2.5 Tangent, Normal and Binormal Vectors
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1 .5. TANGENT, NORMAL AND BINORMAL VECTORS 95.5 Tangent, Normal and Binormal Vectors Three vectors play an important role when studying the motion of an object along a space curve. These vectors are the unit tangent vector, the principal normal vector and the binormal vector. We have already defined the unit tangent vector. In this section, we define the other two vectors. Let us start by reviewing the definition of the unit tangent vector. Definition 59 Unit Tangent Vector) Let C be a smooth curve with position vector r t). The unit tangent vector, denoted T t) is defined to be T t) r t) r t).5. Normal and Binormal Vectors Definition 60 Normal Vector) Let C be a smooth curve with position vector r t). The principal unit normal vector or simply the normal vector, denoted N t) is defined to be: T t) N t) T t).5) The name of this vector suggests that it is normal to something, the question is to what? By definition, T is a unit vector, that is T t). From proposition 5, it follows that T t) T t). Thus, N t) T t). In fact, N t) is a unit vector, perpendicular to T pointing in the direction where the curve is bending. Proposition 6 Let C be a smooth curve with position vector r t). Then, Proof. We know that N t) T t) κ r t) hence T t) N t) κ r t) d T κ ds T t) T and κ t) T t) r t). So, we have T t) T t) T t) κ r t) T t) κ r t)
2 96 CHAPTER. VECTOR FUNCTIONS Earlier, we sat that d T ds T t) r t) hence the second equality. Definition 6 Binormal Vector) Let C be a smooth curve with position vector r t). The binormal vector, denoted B t), is defined to be B t) T t) N t) Since both T t) and N t) are unit vectors and perpendicular, it follows that B t) is also a unit vector. It is perpendicular to both T t) and N t). Example 63 Consider the circular helix r t) cos t, sin t, t. Find the unit tangent, normal and binormal vectors. Unit Tangent: Since T t) r t). r t) r t), we need to compute r t) and r t) sin t, cos t, and Thus r t) sin t + cos t + sin t T t), cos t, Normal: Since T t) N t) T, we need to compute T t) and T t). t) cos t T t), sin t, 0 and T t) cos t + sin t Thus N t) cos t, sin t, 0
3 .5. TANGENT, NORMAL AND BINORMAL VECTORS 97 Figure.7: Helix and the vectors T 0), N 0) and B 0) Binormal: B t) T t) N t) sin t, cos t, B t) sin t, cos t, cos t, sin t, 0 The pictures below figures.7,.8 and.9) show the helix for t [0, π] as well as the three vectors T t), N t) and B t) plotted for various values of t. If the three vectors do not appear to be exactly orthogonal, it is because the scale is not the same in the x, y and z directions..5. Osculating and Normal Planes Definition 64 Osculating and Normal Planes) Let C be a smooth curve with position vector r t). Let P be a point on the curve corresponding to r t 0 ) for some value of t.. The plane through P determined by N t 0 ) and B t 0 ) is called the normal plane of C at P. Note that its normal will be T t).. The plane through P determined by T t 0 ) and N t 0 ) is called the osculating plane of C at P. Note that its normal will be B t).
4 98 CHAPTER. VECTOR FUNCTIONS Figure.8: Helix and the vectors T ), N ) and B ) Figure.9: Helix and the vectors T 4), N 4) and B 4)
5 .5. TANGENT, NORMAL AND BINORMAL VECTORS The osculating circle or the circle of curvature at P is the circle which has the following properties: a) It lies on the osculating plane. b) Has the same tangent at P as C. c) Its radius is κ where κ is computed at P. d) Lies on the side of C where N is pointing. Example 65 Find the normal and osculating planes to the helix given by r t) cos t, sin t, t at the point 0,, π ). Earlier, we found that sin t T t), cos t, and At the point N t) cos t, sin t, 0 sin t B t), cos t, 0,, π ), that is when t π, we have π ) T, 0, π ) N 0,, 0 and π ) B, 0, Normal Plane: It is the plane through, 0,. Thus its equation is 0,, π ) with normal π ) T x 0) + z π ) 0 Multiplying each side by gives or x 0) + z π ) 0 z x π
6 00 CHAPTER. VECTOR FUNCTIONS Osculating Plane: It is the plane through, 0, Thus its equation is x 0) + z π ) 0 0,, π ) with normal π ) B Multiplying each side by gives or x 0) + z π ) 0 x + z π Example 66 Find and graph the osculating circle for the parabola y x at the origin. We need to find r t), N t) and κ. Recall that r t) t, t r t), t At the origin, t 0 hence r 0), 0 i. Using formula.4, we see that κ x) + 4x ) 3 Hence κ 0). Also, N t) T t) T. t) r t) T t) r t) + 4t, t + 4t Thus, T t) 8t + 4t ), + 4t t 8t) ) + 4t ) 8t + 4t ), 8t + 4t )
7 .5. TANGENT, NORMAL AND BINORMAL VECTORS 0 and Thus So, Thus T t) + 4t ) 64t t 4 3t t ) 64t4 + 3t 4 4t + ) + 4t ) + 4t ) N t) 8t +4t ) +4t ), 8t +4t ) +4t ) 4t + 4t, 4t + 4t N 0) 0, j Hence, the osculating circle is a circle of radius. Its center is units from the origin, in the direction of j hence the center is 0, ). Thus, the circle is x + y ) 4. The graph of y x and its osculating circle at the origin are shown in figure 66. y Osculating Circle for y x at 0, 0) x
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