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Why are finding slope and finding area inverse processes?

  1. Jul 9, 2011 #1
    I'm about one week into a Calc II course, and I realized that I have no idea why finding the area under the curve (integrating) would be the inverse operation of finding slope (differentiating). I get that they are opposites, but not why that should be the case... it's not at all intuitive. Is there some sort of intrinsic relationship between slope and area?

    [Sorry if this question is poorly worded; I'm having a hard time putting my confusion into words.]
     
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  3. Jul 9, 2011 #2

    SteamKing

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  4. Jul 9, 2011 #3
    Are you looking for the intuition behind the Fundamental Theorem of Calculus? One informal approach is to think of it as a generalization of distance = rate x time.

    If you are traveling at a certain rate of speed, your total distance covered is your rate of speed, times the length of time you travel.

    Now if you are trying to integrate a function f over an interval [a,b], you can look at f over a tiny little interval, on which the rate of change of f is approximately the derivative f'. If the length of the little interval is dx, then the area under the curve (total distance travelled, amount of work done, etc.) is f'(x) dx. And if you add up all the little intervals, you are taking the integral of f'(x) dx over [a,b] to get the total distance represented by f over that interval.

    Now that's a hand-wavy, imprecise argument, not a proof, and probably not what's in the book. But you can see that if you know your rate of speed over each small part of your trip, you can add up the rate times time for each part of the trip to get the total distance travelled.

    Here's the same idea in a different form. You're driving your car from one point to another. Your passenger is assigned the task of writing down your speed, as indicated on your speedometer, at one-second intervals during the entire trip.

    After you arrive, you look at the list of speeds and you see, well, I was going 50 mph for the first second, then 49mph in the second second, and then 51 mph in the third second, etc. Of course your speed during each second is not constant, you are only looking at one sample point taken during each interval. Yet, using this information, you can get a very good estimate of the total length of your trip. That's because your velocity is well-behaved in each one-second interval: if the sample is 50mph, your speed was reasonably close to that during the entire second. So if you add up all the d = rt calculations over each second, you can estimate the total length of the trip.

    If that helps, good; if not, forget it. In your first week of Calc II you should just pay attention in class, do the homework, do some extra problems for practice, and get through the class. Intuition comes much later, if at all.

    Or as John von Neumann once said: In mathematics you don't understand things. You just get used to them.
     
    Last edited: Jul 9, 2011
  5. Jul 9, 2011 #4

    Stephen Tashi

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    Another way to build intuition about it is to look at material on "The Calculus Of Finite" differences. Look at the relation between summation and "anti-differencing". Notice how it resembles the approximation of derivatives and integrals by finite quantities.

    The basic thought is that if we define
    [tex] \triangle F(x) = \frac{ F(x+h) - F(x)}{h} [/tex] then the Riemann-like sum:

    [tex] h \sum_{k = 0} ^ n \triangle(F(a + kh)) = h (\frac{ F(a+h) - F(a)}{h} + ... \frac{ F(a + nh + h) - F(a+nh)}{h}) [/tex] is a "telescoping sum", which collapses to

    [tex] F(a + nh + h) - F(a) [/tex].

    Think of F(x) as being the "anti-delta" of [itex] \triangle F(x) [/itex] Think of letting n go to infinity while h goes to zero in such a manner that nh + h approaches the upper limit of integration b.
     
  6. Jul 9, 2011 #5
    Minor point but important; the area and slope are not inverses; they're double inverses.

    The integral computes the area. The inverse is the value of the function, not the slope.

    The derivative computes the slope. The inverse is the function, not the area.
     
  7. Jul 10, 2011 #6
    Thanks to everyone who answered, although I don't know how much better it made my gut intuition feel.

    I think this actually makes me feel the most relieved. With pretty much everything in Calculus it's like, "why would THAT be the case?! This makes no sense!" But I took Calc in high school, so Calc I and II at college are basically the second time around for me, and stuff is almost starting to make sense to me, but not quite.

    On a side note, I really wish they taught Calculus (concepts) in elementary school. I feel like just experiencing Calc a little bit when you're young would get rid of the whole "but this doesn't FEEL right" thing that Calculus has going on, at least for me.


    And another question - what happens if you integrate twice? What does that tell you about the function (if anything). f''(x) is concavity, but the second integration of functions isn't really mentioned, as far as I can tell. I don't even know what the notation for it would be. And extra-capital f? Thanks again.
     
  8. Jan 11, 2013 #7
    Is it easy to see the relationship if you dig more on the physics nature of it rather than on the formulation side of it. If you flip it and first see not differentiation(slope) but integration (area) you would realize an area is equal to the multiplication of the sides or ΔY X ΔX.

    Now the definition and study of slope started as way to study the growth different kind of functions (curves) for the same ΔX. the greater the ΔY for the same ΔX the greater the slope. Later formalized through infinitesimal limit calculations slope took its own personality as sort of gradient on infinitesimally small ΔX interval. But as long you consider ΔX an interval no matter how small is then slope will be always ΔY/ΔX = (ΔY X ΔX) / (ΔX X ΔX).

    Hope this help you.
     
  9. Jan 11, 2013 #8

    Mark44

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    Interesting idea, but I'm not sure it would be useful. It takes a certain amount of mathematical maturity to grasp the concepts in calculus. OTOH, if some topics were presented in a highly conceptual way without going into details, these students might gain something from it. On the other, other hand, too many students at the 5th and 6th grade struggle with simple arithmetic, to the point of never learning how to do long division, so there's that.
    If you integrate once, you get area (as one possible application). If you integrate twice, you get volume (again, as one possible application). We run out of handy geometric concepts for triple and higher order integrals, but there are concrete reasons that motivate such integrals. These are usually presented in the 2nd semester or 3rd quarter of college calculus.
     
  10. Jan 11, 2013 #9
    Is [itex]\arcsin\left(x\right)[/itex], then, not an inverse of [itex]\sin\left(x\right)[/itex], since the inverse of [itex]\sin[/itex] is the original number (with some restrictions?) The derivative and the integral are inverse operators, much in the same sense as [itex]\log\left(x\right)[/itex] and [itex]e^x[/itex] are inverses. The derivative of the integral is the original function, the integral of the derivative is the original function (up to a constant.)
     
  11. Jan 13, 2013 #10
    Seems like sort of a dangerous quote, though. I wouldn't take it to mean you shouldn't try to understand. But there are some things that seem initially sort of ugly and non-intuitive and eventually you realize that they aren't that bad. That's what I would like to think he was referring to. Sometimes, that may be just the way it is, but often, it's really the fault of whoever is presenting the material that it isn't very natural. So, that's why I think it's a dangerous quote, although partially true.
     
  12. Jan 13, 2013 #11
    Here's my two cents on this, know that that's all this is:

    Finding Area and Finding Slope really aren't inverses of each other. The confusion is in the difference between definite integrals and indefinite integrals. Indefinite integrals are defined as simply the reverse of differentiation, so of course they have an 'undoing' relationship with deriving.

    Definite Integrals (conceptually, at least) have nothing to do with deriving at all, they are simply the limit of a Riemann sum as was described above me. Indefinite Integrals and Definite Integrals are completely different things, and you can evaluate both slopes and areas under curves without once looking at derivatives or anti-derivatives.

    What's magical about the Fundamental Theorem is that even though Definite Integrals are completely different from differentiation, you are able to evaluate definite integrals by invoking indefinite integrals, which are related to differentiation. Because of that it seems like definite integrals are the opposite of differentiation, but really it is just that they are related to something (indefinite integrals) that by their very definition undo differentiation.
     
  13. Jan 13, 2013 #12
    I should probably say something about this inverse business, although it will probably be a bit abstract for the uninitiated.

    Yes, integration and differentiation are sort of inverse in some sense. Differentiation gives you a mapping from the set, let's say continuously, differentiable functions on R into the set of continuous functions. Strictly speaking, this mapping doesn't have an inverse because it isn't one to one. The problem is, each function plus any constant all map to the same thing. But there is a right inverse, given by the integral from 0 to x. That means you first do the integration map, then differentiation map and you get the identity map on the set of functions. If you try to do it the other way, differentiate first, then integrate, you don't get the identity map because the derivative will kill the constant term. You could say, it's an inverse "up to adding constants".

    Ah, but that's sweeping something under the rug. Why would you make the definition in the first place? An indefinite integral really should be defined as a definite integral where one endpoint is a variable, rather than a constant, not as an anti-derivative by definition. Then, the theorem relates it to taking an anti-derivative.

    The idea is just that the height of a function is the rate of change of its area. If you want to find how much the area goes up by when you increase x by a little bit, you just multiply the height by the change in x and that approximates how much the area goes up by. So, that gives you a little change in the area, A. Then, if the change in x is big, just find what happens when you increase x just a little bit at a time, as I described. Each little change in x gives a little change in A. Then to find total change, you just add up all the little areas. And then you just realize that this is the same as doing Riemann sums. Technically speaking, you need to make some definitions, apply the mean value theorem, take a limit, etc.
     
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