# Thread: Help with planar motion trajectory

1. ## Help with planar motion trajectory

Hi, I have recently been set a set of questions all of which I have been able to do apart from the last part, I just need a starting point if at all possible or some guidance. The question is as follows:

given the 2D trajectory:

r(t) = vector (x(t) y(t))

 = vector (a*cos(omega*t) b*sin(omega*t))

show that r(t) can be expressed as:

r(t)=Re(Re^(i*omega*t)) where Re is the real part and R is the complex amplitude vector (X Y)

The hint is Re(z)=(z + z*)/2 to deduce what X and Y are in terms of a and b.

Ive already plotted the data and deduced that it can be expressed as:

x^2/a^2+y^2/b^2=1

Ive tried different things including working backwards and converting e^(i*omega*t) into the real and imaginary parts (a*cos(omega*t) + i*b*sin(omega*t) etc.) but to no avail. Any help would be much appreciated.





2. ## Nonsensical?

I am interpreting the problem as follows: Given $v(t)=(a\cos(\omega t),b\sin(\omega t))$, find $r=(x,y)$ such that $v(t)=Re[re^{i\omega t}]$

But $Re[re^\theta]$ is just $r\cos\theta$, so v(t) would have to equal $(x,y)\cos(\omega t)=(x\cos(\omega t),y\cos(\omega t))$. So in theory, $x=a, y=b\tan(\omega t)$ could work as a "solution," but I do not think that is what the problem is intending.

3. Many thanks for your reply. That is also the solution I have arrived at which I am happy enough works, I cant see where I would have gone wrong in arriving at it, however the the question states that the solutions to X and Y would be in terms of a and b and be in C (the complex numbers) which I know is technically true, but leads me to believe a different answer is required...I just don't know where I would have gone wrong. Any further help would be appreciated but many thanks for taking your time to reply.

Jamie.

4. Perhaps you can post a photocopy of the question? I think we are missing subtle part of the original framework.

5. Of course, I may have left something out, as it does seem like something is missing. Hopefully the attachment will work! Many thanks again.