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Homework Help: Separable Equations help

  1. Mar 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the y-intercept of the curve that passes through the point (4/9,1) and whose slope at (x,y) is −2/y3.

    2. Relevant equations

    3. The attempt at a solution
    woudnt the y-int be y=-2x/y3 + 5.5 ?
    i got it by plugging the slope into m in the equation y=mx+b
    then plugged in the given points to get b..
    but the question wants a numerical solution :S
  2. jcsd
  3. Mar 27, 2010 #2


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    That's the equation of a straight line, with constant slope, [itex]m[/itex]. The slope you are given isn't constant, it depends on [itex]y[/itex]. Your curve is not a straight line.

    What is the slope of any curve [itex]f(x)[/itex] at the point [itex](x,f(x))[/itex]?...Use that.
  4. Mar 27, 2010 #3


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    That makes no sense whatsoever. I think you know that. They want the y intercept of the curve. I think you have to solve for the curve. Whatever happened to the "separable equations" part of your post title? Shouldn't you separate dy/dx=(-2/y^3) and solve it??
  5. Mar 28, 2010 #4
    okay so i get dy/dx=(-2/y^3)
    ==> y^3 dy = -2dx
    ==> (y^4)/4 = -2x + C
    ==> y= sqrt4(-8x + 4C)

    now what do i do?
  6. Mar 28, 2010 #5


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    They gave you a point that it passes through, so find, the constant C.
  7. Mar 28, 2010 #6
    well if i plug (9/4, 1) in, i get C=4.75
    from there y= sqrt4(-8x + 19)
    now what do i do with this?
  8. Mar 28, 2010 #7


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    So you have a curve passing through the point given, with the slope that they require. What does the question tell you to do with it?
  9. Mar 28, 2010 #8
    i got 2.087
    once i got y = (-8x + 19)^0.25
    y-int is when x=0 so y=19^0.25 = 2.087

    its weird how the slope they gave us dont even help.. its there to make the question scary i guess.
  10. Mar 28, 2010 #9


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    You used the slope to figure out what y was as a function of x. How can you plug in x=0 if you don't know what the function is?
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