Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Algebraic Definition

Consider the following definition for the instantaneous rate of change, $m$, for some value $x$: $$m = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$$

The above definition (usually used as a rigorous definition) does not apply, as the curve is not a function.

Geometric Definition

The following definition is an excerpt from Wikipedia:

In geometry, the tangent line (or simply the tangent) to a plane curve at a given point is the straight line that "just touches" the curve at that point.

Consider the following curve, that for argument's sake is not a function: enter image description here

The above tangent line intersects the curve at two points, meaning Wikipedia's definition is not very rigorous. The above line can be equally thought of as a secant line between the two intersecting points.

Both the algebraic definition and the geometric definition fail to provide a strong definition, leaving to me wonder if a strong defintion exists.

share|improve this question
2  
There is no problem if the curve is locally a function. –  André Nicolas Feb 7 '13 at 1:03
    
@AndréNicolas Would it be possible for the curve to not be a function locally? Or can all curves be locally described as functions to some degree? –  Farhad Yusufali Feb 7 '13 at 1:05
    
The Peano space-filling curve, is an extreme example of such a thing. But this does not really address your main question. –  André Nicolas Feb 7 '13 at 1:35
    
A simple example of a curve that is not a function is $x=y^2$. For each $x > 0$, there are two points $(x,\sqrt x)$ and $(x,-\sqrt x)$ on the curve, which disqualifies it. It is "locally a function" except at $(0,0)$, but even there it has a well-defined tangent. –  TonyK Apr 11 at 21:12
    
By the way, your sketch certainly looks like a function to me! If you rotated it by $90^\circ$ it wouldn't be. (I wanted to make a joke about "for argument's sake", but I couldn't make it work...) –  TonyK Apr 11 at 21:13

2 Answers 2

up vote 1 down vote accepted

Here are two answers, maybe one will be useful to you:

  1. You can parametrize a "smooth" curve in the plane as $(x(t),y(t))$. You can pick numerous parametrizations that will work. For example $(\cos t, \sin t)$ parametrizes a circle, but so does $(\cos t^2, \sin t^2)$ for a different range of $t$. If $x(t)$ and $y(t)$ are functions (even if $y$ is not a function of $x$) then the slope of the curve at a point $t_0$ will be $y'(t_0)/x'(t_0)$. If $x'(t_0) = 0$ then the curve is either vertical or you picked a "bad" parametrization

  2. There is a better geometric definition of a tangent line. If you zoom in on the function $\sin x$ at $x=0$, it appears to be the graph of $y=x$. For instance, if you looked at this graph you might think you were looking at a plot of $y=x$. Roughly speaking, that is why $y=x$ is the tangent line of $\sin x$ at $x=0$. If your function is "smooth", it will always look like a line when you zoom in far enough. The tangent line is the line that you appear to see.

share|improve this answer

An arbitrary curve in the plane is a function $\gamma:[a,b]\to \mathbb R^2$. Such a curve has two component functions $\gamma=(\gamma_1,\gamma_2)$ where $\gamma_i:[a,b]\to \mathbb R$. All of single variable calculus can be recast in terms of such general curves (e.g., tangent lines etc.).

Given a function $f:[a,b] \to \mathbb R$ there is an associated curve $\gamma:[a,b]\to \mathbb R^2$ with $\gamma_1(x)=x$ and $\gamma_2(x)=f(x)$. The image of the curve is the graph of the function and this is how all of the results of calculus for curves will give you back calculus results for single variable functions. Of course, curves are more general than graphs of functions, so I think this answers your question, including instantaneous rate of change and much more.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.