ds dt ds 1 dt 1 dt v v v dt ds and the normal vector is given by N

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1 Normal Vectors and Curvature In the last section, we stated that reparameterization by arc length would help us analyze the twisting and turning of a curve. In this section, we ll see precisely how to do that. The first thing we need to recall, is that a parameterization by arc length yiel velocity vectors of constant length. In fact, we know that a curve that is parameterized by arc length will have unit velocity vectors. Using the same reasoning that we did to prove that spherical curves have tangent vectors orthogonal to their position vectors, we will show that curves parameterized by arc length will have velocity vectors orthogonal to the acceleration vectors. 3 Proof: Suppose ():[, s a b] is a space curve parameterized by arc length. Then 1 1 v v v is the unit tangent vector in the direction of the particle. By the Leibniz Rule for dot product, we know that T() s 1 T(), s T() s 1 2, T( s) 0, T ( s ) 0 We call the unit vector in the direction of, the normal vector, N. The length of the normal vector is caller the curvature and is denoted, κ (kappa). Curvature and Normal Vectors If rt () ( xt (), yt (), zt ()) is a smooth vector-valued function on [a,b], then the curvature of the path traced by r(t) is given by 1 and the normal vector is given by N Since the normal vector is perpendicular to the tangent vector, we need to ask ourselves in which direction is it pointing? Is it pointing away from the direction the curve is turning? Towar the direction of the turn? Or is it orthogonal to the turn and the curve simultaneously?

2 In order to answer this, we need to analyze what the derivative of the tangent vector is measuring? Remember that in single variable calculus, the derivative of a function with respect to time measure the rate of change of the function. In this case, we are taking the derivative of the tangent vector with respect to time, so we are measuring the rate of change of the direction of the tangent vector. T( s1) T( s2) Hence, the difference quotient of the tangent vector of the curve at time, s,, is the s1 s2 difference of the two tangent vectors, divided by the difference in time. The difference vector starts from the tail of T( s1) to the tail end of T( s2). If you draw both these vectors so that their initial point is at the origin, you will see that the limit of the difference vector is orthogonal to the tangent vector. You will further see that the length of the difference vector depen on the quickness of the turn of the tangent vectors. The picture above is the best illustration that I can find. However, as you can see, the velocity vectors are not unit vectors and, therefore, the curve is not parameterized by arc length. However, you can see, that the difference quotient of the velocity vectors is close to orthogonal, and, more importantly, is pointed in the direction that the curve is turning.

3 Formula for Calculating Curvatures If rt () ( xt (), yt (), zt ()) is a smooth curve, then the curvature is given by 1 v v Where T and t is a parameterization other than arc length. v Proof: Suppose r (t) is a smooth curve and s is a parameterization by arc length. Then by definition of curvature, we get 1 1 v Note, the set of equations above is very similar to the set of equations developing the construction of the normal vector. Example: Find the normal vector to the curve traced by rt () (cos( t),sin( t), t) and find the curvature as a function of t. Solution:

4 Example: Find the normal vector and the curvature of the curve r() (2, 2 t cos( t), 2sin( t) t ) for each t. Example: Find the normal vector and the curvature of the curve r() t ( t 2,2t 1,3) t for each t. Example: Find the normal vector and the curvature of the curve rt () (cos(),sin(),1) t t for each t.

5 Exercises: 1) Find T, N and for the plane curves below. a. rt () (, tln(cos( t)) [, ] 2 2 b. rt () ( ln(sec( t)), t) [, ] 2 2 2) Find T, N and for the space curves below. a. rt () ( 3 sin( t),3cos( t), 4t) b. rt () ( cos( t) tsin( t),sin( t) tcos(),3) t 2 3) Show that the parabola y ax has the largest curvature at its vertex and has no minimum curvature. 4) Show that an ellipse has its largest curvature at its major axis and smallest curvature at its minor axis. 5) In the examples above, we found that the curvature of a helix rt () ( acos( t), asin(), t bt) where a and b are non-negative. a) Find the curvature, κ, in terms of a and b. b) What is the largest value of κ can have for a given value of b?