From 8th grade algebra:

Everyone remember slope-intercept form for graphing lines on an X-Y axis? If not, this is it:

y = M*x + b

Where “M” is the slope (rate of change) of the line and “b” is where the line intersects the Y-axis at x=0.

Now, differential Calculus in a nutshell:

Starting with “y = M*x + b” we take the first derivative of “y” with respect to “x”, denoted as such:

(dy/dx)*y = M

or

(dy/dx)*(M*x + b) = M

As you can see, ALL that this non-magical operation does is find the function’s rate of change, “M”. As the equations get more complex, so does the operation, but the result is always the rate of change of the function.

Now on to integral Calculus:

This is more complex to describe, but is no less non-magical and simple in it’s result.

From Wikipedia:

“The definite integral is defined informally to be the area of the region in the xy-plane bounded by the graph of f(x), the x-axis, and the vertical lines x = a and x = b, such that areas above the axis add to the total, and the area below the x axis subtract from the total.”

Explained more simply:

Let’s take a shape we are all familiar with: A square.

Everyone should know how to find the area of a square, right? It’s just length multiplied by width. So, our square will be on the x-y axis, defined initially as a horizontal straight line: y = 5 In this case, our line has no slope. We want to find the area of the square, so we use integral calculus.

The Integral (from 0 to 5 on the x-axis, since this is a square) of y = 5 is:

y = 5*x2 – 5*x1

where x1 = 0 and x2 = 5

Which would be formally:

[Integral from x1=0 to x2=5] y = 5

and then evaluated,

Y = 5*(5) – 5*(0) = 25

Which is clearly the area of a 5×5 square.

Again, as the equations become more complex, so does the operation. As an engineer myself, I can tell you that I use Calculus EVERY DAY. Furthermore, if one experiments a gas and then one finds that the Calculus equations used to model it’s behavior don’t describe the observed behavior, then one has not modeled the system properly and has left some variables unaccounted for in their equations.

Now, to counter Hubbard’s comment on co-change. That is simply a matter of modelling dependent variables using differential equations, the 3rd tier of Calculus.This is merely an extension of differential and integral calculus followed to their logical conclusion. It is a complex system because our physical universe is extremely complex. Determining the rate of change of something that is is dependent on the rate of change of something else (such as time or temperature) is among the primary uses of Calculus.

This is not some belief system or heretical dogma that is open to debate or subject to interpretation, this is mathematics. Without this system of Calculus, NONE of the luxuries we enjoy today, most notably every electronic device in existence, space travel, GPS, flight navigation, etc. would be possible. As a man of faith myself, this is not a challenge to anyone’s beliefs; this is simply an attempt to dissolve ignorance.

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