2 files changed, 217 insertions, 8 deletions
diff --git a/build/website/mathematics/calculus/source/derivative.ms b/build/website/mathematics/calculus/source/derivative.ms
index 00bff8f..6675046 100644
--- a/build/website/mathematics/calculus/source/derivative.ms
+++ b/build/website/mathematics/calculus/source/derivative.ms
@@ -1,4 +1,6 @@
+.EQ
 delim $$
+.EN
 .TL
 The Derivative
 .AU
@@ -8,7 +10,7 @@ Pacific School of Innovation and Inquiry
 
 .NH 1
 Introduction
-.LP
+.PP
 The derivative operator is one of the most important
 concepts in calculus, and one of the fundamental building blocks
 for all advanced studies regarding STEM. Therefore, it is good
@@ -16,14 +18,221 @@ to know the basics behind the subject. I will be introducing them
 from the perspective of infinitesimal number, or the hyperreal number
 system as it is more formally known.
 
-.LP
-Imagine that you are a physicist studying rate of change. You can measure
-the position of a certain object through time, and you want to describe
-the average velocity of the object. Since you know that:
+.PP
+Imagine that you are a physicist studying rate of change. You know that you can measure
+the position of a certain object through time, and that therefore:
 
 .EQ
-delim $$
-v sub average = {\[*D] d(t)} over {\[*D] t}
+v sub average = {d sub final - d sub initial} over {t sub final - t sub initial}
+.EN
+
+If you've taken a grade 11 physics class, you should recall that this is because measuring
+the slope of a line gives you the amount change of d (displacement) in units of t (time), or as
+the math puts it, the ratio between the change in d (displacement) over t (time) is equal to the
+average velocity.
+
+.PP
+As a concrete example, imagine I am walking across a 100m road, with only two directions:
+forwards and backwards. If I start 20m into the road, and after 10 seconds of walking I end
+up 30m into the road, my final displacement would be 30m, my initial displacement would be 20m, and
+the amount of time that elapsed was 10 seconds (my initial amount of time would be 0s into the
+experiment, and my final amount of time would be 10 seconds). Therefore:
+
+.EQ
+v sub average = {30 - 20} over {10 - 0} = 1m/s
+.EN
+
+The above result concludes that on average between the 10 seconds we were walking, we traveled 1m/s.
+Note that this does not mean that we traveled 1m/s all the time; it could be the case that we traveled
+0.5m/s some of the way and 2m/s some other part of the way and it all averaged out to be 1m/s.
+From here we can ask a very important question: is there a way we can find the velocity such that instead
+of giving an average value, we can give an exact value for a specific point in time? Or, another way to
+phrase the question: how to we get the velocity at an instantaneous moment in time instead of over a period
+of time?
+
+.NH 1
+The Derivative
+.PP
+In fact, there is a way. Instead of taking the average over some patch of time, we take the average over
+an
+.I "infinitely small"
+amount of time. In this way, we can calculate the near-instant speed of an object with infinitely small
+inaccuracy.
+
+.PP
+Let's take a simple concrete example of what I am describing. Given a simple function $ d = t sup 2 $ where $ t $ is
+time and $ d $ is displacement, we want to get the instantaneous velocity at a general point
+(time and position respectively) $ (t sub initial, d sub initial) $ with their inifinitely small changes in distance
+and time being \[*D]t and \[*D]d respectively:
+
+.EQ
+d sub final = d sub initial + \[*D]d
+.EN
+.EQ
+t sub final = t sub initial + \[*D]t
+.EN
+.EQ
+v sub average = {d sub final - d sub initial} over {t sub final - t sub initial}
+.EN
+.EQ
+v sub average = {d sub initial + \[*D]d - d sub initial} over {t sub inital + \[*D]t - t sub initial}
+.EN
+.EQ
+v sub average = {\[*D]d} over {\[*D]t}
+.EN
+
+in words, we step through an infinitely small amount of time (\[*D]t) and then we can see what the resulting
+infinitely small step in displacement is (\[*D]d).
+
+.PP
+That might make sense, but how do we actually calculate what $ \[*D]d over \[*D]t $ is?
+We have too many variables, we can't solve for both displacement and time unless we have more information.
+Well, remember,
+we have a function that relates the amount of time that passes to the amount of displacement that happens.
+We can use that in order to get more information about how displacement and time are related so that
+we can express everything in terms of one variable, time. This next section is focused on replacing
+all the displacement variables with time variables, kind of like substitution with systems of linear equations.
+
+.NH 1
+Calculations
+.EQ
+d = t sup 2
+.EN
+.EQ
+d sub initial = t sub initial sup 2
+.EN
+.EQ
+d sub final = t sub final sup 2
+.EN
+
+and all of this must be true by definition. Given this:
+
+.EQ
+d sub final = d sub initial + \[*D]d
+.EN
+
+so we know that:
+
+.EQ
+d sub initial + \[*D]d = t sub final sup 2
+.EN
+
+but we also know that $ t sub final = t sub initial + \[*D]t $, so:
+
+.EQ
+d sub initial + \[*D]d = {(t sub initial + \[*D]t)} sup 2
+.EN
+
+.EQ
+\[*D]d = {(t sub initial + \[*D]t)} sup 2 - d sub initial
+.EN
+
+but remember:
+
+.EQ
+v sub average = {\[*D]d} over {\[*D]t}
+.EN
+
+so:
+
+.EQ
+v sub average = {{(t sub initial + \[*D]t)} sup 2 - d sub initial} over \[*D]t
+.EN
+
+But:
+
+.EQ
+d = t sup 2
+.EN
+
+So:
+
+.EQ
+d sub initial = t sub initial sup 2
+.EN
+
+And, therefore:
+
+.EQ
+v sub average = {{(t sub initial + \[*D]t)} sup 2 - t sub initial sup 2} over \[*D]t
+.EN
+
+So now we can finally do some math in order to figure out what velocity is, where the rest
+should just be easy grade 11 mathematics. We can expand the binomial here:
+
+.EQ
+v sub average = {t sub initial sup 2 + 2t sub initial \[*D]t + \[*D]t sup 2 - t sub initial sup 2} over \[*D]t
+.EN
+
+.EQ
+v sub average = {2t sub initial \[*D]t + \[*D]t sup 2} over \[*D]t
+.EN
+
+.EQ
+v sub average = 2t sub initial + \[*D]t
+.EN
+
+What does this equation mean? It means that this average velocity over an infinitely small period in time is going
+to grow at a rate of $ 2t + \[*D]t $. In other words, for any point in time, the velocity at that point in time will always be
+$ 2t + \[*D]t $. First of all, since \[*D]t is infinitely small, we tend to ignore that term and just say the velocity
+at any given moment in time is going to be $ 2t $. Second of all, instead of being an average velocity, this starts
+becoming more like an instantaneous velocity. So we say:
+
+.EQ
+v = 2t
+.EN
+
+because we used a general point fot $ t sub initial $, we've shown that it works for every point, so we can represent
+the resulting instantaneous velocity as a function. Look at what we've just done: we've taken a function of
+position in terms of time ($ d = t sup 2 $) and turned it into a function of
+.I "velocity"
+in terms of time ($ v = 2t $). This is what people mean when they say taking a derivative. Since velocity is the measurement
+of the change of displacement over time, what people mean when they say to take a derivative of a function is to take
+the rate of change of that function, or how much that function changes over time (or another variable).
+
+.NH 1
+General Form
+
+.PP
+In fact, instead of using a specific function $d = t sup 2$, we can find the general form of a derivative for any function.
+Suppose we know that $d = f(t)$ where $f(t)$ is our general function relating displacement to time. We can do the same
+process as before except more abstractly:
+
+.EQ
+d sub initial = f(t sub {initial})
+.EN
+.EQ
+d sub final = f(t sub {final})
+.EN
+.EQ
+\[*D] = f(t sub initial + \[*D]t) - d sub initial
+.EN
+.EQ
+\[*D] = f(t sub initial + \[*D]t) - f(t sub {initial})
+.EN
+.EQ
+f'(t) = {f(t sub initial + \[*D]t) - f(t sub {initial})} over \[*D]t
 .EN
 
-where \[*D]d(t)
+Where $f'(t)$ is the derivative of $f(t)$. Of course, I skipped a lot of steps, but all the steps that I skipped we already
+went over in another form above. I leave the specific details as an excercise.
+
+.PP
+Of course, we don't just use derivatives for physics, so time sometimes isn't the thing that is changing in our equations.
+The true definition of a derivative is:
+.EQ
+f'(x) = {f(x + \[*D]x) - f(x)} over \[*D]x
+.EN
+where x is a general variable we are measuring the change of. Also, note that the
+.I "initial"
+subscript is removed because there is no final or initial x anymore, there's an instantaneous change.
+
+.NH 1
+Conclusion
+
+.PP
+Think about what we've done. We started with a way to get the average velocity, we came up with a question regarding
+instantaneous velocity, then we used an approach using infinitely small numbers in order to solve the problem. Every
+step of the way started with a less abstract problem, which we then generalized in order to solve more general problems.
+This is the fundamental skill of getting better at anything, not just mathematics. With that, I think this is a good
+end to this article.
diff --git a/build/website/mathematics/calculus/source/derivative.pdf b/build/website/mathematics/calculus/source/derivative.pdf
deleted file mode 100644
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--- a/build/website/mathematics/calculus/source/derivative.pdf
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