Explanation Given info. The path r ( t ) = 〈 5 − t , 21 − t 2 , 3 − t 3 27 〉 Formula used: The vector form of the equation of line is r → = r → 0 + t v → Proof: Position vector is r ( t ) = 〈 5 − t , 21 − t 2 , 3 − t 3 27 〉 r 0 ( t ) = 〈 5 − t 0 , 21 − t 0 2 , 3 − t 0 3 27 〉 Differentiate with respect to t r ' ( t 0 ) = 〈 − 1 , − 2 t 0 , − 3 t 0 2 27 〉 = 〈 − 1 , − 2 t 0 , − t 0 2 3 〉 Now, find time t at which the pilot can hit a target located at the origin. r ( t 0 ) + t v → 0 = r → = 〈 5 − t 0 , 21 − t 0 2 , 3 − t 0 3 27 〉 + t 〈 − 1 , − 2 t 0 , − t 0 2 9 〉 = 〈 0 , 0 , 0 〉 = 〈 5 − t 0 , 21 − t 0 2 , 3 − t 0 3 27 〉 + 〈 − t , − 2 t 0 t , − t t 0 2 9 〉 = 〈 0 , 0 , 0 〉 = 〈 5 − t 0 − t , 21 − t 0 2 − 2 t 0 t , 3 − t 0 3 27 − t t 0 2 9 〉 = 〈 0 , 0 , 0 〉 5 − t 0 − t = 0 t = 5 − t 0 ⋯ ⋯ ( 1 ) 21 − t 0 2 − 2 t 0 t = 0 t = 21 − t 0 2 2 t 0 ⋯ ⋯ ( 2 ) 3 − t 0 3 27 − t t 0 2 9 = ⋯ ⋯ ( 3 ) Substitute t = 5 − t 0 in equation ( 2 ) 5 − t 0 = 21 − t 0 2 2 t 0 10 t 0 − 2 t 0 2 = 21 − t 0 2 t 0 2 − 10 t 0 + 21 = 0 t 0 2 − 7 t 0 − 3 t 0 + 21 = 0 t 0 ( t 0 − 7 ) − 3 ( t 0 − 7 ) = 0 ( t 0 − 3 ) ( t 0 − 7 ) = 0 t

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Author: Rogawski

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Chapter 14.2, Problem 59E

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There is precisely one time t at which the point can hit a target located at the origin. A fighter plane, which can shoot a loser beam straight ahead, travels along the path $r (t) = 〈 5 - t, 21 - t^{2}, 3 - \frac{t^{3}}{27} 〉$ .

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Chapter 14.2, Problem 58E

Chapter 14.2, Problem 60E

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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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There is precisely one time t at which the point can hit a target located at the origin. A fighter plane, which can shoot a loser beam straight ahead, travels along the path r ( t ) = 〈 5 − t , 21 − t 2 , 3 − t 3 27 〉 .

Solution Summary: The author explains that there is precisely one time t at which the point can hit a target located at the origin.

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To show:

There is precisely one time t at which the point can hit a target located at the origin. A fighter plane, which can shoot a loser beam straight ahead, travels along the path $r (t) = 〈 5 - t, 21 - t^{2}, 3 - \frac{t^{3}}{27} 〉$ .

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Chapter 14 Solutions

There is precisely one time t at which the point can hit a target located at the origin. A fighter plane, which can shoot a loser beam straight ahead, travels along the path r ( t ) = 〈 5 − t , 21 − t 2 , 3 − t 3 27 〉 .

Solution Summary: The author explains that there is precisely one time t at which the point can hit a target located at the origin.

Concept explainers

Videos

To show: There is precisely one time t at which the point can hit a target located at the origin. A fighter plane, which can shoot a loser beam straight ahead, travels along the path r(t)=〈5−t,21−t2,3−t327〉.

Want to see the full answer?

Chapter 14 Solutions

To show:

There is precisely one time t at which the point can hit a target located at the origin. A fighter plane, which can shoot a loser beam straight ahead, travels along the path $r (t) = 〈 5 - t, 21 - t^{2}, 3 - \frac{t^{3}}{27} 〉$ .