Velocity of a particle GIVEN position vector!

  • Thread starter vorcil
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  • #1
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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??
 

Answers and Replies

  • #2
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 [tex]\sqrt{4^{2}+0^{2}+3^{2}}[/tex] = 5.
 
  • #3
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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 [tex]\sqrt{4^{2}+0^{2}+3^{2}}[/tex] = 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?
 
  • #4
EDIT: You want the magnitude of v, not r.
 
  • #5
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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?
 
  • #6
You wanted the speed at time t=1.
 
  • #7
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.
 
  • #8
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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)|
 
  • #9
35,285
7,126
v = dr/dt = (4, 0, 3t2)

speed = |v| = [itex]\sqrt{4^2 + 0^2 + 9t^4 }[/itex]
The speed at t = 1 is the value of the radical above at t=1.
 
  • #10
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.
 
  • #11
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(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)?
 
  • #12
395
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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 [tex]\sqrt{4^{2}+0^{2}+3^{2}}[/tex] = 5.


OH RIGHT,
thanks Random variable, XD XD XD :rofl:

Just to confirm, i differentiate r= (4t , 3 , t^3)?????
 
  • #13
35,285
7,126
OH RIGHT,
thanks Random variable, XD XD XD :rofl:

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

Hasn't this been answered in post 9?
 

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