Multivariable Calculus: Conceptual Question on Total Change in a Function

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In summary: That is the flaw in your argument.In summary, the conversation discusses the use of tangent line approximation in single variable calculus and its extension to multivariable calculus using the tangent plane approximation. However, this approach is not valid as it does not take into account the fact that the multivariable function is dependent on both x and y, and the path chosen can affect the results. To accurately evaluate the net change in a multivariable function, a double integral is necessary.
  • #1
cepheid
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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 [itex] y = f(x)[/itex] we have the tangent line approximation:

[tex] \Delta y \approx \frac{df}{dx} \Delta x [/tex]

The relation becomes exact in the limit:

[tex] dy = \frac{df}{dx} dx [/tex]

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

[tex] \int dy = \int_{a}^{b} \frac{df}{dx} dx = \int_{a}^{b} f^{\prime} (x)dx [/tex]

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 [itex] z = f(x, y)[/tex]

[tex] \Delta z \approx \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y [/tex]

In the limit:

[tex] dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy [/tex]

So I just assumed that for a change in [itex] x [/itex] from [itex] a [/itex] to [itex] b [/itex], and a change in [itex] y [/itex] from [itex] c [/itex] to [itex] d [/itex], the total change in [itex] z [/itex] was given by:

[tex] \int_{a}^{b} \frac{\partial f}{\partial x} dx + \int_{c}^{d} \frac{\partial f}{\partial y} dy [/tex]

[tex] = \int_{a}^{b} f_{x} (x,y) dx + \int_{c}^{d} f_{y} (x,y) dy [/tex]

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 [itex] f_{x} [/itex] has [itex] y [/itex] in it, then the first integral cannot be evaluated, and if [itex] f_{y} [/itex] has [itex] x [/itex] in it, then the second integral cannot be evaluated. Also, the region over which [itex] z [/itex] 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 textbook says that the net change in a function [itex] 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 [/itex] 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.
 
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  • #2
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 [tex] dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy [/tex] along a given curve in the plane, then you do get a valid answer.
 
  • #3
"along a given curve in the plane"? An example would be very helpful.
 
  • #4
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
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.
 

1. What is multivariable calculus?

Multivariable calculus is a branch of mathematics that deals with the study of functions of more than one variable. It extends the concepts of single variable calculus to functions with multiple independent variables.

2. What is a conceptual question on total change in a function in multivariable calculus?

A conceptual question on total change in a function in multivariable calculus is a question that asks about the overall change in a function when multiple variables are involved. It may involve understanding the relationship between the variables and how they affect the overall change in the function.

3. How is total change in a function calculated in multivariable calculus?

The total change in a function in multivariable calculus is calculated by taking the partial derivatives of the function with respect to each variable and then finding the magnitude of the resulting vector. This vector represents the overall change in the function.

4. Why is understanding total change in a function important in multivariable calculus?

Understanding total change in a function is important in multivariable calculus because it allows us to analyze the behavior of a function in multiple dimensions. It helps us understand how changes in one variable affect the overall behavior of the function, and how to optimize the function for different scenarios.

5. What are some real-world applications of total change in a function in multivariable calculus?

Total change in a function in multivariable calculus has many real-world applications, such as in economics, physics, and engineering. For example, it can be used to analyze the relationship between multiple variables in a market, or to optimize the design of a structure by understanding how different variables affect its stability and performance.

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