- #1

- 395

- 0

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??

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- Thread starter vorcil
- Start date

- #1

- 395

- 0

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??

- #2

- 116

- 0

- #3

- 395

- 0

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

- 116

- 0

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

- #5

- 395

- 0

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

- 116

- 0

You wanted the speed at time t=1.

- #7

- 116

- 0

- #8

- 395

- 0

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

Mark44

Mentor

- 35,285

- 7,126

speed = |

The speed at t = 1 is the value of the radical above at t=1.

- #10

- 116

- 0

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

- 395

- 0

yeah thanks i know that, but how did you get (4,0,3) from r= (4t , 3 , t^3)?(4,0,3) is a vector with magnitude 5.

- #12

- 395

- 0

OH RIGHT,

thanks Random variable, XD XD XD :rofl:

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

- #13

Mark44

Mentor

- 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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