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Arc length

arc length Let $f(x)$ be class $C^1$. Then the arc length $s$ of a curve $y = f(x)$, where $a \leq x \leq b$ is given by

$\displaystyle s = \int_{a}^{b} \sqrt{1 + (f^{\prime}(x))^{2}} dx $

NOTE Partition $[a,b]$

$\displaystyle \Delta : a = x_{0} < x_{1} < \cdots < x_{n} = b$

Let ${\rm P}_{i}$ be the point $(x_{i},f(x_{i}))$. Then connect the points ${\rm P}_{0},{\rm P}_{1},\ldots,{\rm P}_{n}$ by a straight line to get
${\rm P}_{0}{\rm P}_{1},$ ${\rm P}_{1}{\rm P}_{2}, \ldots, {\rm P}_{n-1}{\rm P}_{n}$.

$\displaystyle \sum_{i=1}^{n}{\rm P}_{i-1}{\rm P}_{i} $

Now letting the norm of length $\vert\Delta\vert$ get smaller, if the Riemann sum converges to $s$, then we say $s$ arc length of $y = f(x)$ for $a \leq x \leq b$.

Smooth Curve
If $f(x)$ is the class $C^1$, the curve of a function $y = f(x)$ is called smooth.

Arc Length Note that the length of the line segment is decomposed with $\Delta x$ and $\Delta y$. Then $\sqrt{(\Delta x)^2 + (\Delta y)^2}$.

Figure 3.26: Arc length
\includegraphics[width=3.5cm]{SOFTFIG-3/arclength.eps}


$\displaystyle {\rm P}_{i-1}{\rm P}_{i}$ $\displaystyle =$ $\displaystyle \sqrt{(x_{i}-x_{i-1})^{2} + (f(x_{i})-f(x_{i-1}))^{2}}$  
  $\displaystyle =$ $\displaystyle \sqrt{1+ \left(\frac{f(x_{i})-f(x_{i-1})}{x_{i}-x_{i-1}}\right)^{2}}(x_{i} - x_{i-1}).$  

Check
$(x_i - x_{i-1})^2 + (f(x_i) - f(x_{i-1}))^2 = (x_i - x_{i-1})^2 (1 + \frac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}})^2$. $\Delta x_i = x_i - x_{i-1}$.

Note that since $f(x)$ is the class $C^{1}$, use the mean value theorem,

$\displaystyle {\rm P}_{i-1}{\rm P}_{i} = \sqrt{1 + (f^{\prime}(\xi))^{2}}\Delta x_{i}, (x_{i-1} < \xi < x_{i}). $

Thus the arc length corresponds to $a \leq x \leq b$ is

$\displaystyle s = \int_{a}^{b}\sqrt{1 + (f^{\prime}(x))^{2}} dx\ensuremath{ \blacksquare}$

Example 3..23   Find the arc length of the following curve.

$\displaystyle x^{2/3} + y^{2/3} = 1$

Figure 3.27: Example3-23
% latex2html id marker 48517 \includegraphics[width=3.5cm]{SOFTFIG-3/reidai3-23_gr1.eps}

SOLUTION Let $x = \cos^{3}{t}, y = \sin^{3}{t} {\rm where} 0 \leq t \leq 2\pi$. Then the small arc length $\Delta l$ is given by

$\displaystyle \Delta l = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\frac{\Delta x}{\Delta t})^2 + (\frac{\Delta y}{\Delta t})^2} \Delta t$

Note that the curve is not smooth at $t = \frac{\pi}{2}$. Thus
$\displaystyle l$ $\displaystyle =$ $\displaystyle 4\int_{0}^{\frac{\pi}{2}}\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$  
  $\displaystyle =$ $\displaystyle 4 \int_{0}^{\frac{\pi}{2}}\sqrt{(-3\cos^{2}{t}\sin{t})^2 + (3\sin^{2}{t}\cos{t})^2} dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2} \sqrt{\cos^4{t}\sin^2{t} + \sin^4{t}\cos^2{t}}dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2} \sqrt{\cos^2{t}\sin^2{t}(\cos^2{t} + \sin^2{t})}dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2} \sqrt{\cos^2{t}\sin^2{t}}dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2}\cos{t}\sin{t}\:dt (u = \sin{t}, du = \cos{t}\:dt)$  
  $\displaystyle =$ $\displaystyle 12 \int_0^1 udu = 12 \left[\frac{u^2}{2}\right]_0^1 = 12\cdot \frac{1}{2} = 6\ensuremath{ \blacksquare}$  

If the curve is hard to represent by $y = f(x)$, then it is better to use parametric representation. For parametric representation $x = f(t), y = g(t)$, if $f'(t), g'(t)$ are continuous on $(a,b)$ and not equal to 0 simultaneously, then the curve is smooth. Since $f'(t) = -3\cos^{2}{t}\sin{t}, g'(t) = 3\sin^{2}{t}\cos{t}$, $f'(t)$ and $g'(t)$ are 0 at $t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ simultaneously. Arc length If the curve of a function is not smooth, then it is not the class $C^{1}$. In this case, we find the arc length of the curve which is smooth.

Figure 3.28: Exercise3-23
% latex2html id marker 48578 \includegraphics[width=3.5cm]{SOFTFIG-3/enshu3-23_gr1.eps}

\begin{displaymath}\left(\begin{array}{l} u = \sin{t}, du = \cos{t}dt\\ \begin{... ...{\pi}{2}\ \hline u & 0 & \to & 1 \end{array}\end{array}\right)\end{displaymath}.

Check
$(x_i - x_{i-1})^2 + (f(x_i) - f(x_{i-1}))^2 = (x_i - x_{i-1})^2 (1 + \frac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}})^2$. $\Delta x_i = x_i - x_{i-1}$.

Note that since $f(x)$ is the class $C^{1}$, use the mean value theorem,

$\displaystyle {\rm P}_{i-1}{\rm P}_{i} = \sqrt{1 + (f^{\prime}(\xi))^{2}}\Delta x_{i}, (x_{i-1} < \xi < x_{i}). $

Thus the arc length corresponds to $a \leq x \leq b$ is

$\displaystyle s = \int_{a}^{b}\sqrt{1 + (f^{\prime}(x))^{2}} dx\ensuremath{ \blacksquare}$

Example 3..24   Find the arc length of the following curve.

$\displaystyle x^{2/3} + y^{2/3} = 1$

Figure 3.29: Example3-23
% latex2html id marker 48605 \includegraphics[width=3.5cm]{SOFTFIG-3/reidai3-23_gr1.eps}

SOLUTION Let $x = \cos^{3}{t}, y = \sin^{3}{t} {\rm where} 0 \leq t \leq 2\pi$. Then the small arc length $\Delta l$ is given by

$\displaystyle \Delta l = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{(\frac{\Delta x}{\Delta t})^2 + (\frac{\Delta y}{\Delta t})^2} \Delta t$

Note that the curve is not smooth at $t = \frac{\pi}{2}$. Thus
$\displaystyle l$ $\displaystyle =$ $\displaystyle 4\int_{0}^{\frac{\pi}{2}}\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$  
  $\displaystyle =$ $\displaystyle 4 \int_{0}^{\frac{\pi}{2}}\sqrt{(-3\cos^{2}{t}\sin{t})^2 + (3\sin^{2}{t}\cos{t})^2} dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2} \sqrt{\cos^4{t}\sin^2{t} + \sin^4{t}\cos^2{t}}dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2} \sqrt{\cos^2{t}\sin^2{t}(\cos^2{t} + \sin^2{t})}dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2} \sqrt{\cos^2{t}\sin^2{t}}dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2}\cos{t}\sin{t}\:dt (u = \sin{t}, du = \cos{t}\:dt)$  
  $\displaystyle =$ $\displaystyle 12 \int_0^1 udu = 12 \left[\frac{u^2}{2}\right]_0^1 = 12\cdot \frac{1}{2} = 6\ensuremath{ \blacksquare}$  

If the curve is hard to represent by $y = f(x)$, then it is better to use parametric representation. For parametric representation $x = f(t), y = g(t)$, if $f'(t), g'(t)$ are continuous on $(a,b)$ and not equal to 0 simultaneously, then the curve is smooth. Since $f'(t) = -3\cos^{2}{t}\sin{t}, g'(t) = 3\sin^{2}{t}\cos{t}$, $f'(t)$ and $g'(t)$ are 0 at $t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ simultaneously. Arc length If the curve of a function is not smooth, then it is not the class $C^{1}$. In this case, we find the arc length of the curve which is smooth.

Figure 3.30: Exercise3-23
% latex2html id marker 48666 \includegraphics[width=3.5cm]{SOFTFIG-3/enshu3-23_gr1.eps}

\begin{displaymath}\left(\begin{array}{l} u = \sin{t}, du = \cos{t}dt\\ \begin{... ...{\pi}{2}\ \hline u & 0 & \to & 1 \end{array}\end{array}\right)\end{displaymath}.

Exercise 3..23   Find the length of following curve.
$\displaystyle{\sqrt{x} + \sqrt{y} = 1}$

SOLUTION Parametrize by $x = \cos^{4}{t}, y = \sin^{4}{t} (0 \leq t \leq \frac{\pi}{2})$.

$\displaystyle \sqrt{x} + \sqrt{y} = \cos^{2}{t} + \sin^{2}{t} = 1.$


$\displaystyle L$ $\displaystyle =$ $\displaystyle \int_{0}^{\pi/2}\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$  
  $\displaystyle =$ $\displaystyle \int_{0}^{\pi/2}\sqrt{(-4\cos^{3}{t}\sin{t})^2 + (4\sin^{3}{t}\cos{t})^2} dt$  
  $\displaystyle =$ $\displaystyle 4\int_{0}^{\pi/2}\sqrt{\cos^{6}{t}\sin^{2}{t} + \sin^{6}{t}\cos^{2}{t}} dt$  
  $\displaystyle =$ $\displaystyle 4\int_{0}^{\pi/2}\sqrt{\cos^{4}{t} + \sin^{4}{t}} \cos{t}\sin{t}dt$  
  $\displaystyle =$ $\displaystyle 4\int_{0}^{\pi/2}\sqrt{(1 - \sin^{2}{t})^2 + \sin^{4}{t}} \sin{t}\cos{t}dt.$  

Now let $u = \sin^{2}{t}$. Then $du = 2\sin{t}\cos{t}dt$ and

\begin{displaymath}\begin{array}{l\vert lll} t&0&\to&\frac{\pi}{2}\ \hline u&0&\to&1 \end{array}\end{displaymath}

Thus
$\displaystyle L$ $\displaystyle =$ $\displaystyle 2\int_{0}^{1}\sqrt{(1 - u)^2 + u^2} du = 2\int_{0}^{1}\sqrt{2u^2 - 2u + 1}du$  
  $\displaystyle =$ $\displaystyle 2\sqrt{2}\int_{0}^{1}\sqrt{(u - \frac{1}{2})^2 + \frac{1}{4}}du.$  

Check
$\sqrt{2u^2 - 2u + 1} = \sqrt{2(u^2 - u + \frac{1}{2}} = \sqrt{2}\sqrt{(u-\frac{... ...-(\frac{1}{2})^2 +\frac{1}{2}} = \sqrt{2}\sqrt{(u-\frac{1}{2})^2 + \frac{1}{4}}$.

Let $w = u - \frac{1}{2}$. Then $dw = du$ and

\begin{displaymath}\begin{array}{l\vert lll} u&0&\to&1\ \hline w&-\frac{1}{2}&\to&\frac{1}{2} \end{array}\end{displaymath}

implies that
$\displaystyle L$ $\displaystyle =$ $\displaystyle 2\sqrt{2}\int_{-1/2}^{1/2}\sqrt{w ^2 + \frac{1}{4}} dw.$  

Now use the following integral formula,

$\displaystyle \int \sqrt{x^2 + a^2} dx = \frac{1}{2}(x\sqrt{x^2 + a^2} + a^2 \log\vert x + \sqrt{x^2 + a^2}\vert) $

Let $x = a\tan{t}$. Then $\sqrt{x^2 + a^2} = \sqrt{a^2(1+\tan^{2}{t})} = \sqrt{a^2 \sec^{2}{t}} = a \sec{t}$. Also, $dx = a\sec^{2}{t}\:dt$,

$\displaystyle \int \sqrt{x^2 + a^2} dx = \int a\sec{t}\cdot a\sec^{2}{t}dt = a^2 \int \sec^{3}{t}dt.$

Now by Exercise3.11, we have

$\displaystyle \int sec^{3}{t}\:dt = \frac{1}{2}\left(\sec{t}\tan{t} + \log\vert sec{t} + \tan{t}\vert\right) + c.$

Thus,
$\displaystyle \int \sqrt{x^2 + a^2} dx$ $\displaystyle =$ $\displaystyle a^2 \int \sec^{3}{t}dt$  
  $\displaystyle =$ $\displaystyle \frac{a^2}{2}\left(\frac{\sqrt{x^2 + a^2}}{a}\frac{x}{a} + \log\vert\frac{\sqrt{x^2 + a^2}}{a} + \frac{x}{a}\vert\right) + c$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}\left(x\sqrt{x^2 + a^2} + a^2 \log\vert\sqrt{x^2 + a^2} + x\vert\right) + c$  


$\displaystyle L$ $\displaystyle =$ $\displaystyle 2\sqrt{2}\cdot \frac{1}{2}\left[(w\sqrt{w^2 + \frac{1}{4}} + \frac{1}{4}\log\vert w + \sqrt{w^2 + \frac{1}{4}}\vert)\right]\mid_{-1/2}^{1/2}$  
  $\displaystyle =$ $\displaystyle \sqrt{2}(\frac{1}{2\sqrt{2}} + \frac{1}{4}\log\vert\frac{1}{2} + ... ...(-\frac{1}{2\sqrt{2}} + \frac{1}{4}\log\vert-\frac{1}{2} + \frac{1}{\sqrt{2}}))$  
  $\displaystyle =$ $\displaystyle \sqrt{2}(\frac{1}{\sqrt{2}} + \frac{1}{4}\log\vert\frac{\frac{1}{... ...t{2}}}) = 1 + \frac{\sqrt{2}}{4}\log\vert\frac{2 + \sqrt{2}}{2 - \sqrt{2}}\vert$  
  $\displaystyle =$ $\displaystyle 1 + \frac{\sqrt{2}}{4}\log\vert\frac{(2+\sqrt{2})^2}{2}\vert = 1 + \frac{\sqrt{2}}{2}\log\vert 1 +\sqrt{2}\vert\ensuremath{ \blacksquare}$  

Exercise A

1.
Find the area of the region bounded by the following curves

(a) $\displaystyle{y = x^{2} , y = x + 2}$ (b) $\displaystyle{y = x^{3}, y = x^{2}}$ (c) $\displaystyle{y = - \sqrt{x}, y = x - 6, y = 0}$

(d) $\displaystyle{y = x^{3} - x, y = 1 - x^{2}}$ (e) $\displaystyle{x+4 = y^{2}, x = 5}$

(f) $\displaystyle{y = 2x, x+y = 9, y = x-1}$

2.
Find the volume of the solid generated by revolving the region about the $x$-axis

(a) $\displaystyle{y = x, y = 0, x = 1}$ (b) $\displaystyle{y = x^{2}, y = 9}$ (c) $\displaystyle{y = \sqrt{x}, y = x^{3}}$

(d) $\displaystyle{y = x^{2} , y = x + 2}$

3.
Find the length of graph

(a) $\displaystyle{y = 2x + 3}$ from $x = 0$ to $x = 2$ (b) $\displaystyle{y = x^{3/2}}$ from $x = 0$ to $x = 44$ (c) $\displaystyle{x(t) = t^{2}, y(t) = 2t}$ from $t = 0$ to $t = \sqrt{3}$

(d) $\displaystyle{r = e^{\theta}}$ from $\theta = 0$ to $\theta = 4\pi$

Exercise B

1.
Find the area of the region bounded by the following curves

(a) $\displaystyle{x = y^{2} , x = 3 -2y^{2}}$ (b) $\displaystyle{x = \cos^{3}{t}, y = \sin^{3}{t}, (0 \leq t \leq \pi)}$ and $x$-axis

2.
Find the volume of the solid generated by the reion about the $x$-axis

(a) $\displaystyle{x^{2}+ (y-2)^{2} \leq 1}$

(b) $\displaystyle{ x = t - \sin{t}, y = 1 - \cos{t}, 0 \leq t \leq 2\pi}$ and $x$-axis

3.
Find the length of graph

(a) $\displaystyle{x^{2/3} + y^{2/3} = 1}$ whole length (b) $\displaystyle{\sqrt{x} + \sqrt{y} = 1}$ whole length (c) $\displaystyle{r = 1+\cos{\theta}}$ whole length

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