Find When Engines Should Be Turned Off for Spaceship to Reach Space Station

[Winzer]

Homework Statement

The position of a spaceship is:
r(t)=(3+t)i +(2+ln(t))j+(7-\frac{4}{t^2+1})k

and the coordinated of the space station are (6,4,9). The captian wants the spaceship to coast into the the space station. When should the engines be turned off?

Homework Equations

r(t)=(3+t)i +(2+ln(t))j+(7-\frac{4}{t^2+1})k

The Attempt at a Solution

Ok the ship coasts(uniform velocity) into the space ship.
So max/min problem right? Find \frac{d^2r}{dx^2} set equall to zero and solve for t right?

 

[Dick]

Nooo. You want r'(t) to be parallel to r(t)-(6,4,9) and pointing in the right direction. Your turn. Why?

 

[Winzer]

So r'(t)=<6,4,9> because it heads in the spacestations direction

 

[Dick]

The sense of your answer is correct. But the direction of the station is <6,4,9>-r(t) from the position of the ship, right? Difference of two positions is the direction.

 

[Winzer]

Ok, so the position vector of the space station is r(t)=<6,4,9>, and r'(t) has to be parallel being r'(t)=<6,4,9> or some scalar multiple.
This means we must solve for t in r'(t) when r'(t)=<6,4,9>?
r&#039;(t)=&lt;1,\frac{1}{t},\frac{4t}{(t^2+1)^2}&gt;

 

[Winzer]

 

[HallsofIvy]

Ok, so the position vector of the space station is r(t)=<6,4,9>, and r'(t) has to be parallel being r'(t)=<6,4,9> or some scalar multiple.
This means we must solve for t in r'(t) when r'(t)=<6,4,9>?
r&#039;(t)=&lt;1,\frac{1}{t},\frac{4t}{(t^2+1)^2}&gt;

You've already been told, twice, that this is wrong. The vector from the ship to the space station is <6, 4, 9>- r(t). You must have r' equal to that.