How Do You Calculate the Length of the Curve from $x=\ln{4}$ to $x=\ln{5}$?

[karush]

$\textsf{12 Find the length of the curve $y=\ln(e^x-1)-\ln( e^x+1 )$
from $x=\ln{4}$ to $x=\ln{5}$}$

ok I thot that curve lengths were done with parametric equations
but didn't know how to convert this

 

[MarkFL]

I would use the formula:

\(\displaystyle s=\int_a^b \sqrt{1+\left(\d{y}{x}\right)^2}\,dx\)

 

[MarkFL]

You could parametrize the curve like so:

\(\displaystyle x(t)=e^t\)

\(\displaystyle y(t)=\ln(x(t)-1)-\ln(x(t)+1)\)

And then use the formula:

\(\displaystyle s=\int_{x(a)}^{x(b)} \sqrt{\left(\d{x}{t}\right)^2+\left(\d{y}{t}\right)^2}\,dt\)

 

[karush]

I would use the formula:

\(\displaystyle s=\int_a^b \sqrt{1+\left(\d{y}{x}\right)^2}\,dx\)

$\displaystyle
\frac{dx}{dy}(\ln(e^x-1)-\ln(e^x+1))
=\frac{2e^x}{e^{2x}-1}$

$\displaystyle s=\int_{\ln 4}^{\ln 5}
\sqrt{1+\left(\frac{2e^x}{e^{2x}-1}\right)^2}\,dx$

this??
got confused on dy/dx

 

[MarkFL]

$\displaystyle
\frac{dx}{dy}(\ln(e^x-1)-\ln(e^x+1))
=\frac{2e^x}{e^{2x}-1}$

$\displaystyle s=\int_{\ln 4}^{\ln 5}
\sqrt{1+\left(\frac{2e^x}{e^{2x}-1}\right)^2}\,dx$

this??
got confused on dy/dx

What you have so far looks good...now combine the terms in the radicand...:)

 

[karush]

$\displaystyle s=\int_{\ln 4}^{\ln 5}
\sqrt{1+\left(\frac{2e^x}{e^{2x}-1}\right)^2}\,dx
= \int_{\ln 4}^{\ln 5}
\sqrt{1-\csch^2(x)} \,dx $

not sure if this was a good alt choice

 

[MarkFL]

$\displaystyle s=\int_{\ln 4}^{\ln 5}
\sqrt{1+\left(\frac{2e^x}{e^{2x}-1}\right)^2}\,dx
= \int_{\ln 4}^{\ln 5}
\sqrt{1-\csch^2(x)} \,dx $

not sure if this was a good alt choice

I don't think those two radicands are equivalent. Can you show you work to justify it?

 

[karush]

I used W|A
this was the input

1+-(\frac{2e^x}{e^{2x}-1})^2

 

[MarkFL]

Have a look here:

W|A - 1+((2e^x)/(e^(2x)-1))^2

However, I would recommend actually doing the algebra instead. :)

 

[karush]

but isn't that just squaring the second term?
there still is a +1

 

[MarkFL]

but isn't that just squaring the second term?
there still is a +1

No, that's the entire radicand...observe:

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