Calculus

Calculus in Latin means pebble or small stone, such that we are working with small pieces. There are two main ideas in calculus, the first being differentiation or finding a derivative (finding a slope, which would equal the rate) and the second being integration, which is the joining (or integration) of the small pieces together to find how much there is.

I remember studying calculus in school and there were so many concepts that never clicked. I could solve the equations, find derivatives, work out the area under the curve, etc. but I didn't see the use of calculus, i.e. the application of calculus. I'm revisiting calculus now because I've been taking part in a biostatistics course offered freely by Coursera, which requires a working knowledge of calculus. The definition of calculus on Wikipedia is as such:

Calculus is the mathematical study of change, in the same way that geometry is the study of shape and algebra is the study of operations and their application to solving equations.

Here's a great introduction to calculus. Again there is this concept that "Calculus is all about changes." They start with an example of trying to estimate the exact speed of a car at a precise moment. Since the formula for speed is:

$latex speed = \frac{distance}{time} $

we can't work out the exact speed because there is no distance and no time at an exact period of time, i.e. $latex t $ would equal zero. The introduction then used another example consisting of a falling body. They use the equation:

$latex d = \frac{1}{2}gt^2 $

where $latex d $ is the distance travelled by an object falling for time $latex t $ and $latex g $ = 10.

At time = 1, an object would have fallen 5 meters and using the speed equation, falling at a speed of 5 m/s. However, this is again not the exact speed at 1 second, just the average speed from 0 to 1 second. And this is where the whole idea of change comes in (and using infinitely small values).

The example created an infinitely small time period called $latex \Delta t $ and then proceeded to working out the difference in distance between t and t + $latex \Delta t $, so that we can calculate the speed. We already worked out that at 1 second, the distance travelled is 5 m. Now at $latex (1 + \Delta t) $ seconds, we can apply the same formula:

$latex d = 5t^2 = 5 \times (1 + \Delta t)^2 = 5 + 10\Delta t + 5(\Delta t)^2. $

Between 1 second and $latex (1 + \Delta t) $ seconds, the distance fallen is:

$latex Change\ in\ d = (5 + 10\Delta t + 5(\Delta t)^2) - 5 = 10\Delta t + 5(\Delta t)^2. $

Now we can work out the speed at $latex \Delta t $, using the distance above:

$latex speed = \frac{10\Delta t + 5(\Delta t)^2}{\Delta t} = 10 + 5\Delta t. $

Now if we shrunk $latex \Delta t $ towards 0, we get speed = 10 m/s.

What just happened? We used algebra to create a small time period just after 1 second and named it $latex \Delta t $. We worked out the distanced travelled from 1 second to $latex \Delta t $ and plugged it back into the speed equation. Then we made $latex \Delta t $, so small that it didn't matter, and arrived at 10 m/s. That was differential calculus, where you cut something into small pieces to find how it changes. Perhaps my school teacher did provide such an example to us (and I wasn't paying attention). All I remember from school was that differentiation is simply the multiplication of the function by the exponent's power and then decreasing it by 1. So if we take the derivative of $latex d = 5t^2 $ we get:

$latex \frac{d}{dt}5t^2 = 10t $

so that at t = 1, we get 10 m/s.

Another example

Here's another gentle introduction to calculus. Firstly, the author defined calculus as:

Calculus does to algebra what algebra did to arithmetic. Arithmetic is about manipulating numbers (addition, multiplication, etc.).
Algebra finds patterns between numbers: $latex a^2 + b^2 = c^2 $ is a famous relationship, describing the sides of a right triangle. Algebra finds entire sets of numbers - if you know a and b, you can find c.
Calculus finds patterns between equations: you can see how one equation (circumference = $latex 2\pi r $) relates to a similar one (area = $latex \pi r^2 $), as you will see below.

Algebra and calculus are a problem-solving duo: calculus finds new equations, and algebra solves them. To demonstrate this, the article then provided a really cool example of using the formula for the circumference of a circle ($latex 2\pi r $) to work out the area of a circle. Please see the link for illustrations.

The idea is that a circle can be made up of smaller circles inside of it. If we used a thick marker and a compass we could draw smaller circles (concentric circles) inside the outer circle until we fill the entire circle; imagine an archery target. Now if we unroll each of these circles and aligned them next to each other from the smallest to largest, we get roughly a triangle. The height of this triangle will be the circumference of the original circle, which is $latex 2\pi r $ and the base of this triangle will be the radius. The area of a triangle is:

$latex \frac{1}{2}bh $

and now we can work out the area of the circle by working out the area of this triangle:

$latex \frac{1}{2}r (2\pi r) = \pi r^2. $

I thought that was really cool; in school I just remembered the two formulas for circumference and area but now I know how they are related. The key idea was that we started with a disc, split it up into smaller circles, and arranged the circles into a triangle so that we could work out the area of a circle. Now imagine if we used a really thin marker to draw the concentric circles to get infinitely smaller lines; we could get a much finer triangle when we unroll each of these circles. This example illustrated how calculus is about breaking bigger things and looking at the smaller parts.

Another example

See also

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