Sec 13.4-sm.dvi - 500 C H A P T E R 14 CALCULUS OF...

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Unformatted text preview: 500 C H A P T E R 14 CALCULUS OF VECTOR-VALUED 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...
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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.

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Sec 13.4-sm.dvi - 500 C H A P T E R 14 CALCULUS OF...

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