Submitted to HEJ Manuscript no.: ANM-010201-A

Length of a curve

Another important geometrical concept associated with a curve leads to an integration. This is the length of arc. To express the length analytically by an integral, in fact, we think of the curve as represented by a function $y=f(x)$ with a continuous derivative $y'$. By the points $a=x_1,x_2,\ldots,x_n=b$ we divide up the interval $a\leq x\leq b$ of the $x$-axis, over which our curve lies, into $n-1$ subintervals of length $\Delta x_{1},\ldots,\Delta x_{n-1}$. In the curve we inscribe a polygon whose vertices lie vertically above these points. The length of the curve is then defined to be the limit of the perimeters of these inscribed polygons, provided that such a limit does exist and is independent of the particular way in which the polygons are chosen. This assumption is called rectifiability. So the total length of the inscribed polygon is given according to Pythagoras theorem by the expression

$$\sum _{i=1}^{n - 1}\,\sqrt{\Delta x_{i}^{2} + \Delta y_{i}^{2... ... + (\frac {\Delta y_{i}}{\Delta x_{i}})^{2}}\,\Delta \,{x_{i}}.$$

But by the mean value theorem of the differential calculus the difference quotient $\Delta y_{i}/\Delta x_{i}$ is equal to $f'(\xi_{i})$, where $\xi_{i}$ is an intermediate value in the interval $\Delta x_{i}$. If we now let $n$ increase beyond all bounds and at the same time let the length of the longest subinterval $\Delta x_i$ tend to zero, then by the definition of integral our expression will tend to the limit

$$\int _{a}^{b}\sqrt{1 + ({\frac {\partial }{\partial x}}\,{\rm y} (x))^{2}}\,dx.$$

We established the following theorem:

Theorem 3.

Every curve $y=f(x)$ for which the derivative $f'(x)$ is continuous is a rectifiable curve, and its length between $x=a$ and $x=b$ ($b\geq a$) is given by the formula $\int _{a}^{b}\sqrt{1 + ({\frac {\partial }{\partial x}}\,{\rm y} (x))^{2}}\,dx$.

Our expression for the length of arc is still subject to the special and artifical assumption that the curve consists of one single-valued branch above the $x$-axis. Parametric representation frees us from this restriction. If a curve of the kind which we have been considering is given in parametric form by the equations $x=x(t),\ y=y(t)$, then by introducing the parameter $t$ in the above expression we obtain the parametric form of the length of arc

$$\int _{\alpha }^{\beta }\sqrt{({\frac {\partial }{\partial t}... ...^{2} + ({\frac {\partial }{\partial t}}\,{\rm y}(t))^{ 2}}\,dt,$$

where $\alpha$ and $\beta$ are the values of $t$ which correspond respectively to the points of the curve $x=a$ and $x=b$.

Excercise 7.1. Give the length of the arc when the curve is expressed in polar coordinates.

EXAMPLE 7.3.

Consider the parabola

$>$ f:=x->1/2*x^2;

$$f := x\rightarrow {\displaystyle \frac {1}{2}} \,x^{2}$$

For its length of arc we immediately obtain the integral

$>$ Int(sqrt(1+x^2),x=a..b);

$${\displaystyle \int _{a}^{b}} \sqrt{1 + x^{2}}\,dx$$

which has the value

$>$ value(%);

$${\displaystyle \frac {1}{2}} \,b\,\sqrt{1 + b^{2}} + {\disp... ... {\displaystyle \frac {1}{2}} \,{\rm ln}(a + \sqrt{1 + a^{2}})$$

EXAMPLE 7.4.

As an example for a motion along a path or trajectory consider the cycloids which arise when a circle rolls along a straight line or another circle. Here we limit ourselves to the simplest case, in which a circle of radius $R$ rolls along the $x$-axis, and we consider a point on its circumference. This point then describes a cycloid. If we choose the origin of the coordinate system and the initial time in such a way that for time $t=0$ the corresponding point of the curve coincides with the origin, we obtain the parametric representation

$$x=R(t-\sin(t)),\ y=R(1-\cos(t))$$

for the cycloid. Here $t$ denotes the angle through which the circle has turned from its original position. From the above equations we obtain at once that

$${\frac {\partial }{\partial t}}\,{\rm x}(t)=R(1-\cos(t)),\ {\frac {\partial }{\partial t}}\,{\rm y}(t) = R\sin(t).$$

Hence the length of the arc is

$>$ Int(sqrt(diff(x(t),t)^2+diff(y(t),t)^2),t=0..alpha)=

$>$ Int(sqrt(2*R^2*(1-cos(t))),t=0..alpha);

$${\displaystyle \int _{0}^{\alpha }} \sqrt{({\frac {\partial ... ...{0}^{\alpha }} \sqrt{ 2}\,\sqrt{R^{2}\,(1 - {\rm cos}(t))}\,dt$$

Since $1-\cos(t)=2\sin(t/2)^2$ the integrand is equal to $2R\sin(t/2)$, hence for $0\leq \alpha\leq 2\pi$ the equation becomes

$>$ Int(2*R*sin(t/2),t=0..alpha);

$${\displaystyle \int _{0}^{\alpha }} 2\,R\,{\rm sin}( {\displaystyle \frac {1}{2}} \,t)\,dt$$

The value of this integral is

$>$ value(%);

$$- 4\,{\rm cos}({\displaystyle \frac {1}{2}} \,\alpha )\,R + 4\,R$$

If we consider the length of arc between two successive cusps we must put $\alpha:=2\,\pi$. Then we have

$>$ eval(subs(alpha=2*Pi,%));

$$8\,R$$

Thus, we obtain that the lenght of arc of the cycloid between successive cusps is equal to four times the diameter of the rolling circle.

Similarly, we calculate the area bounded by one arch of the cycloid and the $x$-axis. If $R=1$ then this area has the form

$>$ plot([t-sin(t),1-cos(t),t=0..2*Pi]);

The area is

$>$ Int(y(t)*diff(x(t),t),t=0..2*Pi)=

$>$ R^2*Int((1-cos(t))^2,t=0..2*Pi);

$${\displaystyle \int _{0}^{2\,\pi }} {\rm y}(t)\,({\frac { \p... ...\displaystyle \int _{0}^{2\,\pi }} (1 - {\rm cos}(t))^{2}\,dt$$

$>$ value(rhs(%));

$$3\,R^{2}\,\pi$$

This area is therefore three times the area of the rolling circle.

Exercise 7.2. Calculate the area bounded by the semicubical parabola $y=x^{\frac{3}{2}}$, the $x$-axis and the lines $x=a$ and $y=b$. Calculate the length of arc of it.

Exercise 7.3. Find the volume and surface area of the torus (or anchor ring) obtained by rotating a circle about a line which does not intersect it.

Exercise 7.4. Find the area of a catenoid, the surface obtained by rotating an arc of the catenary $y=\cosh(x)$ about the $x$-axis.

The possibilities of applications of differential and integral calculus are unbounded. In sciences and engineering mathematical models are developed to aid in the understanding of physical phenomena. These models often yield an equation that contains some derivatives of an unknown function. Such an equation is called differential equation. In order to solve these equations one requires the theory of integration. In this paper we did the first steps towards better understanding the mathematical and real world in which we live.

Submitted to HEJ Manuscript no.: ANM-010201-A