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Slope of a curve and at a point

  1. May 8, 2017 #1
    1. The problem statement, all variables and given/known data
    http://i.imgur.com/In40pGm.png
    In40pGm.png
    Answer: C
    2. Relevant equations
    f'(x)=slope=(y1-y2)/(x1-x2)

    3. The attempt at a solution
    I can't even list a valid formula for that...
    like I tried to integrate f'(x), but f(x) is with y so I don't think I am thinking in the right direction.
    What are the steps in order to get the correct answer?
    Thank you very much.
     
  2. jcsd
  3. May 8, 2017 #2

    Ssnow

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    It is a plane curve? If it is ## y=f(x)## so you can write ##\frac{f'(x)}{f(x)^2}=\frac{1}{x^3}## and integrating both members you will find the equation of ##f(x)## ...
    Ssnow
     
  4. May 8, 2017 #3

    Ray Vickson

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    Solve the differential equation
    $$\frac{dy}{dx} = \frac{y^2}{x^3}$$
    The general solution ##y(x)## will contain an unknown constant ##c##, whose value can be obtained by using the given condition ##y(1) = 1##.
     
  5. May 8, 2017 #4
    I know that (1,1) is for solving the constant. however, i cant find the equation.
    f'(x)=y^2/x^3
    y=f(x)=y^2/x^2*(-1)+c
    This equation seems unreasonable...
    can you help me in this?
    thanks
     
  6. May 8, 2017 #5

    Ssnow

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    You must integrate the differential equation with the separable variables method ..
     
  7. May 8, 2017 #6
    Like what integration method? I cant think of any seems applicable..
    Any suggestion? Thanks
     
  8. May 8, 2017 #7

    vela

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    You have to rewrite the differential equation so all the x's are on one side and all the y's are on the other. Then you can integrate each side.
     
  9. May 8, 2017 #8

    Ssnow

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    Solve the two integrals ##\int \frac{dy}{y^2} =\int \frac{dx}{x^{3}}+c##, you will obtain something as ##y(x)= ......(x)+c##...
    Ssnow
     
  10. May 8, 2017 #9
    how does this come?
    y'=f'(x)=y^2/x^3
    y=f(x)=y^2/x^2*(-2)+c
    1/y=1/x^2*(-2)+c
    you mean like this? or have I calculated anything wrong?
     
  11. May 8, 2017 #10

    Ssnow

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    From ##\frac{dy}{dx}=\frac{y^2}{x^3}## you treat formally ##dx## and ##dy## as certain ''quantities''. Algebraically you can separate ##x## from ##y## obtaining

    ##\frac{dy}{y^2}=\frac{dx}{x^3}##
    after you can integrate both members ...
    see https://en.wikipedia.org/wiki/Separation_of_variables
    Ssnow
     
  12. May 8, 2017 #11
  13. May 8, 2017 #12
    You can't go from 1/y = 1/2x2 to y = 2x2 + C. It must be 1/y = 1/2x2 + C. You add the integration constant in the integration step, not after any subsequent manipulations. Then y = 2x2/(1+2Cx2).
     
  14. May 8, 2017 #13
    thank you for all of your help
    i can finally solved it
     
  15. May 8, 2017 #14

    Ssnow

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    Yes, as @mjc123 said you must put the constant ##C## after the second integration and consider it in all algebraic passages...

    Ssnow
     
  16. May 9, 2017 #15
    Is it a problem of RSM?
     
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