# Thread: Vector Functions & Their Derivatives

1. ## Vector Functions & Their Derivatives

I'm reviewing some of our home work for our midterm this Monday and when I was reading the section on vector functions the part of the directions that kind of lost me is when they said:
"...Write the particle's velocity at that time as the product of its speed and direction."

$
(e^-^t )i +(2 cos3t)j+(3 sin 2t)k , t=0

$

I found the velocity vector $(-e^-^t )i +(-6 sin 3t)j+(-6 cos 2t)k$
and at time t=0 I found that is $-1i + 0j - 6k$
Then found the accelertion vector $(e^-^t )i +(-18 cos3t)j+(-18 sin 2t)k$
and at time t=0 I get $1i-18j+0k$

..My question is : What am I suppose to write down when it asks to write the particles velocity as a product of its speed and direction?

2. Does the problem really say "product" of its speed and direction? I would interpret that as the scalar product of the length of your velocity vector with a unit vector in the direction of motion.

The "length" of $(-e^{-t})i+ (-6 sin(3t))j+ (-6 cos(2t))k$ is $\sqrt{e^{-2t}+ 36sin^2(3t)+ 35cos^2(2t)}$, the particle's speed. Divide each component of the velocity vector by that to get the unit vector in that direction.

Of course, the velocity vector is just the scalar product of that speed and unit vector since you would just be multiplying back in the speed you divided each component by.