# Derivation of vector r(t)

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1. Oct 26, 2014

### GoodTimes

Hello, I am new here, glad I found this Forum :0)
I am german and english is not my mother-tongue, so sorry in advance for any language-related mistakes

1. The problem statement, all variables and given/known data

i don't know how to make a vector arrow on a letter so I declare a v for vector as the following:

A particle moves on a space-curve, described by following position-vector:

2. Relevant equations

3. The attempt at a solution

I think I might differentiate each part, but I am not sure,

I would do it like this

vr'(t)= (2t)vex + (3)vey + (6t² - 8t)vez
so this would be the velocity

and vr''(t) = (2)vex + (0)vey + (12t - 8)vez
would be the acceleration

Last edited by a moderator: May 7, 2017
2. Oct 26, 2014

### Orodruin

Staff Emeritus
Your problem statement does not include a question. Are you simply interested in finding the velocity and acceleration?

The basis vectors (in Cartesian coordinates) are constant so you can just differentiate the prefactors as you have done.

Edit: You can write mathematical notation on Physics Forums using LaTeX commands. Here is a very brief introduction. You can also press Reply on my post and you can see how I have done the following:
$$\vec v(t) = \vec r'(t) = v_x \vec e_x + v_y \vec e_y + v_z \vec e_z$$

3. Oct 26, 2014

### GoodTimes

I am sorry, yes correct, my task is to find the velocity and the acceleration.
does your post mean that the two equations I've done are correct? so i could simply put in a t, for example t=2 and calculate it

4. Oct 26, 2014

### Orodruin

Staff Emeritus
The principle is correct, but one of your derivatives is not. I suggest you double check your computations (alternatively that you typed it in correctly, it may be a typo as well).

5. Oct 26, 2014

### GoodTimes

ahhh it is

$$\vec r'(t) = (2t+1)\vec e_x + (3)\vec e_y + (6t²-8t)\vec e_z$$

6. Oct 26, 2014

### Orodruin

Staff Emeritus
Correct.