Arc Length of a Curve given point in (x,y,z) form $r(t)=\sqrt{2}t$$i$+$\sqrt{2}t$$j$+($1$-$t$$^2$)$k$  Is the function I am considering, and I am finding the arc length. That is not what I am stuck on though, I am asked to find the arc length from $(0,0,1)$ to $(\sqrt{2}$,$\sqrt{2}$, $0$). With the point being in the $($x$,$y$,$z$)$ form, instead of being given a range of values for $t$, how do I find the arc length? 
 A: What one can do is to try and discover the appropriate $t$ values for the given points $(0, 0, 1)$ and $(\sqrt 2, \sqrt 2, 0)$; in this case it's pretty easy, though that certainly isn't always the case.
Anyway, with 
$r(t) = \sqrt 2 t \; \vec i + \sqrt 2 t \; \vec j + (1 - t^2) \; \vec k, \tag 1$
we see that the points
$(0, 0, 1) = r(0), \tag 2$
and 
$(\sqrt 2, \sqrt 2, 0) = r(1); \tag 3$
so we see that $t$ varies 'twixt $0$ and $1$; then we have
$\dot r(t) = \sqrt 2 \; \vec i + \sqrt 2 \; \vec j -2t \; \vec k; \tag 4$
whence
$\vert \dot r(t) \vert = \sqrt{\sqrt 2^2 + \sqrt 2^2 + (-2t)^2} = \sqrt{2 + 2 + 4t^2} = \sqrt{4(1 + t^2)} = 2\sqrt{t^2 + 1}; \tag 5$
the increment of arc-length 'twixt $0$ and $1$ is thus
$\Delta s = \displaystyle 2\int_0^1 \sqrt{t^2 + 1} \; dt, \tag 6$
which I leave to my readers to finalize; the integral is in many tables.
It should of course be observed that we did in fact simply "solve for $t$" in discovering the limits of integration; for exampe, when the $\vec i$ component
$\sqrt 2 t = 0$, $t = 0$ follows.  And so forth and so on.
A: In this cases you have to parametrize your curve. so in this example:
$$x(t)=\sqrt{2} t$$
$$y(t)=\sqrt{2} t$$
$$z(t)=1-t^2$$
So the starting point of the curve is when $t=0$ and the end is $t=1$. These are the limits of integration. A lenght differential  is given by:
$$dr=\sqrt{dx^2+dy^2+dz^2}=\sqrt{ \left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2+\left( \frac{dz}{dt} \right)^2  }dt $$
$$dr=\sqrt{2+2+4t^2} \space dt=2\sqrt{1+t^2} \space dt$$
In general the lenght of a curve is given by:
$$\int_{t_1}^{t_2} \left|\frac{dr}{dt} \right| dt$$
So in this case the integral that reperesents the lenght is:
$$\int_{0}^{1} 2\sqrt{1+t^2} dt$$
