Let $c$ be the path $c(t)$ = $(t, 2sin(t), 3cos(t))$. Find the arc length of $c$ between the to points $(0, 0, 3)$ and $(\pi, 0, -3)$.
I know the formula is the integral of the magnitude of the function but I don't know how to set it up(the bounds). So far, I have:
= $\int{||c'(t)||}dt$
= $\int{||(1 +2cos(t) -3sin(t)||}dt$
= $\int{||1 + 4cos^{2}(t)+9sin^{2}(t)||}$
I'm not sure where to go from here...
On
If you ask Wolfram Alpha, you will get $$\int_0^\pi \sqrt{1+\sin ^2(t)}=2 E(-1)$$ where appears the complete elliptic integral of the second kind.
You can evaluate the result using the series expansion given here $$E(k)=\frac \pi 2 \left(1-\sum_{n=1}^\infty\left[\frac{(2 n-1)\text{!!}}{(2 n)\text{!!}}\right]^2\frac {k^{n}}{2n-1} \right)$$ which, for $k=-1$, would not require too many terms if you are not concerned with too high accuracy. For example, using $\sum_{n=1}^{10}$, would give $1.90894$ while the exact value is $E(-1)\approx 1.91010$ (corresponding to a relative error of $0.06$%).
Your path passes through $(0,0,3)$ when $t=0$ and through $(\pi,0,-3)$ when $t=\pi$. So, the arc length that you're after is equal to$$\int_0^\pi\sqrt{1+4\cos^2t+9\sin^2t}\,\mathrm dt=\int_0^\pi\sqrt{5+5\sin^2t}\,\mathrm dt.$$Can you take it from here?