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Math Help - This is a straight line?

  1. #1
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    This is a straight line?

    Can someone please point out to me how the following reduces to y=mx+b:

    \frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0

    Thanks
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    The only way for a fraction to be zero is when?
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    Let \displaystyle Y = \frac{dy}{dx}, which means \displaystyle \frac{dY}{dx} = \frac{d^2y}{dx^2} and the DE becomes

    \displaystyle \begin{align*} (1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx} &= 0 \\ \int{(1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx}\,dx} &= \int{0\,dx} \\ \int{(1 + Y)^{-\frac{3}{2}}\,dY} &= C_1 \\ \frac{(1 + Y)^{-\frac{1}{2}}}{-\frac{1}{2}} + C_2 &= C_1 \\ -2(1 + Y)^{-\frac{3}{2}} &= C_1 - C_2 \\ (1 + Y)^{-\frac{3}{2}} &= -\frac{1}{2}(C_1 - C_2) \\ 1 + Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} \\ Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1 \end{align*}

    So \displaystyle \frac{dy}{dx} &= C where \displaystyle C = \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1.

    Can you go from here to solve for \displaystyle y?
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Let \displaystyle Y = \frac{dy}{dx}, which means \displaystyle \frac{dY}{dx} = \frac{d^2y}{dx^2} and the DE becomes

    \displaystyle \begin{align*} (1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx} &= 0
    So either [tex]\frac{dY}{dx}= 0[//tex] or (1+ Y)^{-\frac{3}{2}}= 0. But that second equation is impossible.

    That's essentially what Ackbeat said.

     \begin{align*}  \int{(1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx}\,dx} &= \int{0\,dx} \\ \int{(1 + Y)^{-\frac{3}{2}}\,dY} &= C_1 \\ \frac{(1 + Y)^{-\frac{1}{2}}}{-\frac{1}{2}} + C_2 &= C_1 \\ -2(1 + Y)^{-\frac{3}{2}} &= C_1 - C_2 \\ (1 + Y)^{-\frac{3}{2}} &= -\frac{1}{2}(C_1 - C_2) \\ 1 + Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} \\ Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1 \end{align*}

    So \displaystyle \frac{dy}{dx} &= C where \displaystyle C = \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1.

    Can you go from here to solve for \displaystyle y?
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    Is there not a way to delete a post?
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    Quote Originally Posted by rainer View Post
    Can someone please point out to me how the following reduces to y=mx+b:

    \frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0

    Thanks
    A question for the OP. Is it

    \frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0 or \frac{y''(x)}{(1+y'^2(x))^{\frac{3}{2}}}=0

    The first looks odd. The second - standard curvature. Just curious :-)
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    Quote Originally Posted by zg12 View Post
    Is there not a way to delete a post?
    Just report your own post and ask to delete. If it is appropriate, a moderator will delete your post. The report post button is in the lower left corner of your post - looks like a triangle with an exclamation mark inside it.
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    Quote Originally Posted by Danny View Post
    A question for the OP. Is it

    \frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0 or \frac{y''(x)}{(1+y'^2(x))^{\frac{3}{2}}}=0

    The first looks odd. The second - standard curvature. Just curious :-)

    Unless the author has made a type-o, it's the first one, which is the outcome of plugging the arclength formula into the Euler-Lagrange equation.
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    Quote Originally Posted by zg12 View Post
    Is there not a way to delete a post?
    You can click "Edit" and then "Delete Message".
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    Quote Originally Posted by Prove It View Post
    You can click "Edit" and then "Delete Message".
    This works for you, because you are have the MHF Expert badge. There is no way for zg12 directly to delete a post. A moderator must do it for him.
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  11. #11
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    Quote Originally Posted by Ackbeet View Post
    This works for you, because you are have the MHF Expert badge. There is no way for zg12 directly to delete a post. A moderator must do it for him.
    Oops - didn't realise that was only made possible by the badge. Thanks
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  12. #12
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    Quote Originally Posted by rainer View Post
    Unless the author has made a type-o, it's the first one, which is the outcome of plugging the arclength formula into the Euler-Lagrange equation.
    You mean L = \sqrt{1+y'^2}. If so, then it would be the second, wouldn't it?
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