Sec 13.4-sm.dvi

# 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: κ ( t ) = ” T ” ( t ) ” ” r ” ( t ) ” In an arc length parametrization, ” r ” ( t ) ” = 1 for all t , hence the curvature is κ ( t ) = ” T ” ( t ) ” . Using the given information we obtain the following curvature: κ = ” ” 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 κ P = 9? SOLUTION The radius of curvature is the reciprocal of the curvature, hence the radius of curvature at P is: R = 1 κ P = 1 9 Exercises In Exercises 1–6, 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 π t , sin π t , t SOLUTION We compute the derivative vector and its length: r ” ( t ) = ” - π sin π t , π cos π t , 1 ” ” r ” ( t ) ” = (- π sin π t ) 2 + ( π cos π t ) 2 + 1 2 = π 2 ( sin 2 π t + cos 2 π t ) + 1 = π 2 + 1 The unit tangent vector is thus: T ( t ) = r ” ( t ) ” r ” ( t ) ” = 1 π 2 + 1 ” - π sin π...
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Sec 13.4-sm.dvi - 500 C H A P T E R 14 CALCULUS OF...

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