Velocity of a particle GIVEN position vector

[vorcil]

A [article follows the path given by the position vector
r= (4t , 3 , t^3)

what's its' speed at t=1?

no idea how to solve this
i know velocity is the derivative of the position with respect to time

do i just solve, 4, 0, 3t^2, then stick it in?
4 + 3 = 7m/s??

 

[Random Variable]

Velocity is a vector quantity. The particle's velocity at time t is v = (4,0,3t^2). At time t=1, the velocity of the particle is v = (4,0,3). It's speed at time t=1 is $$\sqrt{4^{2}+0^{2}+3^{2}}$$ = 5.

 

[vorcil]

Velocity is a vector. The particles vecolocty at time t is v = (4,0,3t^2). At time t=1, the veclocity of the particle is v = (4,0,3). It's speed at time t=1 is $$\sqrt{4^{2}+0^{2}+3^{2}}$$ = 5.

oh so it's just the magnitude of the differentiated vector?

this was the original equation
r= (4t , 3 , t^3)

so i'd differnetiate it,

(4 , 0 , 3t^2)
and find the |r| or the magnitude of it?

 

[Random Variable]

EDIT: You want the magnitude of v, not r.

 

[vorcil]

Yes. Speed is a scalar.

I find it strange to differentiate the position vector

what happened to the square on the t^3, 3t^2, it became 4,0,3?
what about the t^2? or is the ^2 part of the t, and we forget the t's?

 

[Random Variable]

You wanted the speed at time t=1.

 

[Random Variable]

You're given the particle's postion as a function of time. If you differentiate, then you know the rate and direction at which the particle's position is changing.

 

[vorcil]

You wanted the speed at time t=1.

so how do i make r= (4t , 3 , t^3) become 4,0,3?
4*1t , 3*0t, 1*1^3?

i thought differentiating r= (4t , 3 , t^3) 4t -> 4 ,3 -> 0, t^3 -> 3t^2
then substituting in 1 for t, you get the velocity vector, 4,0,3?
then i'd just get |(4,0,3)|

 

[Mark44]

v = dr/dt = (4, 0, 3t²)

speed = |v| = $\sqrt{4^2 + 0^2 + 9t^4 }$
The speed at t = 1 is the value of the radical above at t=1.

 

[Random Variable]

so how do i make r= (4t , 3 , t^3) become 4,0,3?
4*1t , 3*0t, 1*1^3?

i thought differentiating r= (4t , 3 , t^3) 4t -> 4 ,3 -> 0, t^3 -> 3t^2
then substituting in 1 for t, you get the velocity vector, 4,0,3?
then i'd just get |(4,0,3)|

(4,0,3) is a vector with magnitude 5.

 

[vorcil]

(4,0,3) is a vector with magnitude 5.

yeah thanks i know that, but how did you get (4,0,3) from r= (4t , 3 , t^3)?

 

[vorcil]

Velocity is a vector quantity. The particle's velocity at time t is v = (4,0,3t^2). At time t=1, the velocity of the particle is v = (4,0,3). It's speed at time t=1 is $$\sqrt{4^{2}+0^{2}+3^{2}}$$ = 5.

OH RIGHT,
thanks Random variable, XD XD XD

Just to confirm, i differentiate r= (4t , 3 , t^3)?

 

[Mark44]

OH RIGHT,
thanks Random variable, XD XD XD

Just to confirm, i differentiate r= (4t , 3 , t^3)?

Hasn't this been answered in post 9?