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Math Help - stationary pt. of a curve

  1. #1
    Member maybeline9216's Avatar
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    stationary pt. of a curve

    Help me with this.

    Can you tell me what is your value of x.Thanks
    i dunno where went wrong..
    Attached Thumbnails Attached Thumbnails stationary pt. of a curve-qn10.jpg  
    Last edited by maybeline9216; October 10th 2008 at 03:00 AM.
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  2. #2
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    Can't read the text in the figure.

    To get the stationary point, find the dy/dx and then set that to zero. The x there will give the stationary point.
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  3. #3
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    Quote Originally Posted by maybeline9216 View Post
    Help me with this. Thanks
    Fortunately I have a degree in Egyptology majoring in heiroglyphics. The curve is y = 3 e^{-x/2} + e^{x/2}.

    Solve \frac{dy}{dx} = 0 for x and substitute the value into y:

    0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}

    \Rightarrow 0 = -3 + e^x \Rightarrow x = \ln 3.

    Therefore y = 3 e^{-(\ln 3) /2} + e^{(\ln 3)/2} = 3 = 3 e^{\ln \left(1/\sqrt{3}\right)} + e^{\ln \sqrt{3}} = \frac{3}{\sqrt{3}} + \sqrt{3} = 2 \sqrt{3}.
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  4. #4
    Member maybeline9216's Avatar
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    Quote Originally Posted by ticbol View Post
    Can't read the text in the figure.

    To get the stationary point, find the dy/dx and then set that to zero. The x there will give the stationary point.
    Click the attached file and then click it again to get a better view
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  5. #5
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    Quote Originally Posted by maybeline9216 View Post
    Click the attached file and then click it again to get a better view
    Hmmpph .... So in fact only my mathematical skills were required.
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  6. #6
    Member maybeline9216's Avatar
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    0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}
    can you tell me how to get the value of x from here.


    Its because i tried to use the substitution method to get x

    For eg. Let x=e^x/2
    But i got a different answer.Can you show me another method?

    P.S.My dy/dx is correct
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  7. #7
    Member maybeline9216's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Fortunately I have a degree in Egyptology majoring in heiroglyphics.
    Wow! What is that?? lols
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  8. #8
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    Quote Originally Posted by maybeline9216 View Post
    0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}
    can you tell me how to get the value of x from here.


    Its because i tried to use the substitution method to get x

    For eg. Let x=e^x/2
    But i got a different answer.Can you show me another method?

    P.S.My dy/dx is correct
    Quote Originally Posted by Mr Fantastic

    [snip]
    0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}

    Mr F edit: Multiply both sides by {\color{red}2 e^{x/2}}:

    \Rightarrow 0 = -3 + e^x \Rightarrow x = \ln 3.

    [snip]
    ..
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  9. #9
    Member maybeline9216's Avatar
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    Thanks!!that's so much easier!
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  10. #10
    Member maybeline9216's Avatar
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    3 = 3 e^{\ln \left(1/\sqrt{3}\right)} + e^{\ln \sqrt{3}} = \frac{3}{\sqrt{3}} + \sqrt{3} = 2 \sqrt{3}

    Sorry.... i have another qn, how do u get this(see above) ?

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  11. #11
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    Quote Originally Posted by maybeline9216 View Post
    3 = 3 e^{\ln \left(1/\sqrt{3}\right)} + e^{\ln \sqrt{3}} = \frac{3}{\sqrt{3}} + \sqrt{3} = 2 \sqrt{3}

    Sorry.... i have another qn, how do u get this(see above) ?
    e^{\ln A} = A \, ....
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