Find the time at which the particle has traveled $14$ units If a particle follows the path defined by $$r(t) = (2t^{3/2},2t+1, \sqrt{5} t )$$ and starts at $t=0$, at what time will the particle have traveled a total of $14$ units ?
 A: The formula is a mild generalization of the one for arclength in two dimensions. Take the derivatives, find the sum of their squares, take the square root, and integrate. 
We get "lucky," the sum of the squares of the derivatives is simple, it is $9t+9$. So if we travel from time $0$ to time $\tau$, the path length is
$$\int_0^\tau \sqrt{9t+9}\,dt.$$
Integrate, set the result equal to $14$, and solve for $\tau$. 
A: The distance $d(t)$ traveled between time $t=0$ and time $t=t$ is given by:
$$d^2(t)=|r(t)-r(0)|^2 = t(4t^2+4t+5)$$
Setting
$$d^2(t)=14^2$$ we obtain:
$$t(4t^2+4t+5)=14^2$$
The only real solution of this equation is given by:
$$t=\frac{1}{6}(-2-11\omega^{-1}+\omega)=3.25119...$$
$$\omega=\left(5329+18(87653)^{1/2}\right)^{1/3}$$
A: $$\vec{r}(t)=\left(2t^\frac32,2t+1,\sqrt{5}t\right)$$
You've got a vector of the position $r$ in terms of $t$. What you want to do next is find the velocity vector in terms of $t$, because "14 units" is the area swept by the magnitude of the velocity vector from $t=0$. This is done by simply differentiating:
$$\vec{v}(t)=\left(3t^\frac12,2,\sqrt 5\right)$$
Now you want to find the function for the magnitude of this vector. You probably know how to do that, for vector $\vec{A}=(a, b, c)$, the magnitude is just $\big|\vec{A}\big|=\sqrt{a^2+b^2+c^2}$. Hence:
$$\begin{align}\biggr|\vec{v}(t)\biggr|&=\sqrt{\left(3t^\frac12\right)^2+\left(2\right)^2+\left(\sqrt 5\right)^2} \\ &=\sqrt{9t+4+5}\end{align}$$
Now the distance covered is the integral of that. This is a magnitude so it won't be negative, but you have just velocity, then you'd take the absolute value of it (for distance traveled rather than displacement).
Hence we've got:
$$\int_{t=0}^{t=t_0} \sqrt{9t+9}=14$$
That evaluates to:
$$\frac{1}{13.5}(9t+9)^\frac32\Biggr|_{t=0}^{t=t_0}=14$$
You find that $t_0=3$
