# Finding velocity and acceleration in a vector via differentiation

1. Aug 6, 2012

### Philip Wong

1. The problem statement, all variables and given/known data
Find the velocity and acceleration of a particle with the given position function:

r(t)=<$2cos t$, 3t, $2sin t/t+1$>

3. The attempt at a solution

v(t)=r'(t) dt= <-2sin t, 3, (2cos t/t+1) - (2sin t/(t+1)2)>

a(t)=v'(t) dt = <-2cos t, 0, (4sint t/(t+1)3-(2sint t/(t+1)-(4 cos t/(t+1)2>

therefore a(t) = <-2cos t, 0, 2/(t+1)[(2sint t/(t+1)2-sin t-(2 cos t/(t+1)>

First of all, is this right?
Secondly, can this be further simplify? If yes, can someone please show me how.
Lastly, I don't have a clue on how to solve this two vectors to give a single numerical answer. Or have I misunderstood the concepts, the answer show actually be presented as a vector instead of a single numerical answer?

Phil

2. Aug 6, 2012

### voko

What does $2sin t/t+1$ really mean?

$$\frac {2sin t} {t+1}$$ or

$$2 sin \frac {t} {t+1}$$ or

$$\frac {2sin t} {t} +1$$?

3. Aug 6, 2012

You can't get a numerical value, only a vector. You have to have a t value to plug into the functions you derived to get an exact vector with numerical components. The vector r(t) you start with gives position as a function of t, the derivative r'(t) gives the velocity vector as a function of t (velocity at any time t) and r''(t) gives the acceleration vector as a function of t...

If you had a function like r(t)=<2t,t,t> for example, you'd get r'(t)= <2,1,1> since the derivatives are all constant, but if the components of r(t) had non-constant derivatives then you have a non-constant velocity function like in your example.

4. Aug 10, 2012

### Philip Wong

Adkins Jr: I see what you meant now. thanks for your help!
voko: is the first one. sorry I'm not good using latex, I tried to have the equation type out using it. but somehow it didn't work

5. Aug 11, 2012

### voko

Your solution seems correct. Still, I would like to point out that even if you do not use LaTeX, you could write (sin t)/(t + 1). sin t/t + 1, in the usual convention, means [sin (t/t)] + 1 = [sin 1] + 1.