# Principle of Differentiation and Integration query

• Jul 22nd 2010, 02:02 PM
bugatti79
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
• Jul 22nd 2010, 04:45 PM
Ackbeet
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)?
• Jul 23rd 2010, 12:06 AM
bugatti79
Quote:

Originally Posted by Ackbeet
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
• Jul 23rd 2010, 02:42 AM
Ackbeet
Quote:

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?
• Jul 23rd 2010, 06:34 AM
bugatti79
Quote:

Originally Posted by Ackbeet
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
• Jul 23rd 2010, 07:35 AM
Ackbeet
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.
• Jul 23rd 2010, 12:11 PM
bugatti79
Quote:

Originally Posted by Ackbeet
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
• Jul 23rd 2010, 12:15 PM
Ackbeet
You could do it either way, as long as you remember to put the constant in there somewhere.
• Jul 23rd 2010, 01:58 PM
bugatti79
Principle of Differentiation and Integration query
Quote:

Originally Posted by Ackbeet
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!! (Headbang)

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 (Happy)
Thanks
• Jul 23rd 2010, 02:09 PM
Ackbeet
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:

$\int (5x\frac{dy}{dx}+5y) dx=5\int(x y'(x)+y(x))\,dx
=5\int\frac{d}{dx}(x y(x))\,dx...$
• Jul 25th 2010, 06:02 AM
bugatti79
Principle of Differentiation and Integration query
Quote:

Originally Posted by Ackbeet
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:

$\int (5x\frac{dy}{dx}+5y) dx=5\int(x y'(x)+y(x))\,dx
=5\int\frac{d}{dx}(x y(x))\,dx...$

which just becomes $5xy(x)$...

think that makes sense. Thanks
• Jul 26th 2010, 02:34 AM
Ackbeet
You're very welcome. Have a good one!