Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Find v & a in complex plain

  1. Dec 27, 2009 #1
    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=[tex]\frac{i+2t}{t-i}[/tex]

    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


    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=[tex]\frac{3}{t^{2}+1}[/tex] as this is found using [tex] \frac{dz}{dt}[/tex] but I am asked to verify this using x and y as functions of t.
  2. jcsd
  3. Dec 27, 2009 #2


    User Avatar
    Science Advisor

    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:
    [tex]z=\frac{i+2t}{t-i}\frac{t+i}{t+i}= \frac{2t^2- 1+ i(3t)}{t^2+1}[/tex][/b]
    [tex]= \frac{2t^2-1}{t^2+1}+ i\frac{-t}{t^2+1}[/tex]
    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.
  4. Dec 27, 2009 #3
    I want to verify first: You let [tex]\stackrel{\rightarrow}{z}[/tex] be the position vector on x-y plane where [tex]\stackrel{\rightarrow}{z}=\frac{2t+i}{t-i}[/tex]

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

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

    [tex]\stackrel{\rightarrow}{v}=\frac{\partial\stackrel{\rightarrow}{z}}{\partial t}=\hat{x} \frac{\partial x}{\partial t}+\hat{y}\frac{\partial y}{\partial t}[/tex]

    [tex]\stackrel{\rightarrow}{a}=\frac{\partial\stackrel{\rightarrow}{v}}{\partial t}[/tex]
    Last edited: Dec 28, 2009
  5. Dec 28, 2009 #4
    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?

    [tex] \frac{\partial x}{\partial t}=\frac{(t^{2}+1)(4t)-(2t^{2}-1)(2t)}{(t^{2}+1)^{2}}[/tex]
    [tex] =\frac{4t^{3}+4t-4t^{3}+2t}{(t^{2}+1)^{2}}[/tex]

    [tex] \frac{\partial y}{\partial t}=\frac{(t^{2}+1)(3)-(3t)(2t)}{(t^{2}+1)^{2}}[/tex]
    [tex] =\frac{3t^{2}+3-6t^{2}}{(t^{2}+1)^{2}}[/tex]
    [tex] =\frac{3-3t^{2}}{(t^{2}+1)^{2}}[/tex]
    [tex] =\frac{3(1-t^{2})}{(t^{2}+1)^{2}}[/tex]

    [tex] \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}}[/tex]
  6. Dec 28, 2009 #5
    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).
  7. Dec 28, 2009 #6
    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??
  8. Dec 28, 2009 #7
    [tex]\stackrel{\rightarrow}{v}=\frac{\partial\stackrel{\rightarrow}{z}}{\partial t}=\hat{x} \frac{\partial x}{\partial t}+\hat{y}\frac{\partial y}{\partial t}[/tex]

    [tex]\stackrel{\rightarrow}{a}=\frac{\partial\stackrel{\rightarrow}{v}}{\partial t}[/tex]

    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.
  9. Dec 30, 2009 #8
    My attempt at finding [tex] \frac{\partial x}{\partial t} and \frac{\partial y}{\partial t}[/tex] is above, is this correct?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook