Applications of integrals (1): Length of a curve

As an application of integrals, we consider the length of a curve. Specifically, we consider curves in the 2-dimensional space defined parametrically.

Let $x(t)$ and $y(t)$ be $C^1$ functions defined on an interval containing the closed interval $[a,b]$. If $t$ moves in $[a,b]$, the point $(x(t),y(t))$ on ${\mathbb{R}}^2$ moves smoothly, drawing a curve. Let us denote this curve by $C$. Let $P=(x(a),y(a))$ and $Q=(x(b),y(b))$ be the end points of the curve $C$. We want to measure the ``length'' of the curve $C$. But what is the length of a \emph{curve}, anyway? We do know how to calculate the length of a line segment (Pythagorean theorem). So, let us approximate the curve by line segments. Consider the partition of the closed interval $[a,b]$:

$$\mathrm{\Delta }:a=t_0<t_1<t_2<\cdots <t_{n-1}<t_n=b.$$

Then $P_0=(x(t_0),y(t_0))=P$, $P_1=(x(t_1),y(t_1))$, $P_2=(x(t_2),y(t_2))$, $\cdots$, $P_{n-1}=(x(t_{n-1}),y(t_{n-1}))$, $P_n=(x(t_n),y(t_n))=Q$ are all on $C$. We can approximate the curve $C$ by connecting the line segments $P_0{P}_1$, $P_1{P}_2$, $\cdots$, $P_{n-1}{P}_n$. By adding the lengths of these segments, we can approximate the length $l_{\mathrm{\Delta }}$ of the curve $C$. The length of $P_i{P}_{i+1}$ is given by

$$\sqrt{[x(t_{i+1})-x(t_i){]}^2+[y(t_{i+1})-y(t_i){]}^2}$$

so the approximate length of $C$ is given by

$$\begin{array}{rcl}{l}_{\mathrm{\Delta }}& =& \sum _{i=0}^{n-1}\sqrt{[x(t_{i+1})-x(t_i){]}^2+[y(t_{i+1})-y(t_i){]}^2}\\ & =& \sum _{i=0}^{n-1}\frac{\sqrt{[x(t_{i+1})-x(t_i){]}^2+[y(t_{i+1})-y(t_i){]}^2}}{t_{i+1}-t_i}\cdot (t_{i+1}-t_i)\\ & =& \sum _{i=0}^{n-1}\sqrt{{\left[\frac{x(t_{i+1})-x(t_i)}{t_{i+1}-t_i}\right]}^2+{\left[\frac{y(t_{i+1})-y(t_i)}{t_{i+1}-t_i}\right]}^2}\cdot (t_{i+1}-t_i).\end{array}$$

By the Mean Value Theorem, for each $i=0,1,\cdots ,n-1$, there exist $s_i,s_i^{\prime }\in (t_i,t_{i+1})$ such that

$$\begin{array}{rcl}\frac{d}{dt}x(s_i)& =& \frac{x(t_{i+1})-x(t_i)}{t_{i+1}-t_i},\\ \frac{d}{dt}y(s_i^{\prime })& =& \frac{y(t_{i+1})-y(t_i)}{t_{i+1}-t_i}.\end{array}$$

See also: Mean Value Theorem

Since $y(t)$ is a $C^1$ function, $\frac{d}{dt}y(t)$ is continuous so the difference between $\frac{d}{dt}y(s_i^{\prime })$ and $\frac{d}{dt}y(s_i)$ becomes smaller as $\mathrm{\Delta }$ is more refined. Thus, we have the approximation

$$l_{\mathrm{\Delta }}=\sum _{i=0}^{n-1}\sqrt{{[\frac{d}{dt}x(s_i)]}^2+{[\frac{d}{dt}y(s_i)]}^2}\cdot (t_{i+1}-t_i).$$

Now consider the continuous function $h(t)$ on $[a,b]$ defined by

$$h(t)=\sqrt{{[\frac{d}{dt}x(t)]}^2+{[\frac{d}{dt}y(t)]}^2}.$$

Let us define

$$\begin{array}{rcl}{m}_i& =& min\{h(t)\mid t_i\le t\le t_{i+1}\},\\ M_i& =& max\{h(t)\mid t_i\le t\le t_{i+1}\}.\end{array}$$

Then we have

$$\begin{array}{}\text{(eq:mMlen)}& \sum _{i=0}^{n-1}{m}_i\cdot (t_{i+1}-t_i)\le l_{\mathrm{\Delta }}\le \sum _{i=0}^{n-1}{M}_i\cdot (t_{i+1}-t_i).\end{array}$$

As we refine $\mathrm{\Delta }$, both sides of {eq:mMlen} converge to the same value that is ${\int }_a^{b}h(t)dt$. Thus, it is natural to define the length $l(C)$ of the curve  $C$ by

$$l(C)={\int }_a^b\sqrt{{[\frac{d}{dt}x(t)]}^2+{[\frac{d}{dt}y(t)]}^2}dt.$$

Example. Consider the circle

$$\begin{array}{rcl}x(t)& =& r\cos t,\\ y(t)& =& r\sin t\end{array}$$

where $r>0$ and $0\le t\le 2\pi$. This curve is a circle with length $r$. The length of this curve is

$$\begin{array}{rcl}l(C)& =& {\int }_0^{2\pi }\sqrt{[x^{\prime }(t){]}^2+[y^{\prime }(t){]}^2}dt\\ & =& {\int }_0^{2\pi }\sqrt{r^2{\sin }^{2}t+r^2{\cos }^{2}t}dt\\ & =& {\int }_0^{2\pi }rdt=2\pi r\end{array}$$

as expected. □

Example. Consider the cycloid defined by

$$\begin{array}{rcl}x(t)& =& a(t-\sin t),\\ y(t)& =& a(1-\cos t)\end{array}$$

where $a>0$ and $0\le t\le 2\pi$. See the figure below:

Figure: Cycloid

Since

$$\begin{array}{rcl}{x}^{\prime }(t)& =& a(1-\cos t),\\ y^{\prime }(t)& =& a\sin t,\end{array}$$

we have

$$\begin{array}{rcl}l(C)& =& a{\int }_0^{2\pi }\sqrt{(1-\cos t{)}^2+{\sin }^{2}t}dt\\ & =& a{\int }_0^{2\pi }\sqrt{2(1-\cos t)}dt\\ & =& a{\int }_0^{2\pi }\sqrt{4{\sin }^2\frac{t}{2}}dt\\ & =& 4a{\int }_0^{\pi }\sin udu\\ & =& 8a\end{array}$$

where we have used the formula

$${\sin }^2\frac{t}{2}=\frac{1-\cos t}{2}$$

and the substitution $u=\frac{t}{2}$.

See also: Cycloid (Wikipedia) □