New Notes Blog: `y = x^5/10 + 1/(6x^3) , [2, 5]` Find the arc length of the graph of the function over the indicated interval.

Thursday, 21 May 2015

$\displaystyle{y}=\frac{{{x}}^{{5}}}{{10}}+\frac{{1}}{{{6}{{x}}^{{3}}}},{\left[{2},{5}\right]}$ Find the arc length of the graph of the function over the indicated interval.

Arc length(L) of the function y=f(x) on the interval [a,b] is given by the formula,

$\displaystyle{L}={\int_{{a}}^{{b}}}\sqrt{{{1}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{x}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$ , if y=f(x) and a $\displaystyle\le$ x $\displaystyle\le$ b

Now we have to differentiate the function,

$\displaystyle{y}=\frac{{{x}}^{{5}}}{{10}}+\frac{{1}}{{{6}{{x}}^{{3}}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{{10}}{\left({5}\right)}{{x}}^{{{5}-{1}}}+\frac{{1}}{{6}}{\left(-{3}\right)}{{x}}^{{-{3}-{1}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{x}}^{{4}}}{{2}}-\frac{{1}}{{2}}{{x}}^{{-{4}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{x}}^{{4}}}{{2}}-\frac{{1}}{{{2}{{x}}^{{4}}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{{2}}{\left({{x}}^{{4}}-\frac{{1}}{{{x}}^{{4}}}\right)}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{1}}{{2}}{\left(\frac{{{{x}}^{{8}}-{1}}}{{{x}}^{{4}}}\right)}$

$\displaystyle{L}={\int_{{2}}^{{5}}}\sqrt{{{1}+{{\left(\frac{{{{x}}^{{8}}-{1}}}{{{2}{{x}}^{{4}}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

$\displaystyle{L}={\int_{{2}}^{{5}}}\sqrt{{{1}+\frac{{{{x}}^{{16}}-{2}{{x}}^{{8}}+{1}}}{{{4}{{x}}^{{8}}}}}}{\left.{d}{x}\right.}$

$\displaystyle{L}={\int_{{2}}^{{5}}}\sqrt{{\frac{{{4}{{x}}^{{8}}+{{x}}^{{16}}-{2}{{x}}^{{8}}+{1}}}{{{4}{{x}}^{{8}}}}}}{\left.{d}{x}\right.}$

$\displaystyle{L}={\int_{{2}}^{{5}}}\sqrt{{\frac{{{{x}}^{{16}}+{2}{{x}}^{{8}}+{1}}}{{{4}{{x}}^{{8}}}}}}{\left.{d}{x}\right.}$

$\displaystyle{L}={\int_{{2}}^{{5}}}\sqrt{{{{\left(\frac{{{{x}}^{{8}}+{1}}}{{{2}{{x}}^{{4}}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

$\displaystyle{L}={\int_{{2}}^{{5}}}\frac{{{{x}}^{{8}}+{1}}}{{{2}{{x}}^{{4}}}}{\left.{d}{x}\right.}$

$\displaystyle{L}={\int_{{2}}^{{5}}}{\left(\frac{{{x}}^{{8}}}{{{2}{{x}}^{{4}}}}+\frac{{1}}{{{2}{{x}}^{{4}}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{L}={\int_{{2}}^{{5}}}{\left(\frac{{{x}}^{{4}}}{{2}}+\frac{{1}}{{{2}{{x}}^{{4}}}}\right)}{\left.{d}{x}\right.}$

$\displaystyle{L}={{\left[\frac{{1}}{{2}}{\left(\frac{{{x}}^{{{4}+{1}}}}{{{4}+{1}}}\right)}+\frac{{1}}{{2}}{\left(\frac{{{x}}^{{-{4}+{1}}}}{{-{4}+{1}}}\right)}\right]}_{{2}}^{{5}}}$

$\displaystyle{L}={{\left[\frac{{{x}}^{{5}}}{{10}}-\frac{{1}}{{{6}{{x}}^{{3}}}}\right]}_{{2}}^{{5}}}$

$\displaystyle{L}={\left[\frac{{{5}}^{{5}}}{{10}}-\frac{{1}}{{{6}{{\left({5}\right)}}^{{3}}}}\right]}-{\left[\frac{{{2}}^{{5}}}{{10}}-\frac{{1}}{{{6}{{\left({2}\right)}}^{{3}}}}\right]}$

$\displaystyle{L}={\left[\frac{{3125}}{{10}}-\frac{{1}}{{750}}\right]}-{\left[\frac{{32}}{{10}}-\frac{{1}}{{48}}\right]}$

$\displaystyle{L}={\left[\frac{{{234375}-{1}}}{{750}}\right]}-{\left[\frac{{{768}-{5}}}{{240}}\right]}$

$\displaystyle{L}={\left[\frac{{234374}}{{750}}\right]}-{\left[\frac{{763}}{{240}}\right]}$

$\displaystyle{L}=\frac{{{1874992}-{19075}}}{{6000}}$

$\displaystyle{L}=\frac{{1855917}}{{6000}}$

$\displaystyle{L}={309.3195}$