Page 2 of 2 FirstFirst 12
Results 16 to 21 of 21

Math Help - Algebraic Simplification

  1. #16
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398

    Re: Algebraic Simplification!

    Quote Originally Posted by Umair View Post
    For the equation of x^2-y^2-2x+4y-20=0,

    y'=\frac{dy}{dx}=\frac{1-x}{2-y}

    For d^2y/dx^2, I got the following:
    ...

    (3)   \frac{d^2y}{dx^2}=\frac{-1+\(dy/dx)^2}{2-y}

    SOMEHOW simplifies to:

    (4)   \frac{d^2y}{dx^2}=\frac{x^2-y^2-2x+4y-3}{(2+y)^3} = \frac{17}{(2+y)^3}

    Any idea how (3) simplifies to Middle section of (4)?
    I get d^2y/dx^2=-\frac{x^2-2 x+y^2-4 y+5}{(y-2)^3}\,.

    Which gives: d^2y/dx^2=\frac{2y^2-8 y+25}{(2-y)^3}\,.
    Last edited by Ackbeet; August 5th 2011 at 01:36 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Newbie
    Joined
    Jul 2011
    Posts
    19

    Re: Algebraic Simplification!

    Quote Originally Posted by SammyS View Post
    I get d^2y/dx^2=-\frac{x^2-2 x+y^2-4 y+5}{(y-2)^3}\,.

    Which gives: d^2y/dx^2=\frac{2y^2-8 y+25}{(2-y)^3}\,.
    How did you manage to derive d^2y/dx^2=-\frac{x^2-2 x+y^2-4 y+5}{(y-2)^3}\,?
    Last edited by Ackbeet; August 5th 2011 at 01:37 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Algebraic Simplification!

    You know: \frac{dy}{dx}=\frac{1-x}{2-y} (1)
    and you got:
    \frac{d^2y}{dx^2}=\frac{-1+\left(\frac{dy}{dx}\right)^2}{2-y} (2)

    Substituting (1) in (2) gives:
    \frac{d^2y}{dx^2}=\frac{-1+\left(\frac{1-x}{2-y}\right)^2}{2-y}
    =\frac{-1+\frac{(1-x)^2}{(2-y)^2}}{2-y}
    =\frac{\frac{-1\cdot(2-y)^2+(1-x)^2}{(2-y)^2}}{2-y}
    =\frac{-1(4-4y+y^2)+1-2x+x^2}{(2-y)^3}
    =\frac{-4+4y-y^2+1-2x+x^2}{(2-y)^3}
    =\frac{-y^2+4y+x^2-2x-3}{(2-y)^3}
    Follow Math Help Forum on Facebook and Google+

  4. #19
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2

    Re: Algebraic Simplification

    Quote Originally Posted by Umair View Post
    The fact it may be a circle or not is trivial in the question I think, but it's the "incorrect" equation we need to use.
    That statement makes no sense. If the original problem statement is talking about the equation of a circle centered at (1, -2) and of radius 5, then you are NOT talking about the equation x^2-y^2-2x+4y-20=0, because that simply isn't a circle at all, much less the circle specified.

    However, if you work from the circle equation given by skeeter in post # 10, you don't get to the result indicated in the OP. If you work from the incorrect hyperbola equation, you still don't get to the result indicated in the OP. In summary: you have been asked to do the impossible. There is a mistake in the problem.

    For the equation of x^2-y^2-2x+4y-20=0,

    I got:

    2x -2y\frac{dy}{dx} - 2 +4\frac{dy}{dx} = 0

    which simplifies to: (2-y)\frac{dy}{dx}=1-x

    and finally y'=\frac{dy}{dx}=\frac{1-x}{2-y}

    I'm not sure if this is correct or not ,but it's how I did it... please elaborate
    That's correct for the hyperbola equation.

    Edit:

    For d^2y/dx^2, I got the following:

    (1)   \frac{d}{dx}((2-y)\frac{dy}{dx}+x-1)=0

    (2)   -\left(\frac{dy}{dx}\right)^2+(2-y)\frac{d^2y}{dx^2}+1=0

    (3)   \frac{d^2y}{dx^2}=\frac{-1+\left(dy/dx\right)^2}{2-y}

    SOMEHOW simplifies to:

    (4)   \frac{d^2y}{dx^2}=\frac{x^2-y^2-2x+4y-3}{(2+y)^3} = \frac{17}{(2+y)^3}

    Any idea how (3) simplifies to Middle section of (4)?
    Here you plug in the value for the derivative that you got before:

    \frac{d^2y}{dx^2}=\frac{-1+\left(dy/dx\right)^2}{2-y}=\frac{-1+\left(\frac{1-x}{2-y}\right)^2}{2-y},

    and then you get the common denominator, add the fractions, etc., etc., etc.
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Newbie
    Joined
    Jul 2011
    Posts
    19

    Re: Algebraic Simplification

    Quote Originally Posted by Siron View Post
    You know: \frac{dy}{dx}=\frac{1-x}{2-y} (1)
    and you got:
    \frac{d^2y}{dx^2}=\frac{-1+\left(\frac{dy}{dx}\right)^2}{2-y} (2)

    Substituting (1) in (2) gives:
    \frac{d^2y}{dx^2}=\frac{-1+\left(\frac{1-x}{2-y}\right)^2}{2-y}
    =\frac{-1+\frac{(1-x)^2}{(2-y)^2}}{2-y}
    =\frac{\frac{-1\cdot(2-y)^2+(1-x)^2}{(2-y)^2}}{2-y}
    =\frac{-1(4-4y+y^2)+1-2x+x^2}{(2-y)^3}
    =\frac{-4+4y-y^2+1-2x+x^2}{(2-y)^3}
    =\frac{-y^2+4y+x^2-2x-3}{(2-y)^3}
    Thank you!! This was exactly what I was looking for! However it seems your denominator is (2-y)^3 instead of (2+y)^3

    Although your numerator is exactly what was needed, I think the solutions I got from my professor were wrong lol. No wonder I killed off a few brain cells trying to work this one out, lol. Everyone was getting something different...

    Quote Originally Posted by Ackbeet View Post
    That statement makes no sense. If the original problem statement is talking about the equation of a circle centered at (1, -2) and of radius 5, then you are NOT talking about the equation x^2-y^2-2x+4y-20=0, because that simply isn't a circle at all, much less the circle specified.

    However, if you work from the circle equation given by skeeter in post # 10, you don't get to the result indicated in the OP. If you work from the incorrect hyperbola equation, you still don't get to the result indicated in the OP. In summary: you have been asked to do the impossible. There is a mistake in the problem.



    That's correct for the hyperbola equation.



    Here you plug in the value for the derivative that you got before:

    \frac{d^2y}{dx^2}=\frac{-1+\left(dy/dx\right)^2}{2-y}=\frac{-1+\left(\frac{1-x}{2-y}\right)^2}{2-y},

    and then you get the common denominator, add the fractions, etc., etc., etc.
    I think I discovered quite a few errors in the sheet, lol. Either way what I meant to say was the equation for the hyperbola had to be used.

    More importantly, your expression for dy/dx for the hyperbola equation was different to mine, although you still said I did it correctly?
    Follow Math Help Forum on Facebook and Google+

  6. #21
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2

    Re: Algebraic Simplification

    Quote Originally Posted by Umair View Post
    I think I discovered quite a few errors in the sheet, lol. Either way what I meant to say was the equation for the hyperbola had to be used.

    More importantly, your expression for dy/dx for the hyperbola equation was different to mine, although you still said I did it correctly?
    In my post # 12, multiply top and bottom by -1, thus not changing the fraction, and you get your expression for the derivative.
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. having a problem with algebraic simplification
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 20th 2011, 04:07 AM
  2. Replies: 1
    Last Post: August 25th 2011, 06:09 AM
  3. Algebraic Simplification
    Posted in the Algebra Forum
    Replies: 6
    Last Post: August 7th 2011, 08:04 PM
  4. Algebraic Simplification
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 8th 2008, 01:02 PM
  5. Algebraic Simplification
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 26th 2008, 06:03 PM

Search Tags


/mathhelpforum @mathhelpforum