Explanation Given info: r → ( 2 ) and r → ( − 1 ) for r → ( t ) = ( Sin π 2 t , t 2 , ( t 2 + 1 ) − 1 ) Concept used: Find each component of vector for the given value of variable, then use standard notation to write. Calculation: r → ( 2 ) and r → ( − 1 ) for r → ( t ) = ( Sin π 2 t , t 2 , ( t 2 + 1 ) − 1 ) r x = Sin π 2 t r y = t 2 r z = ( t 2 + 1 ) − 1 r x , r y , and r 2 are the x , y and z component r x ( t = 2 ) = Sin π 2 .2 = Sin π = 0 r y ( t = 2 ) = t 2 putting the value of t = ( 2 ) 2 = 4 r z = ( t 2 + 1 ) − 1 = 1 ( t 2 + 1 ) putting the value of t r z ( t = 2 ) = 1 ( ( 2 ) 2 + 1 ) ...

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Author: Rogawski

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Chapter 14.1, Problem 3E

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Evaluate $\vec{r} (2) and \vec{r} (- 1) for \vec{r} (t) = (Sin \frac{π}{2} t, t^{2}, {(t^{2} + 1)}^{- 1})$

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Chapter 14.1, Problem 2E

Chapter 14.1, Problem 4E

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Chapter 14 Solutions

CALCULUS W/SAPLING ACCESS >IC<

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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Evaluate r → ( 2 ) and r → ( − 1 ) for r → ( t ) = ( Sin π 2 t , t 2 , ( t 2 + 1 ) − 1 )

Solution Summary: The author explains how to calculate r ( 2 ) for t = ( 0, 4, 1

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Evaluate $\vec{r} (2) and \vec{r} (- 1) for \vec{r} (t) = (Sin \frac{π}{2} t, t^{2}, {(t^{2} + 1)}^{- 1})$

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Chapter 14 Solutions

Evaluate r → ( 2 ) and r → ( − 1 ) for r → ( t ) = ( Sin π 2 t , t 2 , ( t 2 + 1 ) − 1 )

Solution Summary: The author explains how to calculate r ( 2 ) for t = ( 0, 4, 1

Videos

Evaluate r→(2) and r→(−1) for r→(t)=(Sinπ2t,t2,(t2+1)−1)

Want to see the full answer?

Chapter 14 Solutions

Evaluate $\vec{r} (2) and \vec{r} (- 1) for \vec{r} (t) = (Sin \frac{π}{2} t, t^{2}, {(t^{2} + 1)}^{- 1})$