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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 $a \leq x \leq b$.


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

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.

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

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
$\displaystyle l$ $\displaystyle =$ $\displaystyle 4\int_{0}^{\frac{\pi}{2}}\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$  
  $\displaystyle =$ $\displaystyle 4 \int_{0}^{\frac{\pi}{2}}\sqrt{(-3\cos^{2}{t}\sin{t})^2 + (3\sin^{2}{t}\cos{t})^2} dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2} \sqrt{\cos^4{t}\sin^2{t} + \sin^4{t}\cos^2{t}}dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2} \sqrt{\cos^2{t}\sin^2{t}(\cos^2{t} + \sin^2{t})}dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2} \sqrt{\cos^2{t}\sin^2{t}}dt$  
  $\displaystyle =$ $\displaystyle 12 \int_0^\frac{\pi}{2}\cos{t}\sin{t}\:dt\ (u = \sin{t}, du = \cos{t}\:dt)$  
  $\displaystyle =$ $\displaystyle 12 \int_0^1 udu = 12 \left[\frac{u^2}{2}\right]_0^1 = 12\cdot \frac{1}{2} = 6\ensuremath{\ \blacksquare}$  

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

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

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


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

\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 =$$\displaystyle 2\sqrt{2}\int_{0}^{1}\sqrt{(u - \frac{1}{2})^2 + \frac{1}{4}}du.$  
Let . Then and $w = u - \frac{1}{2}$$dw = du$

\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 . Then . Also, , $x = a\tan{t}$$\sqrt{x^2 + a^2} = \sqrt{a^2(1+\tan^{2}{t})} = \sqrt{a^2 \sec^{2}{t}} = a \sec{t}$$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?D

(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 -axis?D
$x$

(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?D

from to from to from to

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

from to

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

Exercise B


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

and -axis

(a)$\displaystyle{x = y^{2} , x = 3 -2y^{2}}$(b)$\displaystyle{x = \cos^{3}{t}, y = \sin^{3}{t}, \ (0 \leq t \leq \pi)}$$x$
2.
Find the volume of the solid generated by the reion about the -axis?D
$x$

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

and -axis?D

(b)$\displaystyle{\ x = t - \sin{t}, y = 1 - \cos{t}, \ 0 \leq t \leq 2\pi}$$x$
3.
Find the length of graph?D

whole length whole length whole length

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

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