14.4Ex26-39

# 14.4Ex26-39 - S E C T I O N 14.4 Curvature(ET Section 13.4...

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SECTION 14.4 Curvature (ET Section 13.4) 511 25. In the notation of Exercise 23, assume that a b . Show that b / a 2 κ ( t ) a / b 2 for all t . SOLUTION In Exercise 23 we showed that the curvature of the ellipse r ( t ) = h a cos t , b sin t i is the following function: ( t ) = ab ( b 2 cos 2 t + a 2 sin 2 t ) 3 / 2 Since a b > 0 the quotient becomes greater if we replace a by b in the denominator, and it becomes smaller if we replace b by a in the denominator. We use the identity cos 2 t + sin 2 t = 1 to obtain: ( a 2 cos 2 t + a 2 sin 2 t ) 3 / 2 ( t ) ( b 2 cos 2 t + b 2 sin 2 t ) 3 / 2 ( a 2 ( cos 2 t + sin 2 t )) 3 / 2 ( t ) ( b 2 ( cos 2 t + sin 2 t )) 3 / 2 a 3 = ( a 2 ) 3 / 2 ( t ) ( b 2 ) 3 / 2 = b 3 b a 2 ( t ) a b 2 26. Use Eq. (3) to prove that for a plane curve r ( t ) = h x ( t ), y ( t ) i , ( t ) = | x 0 ( t ) y 00 ( t ) x 00 ( t ) y 0 ( t ) | ( x 0 ( t ) 2 + y 0 ( t ) 2 ) 3 / 2 9 By the formula for curvature we have ( t ) = k r 0 ( t ) × r 00 ( t ) k k r 0 ( t ) k 3 (1) We compute the cross product of r 0 ( t ) = - x 0 ( t ), y 0 ( t ) ® and r 00 ( t ) = - x 00 ( t ), y 00 ( t ) ® . Actually, since the cross product is only deFned for three-dimensional vectors, we will think of these two vectors as follows: r 0 ( t ) = x 0 ( t ) i + y 0 ( t ) j and r 00 ( t ) = x 00 ( t ) i + y 00 ( t ) j . Thus, the cross product is: r 0 ( t ) × r 00 ( t ) = ( x 0 ( t ) i + y 0 ( t ) j ) × ( x 00 ( t ) i + y 00 ( t ) j ) = x 0 ( t ) y 00 ( t ) i × j + y 0 ( t ) x 00 ( t ) j × i = x 0 ( t ) y 00 ( t ) k y 0 ( t ) x 00 ( t ) k = ( x 0 ( t ) y 00 ( t ) y 0 ( t ) x 00 ( t ) ) k We compute the lengths of the following vectors: k r 0 ( t ) × r 00 ( t ) k=k ( x 0 ( t ) y 00 ( t ) y 0 ( t ) x 00 ( t ) ) k k=| x 0 ( t ) y 00 ( t ) y 0 ( t ) x 00 ( t ) | k r 0 ( t ) - x 0 ( t ), y 0 ( t ) ® k= q x 0 ( t ) 2 + y 0 ( t ) 2 Substituting in (1) we get: ( t ) = | x 0 ( t ) y 00 ( t ) y 0 ( t ) x 00 ( t ) | ( x 0 ( t ) 2 + y 0 ( t ) 2 ) 3 / 2 In Exercises 27–30, use Eq. (9) to compute the curvature at the given point. 27. - t 2 , t 3 ® , t = 2 ±or the given parametrization, x ( t ) = t 2 , y ( t ) = t 3 , hence x 0 ( t ) = 2 t x 00 ( t ) = 2 y 0 ( t ) = 3 t 2 y 00 ( t ) = 6 t At the point t = 2wehave x 0 ( 2 ) = 4 , x 00 ( 2 ) = 2 , y 0 ( 2 ) = 3 · 2 2 = 12 , y 00 ( 2 ) = 12

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512 CHAPTER 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) Substituting in Eq. (9) we get κ ( 2 ) = | x 0 ( 2 ) y 00 ( 2 ) x 00 ( 2 ) y 0 ( 2 ) | ( x 0 ( 2 ) 2 + y 0 ( 2 ) 2 ) 3 / 2 = | 4 · 12 2 · 12 | ( 4 2 + 12 2 ) 3 / 2 = 24 160 3 / 2 0 . 012 28. - cosh s , s ® , s = 0 SOLUTION For this parametrization x ( s ) = cosh s , y ( s ) = s , hence x 0 ( s ) = sinh s , x 00 ( s ) = cosh s , y 0 ( s ) = 1, y 00 ( s ) = 0. At the point s = 0wehave x 0 ( 0 ) = sinh 0 = 0 , x 00 ( 0 ) = cosh 0 = 1 , y 0 ( 0 ) = 1 , y 00 ( 0 ) = 0 Substituting in Eq. (9) we obtain the following curvature: ( 0 ) = | x 0 ( 0 ) y 00 ( 0 ) y 0 ( 0 ) x 00 ( 0 ) | ( x 0 ( 0 ) 2 + y 0 ( 0 ) 2 ) 3 / 2 = | 0 · 0 1 · 1 | ( 0 2 + 1 2 ) 3 / 2 = 1 1 = 1 29. - t cos t , sin t ® , t = π We have x ( t ) = t cos t and y ( t ) = sin t , hence: x 0 ( t ) = cos t t sin t x 0 ( ) = cos sin =− 1 x 00 ( t ) sin t ( sin t + t cos t ) 2sin t t cos t x 00 ( ) cos = y 0 ( t ) = cos t y 0 ( ) = cos 1 y 00 ( t ) sin t y 00 ( ) sin = 0 Substituting in Eq. (9) gives the following curvature: ( ) = | x 0 ( ) y 00 ( ) x 00 ( ) y 0 ( ) | ( x 0 ( ) 2 + y 0 ( ) 2 ) 3 / 2 = |− 1 · 0 · ( 1 ) | ( ( 1 ) 2 + ( 1 ) 2 ) 3 / 2 = 2 2 1 . 11 30. - sin 3 s , 2sin4 s ® , s = 2 x ( s ) = sin 3 s , y ( s ) = s . Hence x 0 ( s ) = 3cos3 s x 0 ³ 2 ± = 3cos 3 2 = 0 x 00 ( s ) 9sin3 s x 00 ³ 2 ± 9sin 3 2 = 9 y 0 ( s ) = 8cos4 s y 0 ³ 2 ± = 8cos2 = 8 y 00 ( s ) 32 sin 4 s y 00 ³ 2 ± 32 sin 2 = 0 Substituting in Eq. (9) we get ³ 2 ± = | x 0 ( 2 ) y 00 ( 2 ) x 00 ( 2 ) y 0 ( 2 ) | ³ x 0 ( 2 ) 2 + y 0 ( 2 ) 2 ± 3 / 2 = | 0 · 0 9 · 8 | ( 0 2 + 8 2 ) 3 / 2 = 72 8 3 = 9 64 31. Let s ( t ) = Z t −∞ k r 0 ( u ) k du for the Bernoulli spiral r ( t ) = - e t cos 4 t , e t sin 4 t ® (see Exercise 25 in Section 14.3).
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## This note was uploaded on 02/13/2010 for the course MATH MATH 32A taught by Professor Park during the Fall '09 term at UCLA.

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14.4Ex26-39 - S E C T I O N 14.4 Curvature(ET Section 13.4...

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