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.
 
Sweet! Thank you Dick!