# Find velocity vector from position vector

• Sep 18th 2010, 11:48 AM
downthesun01
Find velocity vector from position vector
I'm reviewing for my calculus exam on Tuesday and am a bit stuck on this problem. I must be differentiating something wrong.

The position vector of a particle is r(t). Find the requested vector.
Find the velocity vector at time (t)=0,

$\displaystyle r(t)=ln(t^{3}-2t^{2}+3)i - (\sqrt{t^{2}+9})j-8cos(t)k$

Derivative of i component:
$\displaystyle (\frac{3t^{2}-4t}{t^{3}-2t^{2}+3})i$

Derivative of j component:
$\displaystyle (\frac{t}{\sqrt{t^{2}+9}})j$

Derivative of j component:
$\displaystyle 8sin(t)k$

$\displaystyle v(t)=(\frac{3t^{2}-4t}{t^{3}-2t^{2}+3})i$-$\displaystyle (\frac{t}{\sqrt{t^{2}+9}})j$+$\displaystyle 8sin(t)k$

Shouldn't v(0)=0 ?

What am I doing wrong?
• Sep 18th 2010, 11:56 AM
Defunkt
It is zero.. All your differentiations are correct.
• Sep 18th 2010, 12:10 PM
downthesun01
Thanks. I guess the professor's practice review has an error. None of his answer choices are 0. Thanks again.