Find the length of the curve
$$y=\int_0^x \sqrt{(1+t^2)^2-1}dt$$ $$0\le x \le 2$$
Studying for an exam with sample finals, and I came across this problem. I know the formula for arc lengths in polar coordinates, so I surmise that $r=1+t^2$ , which I suppose means $\frac{dr}{d\theta}=-1$, but that's as far as I can figure this out, and I'm not even sure of that part. How do I grok this?
On
Well, in general we can write, if:
$$\text{y}_{\space\text{n}}\left(x\right):=\int_\text{n}^x\text{f}\left(t\right)\space\text{d}t\tag1$$
We get, for the arclength:
$$\mathscr{S}\left(\text{a},\text{b}\right)=\int_\text{a}^\text{b}\sqrt{1+\left(\frac{\text{d}\text{y}_{\space\text{n}}\left(x\right)}{\text{d}x}\right)^2}\space\text{d}x=\int_\text{a}^\text{b}\sqrt{1+\text{f}\left(x\right)^2}\space\text{d}x\tag2$$
So, in your example:
$$\mathscr{S}\left(0,2\right)=\int_0^2\sqrt{1+\left(\sqrt{\left(1+x^2\right)^2-1}\right)^2}\space\text{d}x=\int_0^2\left(1+x^2\right)\space\text{d}x=\frac{14}{3}\tag3$$
The formula for arc length is given by $$L = \int_0^2 \sqrt{1+\left(\frac{dy}{dx}\right)^2} \, dx$$ so it's worth trying what happens when you evaluate this. First, consider what $\frac{dy}{dx}$ should be. Using the fundamental theorem of calculus, you can see that $\frac{dy}{dx} = \sqrt{(1+x^2)^2 - 1}$, so $\sqrt{1+\left(\frac{dy}{dx}\right)^2}=1+x^2$. From here, the arc length integral should be easier to evaluate.