Show that $c(t)_{c(t)}$ and the tangent vector to $c$ at $t$ is perpendicular. Let $c(t)=(c_1(t),...,c_n(t)):[0,1] \to \Bbb R^n$ be a differentiable curve in $\Bbb R^n$ such that $|c(t)|=1$ $\forall t \in [0,1]$. Define tangent vector $v$ of $c$ at $t$ as $c_*((e_1)_t)=((c_1)'(t),...,(c_n)'(t))_{c(t)}$. 

Show that $c(t)_{c(t)}$ and the tangent vector to $c$ at $t$ is perpendicular.

Now I was taking dot-product, $c(t).v=c_1(t)c_1'(t)+...+c_n(t)c_n'(t)$, so why it is zero? Can anyone help me with this arguement?


  Please don't tell $0=(T⋅T)′=T′⋅T+T⋅T′=2T⋅T′$.
I don't quite understand your last line. You know $|c(t)| = 1$ for all $t$, in other words $c(t) \cdot c(t) = 1$ for all $t$. The dot product is a bilinear map $\mu : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$; for such a bilinear map, if $f, g : [0,1] \to \mathbb{R}^n$ are two differentiable maps, then the derivative of $[0,1] \to \mathbb{R}$, $t \mapsto \mu(f(t), g(t))$ is given by 
$$\mu(f'(t), g(t)) + \mu(f(t), g'(t)).$$
I advise you to check this directly from the definition of the derivative and of bilinearity, if you're not convinced. Just compute $\frac{1}{h} \bigl( \mu(f(t+h), g(t+h)) - \mu(f(t), g(t)) \bigr)$ cleverly and take the limit as $h \to 0$.

So in your situation, the derivative of $t \mapsto c(t) \cdot c(t)$ is $c'(t) \cdot c(t) + c(t) \cdot c'(t) = 2 c(t) \cdot c'(t)$ by symmetry. But you know that $c(t) \cdot c(t) = 1$ for all $t$, so this derivative is zero and $c(t) \cdot c'(t) = 0$, i.e. $c(t)$ and $c'(t)$ are orthogonal.
I don't quite understand your last line. You know $|c(t)| = 1$ for all $t$, in other words $c(t) \cdot c(t) = 1$ for all $t$. The dot product is a bilinear map $\mu : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$; for such a bilinear map, if $f, g : [0,1] \to \mathbb{R}^n$ are two differentiable maps, then the derivative of $[0,1] \to \mathbb{R}$, $t \mapsto \mu(f(t), g(t))$ is given by 
$$\mu(f'(t), g(t)) + \mu(f(t), g'(t)).$$
I advise you to check this directly from the definition of the derivative and of bilinearity, if you're not convinced. Just compute $\frac{1}{h} \bigl( \mu(f(t+h), g(t+h)) - \mu(f(t), g(t)) \bigr)$ cleverly and take the limit as $h \to 0$.

So in your situation, the derivative of $t \mapsto c(t) \cdot c(t)$ is $c'(t) \cdot c(t) + c(t) \cdot c'(t) = 2 c(t) \cdot c'(t)$ by symmetry. But you know that $c(t) \cdot c(t) = 1$ for all $t$, so this derivative is zero and $c(t) \cdot c'(t) = 0$, i.e. $c(t)$ and $c'(t)$ are orthogonal.