s07wk04 - Solution: First, r ( t ) = < 30 t 4 ,...

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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 ± T when t = 1 .
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Unformatted text preview: Solution: First, r ( t ) = &lt; 30 t 4 , 12 t 2 , 2 &gt; = 2 &lt; 15 t 3 , 6 t 2 , 1 &gt;, and r (1) = 2 &lt; 15 , 6 , 1 &gt;, so the length | r (1) | = 2 225 + 36 + 1 = 2 262 , so T (1) = 1 262 &lt; 15 , 6 , 1 &gt; . Note that nding T ( t ) rst, before setting t = 1 , makes the problem harder; while setting t = 1 in r ( t ) before dierentiating 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 ) = &amp;amp;lt; 30 t 4 ,...

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