# Partial Derivatives

1. Apr 9, 2005

### Reshma

For a function, z = f(x, y)

$$\frac{\partial z}{\partial x} = \lim_{\delta x\rightarrow0} \frac{f(x +\delta x, y) - f(x, y)}{\delta x}$$

$$\frac{\partial z}{\partial y} = \lim_{\delta y\rightarrow0} \frac{f(x, y+\delta y) - f(x, y)}{\delta y}$$

What is partial increament $\delta x, \delta y$?
Wouldn't the function change if only x or y are increased?
What does the partial derivative of a function represent geometrically?
Wouldn't it produce 2 tangents?
Lastly, what are its applications?

2. Apr 9, 2005

### James R

Geometrically:

The function z=f(x,y) defines a surface, where z is the "height" of the surface above the point (x,y) in the x-y plane.

The derivative $\partial z/\partial x$ is the gradient of the tangent line to the surface which lies in the (x,z) plane which passes through the point (x,y,z).

The derivative $\partial z/\partial y$ is the gradient of a similar tangent in the (y,z) plane.

There are many other possible tangents other than these two, depending on which plane you use to "slice" the surface, and there are more general expressions for calculating the gradients using other planes.

3. Apr 11, 2005

### Reshma

First of all thank you for replying!

So if point (x,y) is any point of the scalar field defined by f(x,y) then 'z' is the height?

The gradient is a vector, right?

Last edited: Apr 11, 2005
4. Apr 11, 2005

### Reshma

How are these tangents different from a total differential?

5. Apr 11, 2005

### Reshma

Wouldn't a change in only 'x' or 'y' alter the given function significantly?

6. Apr 11, 2005

### HallsofIvy

Well, that depends upon the function, doesn't it!

f(x,y)= 4 does not change at all no matter how you change either x or y

f(x,y)= 4x changes if you change x but not if you change only y.

f(x,y)= 4y changes if you change y but not if you change only x.

f(x,y)= 1000x+ 0.000001y changes "significantly" if you change only x. It changes but perhaps not "significantly" if you change only y (really depends on what you mean by "significantly").

Now, what are the partial derivative of those functions and how do they tell you about changing or not changing "significantly"?

You may have learned in a basic science course that, if you have a situation in which there are a number of variables, the best thing to do is experiments in which only one variable changes- all others "remaining the same". That's the idea of partial derivatives: what happens if only one variable changes and the other remain the same?

We first, a "tangent" is a line or plane and so not at all like a "total differential"!

Perhaps you meant to ask how partial derivatives are different from the total differential.

You should learn soon after learning partial derivatives that it is possible for a function to have partial derivatives (at a point) even if it is not continuous (and so not "differentiable"). An example is f(x,y)= 0 if xy= 0, 1 if xy is not 0. That has partial derivates (both 0) at (0,0) but is not continuous there.

It is better to think of the gradient vector (have you learned that yet) as "THE derivative". The gradient vector is the vector whose x component is fx and y component fy. The "total differential" is fxdx+ fydy. We can think of that as grad f dot <dx, dy> just like df= (f ') dx in one variable.

7. Apr 12, 2005

### Reshma

Thank you very much for replying!

I have am aware of the idea that a partial derivative deals with only one variable and treats the other variables as constants. But I don't have a clue in what kinds of situations or experiments(in general I mean applications) these are used.
Sorry! Yes, how do total differentials differ from the partial derivatives?
Yes I have learnt of the gradient vector but more as a physical quantity(physics is my undergraduate major). I don't have much mathematical understanding of it. From my understanding, a gradient vector is defined as:
For a given scalar field: f(x,y,z), the gradient vector is:

$$\nabla f = \hat x\frac{\partial f}{\partial x} + \hat y\frac{\partial f}{\partial y} + \hat z\frac{\partial f}{\partial z}$$

How do you interpret a gradient vector as a derivative?

8. Apr 12, 2005

### Palindrom

Given a function in 2 variables, that defines a surface in 3D-space, z=f(x,y), the gradient of f gives you the direction in the x-y plane, so that if you move in that direction, you will modify you height most significantly (it depends wether you going in the direction or the opposite direction of a gradient).

Now, what year are you in now? Did you take a class in QM or thermodynamics, or analytical mechanics?

9. Apr 12, 2005

### James R

The gradient vector at a given point (x,y,z) in space, $\nabla f$, is a vector which points in the direction you would have to "move" for the field f to increase at its maximum possible rate.

10. Apr 12, 2005

### Hurkyl

Staff Emeritus
Or, the gradient vector is what you use to get the differential approximation:

$$f(\vec{r}) \approx f(\vec{r}_0) + \nabla f(\vec{r_0}) \cdot (\vec{r} - \vec{r}_0)$$

11. Apr 13, 2005

### Reshma

I'm in final year now. Whatever vector calculus I know is from my electrodynamics class(I refer Griffiths). Of course this book isn't as rigorous for hardcore calculus. I don't have much mathematical background of gradients..although I'm trying to learn some from this queer book--Differential and Integral Calculus by N. Piskunow.

12. Apr 13, 2005

### Reshma

This is a theorem I found in Piskunow's book. I better quote it.

Can someone give me some more insight on this theorem?

13. Apr 13, 2005

### Reshma

Also, one more question which was left unanswered! How do partial derivatives differ from a total differential? Is there a mathematical relationship between the two?

14. Apr 13, 2005

### Palindrom

Final year is good. More advanced than me.

So let's see.

Let's look at a solution of the wave equation- like a wave function of a particle, or even just en EM wave. This function depends on 2 variables- space&time. (Let's look at a 1D case, where $$$\vec r = \hat x$$$). Therefore, I have 2 partial derivatives- with respect to x, and with respect to t.
Why do I care about the derivative with respect to t? Because it gives me the speed, given x, of the wave at the point x and in time t. But that's not the same as with respect to x, which would give me, at time t, simply the tangent to the wave for any x yoiu chose.
That was a bit forced upon. Let's look at something more interesting.
Let S be the entropy of a closed system. S is a function of 3 independent variables- E,V,N. (Energy, Volume, Number of particles).
Now let's look at the partial derivatives.
With respect to E- do you know what we get? We get the temperature (Well actually 1/T). Now that's getting interesting.
With respect to V- P/T, that is the pressure (over T, but never mind that).
With respect to N- the chemical potential. So you see, the change of the entropy when you change the energy a bit is elegantly dependent of the temperature of the system. Isn't that nice?
By the way, I'm quite sure you know those things, right?

Last but not least- Let's take a mechanical potential, $$$\phi \left( {\vec r} \right)$$$. I'm convinced you know what $$$\nabla \phi \left( {\vec r} \right)$$$ is. Let's try and see it intuitively with the concepts of the gradient as James R gave us- the gradient is the direction in space in which the potential alters most significantly. In what direction would you have to go to make your potential as large as possible? To the direction where your energy would increase the most, right? That is, right against the direction of the force! And that's why: $$$\vec F = - \nabla \phi \left( {\vec r} \right)$$$

I hope that's more or less what you were looking for...

15. Apr 13, 2005

### Palindrom

Well, $$$\frac{{\partial f}}{{\partial \vec S}}$$$ gives you the change of f in the S direction. Seing the change of f is the greatest in the gradient direction, it is only natural to think the change in any other direction would be proportional to how much this direction "has in common" with the gradient's direction.
And just the get some feeling- for example, take the x direction. By the theorem, it would therefore hold that:
$$$gradf \cdot \hat x = \frac{{\partial f}}{{\partial x}}$$$
Well, let's see:
$$$gradf \cdot \hat x = \left( {\hat x\frac{{\partial f}}{{\partial x}} + \hat y\frac{{\partial f}}{{\partial y}} + \hat z\frac{{\partial f}}{{\partial z}}} \right) \cdot \hat x = \frac{{\partial f}}{{\partial x}}$$$
As well expected!

16. Apr 13, 2005

### dextercioby

If u use russian books,i think the best for integral calculus is Fichetnholtz:"Integral Calculus" (3 vols.).

Daniel.

17. Apr 13, 2005

### Reshma

Palindrom, thank you so much for your time. I think I got the answers for most of my doubts.

I don't particularly care for the origin of the book. It is a bit frustrating when maths and physics are taught in a pretty disconnected manner here(in India). Can you recommend a good book on integral calculus especially for physicists?