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The average rate of change can be modeled as a function: $f:D^2 \to \mathbb{R}$ where $D$ is the domain of the primary function in consideration. It maps two variables - the ends of the interval- to the average rate of change of that particular interval.

The derivative is $f$ restricted to the set: $\{(x,x): x \in D\}$, essentially making it a "slice".

Is this a valid interpretation?

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    When you say "3D" and "2D" it's unclear what you mean. From the rest of your question I assume that you mean average rate of change is a function $f:D^2\to \mathbb R$ where $D$ is the domain of whatever other function you're considering, and the derivative is $f$ restricted to the set $\{(x,x):x\in D\}$. But in general it's best to use standard notation, and if you don't know the notation to look it up.2012-07-15
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    @AlexBecker I'll keep that in mind. I specified my definitions of a "2D" and "3D" functions, which should help clarify any previous vagueness.2012-07-15
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    What is an independent/dependent variable? The question is vague.2012-07-15
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    @copper.hat http://en.m.wikipedia.org/wiki/Dependent_and_independent_variables#section_22012-07-15
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    Why would the usage of "independent/dependent variable" be considered vagueness? Upto my knowledge they are terms with concrete mathematical definitions.2012-07-15
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    There is much vagueness in your question. What do you mean by a '3D' function? I have never heard of such a term. Do you mean a function $f: \mathbb{R}->\mathbb{R}^2$ (two 'dependent', one 'independent' variable)?2012-07-15
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    @copper.hat Yes I do. Apologies if the lack of standard notation caused confusion: I have a limited knowledge of representing information through notation.2012-07-15
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    I don't understand what the question is. Are you asking how to define a derivative for a function $f \colon D^2 \to \mathbb{R}$?2012-07-15
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    @AntonioVargas No I am not. I have provided a possible model of the derivative function and average rate of change function and was looking for confirmation of its validity.2012-07-15

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If you have a function $f\colon I\to\mathbb R$ defined on an interval $I$, and you consider its "average rate of change" as a function $g\colon I^2\to\mathbb R$ defined as $$g(x,y) = \frac{f(y)-f(x)}{y-x},$$ hoping to say that $f'(x) = g(x,x)$, then you immediately run into the problem that $g$ is not even defined at $(x,x)$ because you'll be dividing by zero. You'll have to consider the limit of $g(u,v)$ as $(u,v)$ tends to $(x,x)$ instead. That should be fine -- I'd wager that the limit is defined and equal to $f'(x)$ exactly where $f$ is differentiable. But really, you've just replaced the one-variable limit in the definition of the derivative with a two-variable limit (and those are much more complicated), so I don't see this as a very useful interpretation.