Vectors

**dadon** · May 2nd 2006, 01:56 PM

Hi again!

I was doing a vector question until I got stuck! I have done part a) of the question assuming its right I need help on part b) & c)

In b) what is it meant by x-y plane?

c) I remember doing this type of question before where you have to get rp = rq not sure if this is the case for this question.

Hope you guys can help.

Regards

dadon

**ThePerfectHacker** · May 2nd 2006, 02:05 PM

Originally Posted by dadon

Hi again

In b) what is it meant by x-y plane?

That means the graph in terms of x and y.
Plane means flat and x-y means you made a graph as a function of x and y and you end up with a curve which is flat in this plane

**ThePerfectHacker** · May 2nd 2006, 02:12 PM

You have,
$\bold{R}_p(t)=a\cos \omega t \bold{i}+a\sin \omega t \bold{j}$
This is the displacement vector. The velocity vector is its derivative thus,
$\bold{R}'_p(t)=-a\omega \sin \omega t \bold{i}+a\omega \cos \omega t\bold{j}$
To find the acceleration take derivative again,
$\bold{R}''_p(t)=-a\omega^2\cos \omega t \bold{i} -a\omega^2 \sin \omega t\bold{j}$
-------
For the other vector function ya got,
$\bold{R}_q(t)=bt\bold{i}+ct\bold{j}$
$\bold{R}'_q(t)=b\bold{i}+c\bold{j}$
$\bold{R}''_q(t)=\bold{0}$

**ThePerfectHacker** · May 2nd 2006, 02:15 PM

The vector function,
$ \bold{R}_p(t)=a\cos \omega t \bold{i}+a\sin \omega t \bold{j} $
Means, parametrically
$x=a\cos \omega t$
$y=a\sin \omega t$

Umm, I cannot sketch this you did not specify $a,\omega$

The same in vector function,
$\bold{R}_q(t)=bt\bold{i}+ct\bold{j} $
Means, parametrically
$x=bt$
$y=at$
This is a line, but again cannot graph it I need you to specify the values.

**ThePerfectHacker** · May 2nd 2006, 02:19 PM

They collide when,
$ a\cos \omega t \bold{i}+a\sin \omega t \bold{j}=bt\bold{i}+ct\bold{j} $
That happens when,
$a\cos \omega t=bt$
$a\sin \omega t=ct$

Divide equation 2 by equation 1 to get,
$\tan \omega t=\frac{c}{b}$
So the problem reduces to solving this equation.

**dadon** · May 3rd 2006, 01:15 AM

Hi thanks for that theperfecthacker!

How can I use: $ a\cos \omega t=bt $ and $ a\sin \omega t=ct $

To get it in terms of t.

**dadon** · May 3rd 2006, 01:08 PM

OR is it not possible to get it in terms of t?

**ThePerfectHacker** · May 3rd 2006, 01:15 PM

Originally Posted by dadon

OR is it not possible to get it in terms of t?

Look at my previous post,
it says these equations reduce to,
$\tan \omega t= \frac{c}{b}$
Therefore use the arctangent function to get,
$\omega t=\tan^{-1} (c/b)+\pi k$
Thus,
$t=\frac{1}{\omega}\tan^{-1}(c/b)+\frac{\pi k}{\omega}$

**dadon** · May 3rd 2006, 08:45 PM

hi thanks for that!

is it possible to be in terms of a, b and c?

**ThePerfectHacker** · May 4th 2006, 12:37 PM

Originally Posted by dadon

hi thanks for that!

is it possible to be in terms of a, b and c?

Of corse you can write,
$ t=\frac{a}{a\omega}\tan^{-1}(c/b)+\frac{a\pi k}{a\omega} $

**dadon** · May 4th 2006, 01:13 PM

Thank you for spending time to answer my problem.

Kind Regards,

dadon

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