Calculus has a bad reputation for being tough. I don’t know why.
Yes, there are a lot of rules and theorems you’ll need to understand to do well in a calculus class (or on a test, like the AP Calculus exam). But, when you get down to it, there are only two concepts you need to understand to grasp the basics of Calculus: derivatives and integrals. These two concepts describe the inherent relationship between different functions, and they make up the fundamental theorem of calculus. Understand these, and you understand everything.
To get you started, here’s a quick and dirty explanation of what a derivative is, using a simple physics example.
Calculus is all about related functions. You create one function to model the action of something in real life (like the speed of a free-falling person), and there are many functions that are related to it.
To understand the derivative of a function, we need to start with a basic function. Let’s say we want to model the position of a person skydiving from a plane. Let’s say our skydiver jumps out of a plane 10,000 feet (3,048 meters) above the air. We want a function that will tell us, at any given second, how many feet he has traveled – or, conversely, how close he is to hitting the ground and going splat.
To figure this out, we need to know something else – the speed at which the skydiver is falling. We can determine this based on the acceleration caused by gravity. In physics, when you want simple math, it is common to estimate the acceleration caused by gravity at -10 meters per second. At zero seconds, the person is not moving. At one second, the person is moving at 10 meters per second (towards the ground). At two seconds, the person is moving at 20 meters per second. So on and so forth.
I’ll save you the trouble of plotting out a bunch of data points and estimating the function. If we want a function to estimate the position of the skydiver at any given second, we end up with a function like this: -5x^2 + 3,048.
At zero seconds, the skydiver is 3,048 meters above the ground (10,000 feet). As time increases, the distance above the ground decreases. Eventually – around 24 to 25 seconds – the skydivers position will hit 0 meters above the ground. Umm, that is the ground. In other words… splat. Hopefully he pulled the ripcord a long time before that.
[Note: It would take a skydiver longer to hit the ground. That’s because at a certain point, he hits terminal velocity and stops accelerating. This would make our calculus function more complex, so we’ll ignore it for today. The beauty of modeling reality is that you don’t actually have to represent reality… just come close to it!]
So… What is a Derivative?
Now we have a function. Distance Above Ground = -5x^2 + 3,048. So, what is the derivative and what the heck does that mean?
The derivative of a function is a measure of how the value of that function changes. In this case, it’s a measure of how fast the skydiver’s distance above ground is changing.
Wait a minute… we have a special word for that in physics. Velocity! Velocity is a measure of how fast the sky-divers distance is changing. Or, you could measure the distance a car travels, and the derivative of that function would be the cars velocity (speed). Or, you could measure a runner’s distance along a 5 kilometer trail, and the rate of change would be his speed (velocity). You get the picture?
Eventually, you’ll learn how to take a function and determine the derivative. In this case, it’s simply -10x. If you think back to our starting point – the acceleration caused by gravity – this makes perfect sense. Gravity accelerates a falling body at 10 meters per second per second. In other words, multiply 10 m/s by the time and you’ll learn the velocity of a falling body at that time. That’s exactly the function that I just stated (v = -10x).
Are There More Derivatives…?
But wait. Are there more derivatives?
Sure! Calculus is all about many related functions. Not just one or two. In this case, you can take the original function (distance = -5x^2 + 3,048) and determine two derivatives.
The second derivative of the original function (and the regular derivative of the velocity function) will tell us the rate of change of the falling body’s velocity. Or, in normal person speak, it will tell us the acceleration of the falling body.
When you learn how to determine a derivative, you’ll figure out that the derivative of v = -10x is simply a = -10. In other words, the acceleration is constant. No matter what time it is, the falling body is accelerating at -10 meters per second per second. [Note: This is where our model deviates from reality, since a falling human body would eventually hit terminal velocity and stop accelerating.]
This is also a great example of how Calculus operates in simpler types of math. When you learned Geometry, you probably learned all about lines. One of the properties of a line equation or graph is that it has a slope – a measure of how the line changes over time.
Remember all that rise over run crap? Well, in this case, the rise over run (slope) for velocity is -10. In other words, the derivative of the velocity (a linear function) is the same as the slope of that function. Derivatives represent the same idea, but in a more complicated fashion, when you look at higher order functions (like quadratic functions, or polynomials to the degree three or higher).
You can think of derivatives as a way of breaking down or stepping down a function. Usually, they will end somewhere. In this case, there is no derivative of the acceleration function, because the rate of acceleration doesn’t ever change.
The second fundamental idea of calculus – integrals – goes the opposite way. You build up a function. You could start with our acceleration function (a = -10) and find an infinite number of integral or anti-derivative functions. But that’s a topic for another day…