# Conceptual Question

1. Dec 12, 2003

### cepheid

Staff Emeritus
Please don't laugh at the following: I am new to multivariable calculus, and just thought of this on the fly:

In single variable calculus, for a function $y = f(x)$ we have the tangent line approximation:

$$\Delta y \approx \frac{df}{dx} \Delta x$$

The relation becomes exact in the limit:

$$dy = \frac{df}{dx} dx$$

So to find the total change in $y$ over an interval $[a,b]$ we just have:

$$\int dy = \int_{a}^{b} \frac{df}{dx} dx = \int_{a}^{b} f^{\prime} (x)dx$$

I hope that this reasoning is correct. Having assumed that it was, and being very naive, I tried to extend this result to the multivariable case using the tangent plane approximation. For a function $z = f(x, y)[/tex] $$\Delta z \approx \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y$$ In the limit: $$dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$ So I just assumed that for a change in [itex] x$ from $a$ to $b$, and a change in $y$ from $c$ to $d$, the total change in $z$ was given by:

$$\int_{a}^{b} \frac{\partial f}{\partial x} dx + \int_{c}^{d} \frac{\partial f}{\partial y} dy$$

$$= \int_{a}^{b} f_{x} (x,y) dx + \int_{c}^{d} f_{y} (x,y) dy$$

Now, I know that this is dead wrong. I can think of a couple of reasons. For one thing, the curly d’s set off alarm bells: if $f_{x}$ has $y$ in it, then the first integral cannot be evaluated, and if $f_{y}$ has $x$ in it, then the second integral cannot be evaluated. Also, the region over which $z$ is changing is limited to a rectangle. My question is: where did I go wrong in this line of reasoning? I have a vague inkling that the problem is that you cannot do it in two parts using the partials…you need something else to be the “whole” derivative of the multivariable function. My text book says that the net change in a function $z = f(x, y)[/tex] can be evaluated using a double integral, but I’m not sure how they factor in. To me, [itex] \iint$ seems to be something else entirely. I cannot relate it intuitively in my mind to the concept that I’m working with (which is: to get the net change in the function…multiply the infinitesimal change in the independent variable(s) by the infinitesimal rate of change of the function with respect to that variable, and integrate). I would appreciate any help with this.

Last edited: Dec 12, 2003
2. Dec 12, 2003

### HallsofIvy

Staff Emeritus
The difficulty with what you are doing is that you are applying and integration with respect to x to one term and an integration with respect to y to the other. In other words, you are not "doing the same thing to both sides of the equation".

If, on the other hand, you intgrat $$dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$ along a given curve in the plane, then you do get a valid answer.

3. Dec 13, 2003

### cepheid

Staff Emeritus
"along a given curve in the plane"? An example would be very helpful.

4. Jul 23, 2004

### mathwonk

halls of ivys point is that in the plane you cannot integrate only in the x or y directions but must integrate in a slanted direction to get from one point to another.

your integrand df/dx may depend on both y and x, and the integral may depend on the path so you cannot just integrate only over the x values, for example.

by the way your question shows great imagination, and wonderful intuition for how to pursue analogies, and how to pose questions! I predict you will be a research mathematician someday soon, if that interests you.

5. Jul 28, 2004

### njoshi3

Actually, what you are not taking into account is that z(x,y) is a function of both x and y, while you wrote the differential form of z in terms of the partial derivatives of it. The partial derivatives do not come naturally as an straight forward extension of one variable case, since in multivariables ( 2 variable, in this case), there are many directions one can move (in a plane, in this case).

So suppose I start with the point (a,c) and go to (b,d), there is not a unique path to follow and the function z(x,y) may depend upon which path you follow, except in a special case, where z(x,y) is a conservative function ( curl of it is zero), one can give an absolute meaning to the function z(x,y), in all other cases z(x,y) is path dependent.

In the above calculations, you have done, it is okay till the last step, where if some particular path is chosen, y is related to x and hence dy and dx are not independent.