Lengths of Plane Curves - Calculus 2: $\sqrt{1-x^2} ; x=-\frac{1}{2} \to x=\frac{1}{2}$

Question

$$ \sqrt{1-x^2} ; x=-\frac{1}{2} \to x=\frac{1}{2} $$

I am having problems setting this up.

Taking the derivative of $\sqrt{1-x^2}$. Leaves me with: $$ \frac{1}{2}\left(1-x^2 \right)^{-\frac{1}{2}}(-2x) $$

Which further simplifies down to:

$$ \left(1-x^2\right)^{-\frac{1}{2}}(-x) $$

The solutions manual of the book, however skips over this part and gives it as being $\frac{1}{1-x^2}$

Where did the $(-x)$ go? What happened to the square root (or $-\frac{1}{2}$ exponent)? How in the world did they simplify this down to that?

How would you set this up to solve it?

Answer 1

I'm guessing you're using $\sqrt{1+f'(x)^{2}}$ in the arc-length integral. $$\frac{d}{dx}(\sqrt{1-x^{2}})=\frac{-x}{\sqrt{1-x^{2}}}\\1+f'(x)^{2}=1+\left(\frac{-x}{\sqrt{1-x^{2}}}\right)^{2}=\frac{1-x^2+x^2}{1-x^2}=\frac{1}{1-x^{2}}$$

You can then take the sqrt and integrate from $-\frac{1}{2}$ to $\frac{1}{2}$ to find the arc-length.

Answer 2

Well let's try taking the derivative of the $f(x) = \sqrt{1-x^2}$: $$ f'(x) = \frac{1}{2} (1-x^2)^{-1 / 2}(-2x) = \frac{-2x}{2\sqrt{1-x^2}} = \frac{-x}{\sqrt{1-x^2}} $$ Note that we can do the second to last step since the expression: $$ t^{-\alpha} = \frac{1}{t^\alpha} $$ for $t,\alpha \in \mathbb{R}$.

Now in order to find the arclength of $f$ we need to calculate: $$ \sqrt{1 + (f'(x))^2} = \sqrt{1 + \frac{x^2}{1-x^2}} = \sqrt{\frac{1-x^2}{1-x^2} + \frac{x^2}{1-x^2}} = \sqrt{\frac{1}{1-x^2}} = \frac{1}{\sqrt{1-x^2}} $$ which must be what your book is giving you this above expression.

Now we can evaluate the following integral to find the arc length: $$ L = \int_{-1 / 2}^{1 / 2} \sqrt{1 + (f'(x))^2} \, \mathrm{d}x = \int_{-1 / 2}^{1 / 2} \frac{1}{\sqrt{1-x^2}} \, \mathrm{d} x $$ Let me know if you need help evaluating this!