If curve is given with $r=r(t)$ vector parametric equation, then equation of tangent line at point $M_0$ corresponding to $t_0$ parameter value is given by
$R(\tau)=r(t_0)+$ $\tau$ $r^\prime(t_0)$ where $\tau$ is parameter. Now I understand how we are getting this equation but have question concerning to normal line.
Line that is passing through a tangent point and is perpendicular to a tangent line is called normal line.
My book mentions that equation of normal will be $(R(\tau)-r(t_0))r^\prime (t_0)=0$.
Can you explain how this equation is derived?
The normal vector N(t) to a curve $r(t)$ is a vector that is perpendicular to the tangent vector $T(t) = r'(t)$. In two-dimensional curves, it is easy to find the normal vector by rotating the tangent vector by $90^\circ$ either clockwise or anti-clockwise. If T(t) = (T_x, T_y) then
$N(t) = (-T_y, T_x) $ by a rotation of $90^\circ$ anti-clockwise.
In three-dimensions, the normal vector any vector that perpendicular to the tangent vector $T(t) = r'(t)$, and there is not one vector or two vectors that are perpendicular to the tangent vector, there is an infinite number of normal vectors that lie in the plane that is perpendicular to the T(t) and passing through $r(t_0)$. That said, a normal vector that lies in the same plane as $(dT(t))$ (a differential of T(t)) is given by
$N(t) = \dfrac{d\hat{T}}{dt} $
where
$ \hat{T}(t) = \dfrac{ T(t) } { \| T(t) \| } $
is the unit vector along the tangent vector (i.e. it is the unit tangent vector).
As an example, let $r(t) = (t, t^2 , t^3) $
Then $T(t) = (1, 2 t, 3 t^2) $
and $\hat{T}(t) = \dfrac{ (1, 2t, 3 t^2) }{ \sqrt{ 1 + 4 t^2 + 9 t^4 } }$
The normal vector is
$ N(t) = \dfrac{d}{dt} \left( \dfrac{ (1, 2t, 3t^2) }{ \sqrt{1 + 4 t^2 + 9t^4} } \right) $
Differentiating using the product rule, this becomes
$ N(t) = \dfrac{ (0, 2, 6t ) }{\sqrt{1 + 4 t^2 + 9t^4}} + (1, 2t, 3t^2) \left( (-4 t - 18 t^3 ) ( 1 + 4 t^2 + 9 t^4 )^{-\frac{3}{2}} \right) $
Combining terms,
$N(t) = \dfrac{ (0, 2, 6 t) (1 + 4 t^2 + 9t^4) + (1, 2 t, 3 t^2)( -4 t - 18 t^3) } {(1 + 4 t^2 + 9 t^4 ) ^{\frac{3}{2} }} $
And this simplifies further to
$N(t) = \dfrac{( - 4 t - 18 t^3 , - 18 t^4 + 2 , 6 t + 12 t^3 )}{ ( 1 + 4 t^2 + 9 t^4) ^{\frac{3}{2}}} $
As a check, we can find the dot product between $N(t)$ and $T(t)$
$N(t) \cdot T(t) = \dfrac{( - 4 t - 18 t^3 - 36 t^5 + 4 t + 18 t^3 + 36 t^5 )}{(1 + 4 t^2 + 9 t^4)^{\frac{3}{2}}} = 0 $
So, indeed $N(t)$ is perpendicular to $T(t)$.
Finally, the parametric equation of the normal vector is
$ P(\tau) = r(t_0) + \tau N(t_0) $
Note that we can replace the complicated expression for $N(t)$ with simply $ ( - 2 t - 9 t^3 , - 9 t^4 + 1 , 3 t + 6 t^3 ) $ because $N(t)$ is a scalar multiple of it.