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Partial derivative problems

  1. Feb 8, 2005 #1
    Two homework problems I can't get.

    (1) The question is find the first partial derivatives of the function. The problem is that the function in this problem is
    [tex] f(x, y) = \int_{y}^{x} \cos{t^2} dt [/tex]
    The main obstacle is getting past this function. I can't integrate it and neither can my calculator. Is there a way to do this problem without integrating, or how do you integrate this function?

    (2) The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasing as the resistor heats up. Use Ohm's Law, V = IR, to find how the current I is changing at the moment when R = 400[itex]\Omega[/itex], I = 0.08 A, [itex]\frac{dV}{dt}[/itex] = -0.01 V/s, and [itex]\frac{dR}{dt}[/itex] = 0.03 [itex]\Omega[/itex]/s.
     
  2. jcsd
  3. Feb 8, 2005 #2

    dextercioby

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    1.How about applying the fundamental theorem of calculus...??Both parts of it.

    2.Differentiate wrt time the Ohm's law and plug in tht #-s.

    Daniel.
     
  4. Feb 9, 2005 #3

    xanthym

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    {#1}

    NO integration needed here. Use the Fundamental Theorem of Calculus:

    [tex] f(x, y) = \int_{y}^{x} \cos{t^2} dt [/tex]

    [tex] \frac {\partial f(x,y)} {\partial x} = \cos{x^2} [/tex]

    [tex] \frac {\partial f(x,y)} {\partial y} = -\cos{y^2} [/tex]


    Incidentally, in the more general case, we have:

    [tex] f(x, y) = \int_{y}^{g(x)} h(t) dt [/tex]

    [tex] \frac {\partial f(x,y)} {\partial x} = h(g(x))*g'(x) [/tex]


    {#2}

    V = I*R
    -----> I = V/R
    Using the Quotient Rule:
    (dI/dt) = {(dV/dt)*R - V*(dR/dt)}/(R^2)
    (dI/dt) = {(dV/dt)*R - (I*R)*(dR/dt)}/(R^2)
    (dI/dt) = {(dV/dt) - (I)*(dR/dt)}/(R)

    Placing given values into the above equation, we get:
    (dI/dt) = {(-0.01 V/s) - (0.08 A)*(0.03 ohm/s)}/(400 ohm)
    (dI/dt) = (-3.1)x10^(-5) Amps/sec
    ~
     
    Last edited: Feb 9, 2005
  5. Feb 9, 2005 #4

    dextercioby

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    Well,the general idea of this specific forum is not to solve his problems entirely,but to help him solve them by himself.
    I hope u'l get it.

    Daniel.
     
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