s11wk04su

# s11wk04su - v r ( t ) = &amp;amp;lt; 30 t 4 , 12 t 2 , 2...

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Math 23 B. Dodson Week 2a Homework: (continued) 13.1, 13.2 vector functions, derivatives start Week 2b: 13.3 arc length, curvature . 13.4 velocity, acceleration; Problem 13.2.9: Find the derivative of the vector function v r ( t ) = < t 2 , 1 - t, t > . Solution: We just take the derivative of the components, v r ( t ) = < ( t 2 ) , (1 - t ) , ( t ) > = < 2 t, - 1 , 1 2 t >, where ( t 1 2 ) = 1 2 t 1 2 . Week 2 Homework: (continued) 13.1, 13.2 vector functions, derivatives start Week 2b: 13.3 arc length, curvature; 13.4 velocity, acceleration Problem 13.2.17: If v r ( t ) = < 6 t 5 , 4 t 3 , 2 t >, ±nd the unit tangent vector v T when t = 1 .

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2 . Solution: First,
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Unformatted text preview: v r ( t ) = &lt; 30 t 4 , 12 t 2 , 2 &gt; = 2 &lt; 15 t 3 , 6 t 2 , 1 &gt;, and v r (1) = 2 &lt; 15 , 6 , 1 &gt;, so the length | v r (1) | = 2 225 + 36 + 1 = 2 262 , so v T (1) = 1 262 &lt; 15 , 6 , 1 &gt; . Note that fnding v T ( t ) frst, beore setting t = 1 , makes the problem harder; while setting t = 1 in v r ( t ) beore dierentiating changes the derivative to v 0 : the above steps are in exactly the correct order to get the right answer with least computation....
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## s11wk04su - v r ( t ) = &amp;amp;lt; 30 t 4 , 12 t 2 , 2...

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