Tangent and Normal Vectors

Tangent and Normal Vectors MATH 311, Calculus III J. Robert Buchanan Department of Mathematics Fall 2011

Navigation When an observer is traveling along with a moving point, for example the passengers in an airplane, it can be useful to have a right-handed coordinate system travel with the observer.

Navigation When an observer is traveling along with a moving point, for example the passengers in an airplane, it can be useful to have a right-handed coordinate system travel with the observer. Q: How do we create a moving coordinate system?

Navigation When an observer is traveling along with a moving point, for example the passengers in an airplane, it can be useful to have a right-handed coordinate system travel with the observer. Q: How do we create a moving coordinate system? A: We already have the unit tangent vector T(t) = r (t) r (t) which is in the same direction as the motion. We need only two more vectors perpendicular to it.

Principal Unit Normal Vector Definition The principal unit normal vector N(t) is a unit vector having the same direction as T (t) and is defined by N(t) = T (t) T (t).

Principal Unit Normal Vector Definition The principal unit normal vector N(t) is a unit vector having the same direction as T (t) and is defined by Note: N(t) r (t) r (t) N(t) = T (t) T (t).

Direction of N(t) By the Chain Rule, T (t) = dt dt = dt ds ds dt so N(t) = T (t) T (t) = = = 1 κ dt ds dt ds dt ds dt ds ds dt dt ds ds dt (since r (t) = ds dt (since κ = dt ds ) > 0) dt points in the direction T is turning and thus N(t) always ds points to the concave side of the curve.

Illustration T N N N N T T N N T T

Example (1 of 2) Find the unit tangent and principal unit normal vectors for the following vector-valued function. r(t) = t, t 3

Example (1 of 2) Find the unit tangent and principal unit normal vectors for the following vector-valued function. r(t) = t, t 3 T(t) = r (t) r (t) = 1 1 + 9t 4 1, 3t2

Example (1 of 2) Find the unit tangent and principal unit normal vectors for the following vector-valued function. r(t) = t, t 3 T(t) = r (t) r (t) = 1 1 + 9t 4 1, 3t2 T (t) = 6t (1 + 9t 4 ) 3/2 3t2, 1

Example (1 of 2) Find the unit tangent and principal unit normal vectors for the following vector-valued function. r(t) = t, t 3 T(t) = r (t) r (t) = 1 1 + 9t 4 1, 3t2 T (t) = 6t (1 + 9t 4 ) 3/2 3t2, 1 N(t) = T (t) T (t) = 1 1 + 9t 4 3t2, 1 Remark: in V 2, vectors a, b and b, a are always orthogonal.

Example (2 of 2) Find the unit tangent and principal unit normal vectors for the following vector-valued function. r(t) = sin t, cos t, cos t

Example (2 of 2) Find the unit tangent and principal unit normal vectors for the following vector-valued function. r(t) = sin t, cos t, cos t T(t) = r (t) r (t) = 2 3 cos 2t cos t, sin t, sin t

Example (2 of 2) Find the unit tangent and principal unit normal vectors for the following vector-valued function. r(t) = sin t, cos t, cos t T(t) = r (t) r (t) = T (t) = 2 3 cos 2t cos t, sin t, sin t 2 2 2 sin t, cos t, cos t (3 cos 2t) 3/2

Example (2 of 2) Find the unit tangent and principal unit normal vectors for the following vector-valued function. r(t) = sin t, cos t, cos t T(t) = r (t) r (t) = 2 3 cos 2t cos t, sin t, sin t T (t) = 2 2 2 sin t, cos t, cos t (3 cos 2t) 3/2 N(t) = 1 2 sin t, cos t, cos t 3 cos 2t

Binormal Vector So far we have a pair of orthogonal vectors defined along the path of motion. We can get the third orthogonal vector from the cross product.

Binormal Vector So far we have a pair of orthogonal vectors defined along the path of motion. We can get the third orthogonal vector from the cross product. Definition The binormal vector B(t) is B(t) = T(t) N(t). Note: B(t) = T(t) N(t) = T(t) N(t) sin π 2 = 1.

Binormal Vector So far we have a pair of orthogonal vectors defined along the path of motion. We can get the third orthogonal vector from the cross product. Definition The binormal vector B(t) is B(t) = T(t) N(t). Note: B(t) = T(t) N(t) = T(t) N(t) sin π 2 = 1. Definition The triple of unit vectors T(t), N(t), and B(t) forms a moving frame of reference called the TNB frame or moving trihedral.

Illustration T T B T B N N B B N T N N N T T B

Example Find the binormal vector for the vector-valued function, r(t) = sin t, cos t, cos t

Example Find the binormal vector for the vector-valued function, We have already found: r(t) = sin t, cos t, cos t T(t) = 2 3 cos 2t cos t, sin t, sin t 1 N(t) = 2 sin t, cos t, cos t 3 cos 2t

Example Find the binormal vector for the vector-valued function, We have already found: r(t) = sin t, cos t, cos t T(t) = 2 3 cos 2t cos t, sin t, sin t 1 N(t) = 2 sin t, cos t, cos t 3 cos 2t B(t) = 1 2 0, 1, 1

Osculating Circle Definition For a curve C in the xy-plane, defined by the vector-valued function r(t), if the curvature κ 0 when t = t 0, we define the radius of curvature to be 1/κ. The center of curvature is the terminal point of the vector r(t) + 1 κn(t) (when κ 0). The osculating circle or circle of curvature is the circle whose center is the center of curvature and whose radius is the radius of curvature (when κ 0).

Illustration

Example Find the center of curvature and the radius of curvature for the osculating circle to the curve r(t) = t, t 3 when t = 1.

Example Find the center of curvature and the radius of curvature for the osculating circle to the curve r(t) = t, t 3 when t = 1. κ = N(t) = f (t) (1 + [f (t)] 2 ) 3/2 = 6 t (1 + 9t 4 ) 3/2 1 1 + 9t 4 3t2, 1

Example Find the center of curvature and the radius of curvature for the osculating circle to the curve r(t) = t, t 3 when t = 1. κ = N(t) = radius = f (t) (1 + [f (t)] 2 ) 3/2 = 6 t (1 + 9t 4 ) 3/2 1 1 + 9t 4 3t2, 1 1 κ(1) = 5 10 3

Example Find the center of curvature and the radius of curvature for the osculating circle to the curve r(t) = t, t 3 when t = 1. κ = N(t) = f (t) (1 + [f (t)] 2 ) 3/2 = 6 t (1 + 9t 4 ) 3/2 1 1 + 9t 4 3t2, 1 radius = 1 κ(1) = 5 10 3 center = r(1) + 1 κ(1) N(1) = ( 4, 8 ) 3

Tangential and Normal Components of Acceleration The unit tangent and principal unit normal vectors can explain the forces which work to stabilize and destabilize an object as it moves on a path.

Tangential and Normal Components of Acceleration The unit tangent and principal unit normal vectors can explain the forces which work to stabilize and destabilize an object as it moves on a path. The velocity of an object moving along a path described by r(t) is v(t) = r (t) = r (t) T(t) = ds dt T(t).

Tangential and Normal Components of Acceleration The unit tangent and principal unit normal vectors can explain the forces which work to stabilize and destabilize an object as it moves on a path. The velocity of an object moving along a path described by r(t) is v(t) = r (t) = r (t) T(t) = ds dt T(t). The acceleration is given by a(t) = d dt ( ) ds dt T(t) = d 2 s ds T(t) + dt2 dt T (t).

Tangential and Normal Components (continued) a(t) = d 2 s ds T(t) + dt2 dt T (t) = d 2 s ds T(t) + dt2 ds T(t) + 2 = d 2 s dt = d 2 s dt 2 T(t) + = d 2 s dt 2 T(t) + κ dt T (t) N(t) dt ds dt ds dt N(t) ( ) ds 2 dt dt ds N(t) ( ) ds 2 N(t) dt

Tangential and Normal Components (continued) κ ( ds dt a(t) = d 2 ( ) s ds 2 dt 2 T(t) + κ N(t) dt d 2 s : is called the tangential component of dt2 acceleration and is denoted a T = d 2 s dt 2. ) 2 : is called the normal component of acceleration and is denoted a N = κ ( ) ds 2. dt

Tangential and Normal Components (continued) κ ( ds dt a(t) = d 2 ( ) s ds 2 dt 2 T(t) + κ N(t) dt d 2 s : is called the tangential component of dt2 acceleration and is denoted a T = d 2 s dt 2. ) 2 : is called the normal component of acceleration and is denoted a N = κ By Newton s Second Law of Motion ( ) ds 2. dt F(t) = ma(t) = m d 2 ( ) s ds 2 T(t) + mκ N(t). dt2 dt

Example (1 of 2) Express the acceleration of the object traveling along the path described by r(t) = cos t, sin t, sin 3t in terms of the tangential and normal components.

Example (1 of 2) Express the acceleration of the object traveling along the path described by r(t) = cos t, sin t, sin 3t in terms of the tangential and normal components. Strategy: Find ds dt = r (t). Find a T = d 2 s dt 2 = d dt r (t). Since a(t) 2 = a 2 T + a2 N then find a N = a(t) 2 a 2 T.

Example (2 of 2) r(t) = cos t, sin t, sin 3t

Example (2 of 2) r(t) = cos t, sin t, sin 3t Steps: ds dt = 11 2 + 9 cos 6t 2

Example (2 of 2) Steps: ds 11 = dt 2 + 9 cos 6t 2 27 sin 6t a T =. 11 2 2 + 9 2 cos 6t r(t) = cos t, sin t, sin 3t

Example (2 of 2) Steps: ds 11 = dt 2 + 9 cos 6t 2 27 sin 6t a T =. 11 2 2 + 9 2 cos 6t 23 18 cos 6t a N = 2 11 + 9 cos 6t r(t) = cos t, sin t, sin 3t

Illustration 1.0 y 0.5 0.0 0.5 1.0 1.0 0.5 z 0.0 20 10 0.5 1 2 3 4 5 6 1.0 1.0 10 20 0.5 0.0 x 0.5 1.0 Tangential components of acceleration are shown in green. Normal components of acceleration are shown in red.

Application: Finding Curvature a(t) = d 2 s dt 2 T(t) + κ a(t) T(t) = d 2 s T(t) T(t) dt2 }{{} ( ) ds 2 a(t) T(t) = κ dt r (t) r (t) r (t) = κ r (t) 2 r (t) r (t) r (t) 3 = κ =0 ( ) ds 2 N(t) dt +κ ( ) ds 2 N(t) T(t) dt }{{} = B(t)

Application: Finding the Binormal Vector a(t) = d 2 ( ) s ds 2 dt 2 T(t) + κ N(t) dt ( ) ds 2 T(t) a(t) = κ B(t) dt r (t) r (t) r = r (t) r (t) (t) r (t) 3 r (t) 2 B(t) r (t) r (t) r (t) r = B(t) (t)

Kepler s Laws of Planetary Motion 1 Each planet follows an elliptical orbit, with the sun at one focus. 2 The line segment joining the sun to a planet sweeps out equal areas in equal times. 3 If T is the time required for a given planet to make one orbit of the sun and if the length of the major axes of its elliptical orbit is 2a, then T 2 = ka 3 for some constant k.

Kepler s Laws of Planetary Motion 1 Each planet follows an elliptical orbit, with the sun at one focus. 2 The line segment joining the sun to a planet sweeps out equal areas in equal times. 3 If T is the time required for a given planet to make one orbit of the sun and if the length of the major axes of its elliptical orbit is 2a, then T 2 = ka 3 for some constant k. We can derive Kepler s three laws from two of Newton s Laws of Motion.

Assumptions 1 The sun is at the origin and the planet is at the terminal point of vector r(t).

Assumptions 1 The sun is at the origin and the planet is at the terminal point of vector r(t). 2 The gravitational force of attraction has a magnitude which is proportional to the inverse of the square of the distance separating the sun and the planet.

Assumptions 1 The sun is at the origin and the planet is at the terminal point of vector r(t). 2 The gravitational force of attraction has a magnitude which is proportional to the inverse of the square of the distance separating the sun and the planet. F(t) = ma(t) (Newton s 2nd Law) = GmM r(t) r(t) 2 r(t)

Kepler s Laws (continued) Let r = r(t) and note that r(t) is a unit vector, call it u. r(t) which implies F = ma = GmM r 2 u

Kepler s Laws (continued) Let r = r(t) and note that which implies r(t) is a unit vector, call it u. r(t) F = ma = GmM r 2 u a = GM r 2 u. Note: this says the acceleration is always in the opposite direction of r(t), toward the sun.

Kepler s Laws (continued) Consider d dt (r(t) v(t)) = r (t) v(t) + r(t) v (t) = v(t) v(t) + r(t) a(t) = 0 + 0 = 0

Kepler s Laws (continued) Consider d dt (r(t) v(t)) = r (t) v(t) + r(t) v (t) = v(t) v(t) + r(t) a(t) = 0 + 0 = 0 Hence r v = c a constant vector. This implies the orbit of the planet must lie in plane.

Kepler s First Law 1 Each planet follows an elliptical orbit, with the sun at one focus.

Kepler s First Law 1 Each planet follows an elliptical orbit, with the sun at one focus. Assume the planet s orbit is in the xy-plane so that c is parallel to the z-axis.

Kepler s First Law 1 Each planet follows an elliptical orbit, with the sun at one focus. Assume the planet s orbit is in the xy-plane so that c is parallel to the z-axis. Since r(t) = ru, then v(t) = r (t) = d dr dt (ru) = dt u + r du dt

Kepler s First Law (continued) c = r v ( dr = r dt u + r du ) dt = r dr dt u + r r du dt = dr ( dt r u + r 2 = r 2 ( u du dt ) u du dt )

Kepler s First Law (continued) ( a c = a r 2 u du dt = GM r 2 u r 2 = GMu ) ( u du dt ) ( u du dt = GM u du }{{ dt u (u }{{} u ) du } dt =1 =0 = GM du dt )

Kepler s First Law (continued) GM du dt = a c = dv dt c = d (v c) (since c is constant) dt GMu + b = v c (where b is constant) Note: since v c is orthogonal to c then v c is a vector in the xy-plane. Vector u is in the xy-plane, so b must also be in the xy-plane. We can choose b to be parallel to i.

Kepler s First Law (continued) c 2 = c c = (r v) c = r (v c) = ru (GMu + b) = rgm u }{{} u +ru b =1 = rgm + r b cos θ

Kepler s First Law (continued) c 2 = c c = (r v) c = r (v c) = ru (GMu + b) = rgm u }{{} u +ru b =1 = rgm + r b cos θ Remark: Angle θ is the angle between r and i.

Kepler s First Law (continued) Letting c = c and b = b we may solve for r: r = = c 2 GM + b cos θ ed 1 + e cos θ where e = b GM and d = c2 b.

Kepler s First Law (continued) Letting c = c and b = b we may solve for r: r = = c 2 GM + b cos θ ed 1 + e cos θ where e = b GM and d = c2 b. This is the equation of an ellipse in polar coordinates!

Homework Read Section 11.5. Exercises: 1 29 odd, 39, 40