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Math Help - Another differential I can't solve

  1. #1
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    Another differential I can't solve

    The equation is y= b/a*sqrt(a^2 - x^2)

    the solution apparently has a y term in it but I can't figure out where it comes from. Can anyone give me a hand?
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  2. #2
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    Quote Originally Posted by MrJoe2000 View Post
    The equation is y= b/a*sqrt(a^2 - x^2)

    the solution apparently has a y term in it but I can't figure out where it comes from. Can anyone give me a hand?
    A "y" term?

    What is the derivative of \sqrt{a^2 - x^2}?

    The only thing I can think of is that your answer has the y subbed back in to "clean" things up a bit. Your solution is (you can fill in the details)
    y' = - \frac{b}{a} \frac{x}{\sqrt{a^2 - x^2}} = - \frac{b}{a} \frac{x}{y}

    Is that what your final answer is supposed to look like? If not I have no idea what your "y" term in your question means.

    -Dan
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  3. #3
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    Thanks Dan. I got up to where you did but missed that they subbed the y back into the equation. The book answer is -(b^3*x)/(a^2*y)
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  4. #4
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    you are correct Joe. You will get that answer
    substitute
    \sqrt{a^2-x^2}=\frac{ay}{b}
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  5. #5
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    How do you get the cubed and squared terms? I got to the point that Dan did but I'm not sure how to get from there to the final answer.
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  6. #6
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    taking
    \frac{1}{\sqrt{a^2 - x^2}}=\frac{b}{ay}

    y' = - \frac{b}{a} \frac{x}{\sqrt{a^2 - x^2}} = - \frac{bx}{a} \frac{b}{ay}

    so your answer should be
    y=-\frac{b^2x}{a^2y}

    i dont think it is b^3 in your answer
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  7. #7
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    Can you explain why after you obtain the -b/a*x/y answer you continue to substitute in? Why wouldn't I simply stop there?
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  8. #8
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    if you dont substitute then your answer would be
    y' = - \frac{b}{a} \frac{x}{\sqrt{a^2 - x^2}}

    not y' = - \frac{b}{a} \frac{x}{y} i guess Dan made some mistake in substituting!!!
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  9. #9
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    OOh I see I need to substitute in for 1/y not y itself.
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  10. #10
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    im not clear with what you just said?
    you are substiuting to get y in your answer
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  11. #11
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    I understand why you needed to substitute b/(a*y) and not just y.
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  12. #12
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    yes.
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by MrJoe2000 View Post
    Thanks Dan. I got up to where you did but missed that they subbed the y back into the equation. The book answer is -(b^3*x)/(a^2*y)
    Yes, sorry about that. For some reason I decided to use y = \sqrt{a^2 - x^2} in the final substitution.

    I mean, I was just testing you. Yeah, that was it! Uhhhh...I'm gonna go now. (slinks away into a dark alley)

    -Dan
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