Prove that $ \overrightarrow{a}\left(t\right)||\overrightarrow{x}\left(t\right) $ in circular motion This may sound like a physics problem, but I got it in my homework in differential equations course, so Im asking here.
A particle moves on the standard sphere $ \mathbb{S}^{n-1}\subset\mathbb{R}^{n} $. I have to prove that $ \overrightarrow{v}\left(t\right)\perp\overrightarrow{x}\left(t\right) $, and given that $ |\overrightarrow{v}|=\text{const} $ I have to prove that $ \overrightarrow{a}\left(t\right)||\overrightarrow{x}\left(t\right) $.
The first part wasnt hard, if we note that $ 1=\overrightarrow{x}\cdot\overrightarrow{x} $ then $ 0=\left(\overrightarrow{x}\cdot\overrightarrow{x}\right)'=\overrightarrow{v}\cdot\overrightarrow{x}+\overrightarrow{x}\cdot\overrightarrow{v} $ and thus $ \overrightarrow{x}\cdot\overrightarrow{v}=0 $.
But Im not sure about the second part. If $|v|=const $, how can I show that $ a||x $ ?
I tried to write:
$ 0=\frac{d}{dt}\left(|v|^{2}\right)=\frac{d}{dt}\left(\sum_{k=1}^{n-1}v_{k}^{2}\left(t\right)\right)=\sum_{k=1}^{n-1}2v_{k}\left(t\right)a_{k}\left(t\right)=2\overrightarrow{v}\cdot\overrightarrow{a} $
Which shows that $ \overrightarrow{v}\perp\overrightarrow{a} $, But for general $ \mathbb{S}^{n-1} $ it does not promise that $ a||x $ as far as I understand.
Another way that I tried is to write:
$ 0=\left(\overrightarrow{x}\cdot\overrightarrow{v}\right)'=\overrightarrow{x}\cdot\overrightarrow{a}+|v|^{2} $
and thus $ \overrightarrow{x}\cdot\overrightarrow{a}=-|v|^{2} $
But Im not sure how to proceed or if it gives me anything...
Also, Im trying not to make any physical assumptions because its a mathenatical course, so a mathematical solution would be very helpful. Thanks in advance.
 A: You can re-parametrize any (sufficiently smooth) curve on the sphere so that its speed is constant, this is just a parametrization with some multiple of the curve length.
You only get the claim under additional assumptions, like if the particle moves on the sphere mechanically without the influence of any outside forces, only the holonomic constraint forces. Then the mechanical equation reads
$$
\ddot x =\lambda\nabla h(x)
$$
where $\lambda$ can depend on the time and $h(x)=0$ describes the constraint, here for example $h(x)=\frac12(\|x\|^2-1)$. That gives immediately $\nabla h(x)=x$ and thus $a=\ddot x$ parallel to $x$.
One can refine this by computing the time derivatives of $h(x(t))$ to get
$$
0=x·\dot x\\
0=\|\dot x\|^2+x·\ddot x=\|\dot x\|^2+\lambda\\
\dot\lambda=-2\dot x·\ddot x=-2\lambda\dot x·x=0.
$$
This gives more specifically that $\lambda=-\|v\|^2$ is constant, $\|v\|=c_v$, and $a=-c_v^2x$ which is of course still parallel to $x$.
A: Hint.
If $\|\vec v\| = c_0$ then $\|\vec v\|^2 = c_0^2$ and deriving $
\vec a\cdot \vec v = 0$ but if the movement is circular $\|\vec x\|= c_1$ or $\|\vec x\|^2= c_1^2$ and deriving $\vec v\cdot \vec x = 0$ so we arrived to
$$
\cases{\vec a\cdot \vec v = 0\\
\vec v\cdot \vec x = 0
}
$$
NOTE
Also from $\vec v\cdot \vec x = 0$ deriving again $\vec a\cdot \vec v + \|\vec v\|^2=0$ then $\vec a\cdot \vec v \le 0$
A general result can be obtained assuming the material point moves on the sphere without the influence of external forces. In this case the lagrangian can be written as
$$
L = \frac 12 m \|\dot{\vec x} \|^2-\lambda(\| \vec x \|^2-r^2)
$$
with $\lambda$ as a Lagrange multiplier so the movement equations follow
$$
m\ddot{\vec x}+\lambda\vec x = 0
$$
concluding that $\ddot{\vec x} = \vec a$ is parallel to $\vec x$ in the same way as @Lutz Lehmann's answer.
A: Without further restrictions on the motion, it will not be true in general that $\ \vec{a}\,\|
\,\vec{x}\ $. If $\ \vec{x}(t)=\frac{1}{\sqrt{2}}\big(\cos(t)\,\mathbf{i}+\sin(t)\,\mathbf{j}+\mathbf{k}\big)\ $ in $\ \mathbb{R}^3\ $, for instance, then the particle is moving on $\ \mathbb{S}^2\ $, $\ \vec{v}(t)=\frac{1}{\sqrt{2}}\big(-\sin(t)\,\mathbf{i}+\cos(t)\,\mathbf{j}\big)\ $, so $\ \left|\,\vec{v}(t)\,\right|=\frac{1}{\sqrt{2}}\ $ is constant, but $\ \vec{a}(t)=-\frac{1}{\sqrt{2}}\big(\cos(t)\,\mathbf{i}+\sin(t)\,\mathbf{j}\big)\ $ is not parallel to $\ \vec{x}(t)\ $.
Reply to question in comments by OP
If $\ \vec{a}(t)\,\|\,\vec{x}(t)\ $ for all $\ t\ $, then $\ \vec{a}(t)=\ddot{\vec{x}}(t)=\lambda(t)x(t)\ $ for some scalar function $\ \lambda(t)\ $ of $\ t\ $.  Let $\ \vec{u}_1, \vec{u}_2,\dots, \vec{u}_{n-2}\ $ be  a basis for the subspace of $\ \mathbb{R}^n\ $ perpendicular to $\ \vec{v}(0)\ $ and $\ \vec{x}(0)\ $. I'm assuming here that $\ \vec{v}(0)\ne\vec{0}\ $, so $\ \vec{v}(0)\ $ and $\ \vec{x}(0)\ $ must span a space of dimension $2$.
Let $\ y_j(t)=\vec{u}_j\cdot\vec{x}(t)\ $.  Then $\ y_j(t)\ $ must satisfy the linear differential equation $\ \ddot{y}_j(t)=\lambda(t)y_j(t)\ $, and the initial  conditions $\ y_j(0)=\dot{y}_j(0)=0\ $. It follows that $\ y_j(t)=0\ $ for all $\ t\ $, and hence $\ \vec{x}(t)\ $ must remain in the $2$-dimensional subspace spanned by $\ \vec{v}(0)\ $ and $\ \vec{x}(0)\ $.
Therefore, if $\ \vec{a}\,\|\,\vec{x}\ $, and $\ \left|\,\vec{v}\ \right|\ $ is a non-zero constant, then the only possible motions $\ \vec{x}\ $ can have must be of uniform speed around a circle centered on the origin.
A: Your hypotheses are:
$$\begin{cases}
\vec{x}(t) \cdot \vec{x}(t) & = 1\\
\vec{v}(t) \cdot \vec{v}(t) & = V^2  \neq 0 \text{(constant)}
\end{cases},$$
for all $t$.
Using them, you deduced that:
$$\begin{cases}
\vec{x}(t) \cdot \vec{v}(t) & = 0\\
\vec{a}(t) \cdot \vec{v}(t) & = 0\\
%\vec{a}(t) \cdot \vec{x}(t) &= -V^2
\end{cases},$$
for all $t$.
Using the last two equations, one gets:
$$(\vec{a}(t) + \vec{x}(t)) \cdot \vec{v}(t) = 0,$$
for all $t$.
More in general, given two real constants $m$ and $k$, we get:
$$(m\vec{a}(t) + k\vec{x}(t)) \cdot \vec{v}(t) = 0,$$
for all $t$, i.e. any linear combination of $\vec{a}(t)$ and $\vec{x}(t)$ is perpendicular to $v(t).$
Since the previous holds for all $t$, and hence for any value of $\vec{v}(t)$ satisfying the requirements (arbitrariety), we can conclude that:
$$m\vec{a}(t) + k\vec{x}(t) = \vec{0} \Rightarrow \vec{a}(t) = -\frac{k}{m}\vec{x}(t) \Rightarrow \vec{x}(t) \parallel \vec{a}(t).$$
