# Find the minimum speed of a particle

1. Oct 30, 2013

The position function of a particle is given by r(t)=⟨−5t2,−4t,t2+1t⟩.
At what time is the speed minimum?

I found the speed function by taking the magnitude of the velocity which is

lv(t)l = sqrt(108t^2 +4t + 17)

I then took the derivative of the speed to find the critical point. However, I'm always left with a negative time.

d/dt lv(t)l = (104t + 2)/sqrt(104t^2 +4t+17)

The only way to find the minimum is when the numerator is 0. But when I isolate t, it always equals a negative number.

Is my approach correct?

Any help would be appreciated. Thanks!

2. Oct 30, 2013

### Staff: Mentor

Homework problems must be posted in the Homework & Coursework section, not in the technical math sections. I have moved this thread.

3. Oct 30, 2013

### Staff: Mentor

That should be √(104t2 + 4t + 17).

You can find the minimum value of |v(t)| by completing the square inside the radical. I get a negative time. All this denotes is that the speed is smallest at some time before time t = 0.

4. Oct 30, 2013

But it doesn't make sense when t (time) is less than zero.

5. Oct 31, 2013

### HallsofIvy

Staff Emeritus
Why not? t= 0 is just some arbitrarily chosen time. Why couldn't the particle be moving before that?

6. Oct 31, 2013

### pasmith

The other answer is that, if the domain of $r(t)$ is $[0,\infty)$, then since $\|v\|$ is strictly increasing in $t > 0$ the particle attains its minimum speed at $t = 0$.