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

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$\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
 
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I would use the formula:

$$s=\int_a^b \sqrt{1+\left(\d{y}{x}\right)^2}\,dx$$
 
You could parametrize the curve like so:

$$x(t)=e^t$$

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

And then use the formula:

$$s=\int_{x(a)}^{x(b)} \sqrt{\left(\d{x}{t}\right)^2+\left(\d{y}{t}\right)^2}\,dt$$
 
MarkFL said:
I would use the formula:

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

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

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

this??
got confused on dy/dx
 
karush said:
$\displaystyle
\frac{dx}{dy}(\ln(e^x-1)-\ln(e^x+1))
=\frac{2e^x}{e^{2x}-1}$

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

this??
got confused on dy/dx

What you have so far looks good...now combine the terms in the radicand...:)
 
$\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
 
karush said:
$\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?
 
I used W|A
this was the input

1+-(\frac{2e^x}{e^{2x}-1})^2
 
  • #10
but isn't that just squaring the second term?
there still is a +1
 
  • #11
karush said:
but isn't that just squaring the second term?
there still is a +1

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

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

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