# What is the arc length of f(x)= sqrt(x-1)  on x in [1,2] ?

##### 1 Answer
Apr 1, 2016

$- \ln \frac{\sqrt{5} - 2}{4} + \frac{\sqrt{5}}{2} \cong 1.4789$

#### Explanation:

To solve this problem we should use the formula
$L = {\int}_{a}^{b} \sqrt{1 + \left({f}^{'} {\left(x\right)}^{2}\right)} \cdot \mathrm{dx}$

$f \left(x\right) = \sqrt{x - 1}$
${f}^{'} \left(x\right) = \left(\frac{1}{2}\right) {\left(x - 1\right)}^{- \frac{1}{2}}$
${f}^{'} \left(x\right) = \left(\frac{1}{2}\right) {\left(x - 1\right)}^{- 1}$

So we have the indefinite integral that solves the problem:
$F \left(x\right) = \int \sqrt{1 - \frac{1}{4} \left(\frac{1}{x - 1}\right)} \cdot \mathrm{dx} = \frac{1}{2} \int \sqrt{4 - \left(\frac{1}{x - 1}\right)} \cdot \mathrm{dx}$

Lets begin by using a trigonometrical substitution:
$\left(\frac{1}{x - 1}\right) = 4 \cdot {\tan}^{2} y$
$- \frac{\mathrm{dx}}{x - 1} ^ 2 \cdot \mathrm{dx} = 4 \cdot 2 \tan y {\sec}^{2} y \cdot \mathrm{dy}$
$\to - 16 {\tan}^{4} y \cdot \mathrm{dx} = 8 \tan y {\sec}^{2} y \cdot \mathrm{dy}$ => $\mathrm{dx} = - \frac{{\sec}^{2} y}{2 {\tan}^{3} y} \mathrm{dy}$

Then the indefinite integral becomes
$\frac{1}{2} \int \cancel{2} \sec y \left(- \frac{{\sec}^{2} y}{\cancel{2} {\tan}^{3} y}\right) \mathrm{dy} = - \frac{1}{2} \int {\sec}^{3} \frac{y}{{\tan}^{3} y} \mathrm{dy}$
$= - \frac{1}{2} \int \frac{1}{\cancel{{\cos}^{3} y}} \left(\frac{\cancel{{\cos}^{3} y}}{\sin} ^ 3 y\right) \cdot \mathrm{dy} = - \frac{1}{2} \int {\csc}^{3} y \cdot \mathrm{dy}$

We find this last integral in the more complete tables. Anyway, I'll solve it:
$- \frac{1}{2} \int {\csc}^{3} y \cdot \mathrm{dy} = - \frac{1}{2} \int \csc y \cdot \mathrm{dy} - \frac{1}{2} \int \csc y \cdot {\cot}^{2} y \cdot \mathrm{dy}$

I call this last result as expression 1, and I go on to solve its last term:
$\int \csc y \cdot {\cot}^{2} y \cdot \mathrm{dy} = \int \left(\frac{1}{\sin y}\right) \left({\cos}^{2} \frac{y}{\sin} ^ 2 y\right) \mathrm{dy}$
Making
$\sin y = z$
$\cos y \cdot \mathrm{dy} = \mathrm{dz}$
So
$= \int \frac{\sqrt{1 - {z}^{2}}}{z} ^ 3 \mathrm{dz}$
Using the rule

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$
$\to u = {\left(1 - {z}^{2}\right)}^{\frac{1}{2}}$ => $\mathrm{du} = - 2 z \left(\frac{1}{2}\right) {\left(1 - {z}^{2}\right)}^{- \frac{1}{2}} = - z {\left(1 - {z}^{2}\right)}^{- \frac{1}{2}}$
$\to \mathrm{dv} = \frac{\mathrm{dz}}{z} ^ 3$ => $v = - \frac{1}{2 {z}^{2}}$
we get
$= - {\left(1 - {z}^{2}\right)}^{\frac{1}{2}} / \left(2 {z}^{2}\right) - \frac{1}{2} \int \frac{\mathrm{dz}}{z \cdot \sqrt{1 - {z}^{2}}} \mathrm{dz}$

Calling the result above as expression 2 and proceeding to solve its last term
$- \frac{1}{2} \int \frac{\mathrm{dz}}{z \cdot \sqrt{1 - {z}^{2}}} \mathrm{dz} =$
$z = \sin \alpha$
$\mathrm{dz} = \cos \alpha \cdot \mathrm{da} l p h a$
Therefore
$= - \frac{1}{2} \int \frac{\cancel{\cos \alpha}}{\sin \alpha \cdot \cancel{\cos \alpha}} \mathrm{da} l p h a = \frac{1}{2} \int \csc \alpha \cdot \mathrm{da} l p h a$
$= - \frac{1}{2} \cdot \ln | \csc \alpha - \cot \alpha | = - \frac{1}{2} \cdot \ln | \frac{1}{z} - \frac{\sqrt{1 - {z}^{2}}}{z} | = - \frac{1}{2} \cdot \ln | \frac{1 - \sqrt{1 - {z}^{2}}}{z} |$

Returning to expression 2 :
$= - {\left(1 - {z}^{2}\right)}^{\frac{1}{2}} / \left(2 {z}^{2}\right) - \frac{1}{2} \cdot \ln | \frac{1 - \sqrt{1 - {z}^{2}}}{z} |$
But $z = \sin y$
$\to = - \cos \frac{y}{2 {\sin}^{2} y} - \frac{1}{2} \cdot \ln | \frac{1 - \cos y}{\sin} y | = - \frac{1}{2} \cot y \cdot \csc y - \frac{1}{2} \cdot \ln | \csc y - \cot y |$

Returning to expression 1 ::
$= - \frac{1}{2} \ln | \csc y - \cot y | + \frac{1}{4} \cdot \cot y \cdot \csc y + \frac{1}{4} \cdot \ln | \csc y - \cot y |$
$= - \frac{1}{4} \ln | \csc y - \cot y | + \frac{1}{4} \cdot \cot y \cdot \csc y$
Since
${\tan}^{2} y = \frac{1}{4 \left(x - 1\right)}$
we find that
cscy=sqrt(4x-3); cot y=2sqrt(x-1);siny=1/sqrt(4x-3) and cosy=2sqrt((x-1)/(4x-3))

So

$F \left(x\right) = - \frac{1}{4} \cdot \ln | \sqrt{4 x - 3} - 2 \sqrt{x - 1} | + \left(\frac{1}{4}\right) 2 \sqrt{x - 1} \cdot \sqrt{4 x - 3}$
$F \left(x\right) = - \frac{1}{4} \cdot \ln | \sqrt{4 x - 3} - 2 \sqrt{x - 1} | + \frac{1}{2} \cdot \sqrt{\left(x - 1\right) \left(4 x - 3\right)}$

Finally
$L = F \left(x = 3\right) - F \left(x = 1\right)$
$L = - \ln \frac{\sqrt{5} - 2}{4} + \frac{\sqrt{5}}{2} \cong 1.4789$