Finding minimum speed given position vector.

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dial1revenge
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Homework Statement



r(t) = < t^2, 6t, t^2 − 24t >

Using this position vector, find the minimum speed of the particle.

Homework Equations



v(t) = r'(t)

The Attempt at a Solution



I've found similar topics with similar problems but I'm having a hard time figure mine out.

I know you have to take the derivative of r(t), take the magnitude and set it = 0.

r'(t) = v(t) = <2t, 6, 2t - 24>

|v(t)| = sqrt( 4t^2 + 36 + 4t^2 - 96t + 576)

simplified to

8t^2 - 96t + 612 = 0

I guess this where I'm stuck. At probably the simplest part of the problem; factoring. But I'm getting some weird answers and thinking maybe I've done something wrong along the way?

Thanks,
DRV
 
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dial1revenge said:

Homework Statement



r(t) = < t^2, 6t, t^2 − 24t >

Using this position vector, find the minimum speed of the particle.

Homework Equations



v(t) = r'(t)

The Attempt at a Solution



I've found similar topics with similar problems but I'm having a hard time figure mine out.

I know you have to take the derivative of r(t), take the magnitude and set it = 0.

r'(t) = v(t) = <2t, 6, 2t - 24>

|v(t)| = sqrt( 4t^2 + 36 + 4t^2 - 96t + 576)

simplified to

8t^2 - 96t + 612 = 0

I guess this where I'm stuck. At probably the simplest part of the problem; factoring. But I'm getting some weird answers and thinking maybe I've done something wrong along the way?

Thanks,
DRV

You don't want to solve 8t^2 - 96t + 612 = 0. You want to find the minimum of the function f(t)=8t^2 - 96t + 612. Any ideas on doing that?
 
Ahh. So I should take the derivative of my new function v(t) and set that equal to zero?

Then plug the t value into my new function v(t) to find velocity.
 
dial1revenge said:
Ahh. So I should take the derivative of my new function v(t) and set that equal to zero?

Then plug the t value into my new function v(t) to find velocity.

Yes, but you can ignore the sqrt part to begin with. Just find the minimum of the function inside the sqrt.