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.
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.
$\displaystyle y=ln(1-x^2) \Rightarrow \ \frac{dy}{dx}=\frac{-2x}{1-x^2}$
$\displaystyle \int \sqrt{1+\left( \frac{-2x}{1-x^2} \right)^2}dx$
I'll leave you to put in the limits.
I'm in a bit of a rush, but have you tried integrating it using $\displaystyle \frac{-2x}{1-x^2}=sinh(u)$?
Hello, NickytheNine!
It seems you need to brush up on your Algebra . . .
Find the length of the curve: .$\displaystyle y\: = \:\ln(1 - x^2),\quad 0 \leq x \leq \tfrac{1}{7}$
I got the derivative: .$\displaystyle \frac{dy}{dx} \:=\:\frac{-2x}{1-x^2}$
Then i put it in the formula: .$\displaystyle \sqrt{1+\left(\frac{-2x}{1-x^2}\right)^2}$
My problem seems to be simplifying the terms . . .
Under the radical, we have: .$\displaystyle 1 + \frac{4x^2}{(1-x^2)^2} \;=\;\frac{(1-x^2)^2 + 4x^2}{(1-x^2)^2}$
. . . . . $\displaystyle = \;\frac{1 - 2x^2 + x^4 + 4x^2}{(1-x^2)^2} \;=\;\frac{1 + 2x^2 + x^4}{(1-x^2)^2} \;=\;\frac{(1+x^2)^2}{(1-x^2)^2}$
Then: .$\displaystyle \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\sqrt{\frac{(1+x^2)^2}{(1-x^2)^2}} \;=\;\frac{1+x^2}{1-x^2}$
And we must integrate and evaluate: .$\displaystyle \int^{\frac{1}{7}}_0 \frac{1+x^2}{1-x^2}\,dx$
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I recommend long division: . $\displaystyle \int^{\frac{1}{7}}_0\left(-1 + \frac{2}{1-x^2}\right)\,dx $
Got it?