4

$y = \ln (1 − x^2)$ , $ 0 ≤ x ≤ \frac{1}{7}$

Okay, find the derivative: $\frac{1}{1-x^2}$

Set the derivative in this rule:

$\int_{a}^{b}\sqrt{1+\left(\frac{\text{d}y}{\text{d}x}\right)^2}\space\space\text{d}x$

Using this rule $\int_{1}^{\frac{1}{7}}\sqrt{1+\left(\frac{1}{1-x^2}\right)^2}\space\space\text{d}x$

Am I on the right track?

BenCarson2016
  • 443
  • 6
    The derivate is $\dfrac{-2x}{1-x^2}$ – jimbo Nov 08 '15 at 02:56
  • What @jimbo said. You are on the right track. You wish to find the arc-length of the curve, so you are integrating the tiny changes in arc-length along the curve. Just fix your derivative and you'll be fine. – Simon Nov 08 '15 at 02:58

2 Answers2

1

$$\int\sqrt{1+\left(\dfrac{-2x}{1-x^2}\right)^2}dx=\int\sqrt{1+\dfrac{4x^2}{1-2x^2+x^4}}dx=\int\sqrt{\dfrac{1+2x^2+x^4}{1-2x^2+x^4}}=\int\sqrt{\dfrac{(1+x^2)^2}{(1-x^2)^2}}dx=\int\dfrac{1+x^2}{1-x^2}dx=\int \left(-1-\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)dx=-x-\ln(1-x)+\ln(x+1)$$

jimbo
  • 2,156
0

Using inverse function for convenience, the arc length is

$$ \int_0^{\log \frac{48}{49}} \sqrt{ 1-e^x} dx $$

Narasimham
  • 40,495