Notes: `y = ln((e^x + 1)/(e^x - 1)) , [ln2 , ln3]` Find the arc length of the graph of the function over the indicated interval.

Saturday, 18 July 2015

$\displaystyle{y}={\ln{{\left(\frac{{{{e}}^{{x}}+{1}}}{{{{e}}^{{x}}-{1}}}\right)}}},{\left[{\ln{{2}}},{\ln{{3}}}\right]}$ Find the arc length of the graph of the function over the indicated interval.

The arc length of a function is the length of the described curve within some interval on the x-axis (and a corresponding interval on the y-axis). If we were to lay a piece of string over the line of the graph in this window, the graph going from corner to diagonally opposite corner in a curve, the arc length is the length of the piece of string used.

In this case the function to find the arc length of is

$\displaystyle{y}={\ln{{\left(\frac{{{{e}}^{{x}}+{1}}}{{{{e}}^{{x}}-{1}}}\right)}}}$

where $\displaystyle{\ln{{\left({x}\right)}}}$ is the natural logarithm of $\displaystyle{x}$ (the inverse function to $\displaystyle{{e}}^{{{x}}}$ , recall).

The window over which to find the length is the rectangle given by $\displaystyle{\ln{{2}}}\le{x}\le{\ln{{3}}}$ and $\displaystyle{\ln{{2}}}\le{y}\le{\ln{{3}}}$ .

The function for the arc length of a generic function $\displaystyle{y}={f{{\left({x}\right)}}}$ is

$s = int_a^b sqrt(1+((dy)/(dx))^2) \quad dx$

This integrates along the line of the graph, adding infinitesimal sections together where each very small section added is a straight line diagonal over the small window $\displaystyle{x}_{{0}}\le{x}\le{x}_{{0}}+{\left.{d}{x}\right.}$ , $\displaystyle{y}_{{0}}\le{y}\le{y}_{{0}}+{\left.{d}{y}\right.}$ . Though $\displaystyle{y}={f{{\left({x}\right)}}}$ may be a curve, the point is that these windows over which the length is integrated (added up) are so small that the graph is a straight line within them making it a simple thing to add the small (straight line) sections together. This concept is the basis of calculus.

In this example, since $\displaystyle{y}={\ln{{\left(\frac{{{{e}}^{{x}}+{1}}}{{{{e}}^{{x}}-{1}}}\right)}}}={\ln{{\left({{e}}^{{x}}+{1}\right)}}}-{\ln{{\left({{e}}^{{x}}-{1}\right)}}}$ then its derivative $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}$ is given by

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

Now working in steps to find the integrand in the formula for the arc length $\displaystyle{s}$ , we have that

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

and that

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

$\displaystyle=\frac{\sqrt{{{\left({{e}}^{{{2}{x}}}+{2}{{e}}^{{x}}+{1}\right)}{\left({{e}}^{{{2}{x}}}-{2}{{e}}^{{{x}}}+{1}\right)}+{4}{{e}}^{{{2}{x}}}}}}{{{\left({{e}}^{{x}}+{1}\right)}{\left({{e}}^{{x}}-{1}\right)}}}$

$\displaystyle=\frac{\sqrt{{{\left({{e}}^{{{4}{x}}}-{2}{{e}}^{{{3}{x}}}+{{e}}^{{{2}{x}}}+{2}{{e}}^{{{3}{x}}}-{4}{{e}}^{{{2}{x}}}+{2}{{e}}^{{x}}+{{e}}^{{{2}{x}}}-{2}{{e}}^{{x}}+{1}\right)}+{4}{{e}}^{{{2}{x}}}}}}{{{\left({{e}}^{{x}}+{1}\right)}{\left({{e}}^{{x}}-{1}\right)}}}$

Adding up that very long set of exponential terms, finding that most cancel, we finally have that

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

$\displaystyle=\frac{\sqrt{{{{\left({{e}}^{{{2}{x}}}+{1}\right)}}^{{2}}}}}{{{\left({{e}}^{{x}}+{1}\right)}{\left({{e}}^{{x}}-{1}\right)}}}=\frac{{{{e}}^{{{2}{x}}}+{1}}}{{{\left({{e}}^{{x}}+{1}\right)}{\left({{e}}^{{x}}-{1}\right)}}}$

$\displaystyle=\frac{{{{e}}^{{{2}{x}}}+{1}}}{{{{e}}^{{{2}{x}}}-{1}}}$

Integrating the integrand over the required interval, we have that

$s = int_(ln2)^(ln3) ((e^(2x)+1)/(e^(2x)-1)) \quad dx$

$= int_(ln2)^(ln3) (2e^(2x)- e^(2x) + 1)/(e^(2x)-1) \quad dx = int_(ln2)^(ln3) (2e^(2x))/(e^(2x)-1) -1 \quad dx$

$\displaystyle={\ln{{\left({{e}}^{{{2}{x}}}-{1}\right)}}}{{\left|_{{\ln{{2}}}}^{{{\ln{{3}}}}}-{x}\right|}_{{{\ln{{2}}}}}^{{{\ln{{3}}}}}}$

$\displaystyle={\ln{{\left({{e}}^{{{2}{\ln{{3}}}}}-{1}\right)}}}-{\ln{{\left({{e}}^{{{2}{\ln{{2}}}}}-{1}\right)}}}-{\ln{{3}}}+{\ln{{2}}}$

$\displaystyle={\ln{{\left({9}-{1}\right)}}}-{\ln{{\left({4}-{1}\right)}}}-{\ln{{3}}}+{\ln{{2}}}={\ln{{\left({\left(\frac{{8}}{{3}}\right)}{\left(\frac{{2}}{{3}}\right)}\right)}}}$