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Error computing total derivative

  1. Oct 22, 2011 #1
    find the total derivative dz/dt, given:
    [tex]z=(x^2)-8xy-(y^3)[/tex] where [tex]x=3t[/tex] and [tex]y=1-t[/tex]

    my steps look like this, can someone point out where i am going wrong, please?
    [tex]z'=2x-8(x'y+y'x)-3y^2[/tex] where
    [tex]x'=3[/tex] and [tex]y'=-1[/tex]

    [tex]2(3t)-8[3(1-t)+(-1)(3t)]-3(1-t)^2[/tex]
    [tex]=6t-8[3-3t+(-3t)]-3(1-2t+t^2)[/tex]
    [tex]=6t-8(3-6t)-3(1-2t+t^2)[/tex]
    [tex]=6t-24+48t-3+6t-3t^2[/tex]
    [tex]=12t+48t-27-3t^2[/tex]
    [tex]=-3t^2+60t-27[/tex]

    however, the correct answer seems to be:
    [tex]3t^2+60t-21[/tex]

    i'm going nuts trying to figure out where i went wrong. please help.
     
    Last edited: Oct 23, 2011
  2. jcsd
  3. Oct 22, 2011 #2

    LCKurtz

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    Use just tex, not latex in your tex brackets. You want to use the formula:

    [tex]\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt}
    +\frac{\partial z}{\partial y}\frac{dy}{dt}
    [/tex]
     
  4. Oct 23, 2011 #3
    thanks for the response, i corrected the formatting. the formula is very helpful.

    is there any reason my steps don't work? it seems like i'm applying all the basic principles correctly, so i feel stumped.
     
  5. Oct 23, 2011 #4

    lurflurf

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    If

    z=x2-8x y-y3

    then

    z'=2x{x'}-8(x' y+x y'))-3y2{y'}

    you have omitted the terms inside the {}

    or factorize into chain rule form like LCKurtz

    z'=(2x-y)x'-(8x+3y2)y'
     
    Last edited: Oct 23, 2011
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