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Math Help - Curvature

  1. #1
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    Curvature

    Calculate the curvature at the top of the curve of the parabola y=ax^2+bx+c
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by lynch-mob View Post
    Calculate the curvature at the top of the curve of the parabola y=ax^2+bx+c
    \kappa = \frac{|y''|}{(1 + y'^2)^{3/2}}

    In your case, since the slope is zero at the maximum point, so the curvature is 2|a|.

    -Dan

    Edit: I forgot to add the factor of 2 due to taking the derivatives of y, not to mention the absolute value bars. Sorry about that.
    Last edited by topsquark; June 18th 2008 at 03:56 AM.
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    Quote Originally Posted by topsquark View Post
    \kappa = \frac{|y''|}{(1 + y')^{3/2}}

    In your case, since the slope is zero at the maximum point, so the curvature is a.

    -Dan

    Before I start, shouldnīt the formula be \kappa=\frac{|y''|}{(1+y'^2)^\frac{3}{2}}
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  4. #4
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    Quote Originally Posted by lynch-mob View Post
    Before I start, shouldnīt the formula be \kappa=\frac{|y''|}{(1+y'^2)^\frac{3}{2}}
    Yes of course, it was a typo by topsquark
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  5. #5
    Forum Admin topsquark's Avatar
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    I fixed it in my original post. Thanks for the catch, both of you.

    -Dan
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  6. #6
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    help

    Im sorry guys but Iīm getting nowhere. First of all. All I got is the formula but no examples in the literature except for the curvature of a circle. I need some examples. Second is curvature the correct word for this area of calculus. I read somewhere about concavity of a curve, is it the same thing?

    I tried to put the second derivative of the function in the formula but it takes me nowhere.

    Could somebody also explain to me why the curvature at the top of the curve must be a?

    HELP!!
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  7. #7
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    Pathetic mathematician - Curvature

    Please, I really need some help on this one. I canīt find anything useful anywhere on the internet or in my books.

    What is the curvature at the top of the parabola y=ax^2+bx+c

    And how do I calculate it?

    Quote Originally Posted by topsquark View Post
    \kappa = \frac{|y''|}{(1 + y'^2)^{3/2}}

    In your case, since the slope is zero at the maximum point, so the curvature is a.

    -Dan
    How do I know that slope=0 means the curvature is a?

    Hereīs my effort, which probably is totally useless!
    f'(x)=2ax+b
    f''(x)=2a
    so we have:
    \kappa=\frac{|2a|}{(1+(2ax+b)^2)^{\frac{3}{2}}}

    uhhh....uuhhhh??? Is there a need for a calculation at all or do we only need to explain the obvious? (well not obvious to me, thats for sure!)

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  8. #8
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    Quote Originally Posted by lynch-mob View Post
    Please, I really need some help on this one. I canīt find anything useful anywhere on the internet or in my books.

    What is the curvature at the top of the parabola y=ax^2+bx+c

    And how do I calculate it?



    How do I know that slope=0 means the curvature is a?

    Hereīs my effort, which probably is totally useless!
    f'(x)=2ax+b
    f''(x)=2a
    so we have:
    \kappa=\frac{|2a|}{(1+(2ax+b)^2)^{\frac{3}{2}}}

    uhhh....uuhhhh??? Is there a need for a calculation at all or do we only need to explain the obvious? (well not obvious to me, thats for sure!)

    At the top of the parabola y'(x)=0

    RonL
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  9. #9
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    Quote Originally Posted by topsquark View Post
    \kappa = \frac{|y''|}{(1 + y'^2)^{3/2}}

    In your case, since the slope is zero at the maximum point, so the curvature is a.

    -Dan
    |y''|=2|a| not |a|

    RonL
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    |y''|=2|a| not |a|

    RonL
    (sigh) I hate it when that happens!

    Thanks for the catch.

    -Dan
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