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Math Help - A Hard Calculus Problem

  1. #1
    Member iPod's Avatar
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    A Hard Calculus Problem

    The variables t and x are related by t = x+√(x^2+2bx+c) where b and c are constants and
    b^2<c. Show that

    dx/dt=(t-x)/(t+b)

    and hence integrate

    1/√(x^2+2bx+c)

    Verify by direct integration that your result holds also in the case b^2 = c if x+b > 0 but that
    your result does not hold in the case b^2 = c if x + b < 0 .

    [and come someone tell me how to use the math symbols at this site.. i had to do everything in microsoft word ]
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by iPod View Post
    The variables t and x are related by t = x+√(x^2+2bx+c) where b and c are constants and

    b^2<c. Show that

    dx/dt=(t-x)/(t+b)

    and hence integrate

    1/√(x^2+2bx+c)

    Verify by direct integration that your result holds also in the case b^2 = c if x+b > 0 but that
    your result does not hold in the case b^2 = c if x + b < 0 .

    [and come someone tell me how to use the math symbols at this site.. i had to do everything in microsoft word ]
    Differentiate implicitly x with respect to t. This thread does not belong in this sub-forum.
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  3. #3
    Member iPod's Avatar
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    I managed to differentiate, however Im just stuck at the intergration part,
    and doesnt this thread belong here??
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  4. #4
    MHF Contributor
    Grandad's Avatar
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    Hello iPod

    Welcome to Math Help Forum!
    Quote Originally Posted by iPod View Post
    The variables t and x are related by t = x+√(x^2+2bx+c) where b and c are constants and
    b^2<c. Show that

    dx/dt=(t-x)/(t+b)
    [CENTER]
    t = x+\sqrt{x^2+2bx+c}

    \Rightarrow (t-x)^2 = x^2+2bx+c

    \Rightarrow 2(t-x)\left(\frac{dt}{dx}-1\right)=2x+2b

    \Rightarrow \frac{dt}{dx} = 1+\frac{x+b}{t-x}=\frac{t+b}{t-x}

    \Rightarrow \frac{dx}{dt} =\frac{t-x}{t+b}

    and hence integrate

    1/√(x^2+2bx+c)
    t-x = \sqrt{x^2+2bx+c}

    \Rightarrow \frac{dx}{dt}=\frac{\sqrt{x^2+2bx+c}}{t+b}

    \Rightarrow \int \frac{1}{\sqrt{x^2+2bx+c}}\,dx=\int\frac{1}{(t+b)}  \,dt

    = \ln (t+b)+C

    =\ln(b+x+\sqrt{x^2+2bx+c})+C, provided b+x+\sqrt{x^2+2bx+c}>0

    Can you finish up?

    Grandad


    PS
    [and come someone tell me how to use the math symbols at this site.. i had to do everything in microsoft word ]
    The language is called LaTeX, and you'll find some help on this site here: http://www.mathhelpforum.com/math-help/latex-help/

    I also find the Wikipedia site helpful, here: LaTeX/Mathematics - Wikibooks, collection of open-content textbooks

    When you've written some LaTex code, select what you've written and click the \Sigma button on this editor's toolbar, and the editor will insert [tex]...[/tex] tags around the selection. Use 'Preview Changes' to check that it's OK.

    To see any of the code I've used in my solution above, just click-left on any line and the code will appear in a separate pop-up window. Go on, try it now!

    Grandad
    Last edited by Grandad; July 6th 2009 at 12:13 AM. Reason: Add PS
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  5. #5
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    Oh yeah , it is known as Euler's substitution for integration .

    We can solve integrals in this form  \int \frac{ P(x) }{\sqrt{x^2 + 2ax + b}}~dx ~~ using Euler's substitution  t = x + \sqrt{x^2 + ax + b}

     dt = ( 1 + \frac{x + a }{\sqrt{x^2 + 2ax + b}})~dx

     dt = [ \frac{ ( \sqrt{x^2 + 2ax + b}+x) + a}{ \sqrt{x^2 + 2ax + b}}]~dx

    Therefore , the integral becomes

     \int \frac{ P( f(t)) }{ (t+a)}~dx~dt

    Then , what does f(t) refer to ?

    Since

     t = x + \sqrt{x^2 + 2ax + b}
     t^2 - 2tx = 2ax + b

     x = \frac{t^2 -b}{2(a+t)}

     x = f(t) = \frac{t^2 -b}{2(a+t)}



    For more details , you can take a look at http://planetmath.org/encyclopedia/E...tegration.html .

    Euler's substitution is a powerful tools indeed . Make use of it !!
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  6. #6
    Member iPod's Avatar
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    thx

    Thank you very much Grandad and Simple, however I am not very familiar with this type of integration - is there a certain name for this type of integration and also is there a step by step method for it??
    Thank you for your time,
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