diff --git a/.vscode/3blue1brown.code-snippets b/.vscode/3blue1brown.code-snippets index c7402af0..6c5054ee 100644 --- a/.vscode/3blue1brown.code-snippets +++ b/.vscode/3blue1brown.code-snippets @@ -24,7 +24,7 @@ "description": "Insert a figure shortcode into markdown." }, "Question": { - "scope": "markdown", + "scope": "markdow, mdx", "prefix": "question", "body": [ "\n$7\n" @@ -32,7 +32,7 @@ "description": "Insert a question into markdown." }, "Accordion": { - "scope": "markdown", + "scope": "markdown, mdx", "prefix": "accordion", "body": [ "\n$7\n" diff --git a/.vscode/settings.json b/.vscode/settings.json new file mode 100644 index 00000000..04ebd336 --- /dev/null +++ b/.vscode/settings.json @@ -0,0 +1,3 @@ +{ + "editor.wordWrap": "on" +} \ No newline at end of file diff --git a/data/topics.yaml b/data/topics.yaml index 10685512..62aba0d1 100644 --- a/data/topics.yaml +++ b/data/topics.yaml @@ -16,6 +16,7 @@ - area-and-slope - higher-order-derivatives - taylor-series + - 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These ideas seem completely different, so why are they inverses? +description: Derivatives are about slope, and integration is about area. These ideas seem completely different, so why are they inverses? date: 2017-05-06 chapter: 12 +thumbnail: thumbnail-2880-1620.png video: FnJqaIESC2s source: _2017/eoc/chapter9.py credits: - Lesson by Grant Sanderson +- Text adaptation by Kurt Bruns --- + +> We often hear that mathematics consists mainly of "proving theorems." Is a writer's job mainly that of "writing sentences?" +> +> \- Gian-Carlo Rota + +Here, I want to discuss one common type of problem where integration comes up: Finding the average of a continuous variable. + +
+ +This is a useful thing to know in its own right, but what's really neat is that it gives a completely different perspective for why integrals and derivatives are inverses each other. + +## Quick Intuition + +Before jumping into the main line of reasoning, the way this works is to take the area under the curve, and divide it by the width of that region. You might imagine that area as a pile of sand, and if you shake it all down to be flat, i.e. forming a rectangle whose area matches the area under the curve, the height of that rectangle should be the the average height of the function. To calculate that height, you would take the area of the rectangle and divide by its width. + +
+ +This is intuitive, but you might find it worthwhile to also think about how this relates to the more familiar method of averaging finitely many numbers, which is to add them all up and divide by how many there are. A very common feeling is that when you use a sum in a discrete context, you use an integral in a continuous context. But to avoid subtle errors in making this translation, its useful to be able to think exactly how and why the analogy from sums to integrals works in various contexts where it comes up. This question of finding the average value of a function offers a great chance exercise that muscle. + +## Average of sine function + +Take a look at the graph of $\sin(x)$ between $0$ and $\pi$, which is half its period. + +
+ +What is the average height of this graph on that interval? It's not a useless question. All sorts of cyclic phenomena in the world are modeled with sine waves: For example, the number of hours the sun is up per day as a function of which-day-of-the-year-it-is follows a sine wave pattern. + +
+ +If you wanted to predict, say, the average effectiveness of solar panels in summer months vs. winter months, you'd want to be able to answer a question like this: What's the average value of that sine function over half its period. + +
+ +Whereas a case like that will have all sorts of constants mucking up the function, we'll just focus on an unencumbered $\sin(x)$ function, but the substance of the approach would be the same in any application. + +
+ +It's kind of a weird thing to think about, isn't it, the average of a continuous variable. Usually, with averages, we think of a finite number of values, where you add all them up and divide that sum by how many values there are. But there are infinitely many values of $\sin(x)$ between $0$ and $\pi$, and its not like we can add all those numbers and divide by infinity. + +
+ +This sensation actually comes up a lot in math, and is worth remembering, where you have this vague sense that you want to add together infinitely many values associated with a continuum, even though that doesn't really make sense. Almost always, when you get this sense, the key will be to use an integral somehow. And to think through exactly how, a good first step is usually to approximate your situation with some kind of finite sum. + +In this case, imagine sampling a finite number of points, evenly spaced in this range. + +
+ +Since it's a finite sample, you can find the average by adding up all the heights, $\sin(x)$, at each one, and divide that sum by the number of points you sampled, right? + +
+ +And presumably, if the idea of an average height among all infinitely many points is going to make any sense at all, the more points we sample, which would involve adding up more heights, the closer the average of that sample should be to the actual average of the continuous variable. + +
+ + +This should feel at least somewhat related to taking an integral of $\sin(x)$ between $0$ and $\pi$, even if it might not be clear at first exactly how the two ideas will match up. + +$$ +\int_0^\pi \sin (x) d x +$$ + +For that integral, you also think of a sample of inputs on this continuum, but instead of adding the height $\sin(x)$ at each one, and dividing by how many there are, you add up $\sin(x) \cdot dx$ where $dx$ is the spacing between the samples; that is, you're adding little areas, not heights. + +
+ +Technically, the integral is not quite this sum, it's whatever that sum approaches as $dx$ approaches $0$. But it's helpful to reason with respect to one of these finite iterations, where you're adding the areas of some specific number of rectangles. So what you want to do is reframe this expression for the average, this sum of the heights divided by the number of sampled points, in terms of $dx$, the spacing between samples. + +$$ +\text { (Number of samples) } \approx \frac{x_2 - x_1 }{dx} +$$ + +For example, if the spacing between these points is $0.1$, and you know that they range from $0$ to $\pi$, then you can take the length of the interval, $\pi$, and divide it by the length of the space between each sample. If it doesn't go in evenly, you'd round down to the nearest integer, but as an approximation this is fine. + +$$ +\text { (Number of samples }) \approx \frac{\pi}{0.1} \approx 31 +$$ + +If we write the spacing between samples as $dx$, the number of samples is $\frac{\pi}{dx}$. So replacing the denominator with $\frac{\pi}{dx}$ here, you can rearrange, putting the $dx$ up top and distributing. + +
+ +But, think about what it means to distribute that $dx$ up top; it means the terms you're adding all look like $\sin(x) \cdot dx$ for the various inputs $x$ that you're sampling, so that numerator looks exactly like an integral expression. And for larger and larger samples of points, this average approaches the actual integral of $\sin(x)$ between $0$ and $\pi$, all divided by the length of that interval, $\pi$. + +
+ +In other words, the average height of this graph is this area divided by its width. On an intuitive level, and just thinking in terms of units, that feels pretty reasonable, doesn't it? Area divided by with gives you average height. + +
+ +## Computing a result + +Let's actually compute this integral expression. As we saw, last lesson, to compute an integral you need to find an antiderivative of the function inside the integral; some function whose derivative is $\sin(x)$. And, if you're comfortable with trig derivatives, you know the derivative of $\cos(x)$ is $-\sin(x)$. + +
+ +If you negate that, $-\cos(x)$ is the antiderivative of $\sin(x)$. + +
+ +To gut check yourself on that, look at this graph of $-\cos(x)$. At $0$, the slope is $0$, then it increases to some maximum slope at $\pi/2$, then it goes back down to $0$ at $\pi$, and in general its slope does indeed seem to match the height of the sine graph. + +To evaluate the integral of $\sin(x)$ between $0$ and $\pi$, take that antiderivative at the upper bound, and subtract its value at the lower bound. + +$$ +\int_0^\pi \sin (x) d x=(-\cos (\pi))-(-\cos (0)) +$$ + +More visually, that's the difference in the height of this $-\cos(x)$ graph above $\pi$, and its height at $0$, and as you can see, that change in height is exactly $2$. + + +
+ +That's kind of interesting, isn't it? That the area under this sine graph turns out to be exactly $2$. + +
+ +The answer to our average height problem, this integral divided by the width of the region, evidently turns out to be $\frac{2}{\pi}$, which is around $0.64$. + +$$ +\frac{\int_0^\pi \sin (x) d x}{\pi-0}=\frac{(-\cos (\pi))-(-\cos (0))}{\pi-0}=\frac{2}{\pi} \approx 0.64 +$$ + +## Comprehensive Question + +Now is a good time to try your hand at calculating the average value of a function over a range. For example, given the function $f(x)=-x^2+4 x-3$, what is the average value of the function between $1$ and $3$? + +
+ + + +We can start with the idea that we've been building up throughout the lesson that the average height is equal to the area under a range divided by the width of the range. + +$$ +\text { Average height }=\frac{\text { Area }}{\text { Width }} +$$ + +We know that the area is equal to the integral over that range. + +$$ +\text { Average height } = \frac{\displaystyle \int_{x_1}^{x_2} f(x)dx}{x_2 - x_1} +$$ + +And substituting the function and the range into the expression we get the following. + +$$ +\text { Average height }= \frac{\displaystyle \int_{1}^{3} (x^2+4 x-3) dx}{3 -1} +$$ + +To evaluate the integral of $x^2+4 x-3$ from $1$ and $3$, take the antiderivative at the upper bound and subtract its value at the lower bound. + +$$ +\int_{1}^{3} (x^2+4 x-3) dx = F(3) - F(1) +$$ + +Where we know the antiderivative is $F(x)=-\frac{x^3}{3}+2 x^2-3 x$ plus some arbitrary constant that will get cancelled in the subtraction. + +$$ +\begin{align} +\int_{1}^{3} (x^2+4 x-3) dx &= \left(-\frac{(3)^3}{3}+2(3)^2-3(3)\right)-\left(-\frac{(1)^3}{3}+2(1)^2-3(1)\right) \\[1em] +&= (-9+18-9)-\left(-\frac{1}{3}+2-3\right) \\[1em] +&=0-\left(-\frac{4}{3}\right) \\[1em] +&=\frac{4}{3} +\end{align} +$$ + +This means that the area under the function $x^2+4 x-3$ from $1$ and $3$ is equal to $\frac{4}{3}$. + +
+ +And to calculate the average height of this function, we need to divide by the width of the range, and $\frac{4}{3}$ divided by $2$ is equal to $\frac{2}{3}$. + +
+ +This is the answer to our question! The average height of the function $x^2+4 x-3$ from $1$ to $3$ is equal to $\frac{2}{3}$. + +
+ + + +## Another perspective + +I promised at the start that this question of finding the average value of a function offers an alternate perspective on on why integrals and derivatives are inverses of one and other; why the area under one graph is related to the slope of another. + +Notice how finding this average value $\frac{2}{\pi}$ came down to looking at the change in the antiderivative $-\cos(x)$ over the input range, divided by the length of that input range. Another way to think about that fraction is as the rise-over-run slope between the point of the antiderivative graph below zero, and the point of that graph above $\pi$. + +
+ +Now think about why it might make sense that this slope represents the average value of $\sin(x)$ on that region. By definition, $\sin(x)$ is the derivative of this antiderivative graph; it gives the slope of $-\cos(x)$ at every input. So another way to think about the average value $\sin(x)$ is as the average slope over all tangent lines here between $0$ and $\pi$. And from that view, doesn't it make a lot of sense that the average slope of a graph over all its point in a certain range should equal the total slope between the start and end point? + + + +## Generalize this view + +To digest this idea, it helps to see what it looks like for a general function. For any function $f(x)$, if you want to find its average value on some interval, say between $a$ and $b$, what you do is take the integral of $f$ on that interval, divided by the width of the interval. + +
+ +You can think of this as taking the area under the graph divided by the width. Or more accurately, it's the signed area of that graph, since area below the x-axis is counted as negative. + +
+ +And take a moment to remember the connection between this idea of a continuous average and the usual finite notion of an average, where you add up many numbers and divide by how many there are. When you take some sample of points spaced out by $dx$, the number of samples is about the length of the interval divided by $dx$. + +
+ +If you add up the value of $f(x)$ at each sample and divide by the total number of samples, it's the same as adding up the products $f(x) \cdot dx$ and dividing by the width of the entire interval. + +The only difference between that and the integral expression is that the integral asks what happens as $dx$ approaches $0$, but that just corresponds with samples of more and more points that approximate the true average increasingly well. Like any integral, evaluating this comes down to finding an antiderivative of $f(x)$, commonly denoted capital $F(x)$. + +
+ +In particular, what we want is the change to this antiderivative between $a$ and $b$, $F(b) - F(a)$, which you can think of as the change in the height of this new graph between the two bounds. + +
+ +I've conveniently chosen an antiderivative which passes through $0$ at the lower bound here, but keep in mind that you could freely shift this up and down, adding whatever constant you want to it, and it would still be a valid antiderivative. So the solution to the average problem is the change in the height of this new graph divided by the change to its $x$ value between $a$ and $b$. + +$$ +\frac{\displaystyle \int_a^b f(x) d x}{b-a}=\frac{F(b)-F(a)}{b-a} +$$ + +In other words, it's the slope of the antiderivative graph between these endpoints. + +
+ +And again, that should make a lot of sense, because little $f(x)$ gives the slope of a tangent line to this graph at each point, after all it is by definition the derivative of capital $F$. + +Why are antiderivative the key to solving integrals? Well, my favorite intuition is still the one I showed last lesson, but a second perspective is that when you reframe the question of finding the average of a continuous value as finding the average slope of bunch of tangent lines, it lets you see the answer just by the comparing endpoints, rather than having to actually tally up all points in between. + + + +## Summary + +In the last video, I described a sensation that should bring integrals to your mind. Namely, if you feel like the problem you're solving could be approximated by breaking it up somehow, and adding up a large number of small things. Here I want you to come away recognizing a second sensation that should bring integrals to your mind. If there's some idea that you understand in a finite context, and which involves adding up multiple values, like taking the average of a bunch of numbers, and if you want to generalize that idea to apply to an infinite continuous range of values, try seeing if you can phrase things in terms of an integral. It's a feeling that comes up enough that's it's definitely worth remembering. diff --git a/public/content/lessons/2017/area-and-slope/notes.txt b/public/content/lessons/2017/area-and-slope/notes.txt new file mode 100644 index 00000000..55edc469 --- /dev/null +++ b/public/content/lessons/2017/area-and-slope/notes.txt @@ -0,0 +1,32 @@ +Kurt's Notes: + +- Should the summation notation start at n=1? +- It feels fine to have a gut-check section for the antiderivative of sine + +Grant's Notes: + +- This should be two short articles: + - Finding the average value of a function? + - In what sense are slopes and areas inverses? + + - I wonder if there's a way to open with the central question. + - The standard visual intuition for a derivative of a function is that it gives the slope of the graph of that function. The standard visual intuition for a function's integral is that it gives the area under the graph of that function. + + In this series, we've played with other ways to visualize derivatives, and we've already seen one view that helps explain the fundamental theorem of calculus, which explains the sense in which derivatives and integrals are inverses of each other. But in light of the more typical visualization for derivatives, you might wonder, in what sense are slope and area opposites? + + + +- Before jumping into the main line of reasoning, we should have a section on "quick intuition" + - The way this works is to take the area under the curve, and divide it by width of that region. You might imagine that area as a pile of sand, and if you shake it all down to be flat, i.e. forming a rectangle whose area matches the area under the curve, the height of that rectangle should the the average height of the function. That height, of then, is the area of the rectangle divided by its width. + - [Circle the formula] + - This is intuitive, but you might find it worthwhile to also think about how this relates to the more familiar method of averaging finitely many numbers, which is to add them all up and divide by how many there are. + - A very common feeling is that when you use a sum in a discrete context, you use an integral in a continuous context. But to avoid subtle errors in making this translation, its useful to be able to think exactly how and why the analogy from sums to integrals works in various contexts where it comes up. + - This question of finding the average value of a function offers a great chance to exercise that muscle. + + + + + + +- "And take a moment to remember the connection between this idea of a continuous average and the usual finite notion of an average" This can probably be deleted. + - In general, this "generalizing" section can be much more succinct. \ No newline at end of file diff --git a/public/content/lessons/2017/area-and-slope/thumbnail-2880-1620.png b/public/content/lessons/2017/area-and-slope/thumbnail-2880-1620.png new file mode 100644 index 00000000..da89f0b8 --- /dev/null +++ b/public/content/lessons/2017/area-and-slope/thumbnail-2880-1620.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:ef3f4662e3263cc4fc82dcb3598d7794339caf3527b27e956d6d00d103612e28 +size 156819 diff --git a/public/content/lessons/2017/area-and-slope/transcript.txt b/public/content/lessons/2017/area-and-slope/transcript.txt new file mode 100644 index 00000000..9bd3c8ff --- /dev/null +++ b/public/content/lessons/2017/area-and-slope/transcript.txt @@ -0,0 +1,71 @@ +Opening quote +We often hear that mathematics consists mainly of "proving theorems." Is a writer's job mainly that of "writing sentences?" +- Gian-Carlo Rota +Introduction +Goals +Here, I want to discuss one common type of problem where integration comes up: Finding the average of a continuous variable. +This is a useful thing to know in its own right, but what’s really neat is that it gives a completely different perspective for why integrals and derivatives are inverses of one and other. +Average of sine function +Ask question +Take a look at the graph of sin(x) between 0 and pi, which is half its period. What is the average height of this graph on that interval? +It’s not a useless question. All sorts of cyclic phenomena in the world are modeled with sine waves: For example, the number of hours the sun is up per day as a function of which-day-of-the-year-it-is follows a sine wave pattern. So if you wanted to predict, say, the average effectiveness of solar panels in summer months vs. winter months, you’d want to be able to answer a question like this: What’s the average value of that sine function over half its period. +Whereas a case like that will have all sorts of constants mucking up the function, we’ll just focus on a pure unencumbered sin(x) function, but the substance of the approach would be the same in any application. +Weirdness of it +It’s kind of a weird thing to think about, isn’t it, the average of a continuous variable. Usually, with averages, we think of a finite number of values, where you add all them up and divide that sum by how many values there are. +But there are infinitely many values of sin(x) between 0 and pi, and its not like we can add all those numbers and divide by infinity. +This sensation actually comes up a lot in math, and is worth remembering, where you have this vague sense that you want to add together infinitely many values associated with a continuum like this, even though that doesn’t really make sense. +Almost always, when you get this sense, the key will be to use an integral somehow. And to think through exactly how, a good first step is usually to approximate your situation with some kind of finite sum. +Intuition +In this case, imagine sampling a finite number of points, evenly spaced in this range. Since it’s a finite sample, you can find the average by adding up all the heights, sin(x), at each one, and divide that sum by the number of points you sampled, right? +And presumably, if the idea of an average height among all infinitely many points is going to make any sense at all, the more points we sample, which would involve adding up more heights, the closer the average of that sample should be to the actual average of the continuous variable, don’t you think? +This should feel at least somewhat related to taking an integral of sin(x) between 0 and pi, even if it might not be clear at first exactly how the two ideas will match up. +For that integral, you also think of a sample of inputs on this continuum, but instead of adding the height sin(x) at each one, and dividing by how many there are, you add up sin(x)*dx where dx is the spacing between the samples; that is, you’re adding little areas, not heights. +Technically, the integral is not quite this sum, it’s whatever that sum approaches as dx approaches 0. But it’s helpful to reason with respect to one of these finite iterations, where you’re adding the areas of some specific number of rectangles. +So what you want to do is reframe this expression for the average, this sum of the heights divided by the number of sampled points, in terms of dx, the spacing between samples. +If I tell you that the spacing between these points is 0.1, for example, and you know that they range from 0 to pi, can you tell me how many there are? +Well, you can take that length of the interval, pi, and divide it by the length of the space between each sample. If it doesn’t go in evenly, you’d round down to the nearest integer, but as an approximation this is fine. +So if we write the spacing between samples as dx, the number of samples is pi/dx. +So replacing the denominator with pi/dx here, you can rearrange, putting the dx up top and distributing. +But, think about what it means to distribute that dx up top; it means the terms you’re adding all look like sin(x)*dx for the various inputs x that you’re sampling, so that numerator looks exactly like an integral expression. +And for larger and larger samples of points, the average approaches the actual integral of sin(x) between 0 and pi, all divided by the length of that range, pi. +In other words, the average height of this graph is this area divided by its width. On an intuitive level, and just thinking in terms of units, that feels pretty reasonable, doesn’t it? Area divided by with gives you average height. +Solve +So let’s actually compute this expression. +As we saw, last video, to compute an integral you need to find an antiderivative of the function inside the integral; some function whose derivative is sin(x) +And, if you’re comfortable with trig derivatives, you know the derivative of cos(x) is -sin(x), so if you negate that, -cos(x) is the antiderivative of sin(x). +To gut check yourself on that, look at this graph of -cos(x). At 0, the slope is 0, then it increases to some maximum slope at pi/2, then it goes back down to 0 at pi, and in general its slope does indeed seem to match the height of the sine graph. +To evaluate the integral of sin(x) between 0 and pi, take that antiderivative at the upper bound, and subtract its value at the lower bound. +More visually, that’s the difference in the height of this -cos(x) graph above pi, and its height at 0, and as you can see, that change in height is exactly 2. +That’s kind of interesting, isn’t it? That the area under this sine graph turns out to be exactly 2. +So the answer to our average height problem, this integral divided by the width of the region, evidently turns out to be 2/pi, which is around 0.64. +Alternate FTOC view +I promised at the start that this question of finding the average value of a function offers an alternate perspective on on why integrals and derivatives are inverses of one and other; why the area under one graph is related to the slope of another. +Notice how finding this average value 2/pi came down to looking at the change in the antiderivative -cos(x) over the input range, divided by the length of that input range. +Another way to think about that fraction is as the rise-over-run slope between the point of the antiderivative graph below zero, and the point of that graph above pi. +Now think about why it might make sense that this slope represents the average value of sin(x) on that region. +By definition, sin(x) is the derivative of this antiderivative graph; it gives the slope of -cos(x) at every input. +So another way to think about the average value sin(x) is as the average slope over all tangent lines here between 0 and pi. +And from that view, doesn’t it make a lot of sense that the average slope of a graph over all its point in a certain range should equal the total slope between the start and end point? +Generalize +To digest this idea, it helps to see what it looks like for a general function. +For any function f(x), if you want to find its average value on some interval, say between a and b, what you do is take the integral of f on that interval, divided by the width of the interval. +You can think of this as taking the area under the graph divided by the width. Or more accurately, it’s the signed area of that graph, since area below the x-axis is counted as negative. +And take a moment to remember the connection between this idea of a continuous average and the usual finite notion of an average, where you add up many numbers and divide by how many there are. +When you take some sample of points spaced out by dx, the number of samples is about the length of the interval divided by dx. +So if you add up the value of f(x) at each sample and divide by the total number of samples, it’s the same as adding up the products f(x)*dx and dividing by the width of the entire interval. +The only difference between that and the integral expression is that the integral asks what happens as dx approaches 0, but that just corresponds with samples of more and more points that approximate the true average increasingly well. +Like any integral, evaluating this comes down to finding an antiderivative of f(x), commonly denoted capital F(x). +In particular, what we want is the change to this antiderivative between a and b, F(b) - F(a), which you can think of as the change in the height of this new graph between the two bounds. I’ve conveniently chosen an antiderivative which passes through 0 at the lower bound here, but keep in mind that you could freely shift this up and down, adding whatever constant you want to it, and it would still be a valid antiderivative. +So the solution to the average problem is the change in the height of this new graph divided by the change to its x value between a and b. +In other words, it’s the slope of the antiderivative graph between these endpoints. +And again, that should make a lot of sense, because little f(x) gives the slope of a tangent line to this graph at each point, after all it is by definition the derivative of capital F. +So, why are antiderivative the key to solving integrals? Well, my favorite intuition is still the one I showed last video, but a second perspective is that when you reframe the question of finding the average of a continuous value as finding the average slope of bunch of tangent lines, it lets you see the answer just by the comparing endpoints, rather than having to actually tally up all points in between. +General phenomenon of sums being analogous to integrals +Abstract goal +In the last video, I described a sensation that should bring integrals to your mind. Namely, if you feel like the problem you’re solving could be approximated by breaking it up somehow, and adding up a large number of small things. +Here I want you to come away recognizing a second sensation that should bring integrals to your mind. +If there’s some idea that you understand in a finite context, and which involves adding up multiple values, like taking the average of a bunch of numbers, and if you want to generalize that idea to apply to an infinite continuous range of values, try seeing if you can phrase things in terms of an integral. +It’s a feeling that comes up enough that’s it’s definitely worth remembering. +End +… +My thanks, as always, to those making these videos possible. diff --git a/public/content/lessons/2017/derivatives-power-rule/index.mdx b/public/content/lessons/2017/derivatives-power-rule/index.mdx index d7bd30aa..60e38593 100644 --- a/public/content/lessons/2017/derivatives-power-rule/index.mdx +++ b/public/content/lessons/2017/derivatives-power-rule/index.mdx @@ -274,7 +274,7 @@ $$ answer={4} > -The small nudge $dx$ changes the width of the rectangle, but the the area must still be $1$, so the area gained must cancel out with the area lost. This gives us the expression below. +The small nudge $dx$ changes the width of the rectangle, but the area must still be $1$, so the area gained must cancel out with the area lost. This gives us the expression below. $$ dx \frac{1}{x} + x \cdot d \left( \frac{1}{x} \right) = 0 diff --git a/public/content/lessons/2017/essence-of-calculus/_figures.ai b/public/content/lessons/2017/essence-of-calculus/_figures.ai new file mode 100644 index 00000000..0b17be27 --- /dev/null +++ b/public/content/lessons/2017/essence-of-calculus/_figures.ai @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:8ab74f3b86bbf7cb99c1f45b2df5d309ce07c3bad4d10404c41c4dd700641b9a +size 145055 diff --git a/public/content/lessons/2017/essence-of-calculus/index.mdx b/public/content/lessons/2017/essence-of-calculus/index.mdx index 2e17bf6e..b850f118 100644 --- a/public/content/lessons/2017/essence-of-calculus/index.mdx +++ b/public/content/lessons/2017/essence-of-calculus/index.mdx @@ -11,7 +11,7 @@ credits: --- > "The art of doing mathematics is finding that *special case* that contains all the germs of generality." -> – David Hilbert +> \- David Hilbert My goal is for you to come away from this series feeling like you could have invented calculus. That is, cover these core ideas, but in a way that makes clear where they actually come from, and what they really mean, using an all-around visual approach. @@ -28,7 +28,7 @@ In this chapter, I want to show how you might stumble into the core ideas of cal Contemplating this problem and leaving yourself open to generalizing the tools you use along the way can actually lead you to a glimpse of three big ideas in calculus: Integrals, derivatives, and the fact that they are opposites. -But the story of finding the area starts more simply: Just you, and a circle. To be concrete, let’s say the radius is $1$ unit[^1]. If you weren’t worried about the exact area, and just wanted to estimate it, one way you might go about it is to chop the circle into smaller pieces whose areas are easier to approximate, then add up the results. +But the story of finding the area starts more simply: Just you, and a circle. To be concrete, let's say the radius is $1$ unit[^1]. If you weren't worried about the exact area, and just wanted to estimate it, one way you might go about it is to chop the circle into smaller pieces whose areas are easier to approximate, then add up the results. [^1]: This article uses an example circle of radius $1$ where the original video uses a circle of radius $3$. @@ -39,22 +39,22 @@ There are many ways you might go about this, each of which may lead to its own i caption="This figure illustrates different ways to divide the area of the circle with a radius $1$." /> -Math has a tendency to reward you when you respect symmetry, so among the divisions shown above, either the bottom left one dividing into pizza slices or the bottom right which divides into concentric rings will be likely to lead us down a fruitful path. For now, let’s think about the division into concentric rings [^2]. +Math has a tendency to reward you when you respect symmetry, so among the divisions shown above, either the bottom left one dividing into pizza slices or the bottom right which divides into concentric rings will be likely to lead us down a fruitful path. For now, let's think about the division into concentric rings [^2]. [^2]: You may notice that the other division of the circle that respects symmetry is dividing the circle into slices, like a pizza. Challenge question: Can you use this to provide an alternative argument for the area of circle formula? ## Unraveling rings -Focus on just one of those concentric rings, and let’s call its radius $r$, which will be some number between $0$ and $1$. If we can find a nice expression for the area of one such ring, and if we find a nice way to add them all up, it might lead us to an understanding of the full circle’s area. +Focus on just one of those concentric rings, and let's call its radius $r$, which will be some number between $0$ and $1$. If we can find a nice expression for the area of one such ring, and if we find a nice way to add them all up, it might lead us to an understanding of the full circle's area.
-In the spirit of using what will come to be standard calculus notations, let’s call the thickness of one of these rings “$dr$”, which you can read as meaning “the tiny difference in radius from one ring to the next”. In the drawing above, for example, $dr = 0.1$. +In the spirit of using what will come to be standard calculus notations, let's call the thickness of one of these rings “$dr$”, which you can read as meaning “the tiny difference in radius from one ring to the next”. In the drawing above, for example, $dr = 0.1$. -You should know that some people, perhaps even most people, would object to our using the notation $dr$ to represent a specific not-infinitely-small size like this, suggesting instead we use some other notation, like $\Delta r$. Well, it’s my house and my rules, and I have my reasons, which we’ll get into later. For now, I want you to think about $dr$ as simply being some number without worrying about what phrases like “infinitely small” would mean. +You should know that some people, perhaps even most people, would object to our using the notation $dr$ to represent a specific not-infinitely-small size like this, suggesting instead we use some other notation, like $\Delta r$. Well, it's my house and my rules, and I have my reasons, which we'll get into later. For now, I want you to think about $dr$ as simply being some number without worrying about what phrases like “infinitely small” would mean. -If this unwrapped shape is meant to perfectly match the area of the ring, it will be approximately, but not exactly, a rectangle. Because the outer circumference of the ring will be slightly larger than the inner circumference, the bottom of our unwrapped shape will be slightly wider than the top [^3]. However, if we’re comfortable beginning the exploration by only approximating the area of each small piece, we could consider this to be approximately a rectangle with a width of $2 \pi r$ and a height of $dr$. +If this unwrapped shape is meant to perfectly match the area of the ring, it will be approximately, but not exactly, a rectangle. Because the outer circumference of the ring will be slightly larger than the inner circumference, the bottom of our unwrapped shape will be slightly wider than the top [^3]. However, if we're comfortable beginning the exploration by only approximating the area of each small piece, we could consider this to be approximately a rectangle with a width of $2 \pi r$ and a height of $dr$. [^3]: You might hypothesize that the appropriate shape would be a trapezoid, but how would you be sure that those small diagonal edges are straight lines? It's not hard to intuitively justify but to make it rigorous you end up running into awkward questions about what we even mean by "area" in the first place. @@ -86,12 +86,12 @@ The difference between the true area of the ring and the area of the rectangular
However, and this will be a key idea, *this error becomes tiny compared to the overall area when $dr$ is small*. In other words, approximating the area of each ring as $2 \pi r \cdot dr$ is wrong, but as we chop up the circle into finer and finer rings, it becomes less and less wrong [^4]. -[^4]: The astute among you may worry about whether the errors in our approximations accumulate too much. You might say “Sure, sure, the difference between the true area of a ring as $2 \pi r \cdot dr$ is only a small percentage of the area of that ring, and it only gets smaller as $dr$ approaches $0$, I get that. But how do we know that the sum of all these errors doesn’t remain significant as $dr$ approaches $0$?” +[^4]: The astute among you may worry about whether the errors in our approximations accumulate too much. You might say “Sure, sure, the difference between the true area of a ring as $2 \pi r \cdot dr$ is only a small percentage of the area of that ring, and it only gets smaller as $dr$ approaches $0$, I get that. But how do we know that the sum of all these errors doesn't remain significant as $dr$ approaches $0$?” If you wanted to be more precise and rigorous, you would show that the size of this error for each ring is bounded by some constant times $(dr)^2$, while the total number of rings is about $\frac{R}{dr}$, so the total size of the aggregate error is bounded by some constant times $\frac{R}{dr} (dr)^2 = R \cdot dr$. So as $dr$ goes to zero, the full aggregate error will indeed go to zero with it. @@ -102,14 +102,14 @@ However, and this will be a key idea, *this error becomes tiny compared to the o ## Think to Graph -So you’ve broken up the area of this circle into all these rings, and you’re approximating the area of each one as $2 \pi \cdot r \cdot dr$. You might think that actually adding all those areas together will be a nightmare, especially if you’re seeking more accurate approximations with finer and finer divisions of the circle. However, being the bold mathematician that you are, you might have a hunch that taking this process to the utmost extreme may actually make things easier rather than harder. +So you've broken up the area of this circle into all these rings, and you're approximating the area of each one as $2 \pi \cdot r \cdot dr$. You might think that actually adding all those areas together will be a nightmare, especially if you're seeking more accurate approximations with finer and finer divisions of the circle. However, being the bold mathematician that you are, you might have a hunch that taking this process to the utmost extreme may actually make things easier rather than harder. The size of that inner radius for the rings ranges from $0$ for the smallest ring up to $1$ for the largest, spaced out by whatever thickness we chose for $dr$, like $0.1$. Or rather, they range from $0$ up to $1 - dr$, but as $dr$ gets smaller that upper bound looks more and more like $1$.
To help draw what adding those areas looks like, think about adding together your rectangular approximations. A nice way to visualize this is to fit all those rectangles upright, side-by-side standing on a horizontal axis. We can think of this horizontal axis as representing all the values of $r$ ranging from $0$ up to $1$. @@ -119,7 +119,7 @@ Each rectangle has a thickness of $dr$, which is also the spacing between differ
-Again, we’re being approximate here. Each of those rectangles only approximates the area of their corresponding ring from the circle. But remember, the approximation of $2 \pi \cdot r \cdot dr$ for the area of each ring will get less and less wrong as the size of $dr$ gets smaller and smaller. +Again, we're being approximate here. Each of those rectangles only approximates the area of their corresponding ring from the circle. But remember, the approximation of $2 \pi \cdot r \cdot dr$ for the area of each ring will get less and less wrong as the size of $dr$ gets smaller and smaller. -Again, concretely adding up the areas of all these rectangles would be a royal pain, but that’s when you get a crazy thought: Maybe, just maybe, asking what this sum _approaches_ as the choice of $dr$ gets smaller will be easier than ever actually computing the sum for any specific value of $dr$. +Again, concretely adding up the areas of all these rectangles would be a royal pain, but that's when you get a crazy thought: Maybe, just maybe, asking what this sum _approaches_ as the choice of $dr$ gets smaller will be easier than ever actually computing the sum for any specific value of $dr$.
-I’ll talk through the details of examples like this later in the series, but at a high level, many of these types of problems turn out to be equivalent to finding the area under some graph, in much the same way that our circle problem did. +I'll talk through the details of examples like this later in the series, but at a high level, many of these types of problems turn out to be equivalent to finding the area under some graph, in much the same way that our circle problem did. -This happens whenever the quantities that you’re adding up, the ones whose sum approximates your original problem, can be thought of as the area of many thin rectangles sitting side-by-side like this. If finer and finer approximations of your original problem correspond to thinner and thinner rectangles, the original problem will be equivalent to finding the area under some graph. +This happens whenever the quantities that you're adding up, the ones whose sum approximates your original problem, can be thought of as the area of many thin rectangles sitting side-by-side like this. If finer and finer approximations of your original problem correspond to thinner and thinner rectangles, the original problem will be equivalent to finding the area under some graph. -For example, think about two nearby inputs, like $3$ and $3.001$. The change to $x$, then, is $dx = 0.001$. The change $dA$ would be the difference between the mystery function evaluated at $3.001$ and evaluated at $3$. Even though we don’t know what that mystery function is, we do know something about this change, namely that this change divided by $dx = 0.001$ is approximately $3^2$. +For example, think about two nearby inputs, like $3$ and $3.001$. The change to $x$, then, is $dx = 0.001$. The change $dA$ would be the difference between the mystery function evaluated at $3.001$ and evaluated at $3$. Even though we don't know what that mystery function is, we do know something about this change, namely that this change divided by $dx = 0.001$ is approximately $3^2$. $$ \frac{A(3.001)-A(3)}{0.001} \approx 3^{2} @@ -297,20 +299,20 @@ And this relationship between tiny changes to the mystery function and the value image="figure-13.41.png" /> -That doesn’t immediately tell us how to find $A(x)$, but it provides a very strong clue to work with. And there’s nothing special about the graph $x^2$ here. For any function $f(x)$, if we call the area under its graph $A(x)$, then this area function has the property that $dA/dx \approx f(x)$, a slight nudge to the output of $A$ divided by a slight nudge to the input that caused it is about equal to the height of the graph at that point, $f(x)$. Again, that’s an approximation that gets better and better for smaller choices of $dx$. +That doesn't immediately tell us how to find $A(x)$, but it provides a very strong clue to work with. And there's nothing special about the graph $x^2$ here. For any function $f(x)$, if we call the area under its graph $A(x)$, then this area function has the property that $dA/dx \approx f(x)$, a slight nudge to the output of $A$ divided by a slight nudge to the input that caused it is about equal to the height of the graph at that point, $f(x)$. Again, that's an approximation that gets better and better for smaller choices of $dx$.
-Here we’re stumbling onto another big idea from calculus: Derivatives. This ratio $\frac{dA}{dx}$ is called the “derivative of $A$”. Or, more technically, the derivative is whatever value this ratio approaches as $dx$ gets smaller and smaller. +Here we're stumbling onto another big idea from calculus: Derivatives. This ratio $\frac{dA}{dx}$ is called the “derivative of $A$”. Or, more technically, the derivative is whatever value this ratio approaches as $dx$ gets smaller and smaller. -You and I will dive more deeply into the idea of a derivative next chapter, but loosely speaking it’s a measure of how sensitive a function is to small changes in its input. You’ll see as the series goes on that there are many different ways to visualize a derivative, depending on what function you’re looking at and how you think about tiny nudges to its output. +You and I will dive more deeply into the idea of a derivative next chapter, but loosely speaking it's a measure of how sensitive a function is to small changes in its input. You'll see as the series goes on that there are many different ways to visualize a derivative, depending on what function you're looking at and how you think about tiny nudges to its output. ## Fundamental Theorem -We care about derivatives because they help us solve problems, and in our little exploration here we have a slight glimpse of one way they’re used: They are the key to solving integral problems; problems that require finding an area under a curve. When you gain enough familiarity with computing derivatives, you’ll be able to look at a situation like the one below, where you don’t know what a function is, but you do know that its derivative should be $x^2$, and from that reverse engineer what the function must be. +We care about derivatives because they help us solve problems, and in our little exploration here we have a slight glimpse of one way they're used: They are the key to solving integral problems; problems that require finding an area under a curve. When you gain enough familiarity with computing derivatives, you'll be able to look at a situation like the one below, where you don't know what a function is, but you do know that its derivative should be $x^2$, and from that reverse engineer what the function must be. This back and forth between integrals and derivatives, where the derivative of the function for an area under a graph gives you back the function defining the graph itself, is called the *fundamental theorem of calculus*. It ties together the two big ideas of integrals and derivatives, and shows how in some sense, each one is the inverse of the other. @@ -332,7 +334,7 @@ In this chapter I encouraged you to think of $dr$ as the difference in the radiu text="Infinitely small?" /> -The issue with this is that phrases like “infinitely small” and “infinitesimal” run the risk of, well, not actually making any sense. What is an infinitesimal change? Is it a number? If so, which number? It shouldn’t be zero, but it also shouldn’t be anything bigger than zero. If it’s not a number then what is it, exactly? +The issue with this is that phrases like “infinitely small” and “infinitesimal” run the risk of, well, not actually making any sense. What is an infinitesimal change? Is it a number? If so, which number? It shouldn't be zero, but it also shouldn't be anything bigger than zero. If it's not a number then what is it, exactly? In acknowledgment of this conundrum, one common sentiment is to encourage students not to take this notation too seriously. The typical dictum is that derivatives, despite being written as $\frac{df}{dx}$, are not really fractions, and that integrals, written as $\displaystyle \int f(x) dx$, are not really sums, despite the fact that the symbol $\int$ literally derives from the letter "S" for "Sum". Instead, the student is told, each of these constructs has a separate more complicated definition. But! If a student wishes for loose intuition before stepping up their game to a real analysis class, they might find it helpful to think of $dx$ as an infinitely small change in $x$, even if that phrase doesn't _quite_ make sense. It is, after all, only intuition. @@ -344,8 +346,8 @@ In a nutshell, here's the philosophy for this series: When you see expressions w image="figure-3.04.svg" /> -There are more details and rules of thumb for _how_ you replace terms like $df$ and $dx$ with small numbers, and we'll address them as the series unfolds, but that’s the main idea. +There are more details and rules of thumb for _how_ you replace terms like $df$ and $dx$ with small numbers, and we'll address them as the series unfolds, but that's the main idea. I mention all this now so that you know this is not the universal convention; again finite changes are more typically written with the greek letter $\Delta$, as in an expression like $\Delta x = 0.001$. I also want to assure you that the spirit of this notational philosophy is not to compromise rigor for intuition, but to do the exact opposite. -And now, without further philosophical delay, let’s dive into what exactly a derivative is. +And now, without further philosophical delay, let's dive into what exactly a derivative is. diff --git a/public/content/lessons/2017/higher-order-derivatives/figures/displacement-velocity-acceleration-jerk.svg b/public/content/lessons/2017/higher-order-derivatives/figures/displacement-velocity-acceleration-jerk.svg new file mode 100644 index 00000000..496e158d --- /dev/null +++ b/public/content/lessons/2017/higher-order-derivatives/figures/displacement-velocity-acceleration-jerk.svg @@ -0,0 +1,497 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + 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_2017/eoc/footnote.py credits: - Lesson by Grant Sanderson +- Text adaptation by Kurt Bruns --- + +## Introduction + +In the next chapter, about Taylor series, I make frequent reference to higher order derivatives. If you're already comfortable with second derivatives, third derivatives and such, great! Feel free to skip right ahead to that lesson now. + +Somehow I've managed not to bring up higher order derivatives at all so far this series, so for the sake of completeness, I thought I'd give this little footnote to briefly go over them. I'll focus mainly on the second derivative, showing what it looks like in the context of graphs and motion, and leave you to think about the analogies for higher orders. + +## Graphical meaning + +
+ +Given some function $f(x)$, the derivative can be interpreted as the slope of its graph above some input, right? A steep slope means a high value for the derivative, a downward slope means a negative derivative. + +
+ + + +Find the derivative of $f(x)$. + +$$ +\begin{aligned} +\frac{df}{dx} &= \frac{d}{dx} \left( \frac{x^3}{3}-2 x^2+3 x+1 \right) \\ \rule{0pt}{2.25em} +\frac{df}{dx} &= x^2 - 4x +3 +\end{aligned} +$$ + +Substitute $x=2$ into the derivative to calculate the slope at $x=2$. + +$$ +\begin{aligned} +\frac{df}{dx}(2) &= (2)^2 - 4(2) + 3 \\ \rule{0pt}{2.25em} +\frac{df}{dx}(2) &= 4 - 8 + 3 \\ \rule{0pt}{2.25em} +\frac{df}{dx}(2) &= -1 +\end{aligned} +$$ + +Substitute $x=4$ into the derivative to calculate the slope at $x=4$. +$$ +\begin{aligned} +\frac{df}{dx}(4) &= (4)^2 - 4(4) + 3 \\ \rule{0pt}{2.25em} +\frac{df}{dx}(4) &= 16 - 16 + 3 \\ \rule{0pt}{2.25em} +\frac{df}{dx}(4) &= 3 +\end{aligned} +$$ + +The slope at $x=2$ is $-1$ and the slope at $x=4$ is $3$. + + + +The second derivative, whose notation I'll explain in a moment, is the derivative of the derivative, meaning it tells you how that slope is changing. The way to see this at a glance is to think of how the graph of $f(x)$ curves. At points where it curves upward, the slope is increasing, so the second derivative is positive. At points where it curves downward, the slope is decreasing, so the second derivative is negative. + +
+ +For example, let's plot the derivative and second derivative alongside the original function. The plot tells us a lot about how the slope is changing: At $x=1$ the slope is decreasing at a rate of $-1$, at $x=2$ the slope isn't changing at all, and at $x=3$ the slope is increasing at a rate of $1$. + +
+ +## Notation + +As far as notation goes, you could try writing it like this: + +$$ +\frac{d\left(\frac{d f}{d x}\right)}{d x} +$$ + +... indicating some small change to the derivative function divided by some small change to $x$, where as always the use of that letter $d$ suggests that you really want to consider what this ratio approach as $dx$, both $dx$'s in this case, approach $0$. That's pretty awkward and clunky, so the standard is to abbreviate it as $\frac{d^{2}f}{dx^2}$. + +$$ +\frac{d^2 f}{d x^2} +$$ + +It's not terribly important for getting an intuition of the second derivative, but perhaps it's worth showing how you can read this notation. Think of starting at some input to your function, and taking two small steps to the right, each with a size $dx$. I'm choosing rather big steps here so that we'll better see what's going on, but in principle think of them as rather tiny. + + +
+ +The first step causes some change to the function, which I'll call $df_1$, and the second step causes some similar, but possibly slightly different change, which I'll call $df_2$. + +
+ +The difference between these; the change in how the function changes, is what we'll call $d(df)$. + +
+ +You should think of this as really small, typically proportional to the size of $(dx)^2$. So if your choice for $dx$ was $0.01$, you'd expect this $d(df)$ to be proportional to $0.001$. + +$$ +d x=0.01 \Rightarrow(d x)^2=0.0001 +$$ + +And the second derivative is the size of this change to the change, divide by the size of $(dx)^2$. Or, more precisely, it's whatever that ratio approaches as $dx$ approaches $0$. Even though it's not like the letter $d$ is a variable being multiplied by $f$, for the sake of more compact notation you write this as $\frac{d^2f}{dx^2}$, and you don't bother with any parentheses on the bottom. + + + +For higher order derivatives, like the third derivative, fourth derivative and so are are written using the same compact notation. + +
+ +## Physical meaning + +Maybe the most visceral understanding of the second derivative is that it represents acceleration. + +$$ +\frac{d^2 s}{d t^2}(t) \Leftrightarrow \text { Acceleration } +$$ + +Given some movement along a line, suppose you have some function that records distance traveled vs. time, and maybe its graph looks something like this, steadily increasing over time. + +
+ +Then its derivative tells you velocity at each point in time, right? For the example, the graph might look like this bump, increasing to some maximum, then decreasing back to $0$. + +
+ +Then its second derivative tells you the rate of change for velocity, the acceleration at each point in time. + +
+ +In the example, the second derivative is positive for the first half of the journey, which indicates indicates speeding up. That's sensation of being pushed back into your car seat with a constant force. Or rather, having the car seat push you with a constant force. + +
+ + + +The third derivative, and this is not a joke, is called jerk. So if the jerk is not zero, it means the strength of the acceleration itself is changing. + +
+ +## Comprehensive Question + +A Pi Creature is aboard a spaceship and is set to journey into space. The height $s(t)$ of the spaceship from the ground, in meters, as a function of time $t$, in seconds, is given by: + +$$ +s(t)=t^4-4 t^3+6 t^2 +$$ + + + + + + + + + + + + + +If you plot the acceleration of the function $a(t)$, how would you interpret the motion of the spaceship at $t=1$ second? + + + +If you plot the acceleration function $a(t)$, you'll notice that at $t=1$ second, the acceleration dips to be $0$ before increasing again. + +
+ +One explanation is that this as a two-stage rocket where the first stage completes its burn and then there's a momentary pause before the second stage ignites. The momentary standstill in the change of acceleration, or the point where the jerk becomes zero, is when the spaceship is making that transition. + + + +## Next Lesson + +One of the most useful things about higher order derivatives is how they help in approximating functions, which is the topic of the next chapter on Taylor series, so I'll see you there . diff --git a/public/content/lessons/2017/higher-order-derivatives/transcript.txt b/public/content/lessons/2017/higher-order-derivatives/transcript.txt new file mode 100644 index 00000000..e027f312 --- /dev/null +++ b/public/content/lessons/2017/higher-order-derivatives/transcript.txt @@ -0,0 +1,34 @@ +- Introduction + - In the next chapter, about Taylor series, I make frequent reference to higher order derivatives. And, if you’re already comfortable with second derivatives, third derivatives and such, great! Feel free to skip right ahead to the main event now, you won’t hurt my feelings. + - But somehow I’ve managed not to bring up higher order derivatives at all so far this series, so for the sake of completeness, I thought I’d give this little footnote to very briefly go over them. + - I’ll focus mainly on the second derivative, showing what it looks like in the context of graphs and motion, and leave you to think about the analogies for higher orders. +- Graphical meaning + - Given some function f(x), the derivative can be interpreted as the slope of its graph above some input, right? A steep slope means a high value for the derivative, a downward slope means a negative derivative. + - The second derivative, whose notation I’ll explain in a moment, is the derivative of the derivative, meaning it tells you how that slope is changing. + - The way to see this at a glance is to think of how the graph of f(x) curves. + - At points where it curves upward, the slope is increasing, so the second derivative is positive. + - At points where it curves downward, the slope is decreasing, so the second derivative is negative. + - For example, a graph like this has a very positive second derivative at the input 4, since the slope is rapidly increasing around that point, whereas a graph like this still has a positive second derivative at that same point, but it’s smaller, since the slope is increasing only slowly. + - At points where there’s not really any curvature, the second derivative is zero. +- Address notation + - As far as notation goes, you could try writing it like this, indicating some small change to the derivative function divided by some small change to x, where as always the use of that letter d suggests that you really want to consider what this ratio approach as dx, both dx’s in this case, approach 0. + - That’s pretty awkward and clunky, so the standard is to abbreviate it as d2f/dx2. + - It’s not terribly important for getting an intuition of the second derivative, but perhaps it’s worth showing how you can read this notation. + - Think of starting at some input to your function, and taking two small steps to the right, each with a size dx. I’m choosing rather big steps here so that we’ll better see what’s going on, but in principle think of them as rather tiny. + - The first step causes some change to the function, which I’ll call df1, and the second step causes some similar, but possibly slightly different change, which I’ll call df2. + - The difference between these; the change in how the function changes, is what we’ll call d(df). + - You should think of this as really small, typically proportional to the size of (dx)2. + - So if your choice for dx was 0.01, you’d expect this d(df) to be proportional to 0.001. + - And the second derivative is the size of this change to the change, divide by the size of (dx)2. Or, more precisely, it’s whatever that ratio approaches as dx approaches 0. + - Even though it’s not like the letter d is a variable being multiplied by f, for the sake of more compact notation you write this as d2f/dx2, and you don’t bother with any parentheses on the bottom. +- Physical meaning + - Maybe the most visceral understanding of the second derivative is that it represents acceleration. + - Given some movement along a line, suppose you have some function that records distance traveled vs. time, and maybe its graph looks something like this, steadily increasing over time. + - Then its derivative tells you velocity at each point in time, right? For the example, the graph might look like this bump, increasing to some maximum, then decreasing back to 0. + - So its second derivative tells you the rate of change for velocity, the acceleration at each point in time. + - In the example, the second derivative is positive for the first half of the journey, which indicates indicates speeding up. That’s sensation of being pushed back into your car seat with a constant force. Or rather, having the car seat push you with a constant force. + - A negative second derivative indicates slowing down, negative acceleration. + - The third derivative, and this is not a joke, is called jerk. So if the jerk is not zero, it means the strength of the acceleration itself is changing. +- Next video + - One of the most useful things about higher order derivatives is how they help in approximating functions, which is the topic of the next chapter on Taylor series, so I’ll see you there. + diff --git a/public/content/lessons/2017/integration/figures/antiderivative-constant-issue.png b/public/content/lessons/2017/integration/figures/antiderivative-constant-issue.png new file mode 100644 index 00000000..20ef75a5 --- /dev/null +++ b/public/content/lessons/2017/integration/figures/antiderivative-constant-issue.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:f8b0d3186cefa14921a80f0f0d01eef6f5a542b056689200c7a8e71548538284 +size 531973 diff --git a/public/content/lessons/2017/integration/figures/approximation-t-1.png b/public/content/lessons/2017/integration/figures/approximation-t-1.png new file mode 100644 index 00000000..b8ae93a2 --- /dev/null +++ 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adaptation by Kurt Bruns --- + +> "One should never try to prove anything that is not almost obvious" +> +> \- Alexander Grothendieck + +This guy, Grothendieck, is somewhat of a mathematical idol to me. And I just love this quote, don't you? Too often in math we just dive into showing that a certain fact is true with long series of formulas before stepping back and making sure that it feels reasonable, and preferably obvious, at least on an intuitive level. + +In this article, I want to talk about integrals, and the thing that I want to become "almost obvious" is that they are an inverse of derivatives. We'll focus just on one example, which is kind of dual to the example of a moving car that I talked about in chapter 2 of the series, introducing derivatives. Then in the next lesson, we'll see how the idea generalizes into some other contexts. + +## Velocity + +Imagine you're sitting in a car, and you can't see out the window; all you see is the speedometer. At some point, the car starts moving, speeds up, then slows back down to a stop, all over 8 seconds. + +
+ +Is there a nice way to figure out how far you've traveled during that time, based only on your view of the speedometer? Or better yet, find a distance function $s(t)$ that tells you how far you've traveled after any given amount of time, $t$, between 0 and 8 seconds. Let's say you take note of the velocity at each second, and make a plot over time like this. + +
+ +And maybe you find that a nice function to model your velocity over time, in meters per second, is $v(t) = t(8-t)$. + +
+ +## Relationship between velocity and distance + +You might remember, in chapter 2 of this series, we were looking at the opposite situation, where you know a distance function, $s(t)$, and you want to figure out a velocity function from that. I showed how the derivative of your distance vs. time function gives you a velocity vs. time function. + +
+ +So in our current situation, where all we know is the velocity function, it should make sense that finding a distance vs. time function $s(t)$ comes down to asking what function has a derivative $-t^2 + 8t$. This is often described as finding the antiderivative of a function. + +
+ +And indeed, that's what we'll end up doing, and you could even pause and try that right now for the function $v(t)=-t^2 + 8t$. + +
+ +But first, I want to spend the bulk of this article showing how this question is related to finding an area bounded by the velocity graph, because that helps to build an intuition for a whole class of what are called "integral problems" in math and science. + +
+ +To start off, notice this question would be much easier if the car was moving with a constant velocity, right? In that case, you could just multiply the velocity, in meters per second, by the amount of time passed, in seconds, and that gives you the number of meters traveled. + +
+ +Notice that you can visualize that distance as an area, and if visualizing distance as an area seems weird, I'm right there with you. It's just that on this plot, where the horizontal direction has units of seconds and the vertical direction has units of meters/second, units of area very naturally correspond to meters. But what makes our situation hard is that the velocity is not constant, it's incessantly changing at every instant. + +It would even be a lot easier if it only ever changed at a handful of points, maybe staying static for the first second, then suddenly discontinuously jumping to a constant 7 meters per second for the next second, and so on, with discontinuous jumps to portions of constant velocity. + +
+ +That might make it very uncomfortable for the driver, in fact, it's physically impossible, but it would make your calculations a lot more straightforward. You could compute the distance traveled on each interval by multiplying the constant velocity on that interval by the change in time. Then just add them all up. + +
+ +So what we'll do is just approximate our velocity function as if it was constant on a bunch of different intervals. Then, as is common in calculus, we'll see how refining that approximation leads us to something precise. + +
+ +Here, let's make this more concrete with some numbers. Chop up the time axis between 0 and 8 into many small intervals, each with some little width $dt$, like 0.25 seconds. Consider one of these intervals, like the one between $t=1$, and $1.25$. In reality, the car speeds up from 7 m/s to about 8.4 m/s during that time, which you can find by plugging in $t = 1$ and 1.25 to the equation for velocity. + +
+ +We want to approximate the car's motion as if its velocity was constant on this interval. Again, the reason for doing that is that we don't really know how to handle anything other than constant velocity situations[^1]. You could choose this constant to be anything between $7$ and $8.4$, it doesn't really matter. + +[^1]: Now, you may notice a couple improvements of this approximation right away like averaging the two points, using the right point, or even a trapezoid. (if the constant breaking problem is used above, then some note about how the different approximation method doesn't matter in the end + +All that matters is that our sequence of approximations, whatever they are, gets better and better as $dt$ gets smaller and smaller. Treating this car's journey as a bunch of discontinuous jumps in speed between small portions of constant velocity becomes a less wrong reflection of reality as we decrease the time between those jumps. + +So, for convenience, let's just approximate the speed on each interval with whatever the true car's velocity is at the start of the interval; the height of the graph above the left side, which in this case is 7. + +
+ +So on this example interval, according to our approximation, the car moves $(7,\text{m/s})\cdot(0.25,\text{s})$. That's $1.75$ meters, nicely visualized as the area of this thin rectangle. This is a little under the real distance traveled, but not by much. And the same goes for every other interval: The approximated distance is $v(t) \cdot dt$, it's just that you plug in a different value of $t$ for each one, giving a different height for each rectangle. + + + +## Integral Notation + + +I'm going to write out an expression for the sum of the areas of all these rectangles in kind of a funny way. Take this symbol $\int$, which looks like a stretched "S" for sum, then put a $0$ at its bottom and an $8$ at its top to indicate that we're ranging over time steps between $0$ and $8$ seconds. + +
+ +The amount we're adding up at each time step is $v(t) \cdot dt$ which, for each step, can be visualized as the rectangle of area on the graph. + +
+ +Two things are implicit in this notation: First, the value $dt$ plays two roles: not only is it a factor in each quantity we're adding up, it also indicates the spacing between each sampled time step. So when you make $dt$ smaller, even though it decreases the area of each rectangle, it increases the total number of rectangles whose areas we're adding up, because if they are thinner it takes more of them to fill up that space. And second, the reason we don't use the usual sigma notation $\sum$ to indicate a sum is that this expression is technically not any particular sum for any particular choice of $dt$; it's meant to express whatever that sum approaches as $dt$ approaches 0. + +
+ +Remember, smaller choices of $dt$ indicate closer approximations for our original question, *how far does the car go*? So this limiting value for the sum, the area under this curve, gives the precise answer to the question, in full unapproximated precision. + +
+ +Now tell me that's not surprising. We have this pretty complicated idea of approximations that can involve adding up a huge number of very tiny things, and yet the value those approximations approach can be described so simply, as the area under a curve. + +$$ +\underbrace{\int_0^8 v(t) d t}_{d t \rightarrow 0} +$$ + +This expression is called an "integral" of $v(t)$, since it brings all of its values together, it integrates them. + +## Area Under Graphs + + + + +Now you could say, "How does this help!? You've just reframed one hard question, finding how far the car has traveled, into another equally hard problem, finding the area between this graph and the horizontal axis?" And...you'd be right! If the velocity/distance duo was all we cared about, most of this video with all this area under a curve nonsense would be a waste of time. We could just skip straight ahead to figuring out an antiderivative. + +But finding the area between a function's graph and the horizontal axis is somewhat of a common language for many disparate problems that can be broken down and approximated as the sum of a large number of small things. You'll see more next lesson, but for now I'll just say in the abstract that understanding how to interpret and compute the area under a graph is a very general problem-solving tool. + +In fact, the first video of this series already covered the basics of how this works, but now that we have more of a background with derivatives, we can actually take the idea to its completion. + +### Derivative of area function is the function + +For our velocity example, think of this right endpoint as a variable, capital $T$. + +
+ +So we're thinking of this integral of the velocity function between $0$ and $T$, the area under this curve between those two inputs, as a function, where that upper bound is the variable. That area represents the distance the car has traveled after $T$ seconds, right? So this is really a distance vs. time function $s(T)$. + +
+ +Now ask yourself: What is the derivative of that function? On the one hand, a tiny change in distance over a tiny change in time is velocity; that's what velocity means. But there's another way to see it purely in terms of this graph and this area, which generalizes better to other integral problems. + + +
+ +A slight nudge of $dT$ to the input causes that area to increase, some little $ds$ represented by the area of this sliver. The height of that sliver is the height of the graph at that point, $v(T)$, and its width is $dT$. And for small enough $dT$, we can basically consider that sliver to be a rectangle. So the area of that sliver, $ds$, is approximately equal to $v(T) \cdot dT$. + +Because this approximation gets better and better for smaller $dT$, the derivative of the area function $ds/dT$ at this point equals $v(T)$, the value of the velocity function at whatever time we started on. + +
+ +And that's super general, the derivative of any function giving the area under a graph like this is equal to the function for the graph itself. + +
+ +### Find the antiderivative + +So if our velocity function here is $t(8-t)$, what should $s$ be? What function of $t$ has a derivative $t(8-t)$? This is where we actually have to roll up our sleeves and do some math. + +
+ +It's easier to see if we expand this out as $8t - t^2$. Take each part here one at a time: What function has a derivative $8t$? Well, we know that the derivative of $t^2$ is $2t$, so if we just scale that up by $4$, we see that the derivative of $4t^2$ is $8t$. + +And for that second part, what kind of function might have $-t^2$ as its derivative? Using the power rule again, we know that the derivative of a cubic term, $t^3$, gives a squared term, $3t^2$, so if we scale that down by a third, the derivative of $(\frac{1}{3})t^3$ is exactly $t^2$, and making that negative we see that $-(\frac{1}{3})t^3$ has a derivative of $-t^2$. + +Therefore, the antiderivative of our function $8t - t^2$ is $4t^2 - \frac{1}{3}t^3$. + + + + + + + + + +### The +C issue + +But there's a slight issue here: we could add any constant to this function, and its derivative would still be $8t - t^2$. The derivative of a constant is always 0. And if we graph $s(t)$, you can think of this in the sense that moving a graph of a distance function up and down does nothing to affect its slope above each input. So there are actually infinitely many different possible antiderivative functions, all of which look like $4t^2 - (\frac{1}{3})t^3 + C$ for some constant $C$. + +
+ +But there is one piece of information we haven't used yet that lets us zero in on which antiderivative to use: The lower bound on the integral. This integral must be zero when we drag that right endpoint all the way to the left endpoint, right? + +
+ +The distance traveled by the car between 0 seconds and 0 seconds is zero. So as we found, this area as a function of capital $T$ is an antiderivative for the stuff inside, and to choose what constant to add, subtract off the value of that antiderivative function at the lower bound. If you think about it for a moment, that ensures that the integral from the lower bound to itself will indeed be 0. + +$$ +\int_0^T t(8-t) d t=\underbrace{\left(4 T^2-\frac{1}{3} T^3\right)-\left(4(0)^2-\frac{1}{3}(0)^3\right)}_{\text {Cancels when } T=0} +$$ + +As it so happens, when you evaluate the function we have here at $t=0$, you get zero, so in this specific case you don't actually need to subtract off anything. For example, the total distance traveled during the 8 seconds is this expression evaluated at $T=8$, which is $85.33$, minus 0 which is the answer to our original question! But a more typical example would be something like this integral between $1$ and $7$. That's the area pictured here, and it represents the distance traveled between 1 second and 7 seconds. + +
+ +What you'd do is evaluate the antiderivative we found at the top bound, $7$, and subtract off its value at the bottom bound, $1$. Notice, it doesn't matter what antiderivative we choose here; if for some reason it had a constant added to it, like $5$, that constant would cancel out. + + + + + + + + + +## Fundamental theorem + +
+ +More generally, anytime you want to integrate some function - and remember you think of as adding up the values $f(x) \cdot dx$ for inputs in a certain range then asking what that sum approaches as $dx$ approaches $0$ - the first step to evaluating that integral is to find an antiderivative, some other function, "capital $F(x)$", whose derivative is the thing inside the integral. Then the integral equals this antiderivative evaluated at the top bound, minus its value at the bottom bound. This fact is called the "fundamental theorem of calculus". + +
+ +Here's what's crazy about this fact: The integral, the limiting value for the sum of all these thin rectangles, takes into account every single input on the continuum from the lower bound to the upper bound, that's why we use the word "integrate"; it brings them together. And yet, to actually compute it using the antiderivative, you look at only two inputs: the top bound and the bottom bound. It almost feels like cheating! Finding the antiderivative implicitly accounts for all the information needed to add up all the values between the lower bound and upper bound. + +## Quick recap + +There's kind of a lot packed into this whole concept, so let's recap everything that just happened. We wanted to figure out how far a car goes just by looking at the speedometer, and what makes that hard is that the velocity was always changing. + +
+ +If you approximate it to be constant on multiple different intervals, you can figure out how far the car goes on each interval just with multiplication, then add them all up. Adding up those products can be visualized as the sum of the areas of many thin rectangles like this. + +
+ +Better and better approximations of the original problem correspond to collections of rectangles whose aggregate area is closer and closer to being the area under this curve between the start time and end time, so that area under the curve is the precise distance traveled for the true, nowhere-constant velocity function. + +
+ +If you think of this area as function, with a variable right end point, you can deduce that the derivative of that area function must equal the height of the graph at each point. + +
+ +That's the key! So to find a function giving this area, you ask what function has $v(t)$ as its derivative. + +There are actually infinitely many antiderivatives of a given function, since you can always just add some constant without affecting the derivative, so you account for that by subtracting off the value of whatever antiderivative function you choose at the bottom bound. + +### Negative area + +By the way, one important thing to bring up before leaving is the idea of negative area. What if our velocity function was negative at some point? Meaning the car is going backwards. It's still true that the tiny distance traveled $ds$ on a little time interval is about equal to the velocity times the tiny change in time, it's just that the number you'd plug in for velocity would be negative, so that tiny change in distance is negative. + +
+ +In terms of our thin rectangles, if the rectangle goes below the horizontal axis like this, its area represents a bit of distance traveled backwards, so if what you want is to find the distance between the car's start point and end point, you'd want to subtract it. And this is generally true of integrals: Whenever a graph dips below the horizontal axis, that area underneath is counted as negative. What you'll commonly hear is that integrals measure the "signed" area between a graph and the horizontal axis. + + + +If we graph the function $v(t) = 2 t^3-2 t^2$ we can see that the car travels backwards some unknown distance from $0$ to $1$ where it then startings going forward. + +
+ +This is essentially asking the question: at what point in time $T$ in the future does the integral equal $0$? + +$$ +0 = \int_0^T \left( 2 t^3-2 t^2 \right)dt +$$ + +And this is a problem that we can solve! + +$$ +\begin{aligned} +0 &= \int_0^T \left( 2 t^3-2 t^2 \right)dt \\ \rule{0pt}{2.25em} +0 &= \left[ \frac{2}{4} t^4 - \frac{2}{3} t^3 \right]_0^T \\ \rule{0pt}{2.25em} +0 &= \left( \frac{2}{4} (T)^4 - \frac{2}{3} (T)^3 \right) - \left( \frac{2}{4} (0)^4 - \frac{2}{3} (0)^3 \right) \\ \rule{0pt}{2.25em} +0 &= \frac{2}{4} (T)^4 - \frac{2}{3} (T)^3 \\ \rule{0pt}{2.25em} +\frac{2}{3} (T)^3 &= \frac{2}{4} (T)^4 \\ \rule{0pt}{2.25em} +\frac{1}{3} &= \frac{1}{4} T \\ \rule{0pt}{2.25em} +T &= \frac{4}{3} \\ +\end{aligned} +$$ + +If we plot the antiderivative $s(t)=\frac{1}{2} t^4-\frac{2}{3} t^3$ alongside the original function we can see that it is indeed equal to zero at the time $T = \frac{4}{3}$ seconds. + +
+ + + +## Next video + +Next up I'll bring up more contexts where this idea of an integral and the area under curves comes up, along with some other intuitions for the fundamental theorem of calculus. diff --git a/public/content/lessons/2017/integration/potential-questions.md b/public/content/lessons/2017/integration/potential-questions.md new file mode 100644 index 00000000..8263d5de --- /dev/null +++ b/public/content/lessons/2017/integration/potential-questions.md @@ -0,0 +1,41 @@ + + +
+ + + + + + + + + + + + +$ + + diff --git a/public/content/lessons/2017/integration/transcript.txt b/public/content/lessons/2017/integration/transcript.txt new file mode 100644 index 00000000..65a1998d --- /dev/null +++ b/public/content/lessons/2017/integration/transcript.txt @@ -0,0 +1,109 @@ +Opening quote +"One should never try to prove anything that is not almost obvious" +-Alexander Grothendieck +Introduction +This guy, Grothendieck, is somewhat of a mathematical idol to me. And I just love this quote, don"t you? Too often in math we just dive into showing that a certain fact is true with long series of formulas before stepping back and making sure that it feels reasonable, and preferably obvious, at least on an intuitive level. +In this video I want to talk about integrals, and the thing that I want to become "almost obvious" is that they are an inverse of derivatives. +Here, we"ll focus just on one example, which is kind of dual to the example of a moving car that I talked about in chapter 2 of the series, introducing derivatives. +Then in the next video, we"ll see how the idea generalizes into some other contexts. + +Velocity +Pose problem: +Imagine you"re sitting in a car, and you can"t see out the window; all you see is the speedometer. At some point, the car starts moving, speeds up, then slows back down to a stop, all over 8 seconds. +The question is, is there a nice way to figure out how far you"ve traveled during that time, based only on your view of the speedometer? Or better yet, find a distance function s(t) that tells you how far you"ve traveled after any given amount of time, t, between 0 and 8 seconds. +Let"s say you take note of the velocity at each second, and make a plot over time like this... And maybe you find that a nice function to model your velocity over time, in meters per second, is v(t) = t(8-t). +Why we should expect an antiderivative +You might remember, in chapter 2 of this series, we were looking at the opposite situation, where you know a distance function, s(t), and you want to figure out a velocity function from that. +I showed how the derivative of your distance vs. time function gives you a velocity vs. time function, so in our current situation, where all we know is the velocity function, it should make sense that finding a distance vs. time function s(t) comes down to asking what function has a derivative t(8-t). +This is often described as finding the anti-derivative of a function. +And indeed, that"s what we"ll end up doing, and you could even pause and try that right now. But first, I want to spend the bulk of this video showing how this question is related to finding an area bounded by velocity graph, because that helps to build an intuition for a whole class of what are called "integral problems" in math and science. +It"s hard become velocity is always changing. +This question would be much simpler if the car was moving with a constant velocity, right? In that case, you could just multiply the velocity, in meters per second, by the amount of time passed, in seconds, and that gives you the number of meters traveled. +Notice that you can visualize that distance as an area, and if visualizing distance as an area seems weird, I"m right there with you. It"s just that on this plot, where the horizontal direction has units of seconds and the vertical direction has units of meters/second, units of area very naturally correspond to meters. +But what makes our situation hard is that the velocity not constant, it"s incessantly changing at every instant. It would even be a lot easier if it only ever changed at a handful of points, maybe staying static for the first second, then suddenly discontinuously jumping to a constant 7 meters per second for the next second, and so on, with discontinuous jumps to portions of constant velocity. +That might make it very uncomfortable for the driver, in fact, it"s physically impossible, but it would make your calculations a lot more straightforward. +You could compute the distance traveled on each interval by multiplying the constant velocity on that interval by the change in time. Then just add them all up. +So what we"ll do is just approximate our velocity function as if it was constant on a bunch of different intervals. +Then, as is common in calculus, we"ll see how refining that approximation leads us to something precise. + +Show Riemann sum +Here, let"s make this more concrete with some numbers. Chop up the time axis between 0 and 8 into many small intervals, each with some little width dt, like 0.25 seconds. +Consider one of these intervals, like the one between t=1, and 1.25. +In reality the car speeds up from 7 m/s to about 8.4 m/s during that time, which you can find by plugging in t = 1 and 1.25 to the equation for velocity. +We want to approximate the car"s motion as if its velocity was constant on this interval. Again, the reason for doing that is that we don"t really know how to handle anything other than a constant velocity situations. +You could choose this constant to be anything between 7 and 8.4, it doesn"t really matter. +All that matters is that that our sequence of approximations, whatever they are, gets better and better as dt gets smaller and smaller. That treating this car"s journey as a bunch of discontinuous jumps in speed between small portions of constant velocity becomes a less wrong reflection of reality as we decrease the time between those jumps. +So for convenience, let"s just approximate the speed on each interval with whatever the true car"s velocity is at the start of the interval; the height of the graph above the left side, which in this case is 7. +So on this example interval, according to our approximation, the car moves (7 m/s)*(0.25 s). That"s 1.75 meters, nicely visualized as the area of this thin rectangle. This is a little under the real distance traveled, but not by much. +And the same goes for every other interval: The approximated distance is v(t)*dt, it"s just that you plug in a different value of t for each one, giving a different height for each rectangle. +Integral notation +I"m going to write out an expression for the sum of the areas of all these rectangles in kind of a funny way. +Take this symbol, which looks like a stretched "S" for sum, then put a 0 at its bottom and an 8 at its top to indicate that we"re ranging over time steps between 0 and 8 seconds. +And as I said the amount we"re adding up at each time step is v(t)*dt. +Two things are implicit in this notation: First, the value dt plays two roles: not only is it a factor in each quantity we"re adding up, it also indicates the spacing between each sampled time step. So when you make dt smaller, even though it decreases the area of each rectangle here, it increases the total number of rectangles whose areas we"re adding up. +And second, the reason we don"t use the usual sigma notation to indicate a sum is that this expression is technically not any particular sum for any particular choice of dt; it"s whatever that sum approaches as dt approaches 0. +As you can see, what that approaches is the area bounded by this curve and the horizontal axis. +Remember, smaller choices of dt indicate closer approximations for our original question, how far does the car go, right? So this limiting value for the sum, the area under this curve, gives the precise answer to the question, in full unapproximated precision. +Now tell me that"s not surprising. We have this pretty complicated idea of approximations that can involve adding up a huge number of very tiny things, and yet the value those approximates approach can be described so simply, as the area under a curve. +This expression is called an "integral" of v(t), since it brings all of its values together, it integrates them. + +Fundamental theorem of calculus +Why talk about area at all? +Now you could say, "How does this help!?. You"ve just reframed one hard question, finding how far the car has traveled, into another equally hard problem, finding the area between this graph and the horizontal axis?" +And...you"d be right! If the velocity/distance duo was all we cared about, most of this video with all this area under a curve nonsense would be a waste of time. We could just skip straight ahead to figuring out an antiderivative. +But finding the area between a function"s graph and the horizontal axis is somewhat of a common language for many disparate problems that can be broken down and approximated as the sum of a large number of small things. +You"ll see more next video, but for now I"ll just say in the abstract that understanding how interpret and compute the area under a graph is a very general problem-solving tool. +In fact, the first video of this series already covered the basics of how this works, but now that we have more of a background with derivatives, we can actually take the idea to its completion. +Derivative of area function is the function +For our velocity example, think of this right endpoint as a variable, capital T. So we"re thinking of this integral of the velocity function between 0 and T, the area under this curve between those two inputs, as a function, where that upper bound is the variable. +That area represents the distance the car has traveled after T seconds, right? So this is really a distance vs. time function s(T). +Now ask yourself: What is the derivative of that function? +On the one hand, a tiny change in distance over a tiny change in time is velocity; that"s what velocity means. But there"s another way to see it purely in terms of this graph and this area, which generalizes better to other integral problems. +A slight nudge of dT to the input causes that area to increase, some little ds represented by the area of this sliver. +The height of that sliver is the height of the graph at that point, v(T), and its width is dT. +And for small enough dT, we can basically consider that sliver to be a rectangle. So the area of that sliver, ds, is approximately equal to v(T)*dT. +Because this approximation gets better and better for smaller dT, the derivative of the area function ds/dT at this point equals v(T), the value of the velocity function at whatever time we started on. +And that"s super general, the derivative of any function giving the area under a graph like this is equal to the function for the graph itself. +Find the antiderivative +So if our velocity function here is t*(8-t), what should s be? What function of t has a derivative t*(8-t). This is where we actually have to roll up our sleeves and do some math. +It"s easier to see if we expand this out as 8t - t2. +Take each part here one at a time: What function has a derivative 8t? Well, we know that the derivative of t2 is 2t, so if we just scale that up by 4, we see that the derivative of 4t2 is 8t. +And for that second part, what kind of function might have -t2 as its derivative? Using the power rule again, we know that the derivative of a cubic term, t3, gives a squared term, 3t2, so if we scale that down by a third, the derivative of (⅓)t3 is exactly t2, and making that negative we see that -(⅓)t3 has a derivative of -t2. +Therefore, the antiderivative of 8t - t2 is 4t2 - (⅓)t3. + +The +C issue +But there"s a slight issue here: we could add any constant to this function, and its derivative would still be 8t - t2. The derivative of a constant is always 0. +And if we graph s(t), you can think of this in the sense that moving a graph of a distance function up and down does nothing to affect its slope above each input. +So there are actually infinitely many different possible antiderivative functions, all of which look like 4t2 - (⅓)t3 + C for some constant C. +But there is one piece of information we haven"t used yet that let"s us zero in on which antiderivative to use: The lower bound on the integral. +This integral must be zero when we drag that right endpoint all the way to the left endpoint, right? The distance traveled by the car between 0 seconds and 0 seconds is...zero. +So as we found, this area as a function of capital T is an antiderivative for the stuff inside, and to choose what constant to add, subtract off the value of that antiderivative function at the lower bound. +If you think about it for a moment, that ensures that the integral from the lower bound to itself will indeed be 0. +As it so happens, when you evaluate the function we have here at t=0, you get zero, so in this specific case you don"t actually need to subtract off anything. +For example, the total distance traveled during the 8 seconds is this expression evaluated at T=8, which is 85.33, minus 0. +But a more typical example would be something like this integral between 1 and 7. That"s the area pictured here, and it represents the distance traveled between 1 second and 7 seconds. +What you"d do is evaluate the antiderivative we found at the top bound, 7, and subtract off its value at the bottom bound, 1. +Notice, it doesn"t matter what antiderivative we choose here; if for some reason it had a constant added to it, like 5, that constant would cancel out. +Fundamental theorem +More generally, anytime you want to integrate some function –and remember you think of as adding up the values f(x)*dx for inputs in a certain range then asking what that sum approaches as dx approaches 0– the first step is to find an antiderivative, some other function, "capital F(x)", whose derivative is the thing inside the integral. +Then the integral equals this antiderivative evaluated at the top bound, minus its value at the bottom bound. This fact is called the "fundamental theorem of calculus". +Here"s what"s crazy about this fact: The integral, the limiting value for the sum of all these thin rectangles, takes into account every single input on the continuum from the lower bound to the upper bound, that"s why we use the word "integrate"; it brings them together. And yet, to actually compute it using the antiderivative, you look at only two inputs: the top and the bottom. +It almost feels like cheating! Finding the antiderivative implicitly accounts for all the information needed to add up all the values between the lower bound and upper bound. + +Quick recap +There"s kind of a lot packed into this whole concept, so let"s recap everything that just happened, shall we? +We wanted to figure out how far a car goes just by looking at the speedometer, and what makes that hard is that the velocity was always changing. +If you approximate it to be constant on multiple different intervals, you can figure out how far the car goes on each interval just with multiplication, then add them all up. +Adding up those products can be visualized as the sum of the areas of many thin rectangles like this. +Better and better approximations of the original problem correspond to collections of rectangles whose aggregate area is closer and closer to being the area under this curve between the start time and end time, so that area under the curve is the precise distance traveled for the true, nowhere-constant velocity function. +If you think of this area as function, with a variable right end point, you can deduce that the derivative of that area function must equal the height of the graph at each point. That"s the key! So to find a function giving this area, you ask what function has v(t) as its derivative. +There are actually infinitely many antiderivatives of a given function, since you can always just add some constant without affecting the derivative, so you account for that by subtracting off the value of whatever antiderivative function you choose at the bottom bound. +Negative area +By the way, one important thing to bring up before leaving is the idea of negative area. +What if our velocity function was negative at some point? Meaning the car is going backwards. It"s still true that the tiny distance traveled ds on a little time interval is about equal to the velocity times the tiny change in time, it"s just that that the number you"d plug in for velocity would be negative, so that tiny change in distance is negative. +In terms of our thin rectangles, if the rectangle goes below the horizontal axis like this, its area represents a bit of distance traveled backwards, so if what you want is to find the distance between the car"s start point and end point, you"d want to subtract it. +And this is generally true of integrals: Whenever a graph dips below the horizontal axis, that area underneath is counted as negative. +What you"ll commonly hear is that integrals measure the "signed" area between a graph and the horizontal axis. +End +Next video +Next up I"ll bring up more contexts where this idea of an integral and the area under curves comes up, along with some other intuitions for the fundamental theorem of calculus. diff --git a/public/content/lessons/2017/taylor-series-geometric-view/index.mdx b/public/content/lessons/2017/taylor-series-geometric-view/index.mdx index 6b9b7d4a..b3b4a53a 100644 --- a/public/content/lessons/2017/taylor-series-geometric-view/index.mdx +++ b/public/content/lessons/2017/taylor-series-geometric-view/index.mdx @@ -1,6 +1,6 @@ --- title: Taylor series (geometric view) -description: Taylor series are extremely useful in engineering and math, but what are they? This video shows why they're useful, and how to make sense of the formula. +description: A different perspective of Taylor Series that's related to the fundamental theorem of calculus. date: 2017-05-07 chapter: 15 video: 3d6DsjIBzJ4 diff --git a/public/content/lessons/2017/taylor-series/index.mdx b/public/content/lessons/2017/taylor-series/index.mdx index da55a8ac..ad40bf17 100644 --- a/public/content/lessons/2017/taylor-series/index.mdx +++ b/public/content/lessons/2017/taylor-series/index.mdx @@ -3,6 +3,7 @@ title: Taylor series description: Taylor series are extremely useful in engineering and math, but what are they? This video shows why they're useful, and how to make sense of the formula. date: 2017-05-07 chapter: 14 +thumbnail: thumbnail.png video: 3d6DsjIBzJ4 source: _2017/eoc/chapter10.py credits: @@ -46,8 +47,8 @@ First of all, at the input $0$ the value of $\cos(x)$ is $1$, so if our approxim $$ \begin{aligned} -\cos(x) & = c_0 +c_1 x+c_2 x^2 \\ -\cos(0) & = c_0 +c_1 (0)+c_2 (0)^2 \\ +\cos(x) & = c_0 +c_1 x+c_2 x^2 \\ \rule{0pt}{2em} +\cos(0) & = c_0 +c_1 (0)+c_2 (0)^2 \\ \rule{0pt}{2em} c_0 & = 1 \\ \end{aligned} $$ @@ -75,9 +76,9 @@ This is the same as making the derivative of our approximation as close as we ca $$ \begin{aligned} -\frac{d}{d x}(\cos (x)) & =\frac{d}{d x}\left(c_0+c_1 x+c_2 x^2+\right) \\ --\sin (x) & =c_1+2 c_2 x \\ --\sin (0) & =c_1+2 c_2(0) \\ +\frac{d}{d x}(\cos (x)) & =\frac{d}{d x}\left(c_0+c_1 x+c_2 x^2+\right) \\ \rule{0pt}{2em} +-\sin (x) & =c_1+2 c_2 x \\ \rule{0pt}{2em} +-\sin (0) & =c_1+2 c_2(0) \\ \rule{0pt}{2em} c_1 & =0 \end{aligned} $$ @@ -112,15 +113,15 @@ Just as we wanted the derivative of our approximation to match that of cosine, w $$ \begin{aligned} -\frac{d^2}{d x^2}(\cos (x)) & =\frac{d^2}{d x^2}\left(c_0+c_1 x+c_2 x^2+\right) \\ -\frac{d}{d x}(-\sin (x)) & =\frac{d}{d x}\left(c_1+2 c_2 x\right) \\ --\cos (x) & =2 c_2\\ --\cos (0) & =2 c_2\\ +\frac{d^2}{d x^2}(\cos (x)) & =\frac{d^2}{d x^2}\left(c_0+c_1 x+c_2 x^2+\right) \\ \rule{0pt}{2em} +\frac{d}{d x}(-\sin (x)) & =\frac{d}{d x}\left(c_1+2 c_2 x\right) \\ \rule{0pt}{2em} +-\cos (x) & =2 c_2\\ \rule{0pt}{2em} +-\cos (0) & =2 c_2\\ \rule{0pt}{2em} c_2 & =-\frac{1}{2} \end{aligned} $$ -We can compute that at $x=0$, the second derivative of our polynomial is $2c_2$. In fact, that's its second derivative everywhere, it is a constant. To make sure this second derivative matches that of $\cos(x)$, we want it to equal $-1$, which means $c_2 = -½$. +We can compute that at $x=0$, the second derivative of our polynomial is $2c_2$. In fact, that's its second derivative everywhere, it is a constant. To make sure this second derivative matches that of $\cos(x)$, we want it to equal $-1$, which means $c_2 = -\frac{1}{2}$. This locks in a final value for our approximation: $$ @@ -135,7 +136,7 @@ To get a feel for how good this is, let's try it out for $x = 0.1$ $$ \begin{align*} -\cos (0.1) &= \underbrace{0.9950042\ldots}_{\text{true value}} \\ +\cos (0.1) &= \underbrace{0.9950042\ldots}_{\text{true value}} \\ \rule{0pt}{2em} P(0.1) = 1-\frac{1}{2}(0.1)^2 &= \underbrace{0.995}_{\text{approximation}} \\ \end{align*} $$ @@ -163,8 +164,8 @@ If you take the third derivative of a cubic polynomial, anything quadratic or sm $$ \begin{aligned} -\frac{d^3}{d x^3}\left(c_0+c_1 x+c_2 x^2+c_3 x^3\right) & = \frac{d^2}{d x^2}\left(c_1+2 c_2 x+3 c_3 x^2\right) \\ -& = \frac{d}{d x}\left(2 c_2+6 c_3 x\right) \\ +\frac{d^3}{d x^3}\left(c_0+c_1 x+c_2 x^2+c_3 x^3\right) & = \frac{d^2}{d x^2}\left(c_1+2 c_2 x+3 c_3 x^2\right) \\ \rule{0pt}{2em} +& = \frac{d}{d x}\left(2 c_2+6 c_3 x\right) \\ \rule{0pt}{2em} & = 6 c_3 \end{aligned} $$ @@ -183,9 +184,9 @@ When you keep applying the power rule over and over, with those exponents all ho $$ \begin{aligned} -\frac{d^4}{d x^4}\left(c_0+c_1 x+c_2 x^2+c_3 x^3+c_4 x^4 \right) & =\frac{d^3}{d x^3}\left(c_1+2 c_2 x+3 c_3 x^2+4 c_4 x^3\right) \\ -& =\frac{d^2}{d x^2}\left(2 c_2+6 c_3 x+12 c_4 x^2\right) \\ -& =\frac{d}{d x}\left(6 c_3+24 c_4 x\right) \\ +\frac{d^4}{d x^4}\left(c_0+c_1 x+c_2 x^2+c_3 x^3+c_4 x^4 \right) & =\frac{d^3}{d x^3}\left(c_1+2 c_2 x+3 c_3 x^2+4 c_4 x^3\right) \\ \rule{0pt}{2em} +& =\frac{d^2}{d x^2}\left(2 c_2+6 c_3 x+12 c_4 x^2\right) \\ \rule{0pt}{2em} +& =\frac{d}{d x}\left(6 c_3+24 c_4 x\right) \\ \rule{0pt}{2em} & =24 c_4 \end{aligned} $$ diff --git a/public/content/lessons/2017/taylor-series/thumbnail.ai b/public/content/lessons/2017/taylor-series/thumbnail.ai new file mode 100644 index 00000000..364185f0 --- /dev/null +++ b/public/content/lessons/2017/taylor-series/thumbnail.ai @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:54f02295bc1f8bc1f73d16663cc24f16bf1a8d8faadd43a234ab110f7053a5c9 +size 190471 diff --git a/public/content/lessons/2017/taylor-series/thumbnail.png b/public/content/lessons/2017/taylor-series/thumbnail.png new file mode 100644 index 00000000..41b1ffb6 --- /dev/null +++ b/public/content/lessons/2017/taylor-series/thumbnail.png @@ -0,0 +1,3 @@ +version https://git-lfs.github.com/spec/v1 +oid sha256:b5f2590ca5527f7736a2177779d21951dca0cf5db033ea68919618d236c4fd94 +size 87162 diff --git a/public/content/lessons/2017/taylor-series/thumbnail.svg b/public/content/lessons/2017/taylor-series/thumbnail.svg new file mode 100644 index 00000000..231958f5 --- /dev/null +++ b/public/content/lessons/2017/taylor-series/thumbnail.svg @@ -0,0 +1,229 @@ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/public/content/lessons/2018/derivatives-and-transforms/index.mdx b/public/content/lessons/2018/derivatives-and-transforms/index.mdx index 94ba0445..28fe9485 100644 --- a/public/content/lessons/2018/derivatives-and-transforms/index.mdx +++ b/public/content/lessons/2018/derivatives-and-transforms/index.mdx @@ -2,7 +2,7 @@ title: The other way to visualize derivatives description: A visual for derivatives which generalizes more nicely to topics beyond calculus. Thinking of a function as a transformation, the derivative measure how much that function locally stretches or squishes a given region. date: 2018-05-19 -chapter: 15 +chapter: 16 video: CfW845LNObM source: _2018/alt_calc.py credits: