How to find the time needed to traverse the curve $r(t)=(\cos t, \sin t, t)$ from the origin to $8$ meters?

Question

Given the curve $r(t)=(\cos t, \sin t, t)$ and that a body starts moving from $t=0$, after how much time will the body be positioned at the distance of 8 meters from the axis origin $(0,0,0)$?

One can think of 2 points in 3d space: the first point $(0,0,0)$ and the second point $(x,y,z)$ which is the point where the body will be situated after $8$ meters. Then I could find the distance like this: $$ \sqrt{(x-0)^2+(y-0)^2+(z-0)^2}=8 $$

But not only is this an equation with 3 unknowns, I can't find the $t$.

Answer 1

Write $x(t), y(t), z(t)$ for $x,y,z$; these are the three coordinate functions of $r$, i.e., $x(t) = \cos t$. Then go from there.

Answer 2

$|r(t)| = 8 \Leftrightarrow \sqrt{(x(t)-0)^2+(y(t)-0)^2+(z(t)-0)^2}=8 \Leftrightarrow \sqrt{\cos^2(t) + \sin^2(t) + t^2} = 8 \Leftrightarrow \sqrt{1+t^2} = 8 \Leftrightarrow 1 + t^2 = 64 \Leftrightarrow t = \sqrt{63}=3\sqrt{7}$

We reject the mathematically correct solution $t = -\sqrt{63}=-3\sqrt{7}$ because time cannot be negative.

Answer 3

The point lies on a sphere of radius $8$ units, on a top circle cut by a cylinder(of helix) of radius $1$ and cone lateral length (vertex to top circle) $8$ units.

$$ z^2 + 1^2 = 8^2; \, t^2 + 1^2 = 8^2; \, \rightarrow t= \sqrt {63}. $$