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 \([a,b]\):

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

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 \(P_0P_1\), \(P_1P_2\), \(\cdots\), \(P_{n-1}P_{n}\). By adding the lengths of these segments, we can approximate the length \(l_{\Delta}\) 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

\[ \begin{eqnarray} l_{\Delta}&=& \sum_{i=0}^{n-1}\sqrt{[x(t_{i+1}) - x(t_{i})]^2 + [y(t_{i+1}) - y(t_i)]^2}\\ &=& \sum_{i=0}^{n-1}\frac{\sqrt{[x(t_{i+1}) - x(t_{i})]^2 + [y(t_{i+1}) - y(t_i)]^2}}{t_{i+1}-t_{i}}\cdot (t_{i+1}-t_{i})\\ &=& \sum_{i=0}^{n-1}{\sqrt{\left[\frac{x(t_{i+1}) - x(t_{i})}{t_{i+1}-t_{i}}\right]^2 + \left[\frac{y(t_{i+1}) - y(t_i)}{t_{i+1}-t_{i}}\right]^2}}\cdot (t_{i+1}-t_{i}). \end{eqnarray} \]

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

\[ \begin{eqnarray} \frac{d}{dt}x(s_i) &=& \frac{x(t_{i+1}) - x(t_{i})}{t_{i+1}-t_{i}},\\ \frac{d}{dt}y(s_i') &=& \frac{y(t_{i+1}) - y(t_{i})}{t_{i+1}-t_{i}}. \end{eqnarray} \]

See also: Mean Value Theorem

Since \(y(t)\) is a \(C^1\) function, \(\frac{d}{dt}y(t)\) is continuous so the difference between \(\frac{d}{dt}y(s_i')\) and \(\frac{d}{dt}y(s_i)\) becomes smaller as \(\Delta\) is more refined. Thus, we have the approximation

\[l_{\Delta} = \sum_{i=0}^{n-1}\sqrt{\left[\frac{d}{dt}x(s_i)\right]^2 + \left[\frac{d}{dt}y(s_i)\right]^2}\cdot (t_{i+1}-t_{i}).\]

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

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

Let us define

\[ \begin{eqnarray} m_i &=& \min\{h(t) \mid t_i \leq t \leq t_{i+1}\},\\ M_i &=& \max\{h(t) \mid t_i \leq t \leq t_{i+1}\}. \end{eqnarray} \]

Then we have

\[\sum_{i=0}^{n-1}m_i\cdot(t_{i+1}-t_{i}) \leq l_{\Delta} \leq \sum_{i=0}^{n-1}M_i\cdot(t_{i+1}-t_{i}).\tag{eq:mMlen}\]

As we refine \(\Delta\), 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

\[l(C) = \int_a^b\sqrt{\left[\frac{d}{dt}x(t)\right]^2 + \left[\frac{d}{dt}y(t)\right]^2}dt.\]

Example. Consider the circle

\[ \begin{eqnarray} x(t) &=& r\cos t,\\ y(t) &=& r\sin t \end{eqnarray} \]

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

\[ \begin{eqnarray} l(C) &=& \int_0^{2\pi}\sqrt{[x'(t)]^2 + [y'(t)]^2}dt\\ &=&\int_0^{2\pi}\sqrt{r^2\sin^2t + r^2\cos^2t}~ dt\\ &=&\int_0^{2\pi}rdt = 2\pi r \end{eqnarray} \]

as expected. □

Example. Consider the cycloid defined by

\[ \begin{eqnarray} x(t) &=& a(t - \sin t),\\ y(t) &=& a(1 - \cos t) \end{eqnarray} \]

where \(a > 0\) and \(0 \leq t \leq 2\pi\). See the figure below:

Figure: Cycloid

Since

\[\begin{eqnarray} x'(t) &=& a(1 - \cos t),\\ y'(t) &=& a\sin t, \end{eqnarray}\]

we have

\[ \begin{eqnarray} l(C) &=& a\int_0^{2\pi}\sqrt{(1 - \cos t)^2 + \sin^2 t} dt\\ &=&a\int_0^{2\pi}\sqrt{2(1 - \cos t)} dt\\ &=&a\int_0^{2\pi}\sqrt{4\sin^2\frac{t}{2}} dt\\ &=&4a\int_0^{\pi}{\sin u} du\\ &=&8a \end{eqnarray} \]

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) □