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

Monday, November 15, 2010

$\displaystyle{y}={\ln{{\left(\frac{{{{e}}^{{x}}+{1}}}{{{{e}}^{{x}}-{1}}}\right)}}},{\left[{\ln{{2}}},{\ln{{3}}}\right]}$ 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

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

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

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

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

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

$\displaystyle={\ln{{\left(\frac{{16}}{{9}}\right)}}}$ is the final answer

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Monday, November 15, 2010

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

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Monday, November 15, 2010

Find the arc length of the graph of the function over the indicated interval.

No comments:

Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

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

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?