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Determine M(x)

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data
    knowing that -9x^2 - 6y^3 = -10 => [tex]\frac{dy}{dx}[/tex] = [tex]\frac{M(x)}{18y^2}[/tex],

    determine function M.

    i would like to know how i can find M, im not sure where to start
     
    Last edited: Oct 5, 2007
  2. jcsd
  3. Oct 5, 2007 #2

    Dick

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    Differentiate the equation on the left with respect to x. Then solve for dy/dx.
     
  4. Oct 5, 2007 #3
    the equation on the left gives me -18x, what does the left one help me with getting M?
     
  5. Oct 5, 2007 #4

    Dick

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    d/dx(y)=dy/dx, not zero. Use the chain rule to find d/dx(6*y^3).
     
  6. Oct 5, 2007 #5
    i don´t understand very well...is there a way i can put it in mathematica 6 to just get the solution?
     
    Last edited: Oct 5, 2007
  7. Oct 5, 2007 #6

    Dick

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    This isn't difficult. You should be thinking of y as y(x), a function of x. E.g. d/dx(y(x)^2)=2*y(x)*dy(x)/dx. The dy/dx comes from the chain rule. If you want to try to find this in a textbook, look up 'implicit differentiation'.
     
  8. Oct 5, 2007 #7
    i got dy/dx = -x/y^2...now how can i apply to get M? sry for dumb questions :S
     
  9. Oct 5, 2007 #8
    The solution is the following
    [tex]-9x^{2}-6y^{3}=-10\Rightarrow[/tex]
    [tex]-9x^{2}=6y^{3}-10\Rightarrow[/tex]
    [tex]d(-9x^{2})=d(6y^{3}-10)\Rightarrow[/tex]

    [tex]-18xdx=18y^{2}dy[/tex] since [tex]d(-10)=0[/tex] the derivative of a constant is 0

    [tex]\frac{-18xdx}{dx}=18y^{2}\frac{dy}{dx}\Rightarrow[/tex]
    [tex]-18x=18y^{2}\frac{dy}{dx}\Rightarrow[/tex]
    [tex]\frac{dy}{dx}=\frac{-18x}{18y^{2}}\Rightarrow[/tex]
    Since we know that
    [tex]\frac{dy}{dx}=\frac{M(x)}{18y^{2}}\Rightarrow[/tex]
    [tex]M(x)=-18x[/tex]

    You should have known the solution though, was pretty easy, my suggestion is to read more about implicit differentiation since many times can solve problems where other ways fail.
     
  10. Oct 5, 2007 #9

    Dick

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    Good! Multiply by 1=18/18. So your answer could also be written -18x/(18*y^2).
     
  11. Oct 5, 2007 #10

    Dick

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    Do not solve complete problems for posters. That's against forum rules. Help them to solve it themselves.
     
  12. Oct 5, 2007 #11
    sorry wont happen again, i tried to help because she was little lost
     
  13. Oct 5, 2007 #12
    what helped me most was the term 'implicit differentiation', i made a quick search and found some good examples...thanks for the help......he
     
  14. Oct 5, 2007 #13
    i have another one which is -7y*e^10xy = 5 => M(y) / 1 + 10xy, i did the dy/dx to resolve it and i get -70*y^2* e^10xy. where do i get M(y)
     
  15. Oct 6, 2007 #14

    HallsofIvy

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    Do you mean -7ye^(10xy)= 5 and dy/dx= M(y)/(1+ 10xy)? Please use parentheses to make it clearer.
    No, the derivative is NOT -70y^2 e^(10xy). In fact, your dy/dx should not have a "e^(10xy) in it! How about showing exactly what you did?
     
  16. Oct 6, 2007 #15
    [tex]-7\,y{e^{10\,{\it xy}}}=5[/tex] ==> [tex]{\frac {{\it dx}}{{\it dy}}}={\frac {M \left( y \right) }{1+10\,{\it
    xy}}}
    [/tex]


    i did [tex]{\frac {d \left( -7\,{{\it ye}}^{10\,{\it xy}} \right) }{{\it dy}}}={
    \frac {d \left( 5 \right) }{{\it dy}}}
    [/tex]

    is it wrong way ?
     
    Last edited: Oct 7, 2007
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