# Math Help - Arc length of an eqation

1. ## Arc length of an eqation

from x=6 and x=9

I need to find the arc length of this equation:

What I know: The equation for arc length is [integral of sqrt(1 + (f'(x))^2)]

What I'm stuck on: When I plug the equation into the arc length equation, I get lost. The hard part for me is after I find the number under the square root, then I have NO IDEA on how to do the integral under the square root!

2. Originally Posted by johnnycho

I need to find the arc length of this equation:

What I know: The equation for arc length is [integral of sqrt(1 + (f'(x))^2)]

What I'm stuck on: When I plug the equation into the arc length equation, I get lost. The hard part for me is after I find the number under the square root, then I have NO IDEA on how to do the integral under the square root!

Recall the formula

$s=\int_a^b\sqrt{1+[y']^2}dx$

So, we need $y'$

$
y'=3\frac{u'}{u}=3\frac{\frac{2}{9}x}{\frac{x^2}{9 }-1}
=\frac{6x}{x^2-9}$

Now we have

$s=\int_6^9\sqrt{1+\left(\frac{6x}{x^2-9}\right)^2}dx$

Now simplifying

$s=\int_6^9\frac{\sqrt{x^4+18x^2+81}}{\sqrt{x^4-18x^2+81}}dx$

Can you see that these are perfect trinomial squares?

Once you factor them, the radicals will dissapear...

3. Originally Posted by johnnycho

from x=6 and x=9

I need to find the arc length of this equation:

What I know: The equation for arc length is [integral of sqrt(1 + (f'(x))^2)]

What I'm stuck on: When I plug the equation into the arc length equation, I get lost. The hard part for me is after I find the number under the square root, then I have NO IDEA on how to do the integral under the square root!

I'm sorry, I'll edit it right away.
It is : x=6 and x=9
$y = 3\ln \left( {{{\left( {\frac{x}{3}} \right)}^2} - 1} \right) = 3\ln \left( {\frac{{{x^2}}}{9} - 1} \right).$

$y' = \frac{3}{{\frac{{{x^2}}}
{9} - 1}} \cdot {\left( {\frac{{{x^2}}}
{9} - 1} \right)^\prime } = \frac{{27}}
{{{x^2} - 9}} \cdot \frac{{2x}}
{9} = \frac{{6x}}
{{{x^2} - 9}}.$

$\sqrt {1 + {{\left( {\frac{{6x}}
{{{x^2} - 9}}} \right)}^2}} = \sqrt {1 + \frac{{36{x^2}}}
{{{{\left( {{x^2} - 9} \right)}^2}}}} = \sqrt {\frac{{{{\left( {{x^2} - 9} \right)}^2} + 36{x^2}}}
{{{{\left( {{x^2} - 9} \right)}^2}}}} =$

$= \sqrt {\frac{{{x^4} - 18{x^2} + 81 + 36{x^2}}}
{{{{\left( {{x^2} - 9} \right)}^2}}}} = \sqrt {\frac{{{x^4} + 18{x^2} + 81}}
{{{{\left( {{x^2} - 9} \right)}^2}}}} =$

$= \sqrt {\frac{{{{\left( {{x^2} + 9} \right)}^2}}}
{{{{\left( {{x^2} - 9} \right)}^2}}}} = \left| {\frac{{{x^2} + 9}}
{{{x^2} - 9}}} \right| = \left\{ \begin{gathered}
\frac{{{x^2} + 9}}{{{x^2} - 9}}\qquad{\text{if}}\, x \in \left( { - \infty ; \, - 3} \right) \cup \left( {3; \,+ \infty } \right), \hfill \\
- \frac{{{x^2} + 9}}
{{{x^2} - 9}} \quad \, {\text{if}}\, x \in \left( { - 3;\,3} \right). \hfill \\
\end{gathered} \right.$

${L_y} = \int\limits_{{x_1}}^{{x_2}} {\sqrt {1 + {{\left( {y'} \right)}^2}}\,dx} = \int\limits_6^9 {\frac{{{x^2} + 9}}
{{{x^2} - 9}}\,dx} = \ldots$

4. Thanks so much for your help.