This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 500 C H A P T E R 14 CALCULUS OF VECTORVALUED FUNCTIONS (ET CHAPTER 13) 14.4 Curvature (ET Section 13.4) Preliminary Questions 1. What is the unit tangent vector of a line with direction vector v = 2 , 1 , 2 ? SOLUTION A line with direction vector v has the parametrization: r ( t ) = O P + t v hence, since O P and v are constant vectors, we have: r ( t ) = v Therefore, since v = 3, the unit tangent vector is: T ( t ) = r ( t ) r ( t ) = v v = 2 / 3 , 1 / 3 , 2 / 3 2. What is the curvature of a circle of radius 4? SOLUTION The curvature of a circle of radius R is 1 R , hence the curvature of a circle of radius 4 is 1 4 . 3. Which has larger curvature, a circle of radius 2 or a circle of radius 4? SOLUTION The curvature of a circle of radius 2 is 1 2 , and it is larger than the curvature of a circle of radius 4, which is 1 4 . 4. What is the curvature of r ( t ) = 2 + 3 t , 7 t , 5 t ? SOLUTION r ( t ) parametrizes the line 2 , , 5 + t 3 , 7 , 1 , and a line has zero curvature. 5. What is the curvature at a point where T ( s ) = 1 , 2 , 3 in an arc length parametrization r ( s ) ? SOLUTION The curvature is given by the formula: ( t ) = T ( t ) r ( t ) In an arc length parametrization, r ( t ) = 1 for all t , hence the curvature is ( t ) = T ( t ) . Using the given information we obtain the following curvature: = 1 , 2 , 3 = 1 2 + 2 2 + 3 2 = 14 6. What is the radius of curvature of a circle of radius 4? SOLUTION The definition of the osculating circle implies that the osculating circles at the points of a circle, is the circle itself. Therefore, the radius of curvature is the radius of the circle, that is, 4. 7. What is the radius of curvature at P if P = 9? SOLUTION The radius of curvature is the reciprocal of the curvature, hence the radius of curvature at P is: R = 1 P = 1 9 Exercises In Exercises 16, calculate r ( t ) and T ( t ) , and evaluate T ( 1 ) . 1. r ( t ) = 12 t 3 , 18 t 2 , 9 t 4 SOLUTION The derivative vector is: r ( t ) = 36 t 2 , 36 t , 36 t 3 = 36 t 2 , t , t 3 r ( t ) = 36 t 4 + t 2 + t 6 The unit vector is thus: T ( t ) = r ( t ) r ( t ) = 36 t 2 , t , t 3 36 t 4 + t 2 + t 6 = t t , 1 , t 2  t  1 + t 2 + t 4 = t  t  1 + t 2 + t 4 t , 1 , t 2 At t = 1 we have: T ( 1 ) = 1 1 + 1 2 + 1 4 1 , 1 , 1 2 = 1 3 1 , 1 , 1 = 1 3 , 1 3 , 1 3 . S E C T I O N 14.4 Curvature (ET Section 13.4) 501 2. r ( t ) = cos t , sin t , t SOLUTION We compute the derivative vector and its length: r ( t ) =  sin t , cos t , 1 r ( t ) = ( sin t ) 2 + ( cos t ) 2 + 1 2 = 2 ( sin 2 t + cos 2 t ) + 1 = 2 + 1 The unit tangent vector is thus: T ( t ) = r ( t ) r ( t ) = 1 2 + 1  sin...
View
Full
Document
This note was uploaded on 07/15/2009 for the course MATH 210 taught by Professor Hubscher during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 Hubscher

Click to edit the document details