Velocity and Acceleration Vector Problem

[Loppyfoot]

Homework Statement

Find the velocity and acceleration vectors for the following curves. Determine the points where the velocity vector is changing DIRECTION but not changing SPEED,
and the points where the velocity vector is changing SPEED but not DIRECTION.

For:

a. f(t) = (t,t²,t³)
b. (e^-t, e^2t)

Homework Equations

If we want to change the speed but not direction, the acceleration vector must be parallel to the velocity vector.

If we want to change the direction but not the speed, the acceleration vector should be perpendicular to the velocity vector.

The Attempt at a Solution

Ok so here I go.

I will attempt to change the speed but not direction, so I need to do v=ka, and solve for k and then solve for t to get the points, correct?

And if I want to change the direction but not the speed, how should I find the points?

Thanks again, everyone.

 

[Loppyfoot]

Any Ideas?

 

[cepheid]

I think maybe the simplest way of going about this is to note that:

v⋅a = |v||a|

when the vectors are parallel, and:


v⋅a = 0

when they are perpendicular.

 

[HallsofIvy]

You say "so here I go" but you don't go!

What are the velocity and acceleration vectors for those two problems? As cepheid says, the speed changes but not the direction if and only if the velocity and acceleration vectors are parallel and the direction changes but not the speed if and only if they are perpendicular.

 

[Loppyfoot]

So I found out how to solve to change the direction, not the speed, by doing the dot product between vectors v and a.

How would I find the points when I solve for when they're parallel

If v is parallel to a, there must be a constant k that links the two vectors being parallel.

so a=kv, right?

I am confused on the computational portion of this problem..

 

[Mark44]

Have you actually found the velocity and acceleration vectors yet? The velocity is v(t) = f'(t). The acceleration is a(t) = v'(t) = f''(t). To find the derivative of a vector-valued function, simply differentiate component-wise.

And yes, if a and v are in the same direction, then a = kv for some constant k.

 

[Loppyfoot]

COmputationally I feel like i am not getting anywhere.

A=kv

So ,

<0,2,6t> = k<1,2t,3t²>

I get 0=k ; 2=2kt ; 6t=3t²k.

Where should I go from here?

Should I solve for k, because I tried that and I am going in circles...

 

[Mark44]

Maybe it's saying that for this function, there are no points at which the velocity is changing speed but not direction.

 

[Loppyfoot]

any ideas?

Basically, I'm stuck on the problem:

For what values of t (if any) exist when the vector <1,2t,3t^2> is parallel to <0,2,6t>.

I know <0,2,6t>= k <1,2t,3t^2>.

What do I do next?

 

[cepheid]

Well, that's actually three different equations:

0 = k
2 = 2kt
6t = 3kt^2

Since k is non-zero, the first equation can never be satisfied for any value of t. This suggests that v is never parallel to a. This is pretty obvious just by inspection of the two vectors, right? I mean, the acceleration vector has no x-component, meaning that it is always parallel to the yz-plane. In contrast, the velocity vector always has a non-zero x-component, meaning that it never lies in a plane that is parallel to the yz-plane.

EDIT: I just realized that Mark44 already suggested that your results imply that v and a are never parallel. Sorry.

 

[Loppyfoot]

Ok, I understand. Than you cepheid and Mark44 for helping me visualizing this problem!


Thanks again!

Loppyfoot