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Homework Help: Partial derivatives.

  1. Dec 6, 2006 #1

    MathematicalPhysicist

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    im given the function f(r,a,b) and z=rcos(a) y=rsin(a)sin(b) x=rsin(a)cos(b)
    now i need to find the partial derivative of f'_y, without solving r,a,b in terms of x,y,z, what that i got is:
    f'_y=f'_a*a'_y+f'_r*r'_y+f'_b*b'_y
    the answer should include the derivatives of f wrt r,a,b, which i think i did ok, (i only needed to calculate a'_y,b'_y,r'_y).
    but i have another question and it's to solve r,a,b in terms of x,y,z which i did, and then to calculate the above derivative directly by the chain rule.
    i think i need to use here a jacobian, so i defined the implicit functions: F=z-rcos(a) G=y-rsin(a)sin(b) H=x-rsin(a)cos(b), my proble is that in order to calclualte it: i need to know what's f, cause:
    f'_y=-J((F,G,H)/(x,y,z))/J((F,G,H)/(x,f,z))
    the question is how to do it?

    thanks in advance.
     
  2. jcsd
  3. Dec 7, 2006 #2
    cant u use chain rule?
     
  4. Dec 7, 2006 #3

    HallsofIvy

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    You would use the Jacobian in changing variables in integration.

    To differentiate in different variables, use the chain rule.
     
  5. Dec 8, 2006 #4

    MathematicalPhysicist

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    so how would i calclulate the derivatives of:
    f'_a f'_r and f'_b, i mean after i solved for a,r,b in terms of x,y,z:
    f'_y=f'_r*r'_y+f'_a*a'_y+f'_b*b'_y
    and
    f'_r=f'_x*x'_r+f'_y*y'_r+f'_z*z'_r
    and then to substitue f'_r in the first equation and this way also to do with f'_b and f'_a, is this correct?
     
  6. Dec 8, 2006 #5

    HallsofIvy

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    Yes, that is correct.

    And it is easy to see that [itex]r= \sqrt{x^2+ y^2+ z^2}[/itex], [itex]b= arctan(\frac{y}{x})[/itex], and [itex]a= arctan(\frac{\sqrt{x^2+y^2}}{z}[/itex].
     
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