From cd46e2ba935bbe8c3d55dfae410230f37425643e Mon Sep 17 00:00:00 2001 From: Preston Pan Date: Mon, 28 Nov 2022 14:16:23 -0800 Subject: add derivative article --- website/mathematics/calculus/derivative.pdf | Bin 0 -> 33703 bytes website/mathematics/calculus/source/derivative.ms | 225 ++++++++++++++++++++- website/mathematics/calculus/source/derivative.pdf | Bin 22978 -> 0 bytes 3 files changed, 217 insertions(+), 8 deletions(-) create mode 100644 website/mathematics/calculus/derivative.pdf delete mode 100644 website/mathematics/calculus/source/derivative.pdf (limited to 'website/mathematics/calculus') diff --git a/website/mathematics/calculus/derivative.pdf b/website/mathematics/calculus/derivative.pdf new file mode 100644 index 0000000..7101c7a Binary files /dev/null and b/website/mathematics/calculus/derivative.pdf differ diff --git a/website/mathematics/calculus/source/derivative.ms b/website/mathematics/calculus/source/derivative.ms index 00bff8f..6675046 100644 --- a/website/mathematics/calculus/source/derivative.ms +++ b/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/website/mathematics/calculus/source/derivative.pdf b/website/mathematics/calculus/source/derivative.pdf deleted file mode 100644 index f9b1fcb..0000000 Binary files a/website/mathematics/calculus/source/derivative.pdf and /dev/null differ -- cgit