homework8sol - 32A - HOMEWORK 8 SOLUTIONS PROBLEM 3.1. With...

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32A - HOMEWORK 8 SOLUTIONS PROBLEM 3.1. With the parametrization (2.3) we have r ( t )=( R cos t,R sin t )= r 0 ( t )=( R sin t, R cos t )= ⇒k r 0 ( t ) k = R, thus the length of the half circle is Z π 0 k r 0 ( t ) k dt = R Z π 0 dt = πR , and the length of the circle is (not unexpectedly) = 2 πR. With the parametrization (2.5) we have r ( t )=( t, p R 2 t 2 )= r 0 ( t )= ³ 1 , t R 2 t 2 ´ = ⇒k r 0 ( t ) k = R R 2 t 2 , hence Z R R k r 0 ( t ) k dt = R Z R R dt R 2 t 2 = R Z R R d ( t/R ) p 1 ( t/R ) 2 = µ change of variables t/R = u = R Z 1 1 du 1 u 2 = (arcsin1 arcsin( 1)) = R ³ π 2 ³ π 2 ´ = πR , the same as with the other parametrization. PROBLEM 3.2. We have r ( t )=( t cos t,t sin t )= r 0 ( t )=( t sin t + cos t,t cos t + sin t )= ⇒k r 0 ( t ) k = p 1+ t 2 , thus the length is Z 30 0 p 1+ t 2 dt = t 1+ t 2 + arcsinh t 2 ¯ ¯ ¯ t =30 t =0 = 30
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homework8sol - 32A - HOMEWORK 8 SOLUTIONS PROBLEM 3.1. With...

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