Homework Help: Find v & a in complex plain

1. Dec 27, 2009

colgon

1. A particle moves in the (x,y) plane so that its position (x,y) as a function of time is given by:

2. z=$$\frac{i+2t}{t-i}$$

Find the magnitudes of its velocity and its accleration as functions of t by writing z in the x+iy form and so find x and y as functions of t.

3. I think

x=$$\frac{2t^{2}-1}{t^{2}+1}$$
and
y=$$\frac{3t}{t^{2}+1}$$

I am not sure if these are correct and if so how to use them to find v and a. I thought if I could find mod z using these x and y this would give me this distance then divide through by t to get v but when I try this I don't get anything I can divide through by t and be simplifyable in anyway. I know the answers should be v=$$\frac{3}{t^{2}+1}$$ as this is found using $$\frac{dz}{dt}$$ but I am asked to verify this using x and y as functions of t.

2. Dec 27, 2009

HallsofIvy

Instead of just giving your result, please show how you got it- that would relieve of us "guessing" exactly what you did! I presume that the first thing you did was multiply numerator and denominator by the complex conjugate of the denominator:
$$z=\frac{i+2t}{t-i}\frac{t+i}{t+i}= \frac{2t^2- 1+ i(3t)}{t^2+1}$$[/b]
$$= \frac{2t^2-1}{t^2+1}+ i\frac{-t}{t^2+1}$$
Yes, x and y are what you say.

To find the v and a vectors (I presume you mean velocity and acceleration- it would have been better to say so) just differentiate them.

3. Dec 27, 2009

yungman

I want to verify first: You let $$\stackrel{\rightarrow}{z}$$ be the position vector on x-y plane where $$\stackrel{\rightarrow}{z}=\frac{2t+i}{t-i}$$

Then you take $$\frac{2t+i}{t-i}.\frac{t+i}{t+i}=\frac{2t^{2}-1}{t^{2}+1}+i\frac{3t}{t^{2}+1}$$. You call $$x=\frac{2t^{2}-1}{t^{2}+1}$$ and $$y=i\frac{3t}{t^{2}+1}$$.

You don't get v by divid the position vector by t.

$$\stackrel{\rightarrow}{v}=\frac{\partial\stackrel{\rightarrow}{z}}{\partial t}=\hat{x} \frac{\partial x}{\partial t}+\hat{y}\frac{\partial y}{\partial t}$$

$$\stackrel{\rightarrow}{a}=\frac{\partial\stackrel{\rightarrow}{v}}{\partial t}$$

Last edited: Dec 28, 2009
4. Dec 28, 2009

colgon

Thank you for the replies. I didn't really follow them totally. I had a go with the following results. I think I am missing something here?

$$\frac{\partial x}{\partial t}=\frac{(t^{2}+1)(4t)-(2t^{2}-1)(2t)}{(t^{2}+1)^{2}}$$
$$=\frac{4t^{3}+4t-4t^{3}+2t}{(t^{2}+1)^{2}}$$
$$=\frac{6t}{(t^{2}+1)^{2}}$$

$$\frac{\partial y}{\partial t}=\frac{(t^{2}+1)(3)-(3t)(2t)}{(t^{2}+1)^{2}}$$
$$=\frac{3t^{2}+3-6t^{2}}{(t^{2}+1)^{2}}$$
$$=\frac{3-3t^{2}}{(t^{2}+1)^{2}}$$
$$=\frac{3(1-t^{2})}{(t^{2}+1)^{2}}$$

$$\frac{\partial x}{\partial t}+\frac{\partial y}{\partial t}= \frac{6t+3-3t^{2}}{(t^{2}+1)^{2}}=\frac{-3(t^{2}-2t-1)}{(t^{2}+1)^{2}}$$

5. Dec 28, 2009

yungman

What you are showing here is not a complex number and is not the same as what we did.

1) For complex number, you separate the real part and the imaginary part totally. Your denominator is a complex number, so we multiply both the numerator and denomator by the complex conjugate of the denomator to make the denominator a real number.
2) Then you apply partial derivative of the real and imaginary part separately respect to t.

In you case, the denominator is (t-i), so we multiply with (t+i)/(t+i) to get the denominator to (t^2+1).

6. Dec 28, 2009

colgon

I understand about splitting the real and Im parts.

That is where I got x and y from. x being the real part, y being the Im part.

What I don't understand is how to apply partial derivitives to obtain the correct answer??

7. Dec 28, 2009

yungman

$$\stackrel{\rightarrow}{v}=\frac{\partial\stackrel{\rightarrow}{z}}{\partial t}=\hat{x} \frac{\partial x}{\partial t}+\hat{y}\frac{\partial y}{\partial t}$$

$$\stackrel{\rightarrow}{a}=\frac{\partial\stackrel{\rightarrow}{v}}{\partial t}$$

From post #3, you get your x and y. Partial derivative is exactly like taking the derivatives, it just tell you the derivatives is only part of the total derivative.

8. Dec 30, 2009

colgon

My attempt at finding $$\frac{\partial x}{\partial t} and \frac{\partial y}{\partial t}$$ is above, is this correct?