What Resources Help Explore Math Concepts Beyond Calculus 2?

[ranger1716]

I have a question for those of you who may have seen books/resources on this before.

I am just finishing calculus 2 this semester and will be continuing in math next year.

What I would like to know is whether there are any books written for my mathematical level of knowledge that would allow me to explore deeper into why things work. i.e. why do derivatives give velocity, acceleration, etc. Why do integrals work to approximate volume. Am I asking for something that I don't have the math level for, or is there something like this available?

 

[FrogPad]

Have you taken a calculus level physics course? A physical interpretation of derivatives and integration might make more sense after taking one.

I'm not at the level to really speak on this. But after Calc III you can take Advanced Calculus, Real Analysis, Abstract Algebra, and maybe some Complex analysis to really understand why things do what they do. But I would imagine this is not necessary, and just taking some calculus level physics to suppliment the math, should do just fine.

I would think from just Calc II, that you could see why you can use integration to find volumes though, and why derivatives can be used to find the velocity and acceleration, by just thinking of things as infinitesmals.

 

[FrogPad]

Also Strang (awesome MIT professor) has a free calculus book that you can check out. (Maybe your calc book isn't any good).

http://ocw.mit.edu/ans7870/textbooks/Strang/strangtext.htm

 

[matt grime]

Why do you need anything to tell you why the rate of change of displacement with respect to time is velocity? I mean, that is what the definition of velocity is. If I want to find the speed at time t of something what do I do? I try and look at the average speed over some interval containing t, and then I let the interval get smaller and smaller to get a better idea of the 'instantaneous' speed at time t. Well, that *is* taking a derivative.

Integrals *are* defined to be areas.

And if you think about it its reasonably clear the rate of change of area under the graph of f is f (ie if you integrate something then differentiate it you get back to where you started).

 

[HallsofIvy]

Now that you have completed Calculus II it might be a good idea to go back and reread the first sections on limits and the derivative again! Often, after you are more comfortable with the later sections, the first concepts start to make more sense.

 

[matt grime]

Seeing as I said it is clear why the derivative of the integral of f(x) is f(x) I suppose I should justify that in case someone doesn't believe me.

Let's say we've integrated f(t) from 0 to x, call that I(x). What is the integral from 0 to x+h for some small h? Well, it's obviously approximately I(x)+hf(x), that is the area approximately is what we've got already plus a something that is going to be very close to a rectangle of height f(x) and width h (draw a picture). Thus f(x) is I'(x). It's also clear from this description why we require f(x) to be continuous for this to be true, ie so that that claim about being approximately a rectangle is true with error E(h,x) and that this tends to zero as h tends to zero.