# Position Function and Speed

1. Mar 16, 2013

### user8899

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

The position function of a particle is given by r(t) = <2t, 3, -2t+1> where t is the time in seconds. When is the speed a minimum? What is the minimum speed?

2. Relevant equations

v(t) = r'(t)

3. The attempt at a solution

The derivative of r'(t) = 2 + 0 + (-2) = 0. The t is cancelled out so how are we supposed to find t and the minimum speed? I am so confused. Help please?

Last edited by a moderator: Mar 17, 2013
2. Mar 16, 2013

### jbunniii

No, the derivative should be a vector. If $r(t) = \langle a(t), b(t), c(t)\rangle$, then $r'(t) = \langle a'(t), b'(t), c'(t) \rangle$. You don't sum $a'(t)$, $b'(t)$, and $c'(t)$.

3. Mar 16, 2013

### user8899

Ok, so I would just get <2,3,-2>. Then what would I do?

4. Mar 16, 2013

### Ray Vickson

Speed = magnitude of velocity.

5. Mar 16, 2013

### jbunniii

The second component should not be 3.

6. Mar 16, 2013

### user8899

Yea sorry it's <2,0,-2>

7. Mar 16, 2013

### user8899

So the speed is 0? How do we find the time then? That's what I am stuck on.

8. Mar 16, 2013

### Fredrik

Staff Emeritus
No, it's not. He meant the magnitude (i.e. the norm) of the vector, not the magnitude of the sum of its components.

9. Mar 16, 2013

### user8899

I am really confused now. Could you give me some guidance?

So the derivative is velocity right? The magnitude of the vector would be √((2)^2 + (-2)^2)? Then that would be speed which is √8?

Is this right?

Thanks

10. Mar 16, 2013

### Fredrik

Staff Emeritus
Yes, that's right.

11. Mar 16, 2013

### user8899

Thanks, so for time what would I do since when you find the derivative t is no longer there?

12. Mar 16, 2013

### jbunniii

You have shown that the speed is constant, so it achieves both a minimum and a maximum at all times.

13. Mar 17, 2013

### user8899

oh! okay that makes sense! Thank you everybody :)