14.4%20Ex%2032

# 14.4%20Ex%2032 - S E C T I O N 14.4 Curvature(ET Section...

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SECTION 14.4 Curvature (ET Section 13.4) 513 In our case, x ( t ) = e t cos 4 t and y ( t ) = e t sin 4 t . Hence: x 0 ( t ) = e t cos 4 t 4 e t sin 4 t = e t ( cos 4 t 4sin4 t ) x 00 ( t ) = e t ( cos 4 t t ) + e t ( t 16 cos 4 t ) =− e t ( 15 cos 4 t + 8sin4 t ) y 0 ( t ) = e t sin 4 t + 4 e t cos 4 t = e t ( sin 4 t + 4cos4 t ) y 00 ( t ) = e t ( sin 4 t + t ) + e t ( t 16 sin 4 t ) = e t ( 8cos4 t 15 sin 4 t ) We compute the numerator in (1): x 0 ( t ) y 00 ( t ) x 00 ( t ) y 0 ( t ) = e 2 t ( cos 4 t t ) · ( t 15 sin 4 t ) + e 2 t ( 15 cos 4 t + t ) · ( sin 4 t + t ) = e 2 t ( 68 cos 2 4 t + 68 sin 2 4 t ) = 68 e 2 t We compute the denominator in (1): x 0 ( t ) 2 + y 0 ( t ) 2 = e 2 t ( cos 4 t t ) 2 + e 2 t ( sin 4 t + t ) 2 = e 2 t ( cos 2 4 t t sin 4 t + 16 sin 2 4 t + sin 2 4 t + t cos 4 t + 16 cos 2 4 t ) = e 2 t ( cos 2 4 t + sin 2 4 t + 16 ( sin 2 4 t + cos 2 4 t )) = e 2 t ( 1 + 16 · 1 ) = 17 e 2 t (2) Hence ( x 0 ( t ) 2 + y 0 ( t ) 2 ) 3 / 2 = 17 3 / 2 e 3 t Substituting in (2) we have κ ( t ) = 68 e 2 t 17 3 / 2 e 3 t = 4 17 e t R = 17 4 e t (3) On the other hand, by the Fundamental Theorem and (2) we have s 0 ( t ) =k r 0 ( t ) k= q x 0 ( t ) 2 + y 0 ( t ) 2 = p 17 e 2 t = 17 e t We integrate to obtain s ( t ) = Z 17 e t dt =

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14.4%20Ex%2032 - S E C T I O N 14.4 Curvature(ET Section...

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