Deriving Velocity and Acceleration from a Distance Vector in the x-y Plane

[mystmyst]

I don't study in English, so please bare with me if my wording isn't correct.

Homework Statement

The distance vector (this may possibly be displacement vector. I'm not sure) moves in the x-y plane is: \vec{r} = a\sin(\omega t) \hat{x} + b\cos(\omega t) \hat{y}

Find the velocity vector and the acceleration of the body.

The Attempt at a Solution

To find the velocity vector I do the derivative of the distance. But it's a very funny looking equation. How do I do a derivative on this?

Thanks!

 

[ehild]

It is the sum of two terms, both terms are functions of t multiplied by the unit vectors along the x , y axes. Do you know what is the derivative of sin(wt) and cos(wt)?

ehild

 

[mystmyst]

\sin x\prime = \cos x
*does that mean \sin (\omega t)\prime = \cos (\omega t) or do I need to do something special for the (\omega t)
**also, is the unit x-vector it's own fuction? how do i do the derivative of that?

I would assume to use the product rule (fg)\prime= f\prime g + g\prime f for both terms; and that's assuming sinx, cosx and unit vectors x,y are considered functions

\vec{r} = a\sin(\omega t) \hat{x} + b\cos(\omega t) \hat{y}
r\prime = (a\cos(\omega t) \hat{x} + a\sin(\omega t)) + (-b\sin(\omega t) \hat{x} + a\cos(\omega t))

but that's assuming \hat{x}\prime = 1

 

[mystmyst]

so basically, my only question is am I supposed to find the derivative of \hat{x},\hat{y} or does it belong to \cos x like the a does.

 

[ehild]

<br /> \hat{x},\hat{y}<br />

are constant vectors, like a and b are constant numbers. If a function is multiplied by a constant, its derivative is also multiplied with that constant.

As for the derivatives of cos(wt) and sin(wt), they are -w sin(wt) and w (cos(wt), respectively.

ehild

 

[mystmyst]

so because of your help I arrived at this:

at t=0,

\vec{v} = a \omega \hat{x}
\vec{a} = -b \omega^2 \hat{y}

question: How do I find the angle between these two vectors?

Is the omega referring to angular velocity or is it also a constant like a and b?

I would assume 90 degrees since 'v' only contains the x vector and 'a' only contains the y vector.

 

[ehild]

omega is the constant angular velocity.

Yes, the velocity vector and the acceleration vector are perpendicular to each other at t=0.

ehild

 

[mystmyst]

Thanks ehild,

you rock my sox!

where's the thank you button?