Differential geometry - For any vector in $R^n$ there exists a velocity vector $\gamma\prime(t)$ of closed curve $\gamma$ which is orthogonal to it I have the following problem.
Suppose that I have smooth (regular) closed curve $\gamma: [a,b]\rightarrow\mathbb R^n$.
I need to prove that for every vector $\phi\in\mathbb R^n$ there exists point $t\in[a,b]$ such as the value of the velocity speed vector at this point ($\gamma\prime(t))$ is orthogonal to this vector.
My approach (Intuition) was to assume that if it isn't true then for any vector $\gamma=(r_1,...,r_n)$ and for any point $t$ the velocity speed vector at this point (i.e. derivative vector) isn't orthogonal to this vector so the curve isn't turning back to the origin point $\gamma(a)$ and therefore it will not be closed eventually (at point $b$) and this will give me a contradiction to the fact that this is a closed curve, But I'm really struggling with how to convert my approach to formal proof.
 A: I assume
$\phi \ne 0, \tag 0$
since the notion of orthogonality is at best trivial and at worst meaningless when (0) fails.
Consider the function
$f(t) = \phi \cdot \gamma(t); \tag 1$
since $\gamma(t)$ is a closed regular curve on $[a, b]$, a compact interval, the function $f(t)$ attains its extremal values on $[a, b]$; at such a
$\tau \in [a, b] \tag 2$
we have
$\phi  \cdot \gamma'(\tau) = f'(\tau) = 0; \tag 3$
but this is precisely the assertion that $\gamma'(\tau)$ is orthogonal to $\phi$. $OE\Delta$
Note Added in Edit, Saturday 7 November 2020 12:47 PM PST:  The geometry of this situation may perhaps be further illuminated bye scrutinizing the meaning of $f(t)$: if we write $\phi$ in polar form:
$\phi = \vert \phi \vert \mathbf e_\phi, \tag 4$
where
$\mathbf e_\phi \in S^{n - 1} \tag 5$
is the unit vector in the direction of $\phi$, then $f(t)$ becomes
$f(t)= \vert \phi \vert \mathbf e_\phi \cdot \gamma(t); \tag 6$
aside from the factor of $\vert \phi \vert$, the quantity $f(t)$ is the component of $\gamma(t) \in \Bbb R^n$ along $\mathbf e_\phi$; $\vert \phi \vert > 0$ may then be seen as an arbitrary (constant) scaling factor, and there is no loss of information by taking $\vert \phi \vert = 1$; then
$f(t) = \mathbf e_\phi \cdot \gamma(t) \tag 7$
is of itself precisely the component of $\gamma(t)$ along $\mathbf e_\phi$, and this component is extremal at $\tau$ such that
$f'(\tau) = \mathbf e_\phi \cdot \gamma'(\tau) = 0, \tag 8$
that is, where $\gamma'(\tau)$ is normal to $\mathbf e_\phi$; hence we may think of $\gamma'(\tau)$ as tangent to or lying in the hyperplane normal to $\mathbf e_\phi$ passing through $\gamma(\tau)$.
We may in fact find the second derivitive of $f(t)$:
$f''(t) = \mathbf e_\phi \cdot \gamma''(t), \tag9$
which of course provides information aas to the nature of an extremum $\tau$; when
$f''(\tau) > 0, \tag{10}$
then $\tau$ is a local minumum, and so forth and so on..
A more geometric approach is had by parametrizing $\gamma(s)$ by its arc-length $s$; then
$\dot \gamma(s) = T(s), \tag{11}$
the unit tangent vector field to $\gamma(s)$.  Then
$f'(s) = \mathbf e_\phi \cdot \dot \gamma(s) = \mathbf e_\phi \cdot T(s); \tag{12}$
proceeding in this direction, we have
$f''(s) = \mathbf e_\phi \cdot \ddot \gamma(s) = \mathbf e_\phi \cdot \dot T(s) = \mathbf e_\phi \cdot \kappa(s) N(s), \tag{13}$
where we have used the Frenet-Serret relation
$\dot T(s) = \kappa(s) N{s}, \tag{14}$
where
$\kappa(s) > 0 \tag{15}$
is the curvature, and $N(s)$ is the unit normal vector to $\gamma(s)$.  In light of (15) we see that the sign of $f''(s)$ is given by $\mathbf e_\phi \cdot N(s)$, that is,
$f''(s) > 0 \tag{16}$
when $\gamma(s)$ is turning toward the positive side of the hyperplane normal to $\mathbf e_\phi$, which side is in fact characterized by the equation
$X \cdot \mathbf e_\phi - Y \cdot \mathbf e_\phi = (X - Y) \cdot \mathbf e_\phi >  0, \tag{17}$
where $Y \in \Bbb R^n$ lies in the hyperplane itself and $X \in \Bbb R^n$ on the positive side.  (Under these hypotheses on $X$ and $Y$,
End of Note.
