arc lengthintegrationmultivariable-calculus
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…
You can rotate the coordinates so the first point is at the pole and measure $\theta$ from $0$ at that pole to $\pi$ at the other one. The distance to the other point is then $\theta$. The probability distribution of $\theta$ is proportional to $\sin \theta$ as that is the radius of the small circle the second point is on. Your average distance is then $$\frac {\int_0^\pi \theta \sin(\theta) d\theta}{\int_0^\pi \sin(\theta) d\theta}=\frac {\pi}{2}$$ We can see this is true by matching up points at $\theta$ with points at $\pi - \theta$ The average of each pair is $\frac \pi 2$
We have the following situation:
Points $A$ and $B$ are given by their coordinates. So you can calculate their distance. Now, we have an isosceles triangle with given side lengths. We can calculate $\alpha$.
Given $\alpha$ and $r$, the arclength can be calculated.
Best Answer
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?