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Is there really such a thing as a tangent line? Even in the definition of a derivative, you always get:

$$\frac{d}{dx} f(x)= \lim_{h\to 0} (f'(x)+h)$$

Although that h approaches 0, it can never be 0. Think of it this way. If I show you a picture of somebody running and ask you "at what speed is he/she running" you would never be able to tell me. I would have to show you 2 pictures of that runner and tell you the amount of time between the first and the second picture in order for you to tell me the speed. So does a true tangent line really exist? (except maybe in the case of a circle).

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That's not the definition of the derivative. The proper definition is $f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$. –  George V. Williams Apr 16 '13 at 23:23
    
Does a true circle, a true line or a true function exist? Where? It's the same in your case: it all is a matter of definitions and, indeed, pretty important and logical ones. –  DonAntonio Apr 16 '13 at 23:23
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Please define what you mean by "really". If "really" means "really in mathematics", then by definition the tangent line to $y = f(x)$ at a point $c$ exists if and only if $f$ is differentiable at $c$, and this occurs for a very large class of functions, including virtually all of the elementary functions one meets in precalculus mathematics. –  Pete L. Clark Apr 16 '13 at 23:29
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On the other hand, if "really" is meant to signal that your question is a phyiscal/philosophical inquiry, please say so (and we can direct you to the appropriate place to get an answer to that question). For what it's worth, "Is the concept of instantaneous velocity truly physically reasonable, or is it a purely mathematical idealization?" sounds like a good question to me...just not a good math question. –  Pete L. Clark Apr 16 '13 at 23:31
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@Ovi: your last comment indicates that you haven't fully digested the limit concept. A limit as $h \rightarrow 0$ is defined using only nonzero values of $h$, but the limit itself is not in any way an "approximation": it is an "exact" answer. For instance, the tangent line to $y = x^2$ at $x = 1$ is exactly $y = 2x-1$. For instance, one can readily see that this is the unique nonvertical line passing through the parabola at $(1,1)$ which has exactly one intersection point. If the slope were anything other than $2$ -- no matter how close -- there will be two intersection points. –  Pete L. Clark Apr 17 '13 at 4:23

6 Answers 6

up vote 13 down vote accepted

I'm thinking of a number.

I'm not going to tell you what my number is. But I will tell you that $1$ is close to my number. And that $-0.1$ is closer to my number. And that $0.01$ is even closer to my number. And $-0.001$ is even closer still.

Of course, that's not enough information to figure out my number. I will additionally tell you that each of the numbers $(-1)^n 10^{-n}$ (where $n$ ranges over nonnegative integers) get successively closer to my number; each of these number is closer to my number than the one before it.

Surely from this information, you can figure out what my number is. Despite the fact I never told you my number, and none of the numbers in the sequence I mention is equal to my number.

That is what a limit does.

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Thanks, that makes sense. –  Ovi Apr 17 '13 at 19:29

Going with your analogy, first note that in general we define $f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$. Actually this contains more "information" compared to your picture example (You have more than $2$ pictures, in fact you have infinitely many pictures). First, you have one picture of the runner, which is $f$ evaluated at $x$, $f(x)$. You also take into account many pictures of the runner after infinitesimal time intervals after the moment $x$. These pictures are precisely $f(x+h)$ as $h\rightarrow 0$. Making sure that $h\neq 0$ actually helps here, so that you always have more than one picture of your runner.

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I understand what you're saying but my whole point was that even if you had infinitely many pictures closer and closer in time to each other, you could never tell me the speed from just one picture. –  Ovi Apr 17 '13 at 2:22

Here is one way to think about tangent lines.

Let $C$ be any collection of points in the plane. For any point $P\in C$, we say that a line $L$ is quasi-tangent to $C$ at $P$ if for some open ball $U$ of $P$, the following two conditions hold:

  1. $U\cap L\cap C=\{P\}$ or $U\cap L\cap C=U\cap L$
  2. If $U_1$ and $U_2$ are the two connected components of $U\setminus L$, then at least one of $U_1\cap C$ and $U_2\cap C$ are empty.

Notice that there may be infinitely many $L$ that are quasi-tangent to $C$ at $P$. For example, if $C$ is the graph of $y=|x|$, then the point $(0,0)\in C$ has infinitely many quasi-tangent lines.

Now suppose that only one such $L$ exists. Then we say that $L$ is tangent to $C$ at $P$, and we can define the slope of $C$ at $P$ to be that of $L$. So far, we haven't explicitly referenced any limit (although the limit is probably hiding somewhere).

Intuitively, it seems that this definition should be equivalent to the usual definition when $C$ is the graph of some smooth function (see the comment below for a counterexample when the function is differentiable, but not smooth).

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It definitely doesn't hold if $C$ is the graph of a differentiable (but not smooth) function: the condition $U\cap L\cap C = \{P\}$ can fail to be satisfied. Take $f(x) = x^2\sin x$ with ($f(0) = 0$). Then $$ f'(0) = \lim_{h\to 0}\frac{(h)^2\sin\frac{1}{h}}{h} = 0, $$ but there are infinitely many points in $U\cap L\cap C$ for any neighborhood $U$ of $0$. I don't know if it is equivalent in the smooth case, but I think that if it is, the limit would be hiding in the conditions of the open neighborhood $U$ about the point. –  Stahl Apr 16 '13 at 23:46
    
Hi Stahl, sorry to delete my post, only to undelete it. I had to add an extra condition to make sure that the tangent line remains on one "side" of $C$. Your example is clear. I also am unsure if this definition holds for smooth functions, but I can't seem to think of a counterexample. –  Jared Apr 17 '13 at 0:41
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@Jared: smoothness is not enough: you want something like convexity: e.g. $f''$ has constant sign. Nevertheless this is close enough to the truth so that I have made use of it on the first day of freshman calculus to try to give some quick intuition about tangent lines. –  Pete L. Clark Apr 17 '13 at 4:46
    
Of course. I was trying to be too cute and missed some simple counterexamples. It would be interesting to try to rigorously define a tangent line without reference to secants or derivatives or limits. Above was my first attempt. –  Jared Apr 17 '13 at 5:44
    
I don't know a way to give a definition of a tangent line of, say, an arbitrary smooth function, which does not use or amount to the limit concept. (In fact I am skeptical that such a thing is possible, but you shouldn't let that hinder you.) On the other hand, if you assume convexity there is a very nice theory of supporting lines which seems to have all but disappeared from calculus / honors calculus / undergrad real analysis courses. I have reproduced it in $\S$ 7.3.8 of math.uga.edu/~pete/2400full.pdf. –  Pete L. Clark Apr 17 '13 at 15:21

Part of the problem is you're trying to compare mathematics to reality. As a mathematical object of course tangent lines exist, but the question of to what degree real world phenomena may be accurately modeled by continuous/differentiable functions is another. No one even knows if spacetime is discrete, continuous, non-commutative etc. You could also ask the question "Do spheres exist?", and the answer again would depend on whether you mean do they exist as mathematical objects or do they exist in nature. I would answer "yes" to the former and "no" to the latter, but that's just me.

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Well I'm not limiting math to reality, I'm just making an analogy. My whole point was that even if you had infinitely many pictures closer and closer in time to each other, you could never tell me the speed from just one picture. –  Ovi Apr 17 '13 at 2:24

Tangent lines exist as geometric objects and they have slopes. The real question is why does the derivative equal the slope of the tangent line? The answer is that the derivative $f^{'}(x)$ is not the slope of a secant line for a point really close to $x$. It's the limit of the slopes of the secant lines for points arbitrarily close to $x$.

If I say $\lim_{x\to c} f(x) = L$ I don't mean $f(c) = L$. I mean that if you specify any tolerance on L, I can get within that tolerance by choosing a value for $x$ closer to $c$.

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Thinking about the derivative at a point as the slope of the tangent line at that point is simply a way to guide one's intuition via geometry. Further, it is not true that the derivative (or tangent line) always exists. For example, the absolute value function $y =|x| $ does not have a well defined tangent line at $ x = 0$

Lastly, yes in practice, the derivative is often estimated using similar methods as you've stated, but even in the example you gave, one can argue that the speed of the runner at a particular instance exists, so the derivative's existence is not the issue -- the problem is the derivative is hard or impossible to give precisely in some real-world situations, but it can usually be estimated if it exists.

But I should also say, the 'existence' of a mathematical idea in the real world is really a matter of philosophy.

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Well I'm not limiting math to reality, I'm just making an analogy. My whole point was that even if you had infinitely many pictures closer and closer in time to each other, you could never tell me the speed from just one picture. –  Ovi Apr 17 '13 at 2:24

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