I'm trying to find the arc lenght of a curve in $R^3$. I know that that the formula to find this is $\int_a^b \sqrt{f'(t)^2 + g'(t)^2+h'(t)^2}dt$.
But I want to find the arc lenght between the points $(\frac{1}{3},1,5)$ and $(9,5,13)$ of a curve $\vec{C} = \frac{t^3}{3}, 2t-1,t^2+4)$
So I am confused as to what my $a$ and $b$ would be in thi case? as I have a curve in $R^3$. so the interval [a,b] is not a single value? as I have three values, x,y z.
Do you get what I mean?
On
You need to find the $t$-values for which $C$ passes through the given points. It is easy to see that $$C(1)=\left(\frac{1}{3},1,5\right)\quad\mbox{and}\quad C(3)=(9,5,13),$$and since $C$ is injective (because of the first component), there are no ambiguities or other possibilities. The arclength is $$\int_1^3((t^2)^2+2^2+(2t)^2)^{1/2}\,{\rm d}t.$$
I believe a and b are the value of the parameter at the initial point and the final point since the integral integrates over the "speeds" (the magnitude of the derivative) at every value of the parameter between a and b. At $(\frac{1}{3}, 1, 5)$ $t=1$ and at $(9,5,13)$ $t=3$, so $a=1$ and $b=3$.