Prove that a curve lies on a sphere with center the origin. The problem is as follows: "If a curve has the property that the position vector $\vec{r}(t)$ is always perpendicular to the tangent vector $\vec{r}'(t)$, show that the curve lies on a sphere with center the origin" (Stewart 13.2.50).
I have the following proof:
Let $$\vec{r}(t) = \langle x(t), y(t), z(t) \rangle$$ and $$\vec{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle.$$
We know that $\vec{r}(t) \cdot \vec{r}'(t) = 0$. Therefore,
$$x(t)x'(t) + y(t)y'(t) + z(t)z'(t) = 0.$$
We can then integrate both sides.
$$\int x(t)x'(t) + y(t)y'(t) + z(t)z'(t)dt = \int 0dt$$
$$\frac{x(t)^2}{2} + \frac{y(t)^2}{2} + \frac{z(t)^2}{2} = C_1$$
Multiplying both sides by two then gives us the equation of a sphere centered at the origin, proving that the curve lies on a sphere with center the origin:
$$x(t)^2 + y(t)^2 + z(t)^2 = C_2.$$
(This solution is adapted from the solution found here.)
This does, however, leave me with a question. Why do we integrate the expression and how does it work? What is the connection between the dot product of $\vec{r}(t)$ and $\vec{r}'(t)$ and its integral to proving $\vec{r}(t)$'s being a sphere centered at the origin? Thanks in advance.
 A: If a curve lies on a sphere then its position vector $(x,y,z)$ is characterised by $x^2+y^2+z^2=r^2$. If we differentiate this with respect to the curve parameter $t$, we get $x(t)x'(t)+y(t)y'(t)+z(t)z'(t)=0$, so we have that any curve which is on a sphere has a tangent which is always perpendicular to the radius vector. Suppose we have a curve whose tangent is always perpendicular to the radius vector, $\tau\cdot r=0$ then the integral of this expression $\int r'(t)\cdot r(t)dt=\sum r^2(t)=C$ gives us $r^2=C$ so any curve which is perpendicular to the radius must lie on a sphere.
Intuitively, if you have a tangent which is at right angles to a line to the origin then your motion is neither towards or away from the origin. When you go out for a walk or a drive you go towards the horizon but neither go up into space nor down into a mine shaft, but always perpendicular to the earth's surface, so your path is on the earth's surface, regardless of what kind of curve you trace.
A: We have
$$
\begin{aligned}
\textbf{r}(t)\cdot\textbf{r}'(t)&=0 \\
\implies 2\textbf{r}(t)\cdot\textbf{r}'(t)&=0 \\
\implies \frac{d}{dt}(\textbf{r}(t)\cdot\textbf{r}(t))=0 \\
\implies \frac{d}{dt}|\textbf{r}|^2=0 \\
\implies |\textbf{r}|=\text{constant}
\end{aligned}
$$
This is saying the point represented by $\textbf{r}(t)$ is at a constant distance from the origin. Guess what such a set of points looks like...
