Show that,if $T$ represent the unit tangent vector then $\frac {dT}{ds}$is perpendicular to $T$. In the book written by R.M.Khan it is given that since $T$ is unit modulus $T\frac {dT}{ds}=0$ implies $T$ and $\frac {dT}{ds}$ are perpendicular to each other. 
But $\frac {dT}{ds}$ is not a proper vector or in other words a null vector; how can the result be obtained?
Please help to make me understand what actually the statement means.
 A: it means that the original curve, call it $\gamma(s),$ is required to be arc length parametrized, that is $T = \gamma'$ is always of length one, or $T \cdot T = 1.$ The product rule says
$$(V \cdot W)' = V' \cdot W + V \cdot W'.$$
So $$ 0 = (T \cdot T)' =   T' \cdot T + T \cdot T' = 2 T \cdot T'. $$
So $$ T \cdot T' = 0. $$
Whenever $T' \neq 0,$ one defines a scalar function $\kappa$ and another vector field $N$ with $N \cdot N = 1$ by 
$$  T' = \kappa N $$
with some care needed when $T'=0.$
What are $T, N, \kappa$ when
$$ \gamma(s) = (5 \cos s, 5 \sin s)  $$
in the plane?
A: The derivative of $s \mapsto \Vert T(s) \Vert^2$ is $2 T.\frac{dT}{ds}$.
As $\Vert T \Vert^2=1$ is constant, the derivative is equal to zero which leads to the result.
A: Suppose vector a is of constant magnitude that is it has a scalar magnitude of varying direction da/dt=0.
So we relate this idea with vector T with unit modulus that is dT/ds=0.
We can perform a scalar product operation to ensure the vectors are perpendicular. 
T.dT/ds=|T||dT/ds|cos©
Hence |T||dT/ds|cos©=0
   cos©=0 /|T||dT/ds|
   cos©=0
   ©=arccos(0)
   ©=90°
This shows that vector T is perpendicular to dT/ds. 
A: $\mathbf{T} = (T_x,T_y,T_z)$ is the tangent vector to the curve by construction. If we're talking about the motion of a particle along a curve, this points in the direction of the velocity.
$\frac{d\mathbf{T}}{ds} = (\frac{dT_x}{ds},\frac{dT_y}{ds},\frac{dT_z}{ds})$ is how the tangent vector changes as we progress along the curve. This is akin to an acceleration vector. It points inwards to the curvature of the motion.
If $T^2 = c$ where $c$ represents any constant length, the derivative with respect to ds will reveal that $T$ and $\frac{d\mathbf{T}}{ds}$ are orthogonal.
$$T^2 = c$$
$$\frac{d}{ds}T^2 = \frac{d}{ds}c$$
$$\frac{d}{ds}(\mathbf{T}\cdot\mathbf{T}) = 0$$
$$\frac{d\mathbf{T}}{ds}\cdot\mathbf{T} + \mathbf{T}\cdot\frac{d\mathbf{T}}{ds}= 0$$
Note that we applied the product rule to the LHS; the derivative is NOT simply $2T$. 
$$2\frac{d\mathbf{T}}{ds}\cdot\mathbf{T} = 0$$
$$\frac{d\mathbf{T}}{ds}\cdot\mathbf{T} = 0$$
From the definition of the dot product, $\mathbf{a}\cdot\mathbf{b} = ab\cos\theta$, two vectors are orthogonal when the RHS vanishes. Hence our $\mathbf{T}$ and $\frac{d\mathbf{T}}{ds}$ are orthogonal.
Note that this result is not exclusive to the tangent vector. It's true of any vector $\mathbf{v}$ where $v^2 = c$.
