Results 1 to 7 of 7

Math Help - another implicit differentiation question

  1. #1
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161

    another implicit differentiation question

    Here's another:

    If  ax^2 + by^2 = c show that the second derivative is equal to  -ac/b^2y^3

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,831
    Thanks
    1602
    Quote Originally Posted by differentiate View Post
    Here's another:

    If  ax^2 + by^2 = c show that the second derivative is equal to  -ac/b^2y^3

    Thanks
    Impossible.

    Taking the first derivative will eliminate the c. How can the second derivative have a c in it?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,429
    Thanks
    1857
    Quote Originally Posted by Prove It View Post
    Impossible.

    Taking the first derivative will eliminate the c. How can the second derivative have a c in it?
    Because the second derivative will have an "x" in it and eliminating that reintroduces the "c".

    Since ax^2+ by^2= c, 2ax+ 2byy'= 0. Differentiating again, 2a+ 2by'^2+ 2byy"= 0.

    y"= \frac{-a- by'^2}{by}.

    From ax+ byy'= 0, y'= -\frac{ax}{by} and so y'^2= \frac{a^2x^2}{b^2y^2}.

    Then y"= -\frac{a+ \frac{a^2bx^2}{by^2}}{by}
    y"= -\frac{aby^2+ a^2x^2}{b^2y^3}

    From ax^2+ by^2= c, ax^2= c- by^2 so a^2x^2= ac- aby^2.

    aby^2+ a^2x^2= aby^2+ ac- aby^2= ac

    y"= -\frac{ac}{b^2y^3}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Shanks's Avatar
    Joined
    Nov 2009
    From
    BeiJing
    Posts
    374
    2ax+2byy'=0,\text{ thus }y'=\frac{-ax}{by}<br />
2a+2b(y')^2+2byy''=0
    combined with y'=\frac{-ax}{by},gives
    y''=\frac{-ac}{b^2y^3}.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by HallsofIvy View Post
    Because the second derivative will have an "x" in it
    Brilliant\ work\ !

    (implied.....\ the\ 2nd\ derivative\ contains\ (y')^2\ which\ contains\ x^2)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Dec 2008
    From
    Australia
    Posts
    161
    Thanks

    where does  (y')^2 come from?

    I get that you use the product rule, but how is that equal to  (y')^2
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by HallsofIvy View Post
    Since ax^2+ by^2= c, 2ax+ 2byy'= 0. Differentiating again, 2a+ 2by'^2+ 2byy"= 0.
    1st derivative...

    \frac{d}{dx}(ax^2+by^2)=\frac{d}{dx}c

    2ax+b\frac{d}{dx}y^2=0

    2ax+b\frac{dy}{dx}\frac{d}{dy}y^2=0

    2ax+2yb\frac{dy}{dx}=0

    2nd derivative...
    2by is one factor and y' is the other of 2by(y'), or y and 2by' etc.

    2a+2by\frac{d}{dx}\frac{dy}{dx}+\frac{dy}{dx}2b\fr  ac{dy}{dx}=0

    2a+2by\frac{d^2y}{dx^2}+2b(\frac{dy}{dx})^2=0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Implicit Differentiation question
    Posted in the Calculus Forum
    Replies: 11
    Last Post: February 12th 2010, 03:18 AM
  2. Implicit Differentiation Question
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 4th 2009, 05:34 AM
  3. Implicit Differentiation Question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 7th 2008, 07:08 AM
  4. Question about Implicit Differentiation.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2008, 08:28 PM
  5. Question about Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 23rd 2007, 10:34 PM

Search Tags


/mathhelpforum @mathhelpforum