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100hw3soln

# 100hw3soln - Math 100 Homework 3 fall 2010 1 The arc length...

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Math 100 Homework 3 fall 2010 1. The arc length of the given curve is R 1 - 1 || r 0 ( t ) || dt = R 1 - 1 || ( 1 2 , 1 2 1 - t, 1 2 1 + t ) || dt = 1 2 R 1 - 1 1 2 + 1 - t + 1 + tdt = 3 2 R 1 - 1 dt = 3 . 2. Let g ( t ) = R t 0 || r 0 ( s ) || ds = R t 0 || ( e s cos e s , - e s sin e s , 3 e s ) || ds = R t 0 2 e s ds = 2( e t - 1) . So that its inverse is f ( t ) = ln(1 + t/ 2). The reparametrization r ( f ( t )) of r ( t ) is r ( f ( t )) = (sin(1 + t/ 2) , cos(1 + t/ 2) , 3(1 + t/ 2)) , which would be an arc length reparametrization of r ( t ). We may also verify that [ r ( f ( t ))] 0 = ( 1 2 cos(1+ t/ 2) , - 1 2 sin(1+ t/ 2) , 3 2 ) is a unit vector (why?). Therefore r ( f ( t )) = (sin(1 + t/ 2) , cos(1 + t/ 2) , 3(1 + t/ 2)) is an arc length reparametrization of r ( t ). 3. A level surface of f is the collection of ( x, y, z ) so that f ( x, y, z ) = z - x 2 - y 2 = k for some k . That is, those ( x, y, z ) satisfying x 2 + y 2 = z - k which is a circular paraboloid no matter what k is. 4. Let f ( x, y ) = x 3 y 2 x 6 + y 2 . (a) f ( t, mt ) = mt 4 2 t 6 + m 2 t 2 = mt 2 2 t 4 + m 2 0 as t 0 . f ( t, kt 2 ) = kt 5 2 t 6 + k 2 t 4 = kt 2 t 2 + k 2 0 as t 0 .

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