In my last post, I talked about how I want to do research on how to build better calculators. However, I was being a bit misleading. Your basic hand-held calculator can solve all the math problems that most people encounter in their daily lives, so that problem has already been well solved. When I say I want to build better calculators, I mean that I want to build computers that solve harder math problems.

Here's where I come to a dilemma. My research usually tries to solve problems written in the language of calculus, and just the word "calculus" scares most people. Calculus seems to have this mystique of difficulty. For 60% of the population, calculus means "hard math that you never had because you chose a non-calc path in high school." For the other 39% of the population, calculus means "The highest (and hardest) math I know." My high school physics teacher never took calculus. Whenever he talked about it, his voice was filled with equal parts fear and awe, as if calculus were some black art that could rip the physical planes asunder if used improperly.

I want to dispel this myth. Today, I'll try to explain the notation of calculus and how it helps us. Then, I'll explain all of the results of calculus in 5 minutes. The target for this posts is the absolute lay person-- an amateur. I sometimes expect simple geometry, simple high school algebra, and that's about it. So if you don't understand these posts, please let me know in the comments.

First, the motivation. Around the time of Newton, physicists were making all sorts of graphs to describe the motion of planets and objects. After staring at the laboriously hand-drawn graphs, the physicists noticed strange relationships between the slope of graphs to other phenomina. For example, take a look at these graphs of the total distance traveled by a fictional car. The driver begins by punching the gas, slows down to a crawl, accelerates again, then starts backing up.

The top graph shows the total distance traveled by the car in time, and the bottom graph shows the velocity of the car in time. Also, notice that the slope of the line that touches the distance graph is more vertical when the velocity of car is higher. Physicists noticed this, so they invented a notation to describe it. Let f(t) be an abbreviation for "total distance traveled by the car at time t" and let f'(t) be the "slope of the line that just touches the graph at time t", where the apostrophe means "slope of." Using this notation, physicists could then perfectly describe the relationship between distance and velocity: "velocity at time t" = f'(t). It turns that most physics phenomina can be quickly described using this apostrophe notation.

However, is this notation really useful? After all, the whole point of physics is to translate something in the real world into a math equation we can solve. How do we find the slope of the line just touching a curve? The answer lies in this fact, the first tenant of calculus: If you "zoom in" close enough to any graph, the graph looks like a straight line. Therefore, once you "zoom in" you can just measure the slope of the line.

Take the function %$ f(t) = t^2 $%, for example. Say we want to find the slope when zoomed into t=1. The trick is to pick points that are really close together near t=1, then pretend the function is a line and calculate the slope: slope\; at\; f(a) \approx \frac{f(b)-f(a)}{b-a}

so for a=1 and b=1.1, we get: \frac {1.1^2-1}{1.1-1} = \frac{0.21}{0.1} = 2.1

If we zoom in further with b=1.01 and a=1, we get \frac{1.01^2-1}{1.01-1} = \frac{0.0201}{0.01} = 2.01

See the pattern? For a general point, you can prove that for any time t and small jump in the future h,

slope\; at\; f(t) & \approx & \frac{(t+h)^2-t^2}{(t+h)-h}\ &=& \frac{t^2+2ht+h^2-t^2}{h} \ &=& \frac{2ht+h^2}{h} \ &=& 2t+h

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Why do we care about this? First of all, the "Slope of" notation allows us to summarize physics laws. Secondly, we can use calculus to solve maximization problems. For example, assume we have an equation for the height of a thrown football: g(t)=t*t-t. The ball is thrown at t=0, and lands at t=1. When is the ball at it's maximum height? Before calculus, you spend all your time trying to solve equations like ax+b=0. No one ever tells you how to solve this type of problem.

The trick here is to notice on the graph above that we hit our maximum when the slope is equal to zero. Therefore, we need to solve g'(t)=0. Using the same "small h" trick, we can find that g'(t)=2t-1. Therefore, the ball reaches it's maximum height at t=0.5, because g'(0.5)=1-1=0.

In other words, calculus allows us to take a new problem and put it in a form that we already know how to solve.

This new ability is insanely practical. Lets say you work in business, and you have an equation for expected profit versus requested price, or profit=f(price). Using calculus, you can solve for the maximization of this function and find the price that will earn the most money possible!

This, in a nutshell, is the first 3 months of calculus. You spend all your time calculating "slopes", or "derivatives" of many different functions. You learn some tricks that allow you to quickly calculate the "slope" of arbitrary functions. Then, you drill solving 100 maximization problems until you can do it in your sleep. The reason that you had to learn all those complicated simplification procedures in Algebra II was because calculating the "slope" can sometimes involve some tricky simplifications. From a conceptual standpoint though, you can sum up the first three months of calculus with the following ideas:

• The "slope of a graph at a point" is called the derivative at that point.
• If you zoom into any function, it becomes a straight line.
• Once you zoom in, just measure the slope to find the derivative at that point.
• Whenever you are at a maximum or minimum, the slope is zero.

Next post, I'll talk about the other half of calculus: integration. I'll show you why the "slope of" aka derivative operation is magically related to the area under a curve.

In the same vein as my last post, I think I want to use this journal to see if I can remove the barriers between mathematicians and everyone else. Mathematicians are horrible about introducing arbitrary notation: often, two mathematicians who work on the same subject don't even have the same vocabulary. If you're in a graduate mathematics program, you can forget about understanding anything your colleagues are working on without taking months of math classes.

I'll be writing a series of posts in the near future about mathematics and what really makes mathematics work. Today, I'd like to explain calculus in five minutes.

I'm here to dispel this myth. Here's all of the first 3 months of calculus, once you strip out the terms and definitions: If you zoom close enough into a graph, it turns into a straight line.

The truth is, calculus is pretty much just common sense.

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Today, we're going to talk about the other half of calculus: finding the area under a curve. The need for this happens occasionally 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 is 10 meters long. The total volume of metal you need to build the wing is (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 with derivatives aka "slope of" operator, given a function g(x) you can write a math equation for these integrals, do another "small h" trick (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.

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)=1. The amazing thing is that this works for any function and it's derivative.

If you understood all that, you should be in utter disbelief. We said that the ' 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. Which 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, it's the actual problems that are tedious.

Here's a summary of everything you need to know about calculus: