Results 1 to 8 of 8

Math Help - Please check

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    41

    Please check

    can somebody please check the following for me. many thanks

    Find an Expression for DY/DX (I use D as i will use d for partial derivatives)

    When: x^3 + y^3 - 2x^2 y = 0

    i) Assign;

    z = x^3 + y^3 - 2x^2 y = 0 (eqn 1)

    we also have Ez = dz/dx . Ex + dz/dy . Ey ( Where E = Delta/Change)

    ii) Divide Both Sides By Ex

    which then equates to ;

    Ez/Ex = dz/dx + dz/dy . Ey / Ex (eqn 2)

    and when Ex approaches zero eqn1 becomes;

    DZ/DX = dz/dx + dz/dy . DY/DX (eqn 3)

    iii) we can now find the partial derivatives;

    dz/dx = 3x^2 - 4xy

    dz/dy = 3y^2 - 2x^2


    and since z = zero we can now substitue these back into eqn 3 to get;

    0 = 3x^2 - 4xy + (3y^2 - 2x^2) . DY/DX

    - 3x^2 + 4xy = (3y^2 - 2x^2) . DY/DX

    DY/DX = (-3x^2 + 4xy) / (3y^2 - 2x^2)


    Think thats not far off, sorry for changing the script and making it confursing.

    Thanks again
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,831
    Thanks
    1602
    x^3 + y^3 -2x^2y = 0.

    \frac{d}{dx}(x^3 +  y^3 - 2x^2y) = \frac{d}{dx}(0)

    \frac{d}{dx}(x^3) +  \frac{d}{dx}(y^3) -2 \frac{d}{dx}(x^2y) = 0

    3x^2 +  \frac{d}{dy}(y^3)\,\frac{dy}{dx} -2 \left(x^2\,\frac{dy}{dx} +  2xy\right) = 0

    3x^2 + 3y^2\,\frac{dy}{dx} -  2x^2\,\frac{dy}{dx} - 4xy = 0

    \left(3y^2 -  2x^2\right)\frac{dy}{dx} = 4xy - 3x^2

    \frac{dy}{dx} =  \frac{4xy - 3x^2}{3y^2 - 2x^2}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,453
    Thanks
    1868
    There are NO "partial derivatives" here. y is a function of the single variable x.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2008
    Posts
    41
    there has to be, its part of an assignment so the question would not be designed to catch us out. thanks though
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Parton-Bill View Post
    there has to be, its part of an assignment so the question would not be designed to catch us out. thanks though
    And why would you assume that there has to be a partial derivative? The question asks you to find \frac{dy}{dx}, which you can easily do by implicit differentiation. Also note that you got the same answer.

    If you're not sure, ask your instructor - guessing usually doesn't work, and to anyone else trying to solve this it would not seem that there this has anything to do with multi-variable functions, so until you get that solved out we can't help you.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2008
    Posts
    41
    to be clearer, the question states:

    Use partial Differentiation to determine an expression for dy/dx.

    so although it can be solved without partial derivatives, my task is to do so.

    sorry, should have been clearer but thanks for the advice anyway
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,453
    Thanks
    1868
    Quote Originally Posted by Parton-Bill View Post
    to be clearer, the question states:

    Use partial Differentiation to determine an expression for dy/dx.

    so although it can be solved without partial derivatives, my task is to do so.

    sorry, should have been clearer but thanks for the advice anyway
    "Implicit differentiation" of, say f(x,y)= 0, can be written as " \frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}\frac{dy}{dx}= 0" so that \frac{dy}{dx}= \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}
    but since implicit differentiation is usually introduced well before partial derivatives, I would consider that a confusing notation.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Nov 2008
    Posts
    41
    in honesty the entire assignment has been confusing, it was even worse as an exam. alot comes down to what the british term as education, we are taught to jump through hoops, not use best judgment, I suppose judgment comes later when i'm designing building by numbers lol
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: March 12th 2011, 05:05 AM
  2. Can you check please
    Posted in the Statistics Forum
    Replies: 8
    Last Post: February 3rd 2010, 04:42 AM
  3. Check these for me?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 23rd 2009, 10:20 PM
  4. check
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 6th 2009, 01:11 PM
  5. Check IVP
    Posted in the Calculus Forum
    Replies: 14
    Last Post: September 12th 2007, 12:33 PM

Search Tags


/mathhelpforum @mathhelpforum