Find the length of the curve y = $\sqrt{-x(x+1)} - \arctan \sqrt{\frac{-x}{x+1}}$

Question

I used the formula for this example:

$\displaystyle L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$

And I start by computing the derivative:

\begin{align*} y' &= \left(\sqrt{-x^2-x}\right)' - \left(\arctan \sqrt{\frac{-x}{x+1}}\right)' \\ &= \frac{1}{2\sqrt{-x^2-x}} \cdot (-2x-1)-\frac{1}{\left(\sqrt{\frac{-x}{x+1}}\right)^2+1} \cdot \frac {1}{2\sqrt{\frac{-x}{x+1}}} \cdot \frac{-1}{(x+1)^2} \\ &= \frac{-2x-1}{2\sqrt{-x^2-x}} - \frac{1}{{\frac{-x}{x+1}}+1} \cdot \frac {1}{2\sqrt{\frac{-x}{x+1}}} \cdot \frac{-1}{(x+1)^2} \\ &= \frac{-2x-1}{2\sqrt{-x^2-x}}+ \frac{1}{2(x+1)^2 \sqrt{\frac{-x}{x+1}}\left(1-\frac{x}{x+1}\right)} \end{align*}

but I stopped at this moment because it does not seem to me that this derivation was so complicated, how to do it then? Maybe I made a mistake somewhere?

Answer 1

For your actual task it might be beneficial to substitute $z = \sqrt{\tfrac{-x}{x+1}}$. Then your curve is $y = \tfrac z{1+z^2}-\arctan(z)$. Note that you will also have to adapt the boundaries.

Answer 2

I think that you had problems with the simplifications of derivatives. Let $y=A-B$ $$A=\sqrt{-x (x+1)}\implies A'=\frac{-2 x-1}{2 \sqrt{-x (x+1)}}$$ $$B=\tan ^{-1}\left(\sqrt{-\frac{x}{x+1}}\right)\implies B'=\frac{\frac{x}{(x+1)^2}-\frac{1}{x+1}}{2 \sqrt{-\frac{x}{x+1}} \left(1-\frac{x}{x+1}\right)}=\frac{\sqrt{\frac{1}{x+1}-1}}{2 x}$$ $$y'=A'-B'=-\frac{x}{\sqrt{-x (x+1)}}\implies 1+(y')^2=\frac 1 {1+x}$$ I am sure that you can take it from here.