Finding when something is closest to something.

[Sanguinne]

So in the insight 2012 exam 2, they ask me to find when the tugboat is closest to the oil rig. Wouldn't I just do dot product of the position vector and the vector from the oil rig to the position vector? They instead use velocity vector. Why?

thanks

[keltingmeith]

This image will be helpful:

It's a little big

Okay, so let's say we have a particle with position vector . This means that in the diagram above, its movement can be tracked by the red arrow - wherever the red arrow points is where the particle is at that time t. Now, this particle's velocity vector is obviously . Here's where it gets good - the velocity vector is the DERIVATIVE of the position, so the velocity vector is going to point tangent to the curve that the position vector produces. If you draw the velocity vector into the diagram so that it points tangent to the curve (this is the green arrow), you'll see that it's actually perpendicular to the red arrow. In fact, the velocity vector is point in the direction of motion, whereas the position vector points to wherever the particle is, not necessarily in the direction of motion.

So, we see that position and velocity vectors will (in this case, and in a few other cases, too) be perpendicular to each other. The key point to take away from this is for questions like this, draw the situation - then, draw the DIRECTION OF MOTION of your particle (the boat) and compare it to the DIRECTION OF POSITION from its origin (in this case, the oil rig), and see if it looks like they'll be perpendicular at any stage. This is when you take the dot product of the position and velocity vectors.

[psyxwar]

This image will be helpful:

It's a little big

Okay, so let's say we have a particle with position vector . This means that in the diagram above, its movement can be tracked by the red arrow - wherever the red arrow points is where the particle is at that time t. Now, this particle's velocity vector is obviously . Here's where it gets good - the velocity vector is the DERIVATIVE of the position, so the velocity vector is going to point tangent to the curve that the position vector produces. If you draw the velocity vector into the diagram so that it points tangent to the curve (this is the green arrow), you'll see that it's actually perpendicular to the red arrow. In fact, the velocity vector is point in the direction of motion, whereas the position vector points to wherever the particle is, not necessarily in the direction of motion.

So, we see that position and velocity vectors will (in this case, and in a few other cases, too) be perpendicular to each other. The key point to take away from this is for questions like this, draw the situation - then, draw the DIRECTION OF MOTION of your particle (the boat) and compare it to the DIRECTION OF POSITION from its origin (in this case, the oil rig), and see if it looks like they'll be perpendicular at any stage. This is when you take the dot product of the position and velocity vectors.

Hey I'm still not sure why this gets you the minimum distance :S

i've just been finding the minimum of |r(t)| as of late, but dot product seems to be a better way of doing it.

[keltingmeith]

Hey I'm still not sure why this gets you the minimum distance :S

i've just been finding the minimum of |r(t)| as of late, but dot product seems to be a better way of doing it.

So, I just showed that in that situation (note: this isn't true for every situation), the velocity vector and position point perpendicular to each other. The perpendicular distance is always going to be the shortest, which you see intuitively by drawing lines between a point and a line and looking at which one is the shortest.