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Am I doing it right?

  1. Nov 19, 2005 #1
    Hi all,

    we were given some recommended home excercises and since we hadn't been given the right results, I'm curous if I'm doing it right:

    1. Find [itex]\frac{dH}{dt}[/itex], where

    [tex]
    H(t) = sin (3x) - y
    [/tex]

    [tex]
    x = 2t^2 - 3
    [/tex]

    [tex]
    y = \frac{t^2}{2} - 5t + 1
    [/tex]


    This is what I did:

    [tex]
    \frac{dH}{dt} = \frac{\partial H}{\partial x}\frac{du}{dt} + \frac{\partial H}{\partial y}\frac{dv}{dt} = 12t\cos (3x) - t + 5
    [/tex]

    Which seems kind of strange to me. Should I replace x with t? It would be this:

    [tex]
    \frac{dH}{dt} = 12t\cos (6t^2 - 9) - t + 5
    [/tex]

    Is it ok?

    Thank you.
     
    Last edited: Nov 19, 2005
  2. jcsd
  3. Nov 19, 2005 #2

    Fermat

    User Avatar
    Homework Helper

    Yes, it's fine.

    But you should have had dx/dt ,not du/dt, and dy/dt, not dv/dt, in the expression for dH/dt.

    Why did it seem strange to you ?
     
  4. Nov 19, 2005 #3
    I'm sorry, I forgot to mention that I defined

    [tex]
    u(t) = 2t^2 - 3
    [/tex]

    [tex]
    v(t) = \frac{t^2}{2} - 5t + 1
    [/tex]

    So which form is ok? The first or the second?

    I find the first one strange because of the mixing of x and t.
     
  5. Nov 20, 2005 #4

    Fermat

    User Avatar
    Homework Helper

    OK.
    I'm not sure why you defined u and v like that but your mixing of symbols like that is a bit confusing and your differential eqn is wrong the way it is written.
    By that I mean ...

    [tex]\frac{dH}{dt} = \frac{\partial H}{\partial x}\cdot\frac{du}{dt} + \frac{\partial H}{\partial y}\cdot\frac{dv}{dt}[/tex]

    is just plain wrong where it has written,

    [tex]\frac{\partial H}{\partial x}\cdot\frac{du}{dt}[/tex]

    You should have,

    [tex]\frac{\partial H}{\partial x}\cdot\frac{dx}{dt}[/tex]
    or
    [tex]\frac{\partial H}{\partial u}\cdot\frac{du}{dt}[/tex]

    The [tex]\partial u[/tex] should sort of "cancel" with [tex]du[/tex] to give[tex]\frac{\partial H}{dt}[/tex] so that you have [tex]\frac{dH}{dt}[/tex] on one side of the "=" sign and [tex]\frac{\partial H}{dt}[/tex] (twice) on the other side, making both sides consistent with each other.

    So, the form of your differential eqn should be either,

    [tex]\frac{dH}{dt} = \frac{\partial H}{\partial x}\cdot\frac{dx}{dt} + \frac{\partial H}{\partial y}\cdot\frac{dy}{dt}[/tex]
    or
    [tex]\frac{dH}{dt} = \frac{\partial H}{\partial u}\cdot\frac{du}{dt} + \frac{\partial H}{\partial v}\cdot\frac{dv}{dt}[/tex]

    depending upon what definitions (u = f(t) or x = f(t)) you are using.

    You are finding dH/dt, which is the rate at which H varies as t varies. i.e. dH/dt is a function of t, so should be expressed in terms of t.
    You can work out things using x and y,or u and v (sometimes using a substitution can simplify working) but your final result shoud be converted to give an expresion that involves t only.

    HTH
     
    Last edited: Nov 20, 2005
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