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The normal and binormal vectors

We have seen that the derivative r ( t ) of a vector-valued function is a tangent vector to the curve defined by r ( t ) , and the unit tangent vector T ( t ) can be calculated by dividing r ( t ) by its magnitude. When studying motion in three dimensions, two other vectors are useful in describing the motion of a particle along a path in space: the principal unit normal vector    and the binormal vector    .

Definition

Let C be a three-dimensional smooth    curve represented by r over an open interval I. If T ( t ) 0 , then the principal unit normal vector at t is defined to be

N ( t ) = T ( t ) T ( t ) .

The binormal vector at t is defined as

B ( t ) = T ( t ) × N ( t ) ,

where T ( t ) is the unit tangent vector.

Note that, by definition, the binormal vector is orthogonal to both the unit tangent vector and the normal vector. Furthermore, B ( t ) is always a unit vector. This can be shown using the formula for the magnitude of a cross product

B ( t ) = T ( t ) × N ( t ) = T ( t ) N ( t ) sin θ ,

where θ is the angle between T ( t ) and N ( t ) . Since N ( t ) is the derivative of a unit vector, property (vii) of the derivative of a vector-valued function tells us that T ( t ) and N ( t ) are orthogonal to each other, so θ = π / 2 . Furthermore, they are both unit vectors, so their magnitude is 1. Therefore, T ( t ) N ( t ) sin θ = ( 1 ) ( 1 ) sin ( π / 2 ) = 1 and B ( t ) is a unit vector.

The principal unit normal vector can be challenging to calculate because the unit tangent vector involves a quotient, and this quotient often has a square root in the denominator. In the three-dimensional case, finding the cross product of the unit tangent vector and the unit normal vector can be even more cumbersome. Fortunately, we have alternative formulas for finding these two vectors, and they are presented in Motion in Space .

Finding the principal unit normal vector and binormal vector

For each of the following vector-valued functions, find the principal unit normal vector. Then, if possible, find the binormal vector.

  1. r ( t ) = 4 cos t i 4 sin t j
  2. r ( t ) = ( 6 t + 2 ) i + 5 t 2 j 8 t k
  1. This function describes a circle.
    This figure is the graph of a circle centered at the origin with radius of 2. The orientation of the circle is clockwise. It represents the vector-valued function r(t) = 4costi – 4 sintj.
    To find the principal unit normal vector, we first must find the unit tangent vector T ( t ) :
    T ( t ) = r ( t ) r ( t ) = −4 sin t i 4 cos t j ( −4 sin t ) 2 + ( −4 cos t ) 2 = −4 sin t i 4 cos t j 16 sin 2 t + 16 cos 2 t = −4 sin t i 4 cos t j 16 ( sin 2 t + cos 2 t ) = −4 sin t i 4 cos t j 4 = sin t i cos t j .

    Next, we use [link] :
    N ( t ) = T ( t ) T ( t ) = cos t i + sin t j ( cos t ) 2 + ( sin t ) 2 = cos t i + sin t j cos 2 t + sin 2 t = cos t i + sin t j .

    Notice that the unit tangent vector and the principal unit normal vector are orthogonal to each other for all values of t :
    T ( t ) · N ( t ) = sin t , cos t · cos t , sin t = sin t cos t cos t sin t = 0.

    Furthermore, the principal unit normal vector points toward the center of the circle from every point on the circle. Since r ( t ) defines a curve in two dimensions, we cannot calculate the binormal vector.
    This figure is the graph of a circle centered at the origin with radius of 2. The orientation of the circle is clockwise. It represents the vector-valued function r(t) = 4costi – 4 sintj. On the circle in the first quadrant is a vector pointing inward. It is labeled “principal unit normal vector”.
  2. This function looks like this:
    This figure is a curve in 3 dimensions. It is inside of a box. The box represents the first octant. The curve starts at the bottom right of the box and curves through the box in a parabolic curve to the top.
    To find the principal unit normal vector, we first find the unit tangent vector T ( t ) :
    T ( t ) = r ( t ) r ( t ) = 6 i + 10 t j 8 k 6 2 + ( 10 t ) 2 + ( −8 ) 2 = 6 i + 10 t j 8 k 36 + 100 t 2 + 64 = 6 i + 10 t j 8 k 100 ( t 2 + 1 ) = 3 i 5 t j 4 k 5 t 2 + 1 = 3 5 ( t 2 + 1 ) 1 / 2 i t ( t 2 + 1 ) 1 / 2 j 4 5 ( t 2 + 1 ) 1 / 2 k .

    Next, we calculate T ( t ) and T ( t ) :
    T ( t ) = 3 5 ( 1 2 ) ( t 2 + 1 ) 3 / 2 ( 2 t ) i ( ( t 2 + 1 ) 1 / 2 t ( 1 2 ) ( t 2 + 1 ) 3 / 2 ( 2 t ) ) j 4 5 ( 1 2 ) ( t 2 + 1 ) 3 / 2 ( 2 t ) k = 3 t 5 ( t 2 + 1 ) 3 / 2 i 1 ( t 2 + 1 ) 3 / 2 j + 4 t 5 ( t 2 + 1 ) 3 / 2 k T ( t ) = ( 3 t 5 ( t 2 + 1 ) 3 / 2 ) 2 + ( 1 ( t 2 + 1 ) 3 / 2 ) 2 + ( 4 t 5 ( t 2 + 1 ) 3 / 2 ) 2 = 9 t 2 25 ( t 2 + 1 ) 3 + 1 ( t 2 + 1 ) 3 + 16 t 2 25 ( t 2 + 1 ) 3 = 25 t 2 + 25 25 ( t 2 + 1 ) 3 = 1 ( t 2 + 1 ) 2 = 1 t 2 + 1 .

    Therefore, according to [link] :
    N ( t ) = T ( t ) T ( t ) = ( 3 t 5 ( t 2 + 1 ) 3 / 2 i 1 ( t 2 + 1 ) 3 / 2 j + 4 t 5 ( t 2 + 1 ) 3 / 2 k ) ( t 2 + 1 ) = 3 t 5 ( t 2 + 1 ) 1 / 2 i 5 5 ( t 2 + 1 ) 1 / 2 j + 4 t 5 ( t 2 + 1 ) 1 / 2 k = 3 t i + 5 j 4 t k 5 t 2 + 1 .

    Once again, the unit tangent vector and the principal unit normal vector are orthogonal to each other for all values of t :
    T ( t ) · N ( t ) = ( 3 i 5 t j 4 k 5 t 2 + 1 ) · ( 3 t i + 5 j 4 t k 5 t 2 + 1 ) = 3 ( −3 t ) 5 t ( −5 ) 4 ( 4 t ) 5 t 2 + 1 = −9 t + 25 t 16 t 5 t 2 + 1 = 0.

    Last, since r ( t ) represents a three-dimensional curve, we can calculate the binormal vector using [link] :
    B ( t ) = T ( t ) × N ( t ) = | i j k 3 5 t 2 + 1 5 t 5 t 2 + 1 4 5 t 2 + 1 3 t 5 t 2 + 1 5 5 t 2 + 1 4 t 5 t 2 + 1 | = ( ( 5 t 5 t 2 + 1 ) ( 4 t 5 t 2 + 1 ) ( 4 5 t 2 + 1 ) ( 5 5 t 2 + 1 ) ) i ( ( 3 5 t 2 + 1 ) ( 4 t 5 t 2 + 1 ) ( 4 5 t 2 + 1 ) ( 3 t 5 t 2 + 1 ) ) j + ( ( 3 5 t 2 + 1 ) ( 5 5 t 2 + 1 ) ( 5 t 5 t 2 + 1 ) ( 3 t 5 t 2 + 1 ) ) k = ( −20 t 2 20 25 ( t 2 + 1 ) ) i + ( −15 15 t 2 25 ( t 2 + 1 ) ) k = −20 ( t 2 + 1 25 ( t 2 + 1 ) ) i 15 ( t 2 + 1 25 ( t 2 + 1 ) ) k = 4 5 i 3 5 k .
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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