# Introduction to Calculus/Introduction

Before starting this course check the recommendations and resources on the Overview page

## Welcome!

    Many people think calculus is hard. This is a matter of opinion. I think it is not very dificult; then again, I probably wouldn't be writing this if I didn't really like the topic. However, this course should be easy to understand by both people who find it easy and people who find it hard. If you have any questions, just write them at the top of the discussion page, and someone should quickly add the information. Hopefully, with this, the course will not be too difficult to understand.
    This course consists of many lessons. They include this one, the next one on limits, and the third one on using derivatives.

## Why is Calculus necessary?

How fast can you run?
It seems a simple question but let's think:
$speed={\frac {distance\,\ traveled}{time\,\ taken}}$
So if you have run 100 meters in 10 seconds then your speed was 10 metres per second. Simple.
But that assumes you were running at the same speed all the time. Maybe you started off slowly then speed up and then slowed down slightly at the end. Its likely that your speed wasn't constant so all we are calculating is your average speed over the whole distance. The really interesting question is:
How fast am I going now?
Good question. If we are very good at measuring we might be able to find out that you traveled 11 meters in one second. But that's still only your average speed over 1 second. You might actually be slowing down: so your speed when we started to measure was actually slightly faster than 11 metres per second.
You see the problem. By using the formula above we cannot actually tell you how fast you are going now, we can only give you an average.
That's where calculus comes in. Using calculus, we can tell exactly how fast something is traveling right now.
How steep is this slope?
Again the equation that comes to mind is:
$gradient={\frac {vertical\,\ height\,\ gained}{horizontal\,\ distance\,\ traveled}}$
This is great if the slope is the same all the way up. If you are climbing a mountain or driving up a hill in the real world then you run into the same problem. We can make an average measurement but the formula can't tell us exactly how steep the slope is right here. Neither can calculus - unless the distance function is differentiable. Calculus only works when motion is described by a differentiable function.
For a function to be differentiable, it must be continuous and its rate of change over time must be continuous. This is to say that when you start running you cannot go from a standstill to your top speed instantly, you must at some instant in time move at every speed between the two.

## Finding the speed of a falling rock from the distance

Gravity is a very important feature of our everyday world. It also provides a very good basis for an introduction to the first section of calculus.
Say that, for whatever reason, my friend has dropped a rock over a cliff. The rock takes ten seconds to reach the lake below, at which point it makes a very large plop. Now, I wonder; how fast was the rock traveling when it hit the water?
Well, I know a few things about physics. I know that on the surface of the earth, a rock will fall $16t^{2}$ feet or $4.9t^{2}$ meters in t seconds. I also know that the average speed over an interval of time equals:
    ${\frac {Total\,\ distance\,\ traveled\,\ in\,\ the\,\ interval}{Length\,\ of\,\ interval\,\ of\,\ time}}$.
From this, you can say that at $t$ seconds, the distance the rock will have traveled will be $16t^{2}$ feet, and that at $t_{1}$ seconds, the distance the rock will have traveled will be $16t_{1}^{2}$. Next, we say that the length of the interval of time between $t$ and $t_{1}$ will be $t_{1}-t$, and that the total distance traveled during this interval will be $16t_{1}^{2}-16t^{2}$. Make sense? Now, for the grand finale. Let's use our equation for average speed, our distance traveled in an interval, and the length of the interval to create an average speed equation for any point during the rock's fall. The expression for this is:
    ${\frac {16t_{1}^{2}-16t^{2}}{t_{1}-t}}$

### Try-This-Yourself (TTYS) 1

    TTYS 1 will show up a lot during this part of this lesson. This is so you can follow what we are doing, by doing it yourself. First step;
create the average speed formula of a rocket in which the rocket, t seconds after blast off, will be $4t^{3}$ feet off the ground. Use the interval of time between X and Y (Or $t$ and $t_{1}$; it really doesn't matter what variables you use; the people who write calculus textbooks prefer the $t_{1}$ format.)

## Finding the distance traveled by a car from the speed

If we assume a car traveling at a known constant speed, can you guess how long it has passed since you started your stopwatch?. Well..let's try this simple formula,
    $v={\frac {\Delta d}{\Delta t}}\,$
    and so,
    $v\cdot \Delta t=\Delta d\,$
For example if your stopwatch reached 120 seconds and the speed of the car was 60 km/h ( = 60 km/(3600 s) = 1/60 km/s) then, the traveled distance is expected to be:
    1/60 [km/s] × 120 [s] = 2 km!
Now let's assume the car is traveling at different speeds. In this case we will have to record different times on our stopwatch, and make a note of the new speed at every recorded time. If we use the previous formula for calculating individual intervals, then we can get the distance traveled in every change of speed and time. By collecting or summing all these distances, we can get an approximation of the total distance the car has passed. In other words:
    $TotalDistance\approx \Delta d_{1}+\Delta d_{2}+\Delta d_{3}+\cdots +\Delta d_{n}\,$
Where n represents the number of recorded times. In terms of speed or velocity, we get:
    $TotalDistance\approx v_{1}\cdot \Delta t_{1}+v_{2}\cdot \Delta t_{2}+v_{3}\cdot \Delta t_{3}+\cdots +v_{n}\cdot \Delta t_{n}\,$
For example let's assume the following table is the recorded times and velocity changes the car has done:
| Time (s) | 0 | 30 | 60 | 100 | 120 | 160 |
| Speed (m/s) | 30 | 25 | 20 | 25 | 30 | 40 |
by making a new table in terms of the time differences and corresponding distance in that difference, we get:
| Time difference (s) | 30 | 30 | 40 | 20 | 40 |
| Distance in that difference (m) = v·Δt | 30 × 30 = 900 | 25 × 30 = 750 | 20 × 40 = 800 | 25 × 20 = 500 | 30 × 40 = 1200 |
So, the total distance is 900 + 750 + 800 + 500 + 1200 = 4150 m or 4.15 km Note that we ignored the last speed record. Indeed this is just one approximation! We will try the other approximation by ignoring the first speed and check the resulted distance.
| Time difference (s) | 30 | 30 | 40 | 20 | 40 |
| Distance in that difference (m) = v·Δt | 25 × 30 = 750 | 20 × 30 = 600 | 25 × 40 = 1000 | 30 × 20 = 600 | 40 × 40 = 1600 |
And, the total distance will be 750 + 600 + 1000 + 600 + 1600 = 4550 m or 4.55 km.
Can you see that much difference in the two solutions? Now let's take a more accurate approximation by taking the average speed between every two points, recall that average v = 1/2 × (v$_1$ + v$_2$)
| Time difference (s) | 30 | 30 | 40 | 20 | 40 |
| Distance in that difference (m) = v$_{av}$·Δt | 27.5 × 30 = 825 | 22.5 × 30 = 675 | 22.5 × 40 = 900 | 27.5 × 20 = 550 | 35 × 40 = 1400 |
The total distance is 825 + 675 + 900 + 550 + 1400 = 4350 m or 4.35 km.
We find that the three approximations are almost similar. Can you improve these approximations? The answer is yes but by measuring more time differences for the same previous example! The more the time intervals we reduce the more the recorded points we get the more the approximation in the three ways of solution.
NEXT Limits