How Do You Calculate Velocity and Acceleration from Complex Functions?

[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 acceleration 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.

 

[HallsofIvy]

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 acceleration 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.

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.

 

[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.

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

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

 

[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?

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



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

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

 

[yungman]

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?

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



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

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

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).

 

[colgon]

I understand about splitting the real and I am parts.

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

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

 

[yungman]

I understand about splitting the real and I am parts.

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

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

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

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

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.

 

[colgon]

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