Tangent and Normal Vector - (.5). Principal Unit Normal Vector Let C be the curve traced out by the vector-valued function rt vector T t r r t t is the unit tangent vector to the curve C. Now define N t T t T t The vector N t is called the principal unit normal vector. Observe that N t is a unit vector N t is in the same direction as T t N t and T t are perpendicular. The reason is the following. It is known if vt c then. vt c vt vt c xt, yt, zt. We know the v t vt vt v t vt v t 0 vt v t 0 So, vt and v t are perpendicular. Since T t, T t T t 0. N t T t T t T t T t T t T t T t 0 N t dt t N t T t T t dt t, 0. N t will always point to the direction in which T t is turning as arc length increases and point to the concave side of the curve. Example () Let the helix C be traced out by rt cost, sint, t. a. Find the unit tangent and principal unit normal vectors to the curve at anytime t. b. Sketch the helix C and the unit tangent and principal unit normal vectors when t a. and t.
r t sint, cost,, r t 4sin t cos t T t 4sin t cos t sint,cost, 3sin t sint,cost, N t When t, T t 3sin t 3/ T t 3sin t 3/ 3sin t 3/ 3sin t 3/ 9sin t cos t 9sin t cos t sint 3sin t, cost, 3sin t 4cost, 5sint, 3 sint, 3sin t cos t 5sin t 9sin tcos t 9sin t cos t 3sin t 3/ 4cost, 5sint, 3 sint 4cost, 5sint, 3 sint T 5, 0,, N 5 0, 5, 0 0,, 0 When t, T 0,,, N 4 4, 0, 0, 0, 0 4 - -0.5 y 0 t 0.5 - - rt cost, sint, t, t /, -.- t
. Binormal Vector B t T t N t Note that B t is a unit vector. Since T t, N t, and the angle between T t and N t is, B t T t N t sin B t is perpendicular to both T t and N t by definition. TNB frame - a frame of reference formed by vectors T, N and B. basis of the 3-D space (as i, j and k form a basis of the 3-D space). Vectors T, N and B form a 3. Normal Plane and Osculating Plane The plane spanned by vectors N t and B t is called the normal plane. The plane spanned by T t and N t is called the osculating plane. Note that The normal vector to the normal plane ist t and the normal vector to the osculating plane is B t. The osculating circle (or the circle of curvature) is the circle of radius for 0 lying completely in the osculating plane. is called the radius of curvature and the center of curvature is the center of the osculating circle. For example: in a -D plane: http://poncelet.math.nthu.edu.tw/disk3/cabrijava/osc-ell.html in a 3-D space: http://www.lsus.edu/sc/math/rmabry/live3d/helix.htm Example () Find the osculating circle (the equation) of the curve traced out by rt t, t 3 at t. () rt t, t 3, r t t, 3t, r t, t, r,, r, 3, r r t 4t 9t 4, r 3 () Compute the unit tangent vector T t and its derivative T t : T t r r t t, 3t t, 3t t 4t 9t 4 t 4 9t 4 9t, 3t, t 0 T 3, 3, r 3 3
T t 4 9t 0, 3 8t 4 9t 3/, 3t 4 9t 3/ 4 9t 0,3 9t,3t 4 9t 3/ 8t, T 8,, 3/ 3 T 8 3 3/ 3 3 3/ 3 (3) Compute the curvature at t : T r 3 3 3 3, 3 3 (4) Compute the principal normal vector at t : N T T 3 3 8, 3/ 3 3, (5) The center of the osculating circle: w r N, 3 3 3 3,, 3 3,, 3 3,, 3 The equation of the circle of curvature: x y 3 3 3 3 or rt 3 cost, 3 3 sint,0 t 0 The graphs of the curve traced out by rt 8 and the osculating circle. 4 - -0-8 - -4-0 4 - Example Let the helix C be traced out by rt cost, sint, t. Find the radius and center of the osculating circle at t. From Example () we have: rt cost, sint, t, r t sint, cost,, r t 4sin t cos t 4
When t, T t 3sin t 3/ T t 3sin t 3/ N t 9sin t cos t r 4cost, 5sint, 3 sint 9sin t cos t 4cost, 5sint, 3 sint, 0, r, 0,, r 0,,, r T 0,,, T 3/ 4, 0, 0,0,0, T N, 0, 0 T r, radius of the osculating circle: The center of osculating circle: w r N, 0,, 0, 0, 0, 4. Tangential and Normal Components of Acceleration Suppose that the position of an object at the time t can be described by the end point of rt. Then the velocity and acceleration vectors of the object are vt and at, respectively. Recall that vectors T t and N t are orthogonal and form a basis for a -dimensional plane in space. Consider vt r t r t T t T t at v t d N t T t T t d s T t T t T t T t T t N t T t dt at d s T t T t d s T t N t Tangential component of acceleration: a T d s, depending on the rate of change of the speed Normal components of acceleration: a N square Compute a N without computing : 5, depending on the curvature and speed
Fact: if u and v are orthogonal then u v u v by the Pythegorean Theorem. Here T and N are orthogonal and are unit vectors ( T t and N t ). So, at at a N a N at a T since a N 0. Example Find tangential and normal components of acceleration for an object with position vector rt sint, cost, 4t. r t cost, sint, 4, r t cos t sin t 4 4 0 r t 0, d s 0, a T 0 b b a 4 0, a N 0 0 at N t The other way: at sint, cost, 0, at, a N 0 Applications: Force acting on a space curve: F t mat m d s T t m N t