| rob ( @ 2007-09-28 13:42:00 |
| Entry tags: | calculus, math, whatido |
Calculus Tutorial, cont.
Brock, I'd like to thank you for the links and sending attention my way.
Again, if you're confused by any of this, please respond in the comments. I'll edit the post to make things even clearer. I'd like to keep these posts as a reference that I can refer people to later.
Today, we're going to talk about the other half of calculus: finding the area under a curve. You occasionally need this in engineering. Let's say you are an aircraft engineer, and you've figured out an equation g(x) that describes the perfect cross section for an airplane wing. 
Assume for the moment that the wing doesn't taper, and that the wing is 10 meters long. The total volume of metal you need to build the wing equals (base area of cross section)*(length of the wing). Therefore, you want to know what is the area of the cross section under the curve so you can tell your boss how much metal he needs to buy. How do you find the area under this curve?
In calculus, this process is called "integration." Much like computing derivatives, the idea behind integration is simple: To measure the area under any curve, divide it up into a whole bunch of tall and thin rectangles, then add up the area of those rectangles. 
As your rectangles get thinner and thinner, you get a better and better estimate until you get exactly the right answer. Also, notice that the left side of each rectangle touches the function. This is just a convention. I could have chosen the right side, the midpoint, or pretty much any point. As the rectangles get thinner, you get closer to the correct answer regardless of your choice.
Like the derivative operation eventually produces a new function that describes the slope, you can use math to define a function that computes the area under a curve. Given any function g(x), you can write a math equation for these thin rectangles, factor out the small "h" (where h is the width of the rectangle), and get another function G(x) so that the area between "a" and "b" is G(b)-G(a). As an engineer, this helps you because you can now you have a simple expression that gives you the exact area under the curve. Unlike the derivatives, I'm not going to show you this because it's a bit mathy and doesn't give you any intuition. The important thing to remember is that like a derivative, integration is a process you apply to a function g(x) that produces another function G(x), and G(x) allows you to measure the area under a curve.
Here's the surprising part: Derivatives and integration 'undo' each other. In other words, (area of f'(x) from a to b) = f(b)-f(a). Do you remember that we showed yesterday that if f(x)=x^2, then f'(x)=2x? If you wanted to find the area under the graph of f'(x)=2x from 0 to 1, all you need to compute is f(1)-f(0) = 1-0 = 1. This matches our geometric formula: (area of a triangle)=0.5*(base)*(height)=0.5*(1)*(2)
If you understood all that, you should be experiencing utter disbelief. We said that derivative meant "slope of" and that integration meant "area under the curve". How can these two operations 'undo' each other? Why are they even related? What's going on? You don't find out the reason why until your junior or senior year of college as a Math major. Most people never find out. This brings us to our third take home point: Derivatives and integration undo each other because they are Magic.
There you go, that is everything that you learn in a year of calculus. The rest of the time is spent doing endless drills and problems, which you forget in 5 years anyway. If you've gotten this far, you have as good a conceptual understanding of calculus as anyone who has taken 1 year of study in the subject. The ideas behind calculus are simple.
Here's a summary of everything you need to know about calculus to understand the rest of my posts.