MODERN ALCHEMIST

Calculus

There are two aspects of calculus to consider:

Differentiation
Integration

Calculus extracts information from a plotted function. Recall making y=mx + b graphs (or any function where y is equal to a function in terms of x). You may recall referring to y as the "dependent variable" and x as the "independent variable." For these calculations, we will be using t (time) for the independent variable. Dependent variables will include:

d - distance traveled (odometer reading)
v- velocity (speed)
a- acceleration (force applied by the engines of a spacecraft)

For our work with calculus we will use a function to plot the speed, acceleration, and distance traveled by an automobile.

Calculus can convert distance to velocity using differentiation.
Calculus can convert velocity to acceleration using differentiation.
Calculus can convert acceleration to velocity using integration, provided that the velocity at time zero is given.
Calculus can convert velocity to distance traveled using integration, provided that the distance traveled at time zero is given.

Assume that your car travels at an unknown constant speed miles per hour. The digital speedometer does not give readings above 85 mph, but the tachometer gives a steady reading, and the road is flat, so we assume we are traveling at a constant speed. We reach the highway at midnight, and set time=0 for the purposes of our graph.

At time=0.25 hours (15 minutes), we have traveled 25 miles. At time=0.50 hours (30 minutes), we have traveled 50 miles. You plot distance d as a function of time t, and notice a slope (rise over run) of 100 miles per hour.
The slope of a plot of distance vs. time calculates velocity.
Differentiation of the function which gives distance as a function of time produces velocity.
Your graph is a straight line, which makes it easy to determine the function:
Let's assume that we want to confirm that we are traveling at a constant speed.

The following is a known fact: distance traveled ‘d' is the derivative of velocity 'v' with respect to time 't'. We will show the formula below:

Calculus can perform the same operation, using differentiation.
The notation dv/dt is a shorthand for the expression "the derivative for v with respect to t."

A graph is shown below for your convenience.

For this exercise, the velocity was kept constant so that you could readily see that the slope of distance vs. time was equal to velocity.

What if v was not a constant? What if we accelerated and decelerated to change the velocity? Then the graph of distance vs. time would not be a straight line. It would not be possible to determine the slope by visual inspection. With a knowledge of calculus, if the graph fits a function that we can differentiate with our knowledge of calculus, then we can differentiate the function to give a function that calculates velocity as a function of time.

Previously we said that differentiation converted distance to velocity, and that differentiation also converted velocity to acceleration.

We now switch to a problem involving acceleration.

In an elevator, when it begins to travel upward, increasing it's speed, from zero to maybe 1 kilometer per hour or thereabouts, you feel the force, and it makes you feel heavier. Of course, a second later, you feel the same again.

The same principle would apply in a spacecraft. Assume you are currently traveling at a constant velocity. You are floating in the middle of the ship. Your velocity (relative to a fixed point) is the same as the ship, so you stay in the center, floating. Now assume the ships engines turn on for a picosecond, and now the ship is traveling two kph faster than you. You turn, and see the wall of the ship approaching you at 2 kph. The wall hits you (hopefully you aren't hurt too bad) and pushes against you, increasing your velocity by 2 kph and once again, you feel weightless.

Now, what if the ships engines remained on, continuously increasing the speed of the ship? You would remain against the wall, feeling a force similar to gravity. In fact, if the ship accelerated at a force equivalent to the force of gravity, you could stand up on the wall, and it would seem you were back on earth.

Recall that distance has the units of meters, and velocity has the units of meters per second. Acceleration is the change in velocity, and has the units
```
(meters per second)      meters
-------------------  or  ------
      second             second²
```
The gravitational force at sea level on earth is 9.8 m s^-2. Our bodies are adjusted to this force. Below is a graph that plots distance vs. time for a object accelerating at 9.8 m s^-2.

Before we go further into the calculations, now would be a good time to learn how to integrate and differentiate polynomials.

For more in depth information, try the MIT Calculus Index

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Last Revised 01/28/98.
Copyright ©1998 by William L. Dechent. All rights reserved.