When Should the Spaceship's Engines Be Turned Off?

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SUMMARY

The discussion focuses on determining when a spaceship should turn off its engines to coast into a space station located at coordinates (6,4,9). The position function of the spaceship is defined as r(t) = (3+t)i + (2+ln t)j + (7 - 4/(t^2+1))k. The key conclusion is that the engines should be turned off when the velocity vector, derived from r'(t) = i + (1/t)j + (8t/(t^2+1)^2)k, is parallel to the vector from the spaceship's position to the space station. This parallelism indicates that the spaceship will continue on its trajectory towards the station without further propulsion.

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Homework Statement


The position function of a spaceship is

r(t) = (3+t)i + (2+ln t)j + (7 - 4/(t^2+1)k and the coordinates of the space station are (6,4,9). If the spaceship were to "coast" into the space station, when should the engines be turned off?


Homework Equations



The relevant equations are r' = velocity, and r'' = acceleration.

The Attempt at a Solution



I am really not sure how to go about solving this problem. So, this means that r" is zero but then there is still velocity? So, if r' is i + 1/tj + 8t/(t^2+1)^2k. r" = -1/t^2 + (8(t^2+1)^4 - 16t^2(t^2+1))/(t^2+1)^4. But then I am not sure what to do after that. Thanks!
 
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bodensee9 said:

Homework Statement


The position function of a spaceship is

r(t) = (3+t)i + (2+ln t)j + (7 - 4/(t^2+1)k and the coordinates of the space station are (6,4,9). If the spaceship were to "coast" into the space station, when should the engines be turned off?
Am I correct that that r(t) is the position of the spacecraft with the engines on? And with the engines off it will continue in that direction?


Homework Equations



The relevant equations are r' = velocity, and r'' = acceleration.

The Attempt at a Solution



I am really not sure how to go about solving this problem. So, this means that r" is zero but then there is still velocity? So, if r' is i + 1/tj + 8t/(t^2+1)^2k. r" = -1/t^2 + (8(t^2+1)^4 - 16t^2(t^2+1))/(t^2+1)^4. But then I am not sure what to do after that. Thanks!

Yes, with the engines turned of the spaceship will still have non zero velocity. Look at the velocity vector for each t (the derivative of r) as well as the vector from r(t) directly to the space station. When those vectors are parallel, turn off the engines will cause the spaceship to continue to the space station.
 
So do you mean that if the vector formed by subtracting r(t) from (6,4,9) needs to be parallel to r'? So this means that <3-t, 2-ln t, 2+4/t^2+1> needs to be a multiple of <i, 1/tj, 8t/(t^2+1)^2k?

But how do you go about solving for that, because you have 3-t must be a multiple of i, and 2-ln t must be a multiple of 1/t? So does that mean you try to make ln t go away? Thanks.
 

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