MIT18_024S11_soln-pset4

MIT18_024S11_soln-pset4 - Solutions for PSet 4 1. (B63:3) (...

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Unformatted text preview: Solutions for PSet 4 1. (B63:3) ( t, t cos ) for < t 1 , f ( t ) = t (0 , 0) for t = 0 . f : 2 R R , thus it is continuous if both f 1 ( t ) = t and t cos f 2 ( t ) = for < t 1 , t for t = 0 . are continuous. f 1 is clearly continuous at any point t , and so is f 2 at any point t = 0. For t = we have to check that if t n 0, then f 2 ( t n ) f 2 (0) = 0: | f 2 ( t n ) | = | t n cos t n t n | | | thus it tends to too. Further, note that both f 1 ( t ) and f 2 ( t ) do not have any self intersections and are therefore simple curves. Because f ( t ) is simple and continuous over a continuous stretch of t , we can assess whether f ( t ) has a finite arc length (is rectifiable) by: (a) partitioning the space of t into n discrete blocks defined by vertices t 1 , t 2 , . . . t n . (b) defining a polygonal arc connecting points ( t 1 , f ( t 1 )), ( t 2 , f ( t 2 )), . . . ( t n , f ( t n ))- this represents a sampled approximation of f ( t ) (c) considering the limit as n of the length of this polygonal arc For n = 5, the partition of t is defined by the collection of points...
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MIT18_024S11_soln-pset4 - Solutions for PSet 4 1. (B63:3) (...

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