anonymous
  • anonymous
Find the arc length of the graph of the function from x=0 to x=2. F(x) = ln(x+1).
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
hello
anonymous
  • anonymous
sjin, the arc length of a function is given by\[s=\int\limits_{x_1}^{x_2}\sqrt{1+(y')^2}dx\]after considering a differential of arc length using Pythagoras' Theorem. The problem here will be with the integration.
anonymous
  • anonymous
\[y=\log (x+1) \rightarrow y'=\frac{1}{x+1}\]so\[s=\int\limits_{0}^{2}\sqrt{1+\left( \frac{1}{x+1} \right)^2}dx=\int\limits_{0}^{2}\frac{\sqrt{(x+1)^2+1}}{x+1}dx\]

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anonymous
  • anonymous
You're going to have to make a few substitutions along the way to solve this. You should start with \[u=x+1\]to obtain\[\int\limits_{u_1}^{u_2}\frac{\sqrt{u^2+1}}{u}du\]then\[w=u^2+1\]to obtain\[\frac{1}{2}\int\limits_{w_1}^{w_2}\frac{w^{1/2}}{w-1}dw\]and finally, \[\zeta = w^{1/2}\]to obtain\[\int\limits_{\zeta_1}^{\zeta_2}\frac{\zeta^2}{\zeta^2-1}d \zeta\]The last integrand may be written\[\frac{\zeta^2}{\zeta^2-1}=\frac{\zeta^2-1+1}{\zeta^2 -1}=1+\frac{1}{\zeta^2-1}=1+\frac{1}{2}\frac{1}{\zeta -1}-\frac{1}{2}\frac{1}{\zeta +1}\]
anonymous
  • anonymous
I would have banged out a sinh sub, but I haven't looked too closely.
anonymous
  • anonymous
After solving the final integral form, and back-substituting, you obtain,\[s=\sqrt{(x+1)^2+1}+\frac{1}{2}\log \frac{\sqrt{(x+2)^2+1}-1}{\sqrt{(x+1)^2+1}+1}|_0^2\]
anonymous
  • anonymous
with\[s=\sqrt{10}-\sqrt{2}+\frac{1}{2}\log \frac{\sqrt{10}-1}{\sqrt{10}+1}-\frac{1}{2}\log \frac{\sqrt{2}-1}{\sqrt{2}+1}\]
anonymous
  • anonymous
As INewton mentioned, there are other avenues for substitution.
anonymous
  • anonymous
thank you so much!
anonymous
  • anonymous
np
anonymous
  • anonymous
Indeed, I'm pretty impressed you typed all that LaTeX on what is a pretty poor engine on this site.
anonymous
  • anonymous
I'm impressed too. I feel exhausted.

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