4x^2}\, dx \end{align}

In this chapter, we learned that

$\int \sqrt{1 + a^{2} x^{2}} d x = \frac{x}{2} \sqrt{1 + a^{2} x^{2}} + \frac{1}{2 a} \ln | a x + \sqrt{1 + a^{2} x^{2}} | + C$

Here $a = 2$ . Therefore

\begin{align} s & =\int_{0}^{2} \sqrt{1+4 x^{2}}\, dx \\ & =\left[\frac{x}{2} \sqrt{1+4 x^{2}}+\frac{1}{4} \ln \left(2 x+\sqrt{1+4 x^{2}}\right)\right]_{0}^{2}\\ & =\sqrt{17}+\frac{1}{4} \ln (4+\sqrt{17})-0-\frac{1}{4} \ln 1 \\ & =\sqrt{17}+\frac{1}{4} \ln (4+\sqrt{17}) \\ & \approx 4.64678 \end{align}

Exercise 22.5. Find the length of the curve $y = m x^{2}$ between the two points for which $x = 0$ and $x = \frac{1}{2 m}$ .

Answer

$s = \frac{0.57}{m}$ .

Solution

$y = m x^{2} \Rightarrow \frac{d y}{d x} = 2 m x$

\begin{align} s & =\int_{0}^{\frac{1}{2 m}} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \\ & =\int_{0}^{\frac{1}{2 m}} \sqrt{1+4 m^{2} x^{2}} d x \end{align}

Again, using the formula $\int \sqrt{1 + a^{2} x^{2}} d x = \frac{x}{2} \sqrt{1 + a^{2} x^{2}} + \frac{1}{2 a} \ln | a x + \sqrt{1 + a^{2} x^{2}} | + C,$ we get \begin{align} s & =\left[\frac{x}{2} \sqrt{1+4 m^{2} x^{2}}+\frac{1}{4 m} \ln \left|2 m x+\sqrt{1+4 m^{2} x^{2}}\right|\right]_{0}^{\frac{1}{2 m}} \\ & =\frac{1}{4 m} \sqrt{1+4 m^{2} \frac{1}{4 m^{2}}}+\frac{1}{4 m} \ln \left(1+\sqrt{1+4 m^{2} \frac{1}{4 m^{2}}}\right)-0 \\ & =\frac{\sqrt{2}}{4 m}+\frac{\ln (1+\sqrt{2})}{4 m} \approx \frac{0.573897}{m} \end{align}

Exercise 22.6. Find the length of the curves $r = a \cos θ$ and $r = a \sin θ$ between $θ = θ_{1}$ and $θ = θ_{2}$ .

Answer

$s = a (θ_{2} - θ_{1})$ .

Solution

$r = a \cos θ \Rightarrow \frac{d r}{d θ} = - a \sin θ$

\begin{align} s & =\int_{\theta_{1}}^{\theta_{2}} \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}} d \theta \\ & =\int_{\theta_{1}}^{\theta_{2}} \sqrt{a^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} d \theta \\ & =|a| \int_{\theta_{1}}^{\theta_{2}} d \theta \\ & =|a|\left(\theta_{2}-\theta_{1}\right) \end{align}

Similarly $r = a \sin θ \Rightarrow \frac{d r}{d θ} = a \cos θ$

\begin{align} s & =\int_{\theta_{1}}^{\theta_{2}} \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}} d \theta \\ & =\int_{\theta_{1}}^{\theta_{2}} \sqrt{a^{2} \sin ^{2} \theta+a^{2} \cos ^{2} \theta} d \theta \\ & =|a| \int_{\theta_{1}}^{\theta_{2}} d \theta \\ & =|a|\left(\theta_{2}-\theta_{1}\right) \end{align}

Exercise 22.7. Find the length of the curve $r = a \sec θ$ .

Answer

$s = \sqrt{r^{2} - a^{2}}$ .

Solution

$r = a \sec θ$

To calculate $\frac{d r}{d θ}$ , we need to differentiate $\sec θ$ . In one of the exercises of chapter 15, we have differentiated it. However, if you do not recall the result, we can derive it again using the Quotient Rule:

\begin{align} \frac{d(\sec \theta)}{d \theta} & =\frac{d}{d \theta}\left(\frac{1}{\cos \theta}\right) \\ & =\frac{0 \times \cos \theta-(-\sin \theta)}{\cos ^{2} \theta} \\ & =\frac{\sin \theta}{\cos ^{2} \theta}=\frac{\sin \theta}{\cos \theta} \frac{1}{\cos \theta} \\ & =\tan \theta \sec \theta \end{align} Therefore, $r = a \sec θ \Rightarrow \frac{d r}{d θ} = a \sec θ \tan θ$

\begin{align} & s=\int \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}}\ d \theta \\ & =\int \sqrt{a^{2} \sec ^{2} \theta+a^{2} \sec ^{2} \theta \tan ^{2} \theta}\ d \theta \\ & =\int|a \sec \theta| \sqrt{1+\tan ^{2} \theta}\ d \theta \end{align}

since $1 + \tan^{2} θ = \sec^{2} θ$

\begin{align} s & =\int|a \sec \theta| \sqrt{\sec ^{2} \theta}\,d \theta \\ & =\int a \sec ^{2} \theta\, d\theta \end{align}

(suppose $a > 0$ )

But $\frac{d (\tan θ)}{d θ} = \sec^{2} θ$ , hence

$s = [a \tan θ]_{θ_{1}}^{θ_{2}} = a (\tan θ_{2} - \tan θ_{1})$

since $\tan^{2} θ = \sec^{2} θ - 1$ or

$\tan θ = \sqrt{\sec^{2} θ - 1}$

the result may be written as

\begin{align} s & =a \sqrt{\sec ^{2} \theta-1} \\ & =\sqrt{a^{2} \sec ^{2} \theta-a^{2}} \\ & =\sqrt{r^{2}-a^{2}} \end{align} Here we have assumed $θ_{1} = 0$ .

Remark: Let’s write the equation $r = a \sec θ$ in polar coordinates. Since $\sec θ = \frac{1}{\cos θ}$ , we have $r = \frac{a}{\cos θ}$ Multiplying both sides by $\cos θ$ (when it is not equal to $0$ ) gives $r \cos θ = a$ But $r \cos θ = x$ . Therefore, we want to find the length of the vertical line $x = a$ between $θ_{1} = 0$ to $θ$ . Now we can say that the above formulas make sense.

Exercise 22.8. Find the length of the arc of the curve $y^{2} = 4 a x$ between $x = 0$ and $x = a$ .

Answer

$s = \int_{0}^{a} \sqrt{1 + \frac{a}{x}} d x$ and $s = a \sqrt{2} + a \ln (1 + \sqrt{2})$ .

Solution

For this problem, we use the formula

$s = \int \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y$

$y^{2} = 4 a x \Rightarrow 2 y d y = 4 a d x \Rightarrow \frac{d x}{d y} = \frac{1}{2 a} y$

When $x = 0, y = 0$ and when $x = a, y = 2 a$

(We work with the upper branch $y > 0$ )

Therefore

$s = \int_{0}^{2 a} \sqrt{1 + \frac{1}{4 a^{2}} y^{2}} d y$

Again, since $\int \sqrt{1 + a^{2} x^{2}} d x = \frac{x}{2} \sqrt{1 + a^{2} x^{2}} + \frac{1}{2 a} \ln | a x + \sqrt{1 + a^{2} x^{2}} | + C$ We have

\begin{align} s & =\left[\frac{y}{2} \sqrt{1+\frac{y^{2}}{4 a^{2}}}+a \ln \left(\frac{y}{2 a}+\sqrt{1+\frac{y^{2}}{4 a^{2}}}\right)\right]_{0}^{2 a} \\ & =a \sqrt{2}+a \ln (1+\sqrt{2})-a \ln 1 \\ & =a[\sqrt{2}+\ln (1+\sqrt{2})] \approx 2.29559 a \end{align}

Exercise 22.9. Find the length of the are of the curve $y = x (\frac{x}{2} - 1)$ between $x = 0$ and $x = 4$ .

Answer

$s = \frac{x - 1}{2} \sqrt{(x - 1)^{2} + 1} + \frac{1}{2} \ln {(x - 1) + \sqrt{(x - 1)^{2} + 1}}$ and $s \approx 6.80$ .

Solution

We first differentiate $y$ with respect to $x$ : $y = x (\frac{x}{2} - 1) = \frac{x^{2}}{2} - x \Rightarrow \frac{d y}{d x} = x - 1$

The length $s$ of the curve is given by the integral: \begin{align} s & =\int_{0}^{4} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \\ & =\int_{0}^{4} \sqrt{1+(x-1)^{2}} d x \end{align}

To evaluate this integral, we make a substitution $u = x - 1$ and apply the following formula:

$\int \sqrt{1 + a^{2} u^{2}} = \frac{u}{2} \sqrt{1 + a^{2} u^{2}} + \frac{1}{2 a} \ln | a u + \sqrt{1 + a^{2} u^{2}} | + C .$

We substitute $a = 1$ and $u = x - 1$ in the above formula:

\begin{align} s & =\left[\frac{(x-1)}{2} \sqrt{1+(x-1)^{2}}+\frac{1}{2} \ln \left|(x-1)+\sqrt{1+(x-1)^{2}}\right| \right]_{0}^{4} \\ & =\left[\frac{3}{2} \sqrt{10}+\frac{1}{2} \ln (3+\sqrt{10})\right]-\left[\frac{-1}{2} \sqrt{2}+\frac{1}{2} \ln (-1+\sqrt{2})\right] \\ & \approx 6.80043 \end{align}

Exercise 22.10. Find the length of the arc of the curve $y = e^{x}$ between $x = 0$ and $x = 1$ .
(Note. This curve is in rectangular coordinates, and is not the same curve as the logarithmic spiral $r = e^{θ}$ which is in polar coordinates. The two equations are similar, but the curves are quite different.)

Answer

$s = \int_{1}^{e} \frac{\sqrt{1 + y^{2}}}{y} d y$ . Make a substitution $1 + y^{2} = u^{2}$ to get $s = \sqrt{1 + y^{2}} + \frac{1}{2} \ln \frac{\sqrt{1 + y^{2}} - 1}{\sqrt{y^{2} + 1} + 1}$ and $s \approx 2.00$ .

Solution

Method (a) $y = e^{x} \Rightarrow \frac{d y}{d x} = e^{x}$

\begin{align} s & =\int_{0}^{1} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \\ & =\int_{0}^{1} \sqrt{1+e^{2 x}} d x \end{align}

Let $u = \sqrt{1 + e^{2 x}}$ or $u^{2} = 1 + e^{2 x}$ . Then

$2 u d u = 2 e^{2 x} d x$

$d x = \frac{u}{e^{2 x}} d u$ and \begin{align} \int \sqrt{1+e^{2 x}} d x & =\int u \frac{u}{e^{2 x}}\ d u \\ & =\int \frac{u^{2}}{u^{2}-1}\ d u \\ & =\int \frac{u^{2}-1+1}{u^{2}-1}\ d u \\ & =\int\left(1+\frac{1}{u^{2}-1}\right)\ du. \end{align}

Since $u^{2} - 1 = (u - 1) (u + 1)$ We can write

$\frac{1}{u^{2} - 1} = \frac{A}{u - 1} + \frac{B}{u + 1}$

After some manipulations, we get

$\frac{1}{u^{2} - 1} = \frac{\frac{1}{2}}{u - 1} - \frac{\frac{1}{2}}{u + 1} .$

Hence

\begin{align} \int\left(1+\frac{1}{u^{2}-1}\right) d u & =\int\left(1+\frac{1}{2} \frac{1}{u-1}-\frac{1}{2} \frac{1}{u+1}\right) d u \\ & =u+\frac{1}{2} \ln \left|\frac{u-1}{u+1}\right|+C \\ & =\sqrt{1+e^{2 x}}+\frac{1}{2} \ln \frac{\sqrt{1+e^{2 x}}-1}{\sqrt{1+e^{2 x}}+1}+C \end{align} and \begin{align} s&=\int_{0}^{1} \sqrt{1+e^{2 x}}\ d x\\ &=\left[\sqrt{1+e^{2 x}}+\frac{1}{2} \ln \frac{\sqrt{1+e^{2 x}}-1}{\sqrt{1+e^{2 x}}+1}\right]_{0}^{1} \\ & =\left(\sqrt{1+e^{2}}+\frac{1}{2} \ln \frac{\sqrt{1+e^{2}}-1}{\sqrt{1+e^{2}}+1}\right)-\left(\sqrt{2}+\frac{1}{2} \ln \frac{\sqrt{2}-1}{\sqrt{2}+1}\right) \\ & \approx 2.0035 \end{align}

Method (b) $y = e^{x} \Leftrightarrow x = \ln y$ $\Rightarrow \frac{d x}{d y} = \frac{1}{y}$

$d s = \sqrt{(d x)^{2} + (d y)^{2}} = \sqrt{{(\frac{d x}{d y})}^{2} + 1}$

When $x = 0$ , $y = 1$

When $x = 1$ , $y = e$ .

Therefore, $s = \int_{1}^{e} \sqrt{\frac{1}{y^{2}} + 1} d y = \int_{1}^{e} \frac{\sqrt{1 + y^{2}}}{y} d y$ Make a substitution: $1 + y^{2} = u^{2}$ then $2 y d y = 2 u d u \Leftrightarrow d y = \frac{u}{y} d u$ So \begin{align} \int \frac{\sqrt{1+y^2}}{y}dy&=\int \frac{u}{y}\frac{u\, du}{y}\\ &=\int \frac{u^2}{y^2} du\\ &=\int \frac{u^2}{u^2-1}du\\ &=\int \frac{u^2-1+1}{u^2-1} du\\ &=\int \left(1+\frac{1}{u^2-1} \right)du \end{align} Using partial fractions: $\frac{1}{u^{2} - 1} = \frac{1}{(u - 1) (u + 1)} = \frac{1}{2} (\frac{1}{u - 1} - \frac{1}{u + 1})$ Hence \begin{align} \int \frac{\sqrt{y^2+1}}{y}dy&=\int \left(1+\frac{1}{u^2-1} \right)du\\ &=\int \left(1+\frac{1}{2(u-1)}-\frac{1}{2(u+1)}\right)du\\ &=u+\frac{1}{2}\ln|u-1|-\frac{1}{2}\ln|u+1|+C\\ &=u+\frac{1}{2}\ln\left|\frac{u-1}{u+1}\right|+C\\ &=\sqrt{1+y^2}+\frac{1}{2}\ln\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}+C \end{align}

Finally \begin{align} s&=\int_1^e \frac{\sqrt{1+y^2}}{y}dy\\ &=\left[\sqrt{1+y^2}+\frac{1}{2}\ln\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}\right]_1^e\\ &=\left(\sqrt{1+e^{2}}+\frac{1}{2} \ln \frac{\sqrt{1+e^{2}}-1}{\sqrt{1+e^{2}}+1}\right)-\left(\sqrt{2}+\frac{1}{2} \ln \frac{\sqrt{2}-1}{\sqrt{2}+1}\right) \\ & \approx 2.0035 \end{align} as before.

Exercise 22.11. A curve is such that the coordinates of a point on it are $x = a (θ - \sin θ)$ and $y = a (1 - \cos θ), θ$ being a certain angle which varies between 0 and $2 π$ . Find the length of the curve. (It is called a cycloid.)

Answer

$s = 2 a \int \sin \frac{θ}{2} d θ$ and $s = 8 a$ .

Solution

$x = a (θ - \sin θ) y = a (1 - \cos θ)$

We start with

$d s = \sqrt{(d x)^{2} + (d y)^{2}}$ then

$\frac{d s}{d θ} = \sqrt{{(\frac{d x}{d θ})}^{2} + {(\frac{d y}{d θ})}^{2}}$ and $s = \int \sqrt{{(\frac{d x}{d θ})}^{2} + {(\frac{d y}{d θ})}^{2}} d θ$ Here \begin{align} & x=a(\theta-\sin \theta) \Rightarrow \frac{d x}{d \theta}=a(1-\cos \theta) \\ & y=a(1-\cos \theta) \Rightarrow \frac{d y}{d x}=a \sin \theta \end{align} and

\begin{align} s & =\int_{0}^{2 \pi} \sqrt{a^{2}(1-\cos \theta)^{2}+a^{2} \sin ^{2} \theta}\ d \theta \\ & =a \int_{0}^{2 \pi} \sqrt{1-2 \cos \theta+\cos ^{2} \theta+\sin ^{2} \theta}\ d \theta \\ & =a \int_{0}^{2 \pi} \sqrt{2} \sqrt{1-\cos \theta} d \theta \end{align}

Since $1 - \cos θ = 2 \sin^{2} \frac{θ}{2}$ , we get

\begin{align} s & =a \int_{0}^{2 \pi} \sqrt{2} \sqrt{2 \sin ^{2} \frac{\theta}{2}} d \theta \\ & =2 a \int_{0}^{2 \pi}\left|\sin \frac{\theta}{2}\right| d \theta \end{align}

For $0 < θ < 2 π, \sin \frac{θ}{2} > 0$ . Therefore

\begin{align} s & =2 a \int_{0}^{2 \pi} \sin \frac{\theta}{2} d \theta \\ & =2 a \int_{0}^{2 \pi} 2 \sin \frac{\theta}{2} d\left(\frac{\theta}{2}\right) \\ & =4 a\left[-\cos \frac{\theta}{2}\right]_{0}^{2 \pi} \\ & =8 a \end{align}

Exercise 22.12. Find the length of the curve $y^{2} = m x$ between the two points for which $x = 0$ and $x = \frac{m}{4}$ .

Answer

$s = \frac{m}{4} \sqrt{2} + \frac{m}{4} \ln (1 + \sqrt{2})$ .

Solution

$x = \frac{1}{m} y^{2} \Rightarrow \frac{d x}{d y} = \frac{2}{m} y$

\begin{align} ds&=\sqrt{(dx)^2+(dy)^2}=\sqrt{\left(\frac{dx}{dy}\right)^2+1}\,dy\\ &=\sqrt{\frac{4}{m^2}y^2+1}\,dy \end{align}

When $x = 0$ , $y = 0$

When $x = \frac{m}{4}$ , $y = \frac{m}{2}$

Hence,

$s = \int_{0}^{\frac{m}{2}} \sqrt{\frac{4}{m^{2}} y^{2} + 1} d y$

Recall that (see Example 22.7) $\int \sqrt{1 + a^{2} x^{2}} d x = \frac{x}{2} \sqrt{1 + a^{2} x^{2}} + \frac{1}{2 a} \ln (a x + \sqrt{1 + a^{2} x^{2}})$

Therefore,

$\sqrt{1 + \frac{4}{m^{2}} y^{2}} d y = \frac{y}{2} \sqrt{1 + \frac{4 y^{2}}{m^{2}}} + \frac{m}{4} \ln (\frac{2 y}{m} + \sqrt{1 + \frac{4 y^{2}}{m^{2}}})$

and

\begin{align} s&=\int_0^\frac{m}{2} \sqrt{\frac{4}{m^2}y^2+1}\,dy\\ &=\left[ \frac{y}{2}\sqrt{1+\frac{4y^2}{m^2}}+\frac{m}{4}\ln\left(\frac{2y}{m}+\sqrt{1+\frac{4y^2}{m^2} }\right)\right]_0^\frac{m}{2}\\ &=\frac{m}{4}\sqrt{2}+\frac{m}{4}\ln\left(1+\sqrt{2}\right) \end{align}

Exercise 22.13. Find the expression for the length of an arc of the curve $y^{2} = \frac{x^{3}}{a}$ .

Answer

$s = \frac{8 a}{27} {1 + (\frac{9 x}{4 a})}^{\frac{3}{2}}$ .

Solution

$y^{2} = \frac{x^{3}}{a} \Rightarrow y = \frac{1}{\sqrt{a}} x^{3 / 2} (assuming y > 0)$

$\frac{d y}{d x} = \frac{3}{2 \sqrt{a}} x^{\frac{1}{2}}$

\begin{align} s&=\int \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}\ d x &=\int \sqrt{1+\frac{9}{4 a} x}\ d x \end{align}

Let $u = 1 + \frac{9}{4 a} x$ . Then $d u = \frac{9}{4 a} d x$ or

$d x = \frac{4 a}{9} d u$ and \begin{align} s & =\int \sqrt{u}\left(\frac{4 a}{9} d u\right)\\ &=\frac{4 a}{9} \int u^{\frac{1}{2}} d u \\ & =\frac{4 a}{9}\cdot \frac{2}{3} u^{\frac{3}{2}} \\ & =\left[\frac{8 a}{27}\left(1+\frac{9}{4 a} x\right)^{3 / 2}\right]_{x_{1}}^{x_{2}} \end{align}

Exercise 22.14. Find the length of the curve $y^{2} = 8 x^{3}$ between the two points for which $x = 1$ and $x = 2$ .

Answer

$s \approx 5.27$ .

Solution

$y^{2} = 8 x^{3}$

In the previous exercise, we showed that the arc length of $y^{2} = \frac{x^{3}}{a}$ between $x = x_{1}$ and $x = x_{2}$ is given by

${[\frac{8 a}{27} {(1 + \frac{9}{4 a} x)}^{\frac{3}{2}}]}_{x_{1}}^{x_{2}}$ In this exercise $a = \frac{1}{8}, x_{1} = 1$ and $x_{2} = 2$ . Therefore, the arc length is given by

\begin{align} & {\left[\frac{8 \times \frac{1}{8}}{27}\left(1+\frac{9}{4 \times \frac{1}{8}} x\right)^{\frac{3}{2}}\right]_{x=1}^{x=2}} \\ & =\frac{1}{27}(1+18 \times 2)^{\frac{3}{2}}-\frac{1}{27}(1+18 \times 1)^{\frac{3}{2}} \\ & \approx 5.26826 \end{align}

Exercise 22.15. Find the length of the curve $y^{\frac{2}{3}} + x^{\frac{2}{3}} = a^{\frac{2}{3}}$ between $x = 0$ and $x = a$ .

Answer

$s = \frac{3 a}{2}$ .

Solution

$y^{\frac{2}{3}} + x^{\frac{2}{3}} = a^{\frac{2}{3}} \Rightarrow y^{\frac{2}{3}} = a^{\frac{2}{3}} - x^{\frac{2}{3}}$ and $y = {(a^{\frac{2}{3}} - x^{\frac{2}{3}})}^{\frac{3}{2}}$

To find $\frac{d y}{d x}$ , let $u = a^{\frac{2}{3}} - x^{\frac{2}{3}}$ . Then $y = u^{\frac{3}{2}}$

\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =\left(\frac{3}{2} u^{\frac{1}{2}}\right)\left(-\frac{2}{3} x^{\frac{-1}{3}}\right) \\ & =-\frac{u^{\frac{1}{2}}}{x^{\frac{1}{3}}}=-\frac{\left(a^{\frac{2}{3}}-x^{\frac{2}{3}}\right)^{\frac{1}{2}}}{x^{\frac{1}{3}}} \end{align} and \begin{align} s&=\int_{0}^{a} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \\ & =\int_{0}^{a} \sqrt{1+\frac{a^{\frac{2}{3}}-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}} d x \\ & =\int_{0}^{a} \sqrt{\frac{x^{\frac{2}{3}}+a^{\frac{2}{3}}-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}} d x \\ & =\int_{0}^{a} \frac{a^{\frac{1}{3}}}{x^{\frac{1}{3}}} d x \\ & =a^{\frac{1}{3}} \int_{0}^{a} x^{-\frac{1}{3}} d x \\ & =\left[a^{\frac{1}{3}} \times \frac{3}{2} x^{\frac{2}{3}}\right]_{0}^{a} \\ & =\frac{3}{2} a \end{align}

Exercise 22.16. Find the length of the curve $r = a (1 - \cos θ)$ between $θ = 0$ and $θ = π$ .

Answer

$4 a$ .

Solution

$r = a (1 - \cos θ) \Rightarrow \frac{d r}{d θ} = a \sin θ$ and \begin{align} s & =\int_{0}^{\pi} \sqrt{r^{2}+\left(\frac{d r}{d x}\right)^{2}}\ dx \\ & =\int_{0}^{\pi} \sqrt{a^{2}(1-\cos \theta)^{2}+a^{2} \sin ^{2} \theta}\ d\theta \\ & =a \int_{0}^{\pi} \sqrt{1-2 \cos \theta+\cos ^{2} \theta+\sin ^{2} \theta}\ d\theta \end{align}

Since $\cos^{2} θ + \sin^{2} θ = 1$ , then

$s = a \int_{0}^{π} \sqrt{2 - 2 \cos θ} d θ = a \sqrt{2} \int_{0}^{π} \sqrt{1 - \cos θ} d θ$

Since $1 - \cos θ = 2 \sin^{2} \frac{θ}{2}$ , we get

\begin{align} s & =a \sqrt{2} \int_{0}^{\pi} \sqrt{2 \sin ^{2} \frac{\theta}{2}}\ d\theta \\ & =2 a \int_{0}^{\pi} \sin \frac{\theta}{2}\ d \theta \\ & =2 a \int_{0}^{\pi} 2 \sin \frac{\theta}{2}\ d\left(\frac{\theta}{2}\right) \\ & =4 a\left[-\cos \frac{\theta}{2}\right]_{0}^{\pi} \\ & =4 a \end{align}

An alternative substitution is to put $a x = \sinh t$ . Then $a d x = \cosh t d t$ and \begin{align} \int \frac{dx}{\sqrt{1+a^2x^2}}=\int \frac{\frac{1}{a}\cosh t}{\sqrt{1+\sinh^2 t}}dt=\frac{1}{a}\int\frac{\cosh t}{\cosh t}dt=\frac{1}{a}t+C \end{align} But $t = arcsinh a x$ or $t = \sinh^{- 1} (a x)$ . So the integral is $\frac{1}{a} arcsinh a x + C$ . Although this is a totally acceptable answer, we can express the result in terms of logarithms by recalling that $arcsinh A = \ln (A + \sqrt{1 + A^{2}})$ (see here). Hence, $\int \frac{d x}{\sqrt{1 + a^{2} x^{2}}} = \frac{1}{a} \ln (a x + \sqrt{1 + a^{2} x^{2}}) + C,$ as before.↩︎

Arc Length

Exercises

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