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>?
$$r'(t)=<1,\frac{1}{t},\frac{4t}{(t^2+1)^2}>$$

 

[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>?
$$r'(t)=<1,\frac{1}{t},\frac{4t}{(t^2+1)^2}>$$

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.