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Unformatted text preview: Solution: First, ± r ± ( t ) = < 30 t 4 , 12 t 2 , 2 > = 2 < 15 t 3 , 6 t 2 , 1 >, and ± r ± (1) = 2 < 15 , 6 , 1 >, so the length  ± r ± (1)  = 2 √ 225 + 36 + 1 = 2 √ 262 , so ± T (1) = 1 √ 262 < 15 , 6 , 1 > . Note that ﬁnding ± T ( t ) ﬁrst, before setting t = 1 , makes the problem harder; while setting t = 1 in ± r ( t ) before diﬀerentiating changes the derivative to ± 0 : the above steps are in exactly the correct order to get the right answer with least computation....
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 Spring '06
 YUKICH
 Arc Length, Derivative

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