Answer
(a) The unit vector is perpendicular to the tangent vector of the curve , so a formula for the unit vector is $N(t)=\frac{T'(t)}{|T'(t)|}$. The binormal vector is perpendicular to both $T(t)$ and N(t) thus, it is the cross product of the two vectors: $B(t)=T(t) \times N(t)$
(b) The normal plane of a curve $C$ is the plane formed by the normal and binormal vectors $N$ and $B$ at a point $P$ on $C$ . The oscillating plane of a curve C is the plane formed by the unit normal vectors $T$ and $N$ , this is the plane that comes closet to containing the part of the curve near $P$. The oscillating circle a of a curve C at P is the circle that lies in the oscillating plane of a curve C that has the same tangent as C at P and the radius $ \rho=\frac{1}{\kappa}$.
Work Step by Step
(a) The unit vector is perpendicular to the tangent vector of the curve , so a formula for the unit vector is $N(t)=\frac{T'(t)}{|T'(t)|}$. The binormal vector is perpendicular to both $T(t)$ and N(t) thus, it is the cross product of the two vectors: $B(t)=T(t) \times N(t)$
(b) The normal plane of a curve $C$ is the plane formed by the normal and binormal vectors $N$ and $B$ at a point $P$ on $C$ . The oscillating plane of a curve C is the plane formed by the unit normal vectors $T$ and $N$ , this is the plane that comes closet to containing the part of the curve near $P$. The oscillating circle a of a curve C at P is the circle that lies in the oscillating plane of a curve C that has the same tangent as C at P and the radius $ \rho=\frac{1}{\kappa}$.
