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# s07wk04 - Solution First ± r ± t =< 30 t 4 12 t 2 2>...

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Math 23 B. Dodson Week 4 Homework: 13.1, 13.2 vector functions, derivatives . [due Friday, Feb 9] start Week 5: 13.3 arc length, curvature . 13.4 velocity, acceleration; . 14.1 functions of several variables Problem 13.2.9: Find the derivative of the vector function r ( t ) = < t 2 , 1 - t, t > . Solution: We just take the derivative of the components, r ( t ) = < ( t 2 ) , (1 - t ) , ( t ) > = < 2 t, - 1 , 1 2 t >, where ( t 1 2 ) = 1 2 t - 1 2 . Week 4 Homework: 13.1, 13.2 vector functions, derivatives start Week 5: 13.3 arc length, curvature; 13.4 velocity, acceleration Problem 13.2.17: If r ( t ) = < 6 t 5 , 4 t 3 , 2 t >, find the unit tangent vector

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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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s07wk04 - Solution First ± r ± t =< 30 t 4 12 t 2 2>...

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