Length of the Curve

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• April 17th 2009, 01:59 PM
NickytheNine
Length of the Curve
Find the Length of The Curve
y = ln(1 - x2)

0 < x < (1/7)

I got the derivative: (-2x)/(1-x^2)

Then i put it in the formula : sqr*(1+((-2x)/(1-x^2))^2)

My problem seems to be simplifying the terms to cancel the sqr* before integrating.
• April 17th 2009, 02:03 PM
Showcase_22
$y=ln(1-x^2) \Rightarrow \ \frac{dy}{dx}=\frac{-2x}{1-x^2}$

$\int \sqrt{1+\left( \frac{-2x}{1-x^2} \right)^2}dx$

I'll leave you to put in the limits.

I'm in a bit of a rush, but have you tried integrating it using $\frac{-2x}{1-x^2}=sinh(u)$?
• April 17th 2009, 02:07 PM
NickytheNine
0 < x < (1/7)
• April 17th 2009, 02:20 PM
NickytheNine
i dont really see how you use a hyperbolic function to make it possible to integrate
• April 17th 2009, 04:18 PM
Soroban
Hello, NickytheNine!

It seems you need to brush up on your Algebra . . .

Quote:

Find the length of the curve: . $y\: = \:\ln(1 - x^2),\quad 0 \leq x \leq \tfrac{1}{7}$

I got the derivative: . $\frac{dy}{dx} \:=\:\frac{-2x}{1-x^2}$

Then i put it in the formula: . $\sqrt{1+\left(\frac{-2x}{1-x^2}\right)^2}$

My problem seems to be simplifying the terms . . .

Under the radical, we have: . $1 + \frac{4x^2}{(1-x^2)^2} \;=\;\frac{(1-x^2)^2 + 4x^2}{(1-x^2)^2}$

. . . . . $= \;\frac{1 - 2x^2 + x^4 + 4x^2}{(1-x^2)^2} \;=\;\frac{1 + 2x^2 + x^4}{(1-x^2)^2} \;=\;\frac{(1+x^2)^2}{(1-x^2)^2}$

Then: . $\sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\sqrt{\frac{(1+x^2)^2}{(1-x^2)^2}} \;=\;\frac{1+x^2}{1-x^2}$

And we must integrate and evaluate: . $\int^{\frac{1}{7}}_0 \frac{1+x^2}{1-x^2}\,dx$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I recommend long division: . $\int^{\frac{1}{7}}_0\left(-1 + \frac{2}{1-x^2}\right)\,dx$

Got it?

• April 19th 2009, 02:28 PM
NickytheNine
thank you very much, i just got to work on my algebra i guess lol