Notes: `y=lnx , [1,5]` Find the arc length of the curve over the given interval.

Arc length of curve can be denoted as " $\displaystyle{S}$ ". We can determine it by using integral formula on a closed interval [a,b] as: $\displaystyle{S}={\int_{{a}}^{{b}}}{d}{s}$

where:

$\displaystyle{d}{s}=\sqrt{{{1}+{{\left(\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$ $\displaystyle{\quad\text{if}\quad}{y}={f{{\left({x}\right)}}}$

$\displaystyle{d}{s}=\sqrt{{{1}+{{\left(\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}\right)}}^{{2}}}}{\left.{d}{y}\right.}{\quad\text{if}\quad}{x}={h}{\left({y}\right)}$

$\displaystyle{a}$ = lower boundary of the closed interval

$\displaystyle{b}$ =upper boundary of the closed interval

From the given problem: $\displaystyle{y}={\ln{{\left({x}\right)}}},{\left[{1},{5}\right]}$ , we determine that the boundary values are:

...

Arc length of curve can be denoted as " $\displaystyle{S}$ ". We can determine it by using integral formula on a closed interval [a,b] as: $\displaystyle{S}={\int_{{a}}^{{b}}}{d}{s}$

where:

$\displaystyle{d}{s}=\sqrt{{{1}+{{\left(\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$ $\displaystyle{\quad\text{if}\quad}{y}={f{{\left({x}\right)}}}$

$\displaystyle{d}{s}=\sqrt{{{1}+{{\left(\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}\right)}}^{{2}}}}{\left.{d}{y}\right.}{\quad\text{if}\quad}{x}={h}{\left({y}\right)}$

$\displaystyle{a}$ = lower boundary of the closed interval

$\displaystyle{b}$ =upper boundary of the closed interval

From the given problem: $\displaystyle{y}={\ln{{\left({x}\right)}}},{\left[{1},{5}\right]}$ , we determine that the boundary values are:

$\displaystyle{a}={1}$ and $\displaystyle{b}={5}$

Note that $\displaystyle{y}={\ln{{\left({x}\right)}}}$ follows $\displaystyle{y}={f{{\left({x}\right)}}}$ then the formula we will follow can be expressed as $\displaystyle{S}={\int_{{a}}^{{b}}}\sqrt{{{1}+{{\left(\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

For the derivative of $\displaystyle{y}$ or $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ , we apply the derivative formula for logarithm:

$\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{y}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\ln{{\left({x}\right)}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{1}}{{x}}$

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

Plug-in the values on integral formula for arc length of a curve, we get:

$\displaystyle{S}={\int_{{1}}^{{5}}}\sqrt{{{1}+\frac{{1}}{{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

Let $\displaystyle{1}=\frac{{{x}}^{{2}}}{{{x}}^{{2}}}$ then we get:

$\displaystyle{S}={\int_{{1}}^{{5}}}\sqrt{{\frac{{{x}}^{{2}}}{{x}}+\frac{{1}}{{{x}}^{{2}}}}}{\left.{d}{x}\right.}$

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

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

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

From the integration table, we follow the formula for rational function with roots:

$\displaystyle\int\frac{\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}}{{x}}{\left.{d}{x}\right.}=\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}-{a}\cdot{\ln}{\left|\frac{{{a}+\sqrt{{{{x}}^{{2}}+{{a}}^{{2}}}}}}{{x}}\right|}$ .

Applying the integral formula with a^2=1 then a=1, we get:

$\displaystyle{\int_{{1}}^{{5}}}\frac{\sqrt{{{{x}}^{{2}}+{1}}}}{{x}}{\left.{d}{x}\right.}={\left[\sqrt{{{{x}}^{{2}}+{1}}}-{1}\cdot{\ln}{\left|\frac{{{1}+\sqrt{{{{x}}^{{2}}+{1}}}}}{{x}}\right|}\right]}{{\mid}_{{1}}^{{5}}}$

$\displaystyle={\left[\sqrt{{{{x}}^{{2}}+{1}}}-{\ln}{\left|\frac{{{1}+\sqrt{{{{x}}^{{2}}+{1}}}}}{{x}}\right|}\right]}{{\mid}_{{1}}^{{5}}}$

Apply the definite integral formula: $\displaystyle{F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ .

$\displaystyle{\left[\sqrt{{{{x}}^{{2}}+{1}}}-{\ln}{\left|\frac{{{1}+\sqrt{{{{x}}^{{2}}+{1}}}}}{{x}}\right|}\right]}{{\mid}_{{1}}^{{5}}}$

$\displaystyle={\left[\sqrt{{{{5}}^{{2}}+{1}}}-{\ln}{\left|\frac{{{1}+\sqrt{{{{5}}^{{2}}+{1}}}}}{{5}}\right|}\right]}-{\left[\sqrt{{{{1}}^{{2}}+{1}}}-{\ln}{\left|\frac{{{1}+\sqrt{{{{1}}^{{2}}+{1}}}}}{{1}}\right|}\right]}$

$\displaystyle={\left[\sqrt{{{25}+{1}}}-{\ln}{\left|\frac{{{1}+\sqrt{{{25}+{1}}}}}{{5}}\right|}\right]}-{\left[\sqrt{{{1}+{1}}}-{\ln}{\left|\frac{{{1}+\sqrt{{{1}+{1}}}}}{{1}}\right|}\right]}$

$\displaystyle={\left[\sqrt{{{26}}}-{\ln}{\left|\frac{{{1}+\sqrt{{{26}}}}}{{5}}\right|}\right]}-{\left[\sqrt{{{2}}}-{\ln}{\left|{1}+\sqrt{{{2}}}\right|}\right]}$

$\displaystyle=\sqrt{{{26}}}-{\ln}{\left|\frac{{{1}+\sqrt{{{26}}}}}{{5}}\right|}-\sqrt{{{2}}}+{\ln}{\left|{1}+\sqrt{{{2}}}\right|}$

Apply logarithm property: $\displaystyle{\ln{{\left({x}\right)}}}-{\ln{{\left({y}\right)}}}={\ln{{\left(\frac{{x}}{{y}}\right)}}}$ .

$\displaystyle{S}=\sqrt{{{26}}}-\sqrt{{{2}}}+{\ln}{\left|{1}+\sqrt{{{2}}}\right|}-{\ln}{\left|\frac{{{1}+\sqrt{{{26}}}}}{{5}}\right|}$

$\displaystyle{S}=\sqrt{{{26}}}-\sqrt{{{2}}}+{\ln}{\left|\frac{{{1}+\sqrt{{{2}}}}}{{{\left(\frac{{{1}+\sqrt{{{26}}}}}{{5}}\right)}}}\right|}$

$\displaystyle{S}=\sqrt{{{26}}}-\sqrt{{{2}}}+{\ln}{\left|\frac{{{5}\cdot{\left({1}+\sqrt{{{2}}}\right)}}}{{{1}+\sqrt{{{26}}}}}\right|}$

$\displaystyle{S}=\sqrt{{{26}}}-\sqrt{{{2}}}+{\ln}{\left|\frac{{{5}+{5}\sqrt{{{2}}}}}{{{1}+\sqrt{{{26}}}}}\right|}$

$\displaystyle{S}\approx{4.37}$

Notes

Monday, 10 July 2017

$\displaystyle{y}={\ln{{x}}},{\left[{1},{5}\right]}$ Find the arc length of the curve over the given interval.

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How are race, gender, and class addressed in Oliver Optic's Rich and Humble?

Monday, 10 July 2017

Find the arc length of the curve over the given interval.

No comments:

Post a Comment

How are race, gender, and class addressed in Oliver Optic&#39;s Rich and Humble?

$\displaystyle{y}={\ln{{x}}},{\left[{1},{5}\right]}$ Find the arc length of the curve over the given interval.

How are race, gender, and class addressed in Oliver Optic's Rich and Humble?