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Was my test wrong?

  1. Nov 3, 2004 #1

    rpc

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    One of the problems on my AP Calc test:

    The point (1,9) lies on the graph of an equation y=f(x) for which dy/dx = 4x*y^(1/2) where x> or = to 0 and y > or = 0

    When x=0 y=?

    Seperation of variables:

    dy/y^(1/2) = 4x dx

    Integrate :

    2 y^(1/2) = 2x^2 +C now, if you do C now:

    2 * (9)^(1/2) = 2 (1)^2 + C
    6 = 2 + C
    C= 4 ,

    plug in 0 for X and 4 for C:

    2 y^(1/2) = 2 (0)^2 + 4

    y^1/2 = 2
    y = 4 when x =0 <--- thats what the answer key said/ teacher marked


    What I did:

    Seperation of variables:

    dy/y^(1/2) = 4x dx

    Integrate :

    2 y^(1/2) = 2x^2

    solve for Y

    y^(1/2) = x^2
    y = x^4 + C

    solve for C

    9 = (1)^4 + C
    C = 8

    y = x^4 + 8
    Solve for y(0):
    y=8 <--- thats the answer I got, and its a multiple choice question

    Since it satisfies the initial condition (1,9) and the derivitive, then 8 is a correct answer, right?

    If you take the deriv of where I got y = x^4 from, y^(1/2) = x^2, you still get dy/dx = 4x*y^(1/2), and I merely simplified the equation to put it in terms of Y like it says in the intro: "The point (1,9) lies on the graph of an equation y=f(x)"

    Any thoughts? -Thanks
     
  2. jcsd
  3. Nov 3, 2004 #2
    Have you actually checked that your solution satisfies dy/dx = 4x*y^(1/2) for all x >= 0? (It doesn't btw).

    The problem lies between these two steps:

    You forgot the constant of integration. You can't "add it in later" like you did now. Try it and you'll see why.
     
  4. Nov 3, 2004 #3

    JasonRox

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    Homework Helper
    Gold Member

    Don't you have to add C first?

    You should know better than that. ;)

    Like Muzza said, you can't add it later.
     
  5. Nov 3, 2004 #4

    rpc

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    But, the deriv is not unique to 1 equation,

    and isnt the C somewhat arbitrary

    for example:

    Seperation of variables:

    dy/y^(1/2) = 4x dx

    Integrate :

    2 y^(1/2) = 2x^2 +C1

    Simplify

    y^(1/2) = x^2 + C2 <--- and the deriv of this is still dy/dx = 4x*y^(1/2), then all I did was solve for Y

    solve for Y
    y = x^4 + C3

    What am I doing wrong here?, or do you have to solve for C before you can simplify?
     
    Last edited: Nov 3, 2004
  6. Nov 3, 2004 #5

    StatusX

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    Homework Helper

    come on, this is easy. howd you get this second equation? by squaring both sides? try it again.
     
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