Results 1 to 12 of 12

Math Help - Principle of Differentiation and Integration query

  1. #1
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461

    Principle of Differentiation and Integration query

    Hi Folks,

    if we have an implicit function z=x^2-4y^2+5xy-6=0 eqn1

    then we can find the standard derivative \frac{dz}{dx}=2x-8y\frac{dy}{dx}+5x\frac{dy}{dx}+5y eqn 2 and

    the partial derivative \frac{\partial z}{\partial x}=2x+5y eqn3

    However, I am trying to grasp the idea of having a standard/total derivation and partial derivation of functions as above yet we dont have standard or partial integration (ie the reverse). How is that?

    If we integrate back eqn2 wrt to x we get eqn1 with a constant. This constant can be determined if BC's or IC's are known (ie 6) etc...but

    if we integrate eqn3 wrt x we can get eqn1 with a constant also but we lose the -4y^2 term. How does one get this term back?

    I dont think I understand this fully. Can anyone shed some light?

    Thanks
    Bugatti79
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    I'm puzzled by your definition of the implicit function. Is z = 0, period? Or is there a typo in there? And are you trying to think of z = z(x,y)?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Ackbeet View Post
    I'm puzzled by your definition of the implicit function. Is z = 0, period? Or is there a typo in there? And are you trying to think of z = z(x,y)?
    Sorry, that is a typo. It should read z=x^2-4y^2+5xy-6 and yes z is function of (x,y).

    Thanks
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    if we integrate eqn3 wrt x we can get eqn1 with a constant also but we lose the term. How does one get this term back?
    Actually, in that case, your "constant" of integration with respect x is an unknown function of y, which you can then determine sometimes by integrating w.r.t. y and comparing to the original function. Everything here depends on knowing what is a function of what. In the first case, you're treating the function z like this: z(x,y(x)). In the second case, you have z(x,y), and y is not treated like a function of x.

    So in reality, there are two different kinds of integration, corresponding to your total and partial derivatives. The difference will play out in the "constants" of integration. Make sense?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Ackbeet View Post
    Everything here depends on knowing what is a function of what. In the first case, you're treating the function z like this: z(x,y(x)). In the second case, you have z(x,y), and y is not treated like a function of x.
    The difference will play out in the "constants" of integration. Make sense?

    So in the first case the y(x) you are referring to is the -8y\frac{dy}{dx} term?
    If I understand correctly then, the method used to determine the constants will reflect the nature of integration ie partial or total?
    I have never come across a problem like the first case to determine the constants. Can an example of this be provided?

    Thanks for your explanation. Makes it a lot clearer.
    bugatti79
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    I'm saying that if I were given

    \frac{dz}{dx}=2x-8y\frac{dy}{dx}+5x\frac{dy}{dx}+5y,

    and told to integrate with respect to x, I would produce the following:

    z(x,y(x))=x^2 + 5\,x\,y(x) - 4\,y^{2}(x)+C.

    If I were given \frac{\partial z}{\partial x}=2x+5y and told to integrate with respect to x, I would produce the following:

    z(x,y)=x^{2}+5\,x\,y+g(y),

    where g(y) is an unknown function of y. To find that function, I would need more information. If given the correct information, I could recover the - 4\,y^{2} of the original equation.

    You see how this works? In the second case, I have to include an unknown function of y, because in taking the partial derivative of the resulting integral with respect to x, I would eliminate that whole term.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Ackbeet View Post
    I'm saying that if I were given


    You see how this works? In the second case, I have to include an unknown function of y, because in taking the partial derivative of the resulting integral with respect to x, I would eliminate that whole term.

    Would you write g(y) + C or would the C be automatically included in the former?

    But now I understand it very clearly. thanks very much Ackbeet

    Bugatti79
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    You could do it either way, as long as you remember to put the constant in there somewhere.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461

    Principle of Differentiation and Integration query

    Quote Originally Posted by Ackbeet View Post
    I'm saying that if I were given

    \frac{dz}{dx}=2x-8y\frac{dy}{dx}+5x\frac{dy}{dx}+5y,

    and told to integrate with respect to x, I would produce the following:

    z(x,y(x))=x^2 + 5\,x\,y(x) - 4\,y^{2}(x)+C.
    Just when I thought I understood!!

    Integrating dz/dx, I get for each term

    \int2x dx = x^2

    \int-8y\frac{dy}{dx} dx = \int -8y dy = -4y^2 the dx's cancel

    How does the third term go to 0 because I get
    \int 5x\frac{dy}{dx} dx= \int5x dy = 5xy the dx's cancel

    \int 5y dx= 5xy

    All of these term plus a constant etc. However, I now have two '5xy' terms. Should only have one. What have i done wrong?
    I feel stupid now
    Thanks
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    The third term doesn't go to zero. I think you're going to see some interaction between terms there. You have to combine the last two terms, I think:

    \int (5x\frac{dy}{dx}+5y) dx=5\int(x y'(x)+y(x))\,dx<br />
=5\int\frac{d}{dx}(x y(x))\,dx...
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461

    Principle of Differentiation and Integration query

    Quote Originally Posted by Ackbeet View Post
    The third term doesn't go to zero. I think you're going to see some interaction between terms there. You have to combine the last two terms, I think:

    \int (5x\frac{dy}{dx}+5y) dx=5\int(x y'(x)+y(x))\,dx<br />
=5\int\frac{d}{dx}(x y(x))\,dx...
    which just becomes 5xy(x)...

    think that makes sense. Thanks
    Follow Math Help Forum on Facebook and Google+

  12. #12
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    You're very welcome. Have a good one!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] PDE's - Implicit Differentiation Query
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: January 15th 2011, 08:56 AM
  2. [SOLVED] 2nd order differentiation query
    Posted in the Calculus Forum
    Replies: 8
    Last Post: December 5th 2010, 10:46 AM
  3. Integration Query
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 16th 2010, 12:52 PM
  4. Integration Query
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 8th 2008, 02:09 PM
  5. Integration Query
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 30th 2008, 07:17 AM

/mathhelpforum @mathhelpforum