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Transcript
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Hey everyone, Grant here.
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This is the first video in a series on the essence of calculus,
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and I'll be publishing the following videos once per day for the next 10 days.
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The goal here, as the name suggests, is to really get
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the heart of the subject out in one binge-watchable set.
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But with a topic that's as broad as calculus, there's a lot of things that can mean,
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so here's what I have in mind specifically.
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Calculus has a lot of rules and formulas which
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are often presented as things to be memorized.
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Lots of derivative formulas, the product rule, the chain rule,
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implicit differentiation, the fact that integrals and derivatives are opposite,
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Taylor series, just a lot of things like that.
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And my goal is for you to come away feeling like
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you could have invented calculus yourself.
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That is, cover all those core ideas, but in a way that makes clear where they
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actually come from, and what they really mean, using an all-around visual approach.
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Inventing math is no joke, and there is a difference between being
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told why something's true, and actually generating it from scratch.
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But at all points, I want you to think to yourself, if you were an early mathematician,
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pondering these ideas and drawing out the right diagrams,
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does it feel reasonable that you could have stumbled across these truths yourself?
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In this initial video, I want to show how you might stumble into the core ideas of
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calculus by thinking very deeply about one specific bit of geometry,
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the area of a circle.
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Maybe you know that this is pi times its radius squared, but why?
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Is there a nice way to think about where this formula comes from?
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Well, contemplating this problem and leaving yourself open to exploring the
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interesting thoughts that come about can actually lead you to a glimpse of three
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big ideas in calculus, integrals, derivatives, and the fact that they're opposites.
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But the story starts more simply, just you and a circle, let's say with radius 3.
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You're trying to figure out its area, and after going through a lot of
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paper trying different ways to chop up and rearrange the pieces of that area,
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many of which might lead to their own interesting observations,
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maybe you try out the idea of slicing up the circle into many concentric rings.
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This should seem promising because it respects the symmetry of the circle,
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and math has a tendency to reward you when you respect its symmetries.
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Let's take one of those rings, which has some inner radius r that's between 0 and 3.
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If we can find a nice expression for the area of each ring like this one,
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and if we have a nice way to add them all up,
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it might lead us to an understanding of the full circle's area.
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Maybe you start by imagining straightening out this ring.
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And you could try thinking through exactly what this new shape is and what its
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area should be, but for simplicity, let's just approximate it as a rectangle.
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The width of that rectangle is the circumference of the original ring,
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which is 2 pi times r, right?
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I mean, that's essentially the definition of pi.
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And its thickness?
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Well, that depends on how finely you chopped up the circle in the first place,
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which was kind of arbitrary.
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In the spirit of using what will come to be standard calculus notation,
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let's call that thickness dr for a tiny difference in the radius from one ring to
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the next.
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Maybe you think of it as something like 0.1.
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So approximating this unwrapped ring as a thin rectangle,
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its area is 2 pi times r, the radius, times dr, the little thickness.
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And even though that's not perfect, for smaller and smaller choices of dr,
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this is actually going to be a better and better approximation for that area,
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since the top and the bottom sides of this shape are going to get closer and closer to
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being exactly the same length.
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So let's just move forward with this approximation,
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keeping in the back of our minds that it's slightly wrong,
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but it's going to become more accurate for smaller and smaller choices of dr.
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That is, if we slice up the circle into thinner and thinner rings.
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So just to sum up where we are, you've broken up the area of the circle into
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all of these rings, and you're approximating the area of each one of those as
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2 pi times its radius times dr, where the specific value for that inner radius
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ranges from 0 for the smallest ring up to just under 3 for the biggest ring,
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spaced out by whatever the thickness is that you choose for dr, something like 0.1.
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And notice that the spacing between the values here corresponds to the
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thickness dr of each ring, the difference in radius from one ring to the next.
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In fact, a nice way to think about the rectangles approximating each
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ring's area is to fit them all upright side by side along this axis.
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Each one has a thickness dr, which is why they fit so snugly right there together,
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and the height of any one of these rectangles sitting above some specific value of r,
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like 0.6, is exactly 2 pi times that value.
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That's the circumference of the corresponding ring that this rectangle approximates.
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Pictures like this 2 pi r can get tall for the screen,
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I mean 2 times pi times 3 is around 19, so let's just throw up a y axis that's
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scaled a little differently so that we can actually fit all of these rectangles
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on the screen.
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A nice way to think about this setup is to draw the graph of 2 pi r,
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which is a straight line that has a slope 2 pi.
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Each of these rectangles extends up to the point where it just barely touches that graph.
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Again, we're being approximate here.
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Each of these rectangles only approximates the
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area of the corresponding ring from the circle.
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But remember, that approximation, 2 pi r times dr,
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gets less and less wrong as the size of dr gets smaller and smaller.
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And this has a very beautiful meaning when we're
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looking at the sum of the areas of all those rectangles.
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For smaller and smaller choices of dr, you might at first
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think that turns the problem into a monstrously large sum.
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I mean, there's many many rectangles to consider,
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and the decimal precision of each one of their areas is going to be an
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absolute nightmare.
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But notice, all of their areas in aggregate just looks like the area under a graph.
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And that portion under the graph is just a triangle,
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a triangle with a base of 3 and a height that's 2 pi times 3.
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So its area, 1 half base times height, works out to be exactly pi times 3 squared.
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Or if the radius of our original circle was some other value,
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capital R, that area comes out to be pi times r squared.
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And that's the formula for the area of a circle.
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It doesn't matter who you are or what you typically think of math,
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that right there is a beautiful argument.
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But if you want to think like a mathematician here,
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you don't just care about finding the answer,
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you care about developing general problem-solving tools and techniques.
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So take a moment to meditate on what exactly just happened and why it worked,
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because the way we transitioned from something approximate to something
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precise is actually pretty subtle and cuts deep to what calculus is all about.
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You had this problem that could be approximated with the sum of many small numbers,
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each of which looked like 2 pi r times dr, for values of r ranging between 0 and 3.
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Remember, the small number dr here represents our choice for the thickness of each ring,
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for example 0.1.
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And there are two important things to note here.
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First of all, not only is dr a factor in the quantities we're adding up,
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2 pi r times dr, it also gives the spacing between the different values of r.
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And secondly, the smaller our choice for dr, the better the approximation.
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Adding all of those numbers could be seen in a different,
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pretty clever way as adding the areas of many thin rectangles
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sitting underneath a graph, the graph of the function 2 pi r in this case.
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Then, and this is key, by considering smaller and smaller choices for dr,
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corresponding to better and better approximations of the original problem, the sum,
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thought of as the aggregate area of those rectangles,
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approaches the area under the graph.
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And because of that, you can conclude that the answer to the original question,
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in full unapproximated precision, is exactly the same as the area underneath this graph.
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A lot of other hard problems in math and science can be broken down and
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approximated as the sum of many small quantities,
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like figuring out how far a car has traveled based on its velocity at each point in time.
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In a case like that, you might range through many different points in time,
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and at each one multiply the velocity at that time times a tiny change in time, dt,
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which would give the corresponding little bit of distance traveled during that little
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time.
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I'll talk through the details of examples like this later in the series,
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but at a high level many of these types of problems turn out to be equivalent
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to finding the area under some graph, in much the same way that our circle problem did.
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This happens whenever the quantities you're adding up,
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the one whose sum approximates the original problem,
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can be thought of as the areas of many thin rectangles sitting side by side.
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If finer and finer approximations of the original problem correspond to thinner and
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thinner rings, then the original problem is equivalent to finding the area under some
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graph.
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Again, this is an idea we'll see in more detail later in the series,
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so don't worry if it's not 100% clear right now.
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The point now is that you, as the mathematician having just
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solved a problem by reframing it as the area under a graph,
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might start thinking about how to find the areas under other graphs.
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We were lucky in the circle problem that the relevant area turned out to be a triangle,
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but imagine instead something like a parabola, the graph of x2.
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What's the area underneath that curve, say between
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the values of x equals 0 and x equals 3?
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Well, it's hard to think about, right?
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And let me reframe that question in a slightly different way.
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We'll fix that left endpoint in place at 0, and let the right endpoint vary.
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Are you able to find a function, a of x, that gives
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you the area under this parabola between 0 and x?
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A function a of x like this is called an integral of x2.
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Calculus holds within it the tools to figure out what an integral like this is,
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but right now it's just a mystery function to us.
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We know it gives the area under the graph of x2 between some fixed left
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point and some variable right point, but we don't know what it is.
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And again, the reason we care about this kind of question is not just for
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the sake of asking hard geometry questions, it's because many practical
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problems that can be approximated by adding up a large number of small
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things can be reframed as a question about an area under a certain graph.
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I'll tell you right now that finding this area, this integral function,
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is genuinely hard, and whenever you come across a genuinely hard question in math,
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a good policy is to not try too hard to get at the answer directly,
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since usually you just end up banging your head against a wall.
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Instead, play around with the idea, with no particular goal in mind.
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Spend some time building up familiarity with the interplay between the function
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defining the graph, in this case x2, and the function giving the area.
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In that playful spirit, if you're lucky, here's something you might notice.
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When you slightly increase x by some tiny nudge dx, look at the resulting change in area,
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represented with this sliver I'm going to call da for a tiny difference in area.
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That sliver can be pretty well approximated with a rectangle,
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one whose height is x2 and whose width is dx.
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And the smaller the size of that nudge dx, the
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more that sliver actually looks like a rectangle.
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This gives us an interesting way to think about how a of x is related to x2.
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A change to the output of a, this little da, is about equal to x2,
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where x is whatever input you started at, times dx,
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the little nudge to the input that caused a to change.
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Or rearranged, da divided by dx, the ratio of a tiny change in a to the tiny
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change in x that caused it, is approximately whatever x2 is at that point.
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And that's an approximation that should get better
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and better for smaller and smaller choices of dx.
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In other words, we don't know what a of x is, that remains a mystery.
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But we do know a property that this mystery function must have.
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When you look at two nearby points, for example 3 and 3.001,
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consider the change to the output of a between those two points,
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the difference between the mystery function evaluated at 3.001 and 3.001.
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That change, divided by the difference in the input values, which in this case is 0.001,
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should be about equal to the value of x2 for the starting input, in this case 3 squared.
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And this relationship between tiny changes to the mystery function
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and the values of x2 itself is true at all inputs, not just 3.
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That doesn't immediately tell us how to find a of x,
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but it provides a very strong clue that we can work with.
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And there's nothing special about the graph x2 here.
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Any function defined as the area under some graph has this property,
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that da divided by dx, a slight nudge to the output of a divided by a slight
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nudge to the input that caused it, is about equal to the height of the graph at
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that point.
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Again, that's an approximation that gets better and better for smaller choices of dx.
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And here, we're stumbling into another big idea from calculus, derivatives.
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This ratio da divided by dx is called the derivative of a, or more technically,
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the derivative is whatever this ratio approaches as dx gets smaller and smaller.
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I'll dive much more deeply into the idea of a derivative in the next video,
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but loosely speaking it's a measure of how sensitive a function is to small changes in
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its input.
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You'll see as the series goes on that there are many ways you can visualize a derivative,
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depending on what function you're looking at and how you think
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about tiny nudges to its output.
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We care about derivatives because they help us solve problems,
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and in our little exploration here, we already have a glimpse of one way they're used.
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They are the key to solving integral questions,
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problems that require finding the area under a curve.
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Once you gain enough familiarity with computing derivatives,
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you'll be able to look at a situation like this one where you don't know what a function
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is, but you do know that its derivative should be x2,
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and from that reverse engineer what the function must be.
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This back and forth between integrals and derivatives,
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where the derivative of a function for the area under a graph gives you
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back the function defining the graph itself, is called the fundamental
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theorem of calculus.
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It ties together the two big ideas of integrals and derivatives,
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and shows how each one is an inverse of the other.
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All of this is only a high-level view, just a peek
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at some of the core ideas that emerge in calculus.
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And what follows in this series are the details, for derivatives and integrals and more.
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At all points, I want you to feel that you could have invented calculus yourself,
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that if you drew the right pictures and played with each idea in just the right way,
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these formulas and rules and constructs that are presented could have just
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as easily popped out naturally from your own explorations.
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And before you go, it would feel wrong not to give the people who supported this
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series on Patreon a well-deserved thanks, both for their financial backing as
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well as for the suggestions they gave while the series was being developed.
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You see, supporters got early access to the videos as I made them,
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and they'll continue to get early access for future essence-of type series.
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And as a thanks to the community, I keep ads off of new videos for their first month.
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I'm still astounded that I can spend time working on videos like these,
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and in a very direct way, you are the one to thank for that.
— end of transcript —
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