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The question: Find length of curve defined by $\displaystyle y=2\ln\left[\left(\frac{x}{2}\right)^2-1\right] $ from $x=4$ to $x=6$

Here is the work I have done, but I seem to keep getting it wrong. Are there any suggestions as to what else I can do or what I am doing wrong?


$$\frac{dy}{dx} = 2 \ln\left[\left(\frac x2\right)^2-1\right]$$ Then I used the substitution $\displaystyle-u = \left(\frac x2\right)^2 - 1$. The answer I got was $2 \frac 1u \frac x2$. Substituting back in we get $$2 \frac{1}{\frac{x^2}4 - 1} \frac x2 = \frac{4x}{x^2 - 4}$$

Using this arclenght formula I get $$L = \int_4^6 \sqrt{1 + \left(\frac{4x}{x^2 - 4}\right)^2}dx = \int_4^6 \sqrt{1 + \frac{16x^2}{(x^2 - 4)^2}}dx = \int_4^6 1 dx + \int_4^6 \frac{4x}{x^2-4}dx = 1.66 $$

But the correct answer is $2.81$

What did I do wrong?

SIDENOTE: Hi I have posed as another user with the same name, but my computer signed me off and I'm not really sure how to get that account back (cannot remember password to that email!)? If anyone has any suggestions or should I just start this account? I am also not familiar with how to add the proper notation, if someone can point me out to that. Thank you.

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  • $\begingroup$ You can use for formatting meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – kingW3
    Dec 12, 2014 at 20:14
  • $\begingroup$ this is unreadable. So either write a beter copy or type in latex codecogs.com/latex/eqneditor.php $\endgroup$
    – TKM
    Dec 12, 2014 at 20:19
  • $\begingroup$ Trying to edit now... might take a while then. Thanks $\endgroup$
    – Kris
    Dec 12, 2014 at 20:40
  • $\begingroup$ Hi, sorry to bother, but I'm not sure what I did wrong with the MathJax?!?! I do not get along with computers! $\endgroup$
    – Kris
    Dec 12, 2014 at 21:05
  • $\begingroup$ You have to wrap your math commands around the dollar sign :) so write the dollar symbol, then your commands, then again the dollar symbol. $\endgroup$
    – Ant
    Dec 12, 2014 at 21:08

1 Answer 1

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The problem is that $$\sqrt{a + b} \neq \sqrt{a} + \sqrt{b} $$

You have used this in your post but it is wrong; the square root is not a linear operator!


For example, $4 = \sqrt{16} = \sqrt{12 + 4} \neq \sqrt{12} + 2$

Also, think of all the theorems that would become meaningless; Pythagora, for example; $c = \sqrt{a^2 + b^2}$ would become $c = a + b$. But the hypothenuses in a right triangle is not the sum of the other two sides!

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