Category Archives: Math 100

My Teaching

I am currently not teaching anything. Though, I am supervising a few students at the University of Warwick, I have no teaching responsibilities here.

In fall 2016, I taught Math 100 (second semester calculus, starting with integration by parts and going through sequences and series) at Brown University. Here are my concluding remarks.

In spring 2016, I designed and taught Math 42 (elementary number theory) at Brown University. My students were exceptional — check out a showcase of some of their final projects. Here are my concluding remarks.

In fall 2014, I taught Math 170 (advanced placement second semester calculus) at Brown University.

I taught number theory in the Summer@Brown program for high school students in the summers of 2013-2015.

I taught a privately requested course in precalculus in the summer of 2013.

I have served as a TA (many, many, many times) for

• Math 90 (first semester calculus) at Brown University
• Math 100 (second semester calculus) at Brown University
• Math 1501 (first semester calculus) at Georgia Tech
• Math 1502 (second semester calculus, starting with sequences and series but also with 7 weeks of linear algebra) at Georgia Tech
• Math 2401 (multivariable calculus) at Georgia Tech (there’s essentially no content on this site about this – this was just before I began to maintain a website)

I sometimes tutor at Brown (but not limited to Brown students) and around Boston, on a wide variety of topics (not just the ordinary, boring ones). I charge $80/hour, but I am not currently looking for tutees. Below, you can find my most recent posts tagged under “Teaching”. Posted in Brown University, Math 100, Teaching | Leave a comment Math 100 Fall 2016: Concluding Remarks It is that time of year. Classes are over. Campus is emptying. Soon it will be mostly emptiness, snow, and grad students (who of course never leave). I like to take some time to reflect on the course. How did it go? What went well and what didn’t work out? And now that all the numbers are in, we can examine course trends and data. Since numbers are direct and graphs are pretty, let’s look at the numbers first. Math 100 grades at a glance Let’s get an understanding of the distribution of grades in the course, all at once. These are classic box plots. The center line of each box denotes the median. The left and right ends of the box indicate the 1st and 3rd quartiles. As a quick reminder, the 1st quartile is the point where 25% of students received that grade or lower. The 3rd quartile is the point where 75% of students received that grade or lower. So within each box lies 50% of the course. Each box has two arms (or “whiskers”) extending out, indicating the other grades of students. Points that are plotted separately are statistical outliers, which means that they are$1.5 \cdot (Q_3 – Q_1)$higher than$Q_3$or lower than$Q_1$(where$Q_1$denotes the first quartile and$Q_3$indicates the third quartile). A bit more information about the distribution itself can be seen in the following graph. Within each blob, you’ll notice an embedded box-and-whisker graph. The white dots indicate the medians, and the thicker black parts indicate the central 50% of the grade. The width of the colored blobs roughly indicate how many students scored within that region. [As an aside, each blob actually has the same area, so the area is a meaningful data point]. Posted in Brown University, Math 100, Mathematics, Teaching | Tagged , , , , | 1 Comment Computing pi with tools from Calculus Computing$\pi$This note was originally written in the context of my fall Math 100 class at Brown University. It is also available as a pdf note. While investigating Taylor series, we proved that \label{eq:base} \frac{\pi}{4} = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} + \frac{1}{9} + \cdots Let’s remind ourselves how. Begin with the geometric series \frac{1}{1 + x^2} = 1 – x^2 + x^4 – x^6 + x^8 + \cdots = \sum_{n = 0}^\infty (-1)^n x^{2n}. \notag (We showed that this has interval of convergence$\lvert x \rvert < 1$). Integrating this geometric series yields \int_0^x \frac{1}{1 + t^2} dt = x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7} + \cdots = \sum_{n = 0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}. \notag Note that this has interval of convergence$-1 < x \leq 1$. We also recognize this integral as \int_0^x \frac{1}{1 + t^2} dt = \text{arctan}(x), \notag one of the common integrals arising from trigonometric substitution. Putting these together, we find that \text{arctan}(x) = x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7} + \cdots = \sum_{n = 0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}. \notag As$x = 1$is within the interval of convergence, we can substitute$x = 1$into the series to find the representation \text{arctan}(1) = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} + \cdots = \sum_{n = 0}^\infty (-1)^n \frac{1}{2n+1}. \notag Since$\text{arctan}(1) = \frac{\pi}{4}$, this gives the representation for$\pi/4$given in \eqref{eq:base}. However, since$x=1$was at the very edge of the interval of convergence, this series converges very, very slowly. For instance, using the first$50$terms gives the approximation \pi \approx 3.121594652591011. \notag The expansion of$\pi$is actually \pi = 3.141592653589793238462\ldots \notag So the first$50$terms of \eqref{eq:base} gives two digits of accuracy. That’s not very good. I think it is very natural to ask: can we do better? This series converges slowly — can we find one that converges more quickly? | | 1 Comment Series Convergence Tests with Prototypical Examples This is a note written for my Fall 2016 Math 100 class at Brown University. We are currently learning about various tests for determining whether series converge or diverge. In this note, we collect these tests together in a single document. We give a brief description of each test, some indicators of when each test would be good to use, and give a prototypical example for each. Note that we do justify any of these tests here — we’ve discussed that extensively in class. [But if something is unclear, send me an email or head to my office hours]. This is here to remind us of the variety of the various tests of convergence. A copy of just the statements of the tests, put together, can be found here. A pdf copy of this whole post can be found here. In order, we discuss the following tests: 1. The$n$th term test, also called the basic divergence test 2. Recognizing an alternating series 3. Recognizing a geometric series 4. Recognizing a telescoping series 5. The Integral Test 6. P-series 7. Direct (or basic) comparison 8. Limit comparison 9. The ratio test 10. The root test The$n$th term test Statement Suppose we are looking at$\sum_{n = 1}^\infty a_n$and \lim_{n \to \infty} a_n \neq 0. \notag Then$\sum_{n = 1}^\infty a_n$does not converge. When to use it When applicable, the$n$th term test for divergence is usually the easiest and quickest way to confirm that a series diverges. When first considering a series, it’s a good idea to think about whether the terms go to zero or not. But remember that if the limit of the individual terms is zero, then it is necessary to think harder about whether the series converges or diverges. Example Each of the series \sum_{n = 1}^\infty \frac{n+1}{2n + 4}, \quad \sum_{n = 1}^\infty \cos n, \quad \sum_{n = 1}^\infty \sqrt{n} \notag diverges since their limits are not$0$. Recognizing alternating series Statement Suppose$\sum_{n = 1}^\infty (-1)^n a_n$is a series where 1.$a_n \geq 0$, 2.$a_n$is decreasing, and 3.$\lim_{n \to \infty} a_n = 0$. Then$\sum_{n = 1}^\infty (-1)^n a_n$converges. Stated differently, if the terms are alternating sign, decreasing in absolute size, and converging to zero, then the series converges. When to use it The key is in the name — if the series is alternating, then this is the goto idea of analysis. Note that if the terms of a series are alternating and decreasing, but the terms do not go to zero, then the series diverges by the$n$th term test. Example Suppose we are looking at the series \sum_{n = 1}^\infty \frac{(-1)^n}{\log(n+1)} = \frac{-1}{\log 2} + \frac{1}{\log 3} + \frac{-1}{\log 4} + \cdots \notag The terms are alternating. The sizes of the terms are$\frac{1}{\log (n+1)}, and these are decreasing. Finally, \lim_{n \to \infty} \frac{1}{\log(n+1)} = 0. \notag Thus the alternating series test applies and shows that this series converges. Posted in Brown University, Math 100, Mathematics, Teaching | | Leave a comment Math 100: Completing the partial fractions example from class An Unfinished Example At the end of class today, someone asked if we could do another example of a partial fractions integral involving an irreducible quadratic. We decided to look at the integral $$\int \frac{1}{(x^2 + 4)(x+1)}dx.$$ Notice that ${x^2 + 4}$ is an irreducible quadratic polynomial. So when setting up the partial fraction decomposition, we treat the ${x^2 + 4}$ term as a whole. So we seek to find a decomposition of the form $$\frac{1}{(x^2 + 4)(x+1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 4}.$$ Now that we have the decomposition set up, we need to solve for ${A,B,}$ and ${C}$ using whatever methods we feel most comfortable with. Multiplying through by ${(x^2 + 4)(x+1)}$ leads to $$1 = A(x^2 + 4) + (Bx + C)(x+1) = (A + B)x^2 + (B + C)x + (4A + C).$$ Matching up coefficients leads to the system of equations \begin{align} 0 &= A + B \\ 0 &= B + C \\ 1 &= 4A + C. \end{align} So we learn that ${A = -B = C}$, and ${A = 1/5}$. So ${B = -1/5}$ and ${C = 1/5}$. Together, this means that $$\frac{1}{(x^2 + 4)(x+1)} = \frac{1}{5}\frac{1}{x+1} + \frac{1}{5} \frac{-x + 1}{x^2 + 4}.$$ Recall that if you wanted to, you could check this decomposition by finding a common denominator and checking through. Now that we have performed the decomposition, we can return to the integral. We now have that $$\int \frac{1}{(x^2 + 4)(x+1)}dx = \underbrace{\int \frac{1}{5}\frac{1}{x+1}dx}_ {\text{first integral}} + \underbrace{\int \frac{1}{5} \frac{-x + 1}{x^2 + 4} dx.}_ {\text{second integral}}$$ We can handle both of the integrals on the right hand side. The first integral is $$\frac{1}{5} \int \frac{1}{x+1} dx = \frac{1}{5} \ln (x+1) + C.$$ The second integral is a bit more complicated. It’s good to see if there is a simple ${u}$-substition, since there is an ${x}$ in the numerator and an ${x^2}$ in the denominator. But unfortunately, this integral needs to be further broken into two pieces that we know how to handle separately. $$\frac{1}{5} \int \frac{-x + 1}{x^2 + 4} dx = \underbrace{\frac{-1}{5} \int \frac{x}{x^2 + 4}dx}_ {\text{first piece}} + \underbrace{\frac{1}{5} \int \frac{1}{x^2 + 4}dx.}_ {\text{second piece}}$$ The first piece is now a ${u}$-substitution problem with ${u = x^2 + 4}$. Then ${du = 2x dx}$, and so $$\frac{-1}{5} \int \frac{x}{x^2 + 4}dx = \frac{-1}{10} \int \frac{du}{u} = \frac{-1}{10} \ln u + C = \frac{-1}{10} \ln (x^2 + 4) + C.$$ The second piece is one of the classic trig substitions. So we draw a triangle. In this triangle, thinking of the bottom-left angle as ${\theta}$ (sorry, I forgot to label it), then we have that ${2\tan \theta = x}$ so that ${2 \sec^2 \theta d \theta = dx}$. We can express the so-called hard part of the triangle by ${2\sec \theta = \sqrt{x^2 + 4}}$. Going back to our integral, we can think of ${x^2 + 4}$ as ${(\sqrt{x^2 + 4})^2}$ so that ${x^2 + 4 = (2 \sec \theta)^2 = 4 \sec^2 \theta}$. We can now write our integral as $$\frac{1}{5} \int \frac{1}{x^2 + 4}dx = \frac{1}{5} \int \frac{1}{4 \sec^2 \theta} 2 \sec^2 \theta d \theta = \frac{1}{5} \int \frac{1}{2} d\theta = \frac{1}{10} \theta.$$ As ${2 \tan \theta = x}$, we have that ${\theta = \text{arctan}(x/2)}$. Inserting this into our expression, we have $$\frac{1}{10} \int \frac{1}{x^2 + 4} dx = \frac{1}{10} \text{arctan}(x/2) + C.$$ Combining the first integral and the first and second parts of the second integral together (and combining all the constants ${C}$ into a single constant, which we also denote by ${C}$), we reach the final expression $$\int \frac{1}{(x^2 + 4)(x + 1)} dx = \frac{1}{5} \ln (x+1) – \frac{1}{10} \ln(x^2 + 4) + \frac{1}{10} \text{arctan}(x/2) + C.$$ And this is the answer. Other Notes If you have any questions or concerns, please let me know. As a reminder, I have office hours on Tuesday from 9:30–11:30 (or perhaps noon) in my office, and I highly recommend attending the Math Resource Center in the Kassar House from 8pm-10pm, offered Monday-Thursday. [Especially on Tuesday and Thursdays, when there tend to be fewer people there]. On my course page, I have linked to two additional resources. One is to Paul’s Online Math notes for partial fraction decomposition (which I think is quite a good resource). The other is to the Khan Academy for some additional worked through examples on polynomial long division, in case you wanted to see more worked examples. This note can also be found on my website, or in pdf form. Good luck, and I’ll see you in class. Posted in Math 100, Mathematics, Teaching | | Leave a comment Math 100 Fall 2013: Concluding Remarks This is a post written towards my students in Calc II, Math 100 at Brown University, fall 2013. There will be many asides, written in italics. They are to serve as clarifications of method or true asides, to be digested or passed over. The semester is over. All the grades are in and known, fall 2013 draws to a close. As you know, I’m interested in the statistics behind the course. I’d mentioned my previous analysis about the extremely high correlation between first midterm and final grade (much higher than I would have thought!). Let’s reveal the statistics and distribution of this course, below the fold. Posted in Brown University, Math 100, Mathematics, Teaching | | 2 Comments Math 100: Before second midterm You have a midterm next week, and it’s not going to be a cakewalk. As requested, I’m uploading the last five weeks’ worth of worksheets, with (my) solutions. A comment on the solutions: not everything is presented in full detail, but most things are presented with most detail (except for the occasional one that is far far beyond what we actually expect you to be able to do). If you have any questions about anything, let me know. Even better, ask it here – maybe others have the same questions too. Without further ado – And since we were unable to go over the quiz in my afternoon recitation today, I’m attaching a worked solution to the quiz as well. Again, let me know if you have any questions. I will still have my office hours on Tuesday from 2:30-4:30pm in my office (I’m aware that this happens to be immediately before the exam – status not by design). And I’ll be more or less responsive by email. Study study study! Posted in Brown University, Math 100, Mathematics | | Leave a comment Math 100: Week 4 This is a post for my math 100 calculus class of fall 2013. In this post, I give the 4th week’s recitation worksheet (no solutions yet – I’m still writing them up). More pertinently, we will also go over the most recent quiz and common mistakes. Trig substitution, it turns out, is not so easy. Before we hop into the details, I’d like to encourage you all to avail of each other, your professor, your ta, and the MRC in preparation for the first midterm (next week!). 1. The quiz There were two versions of the quiz this week, but they were very similar. Both asked about a particular trig substitution $\displaystyle \int_3^6 \sqrt{36 – x^2} \mathrm{d} x$ And the other was $\displaystyle \int_{2\sqrt 2}^4 \sqrt{16 – x^2} \mathrm{d}x.$ They are very similar, so I’m only going to go over one of them. I’ll go over the first one. We know we are to use trig substitution. I see two ways to proceed: either draw a reference triangle (which I recommend), or think through the Pythagorean trig identities until you find the one that works here (which I don’t recommend). We see a ${\sqrt{36 – x^2}}$, and this is hard to deal with. Let’s draw a right triangle that has ${\sqrt{36 – x^2}}$ as a side. I’ve drawn one below. (Not fancy, but I need a better light). In this picture, note that ${\sin \theta = \frac{x}{6}}$, or that ${x = 6 \sin \theta}$, and that ${\sqrt{36 – x^2} = 6 \cos \theta}$. If we substitute ${x = 6 \sin \theta}$ in our integral, this means that we can replace our ${\sqrt{36 – x^2}}$ with ${6 \cos \theta}$. But this is a substitution, so we need to think about ${\mathrm{d} x}$ too. Here, ${x = 6 \sin \theta}$ means that ${\mathrm{d}x = 6 \cos \theta}$. Some people used the wrong trig substitution, meaning they used ${x = \tan \theta}$ or ${x = \sec \theta}$, and got stuck. It’s okay to get stuck, but if you notice that something isn’t working, it’s better to try something else than to stare at the paper for 10 minutes. Other people use ${x = 6 \cos \theta}$, which is perfectly doable and parallel to what I write below. Another common error was people forgetting about the ${\mathrm{d}x}$ term entirely. But it’s important!. Substituting these into our integral gives $\displaystyle \int_{?}^{??} 36 \cos^2 (\theta) \mathrm{d}\theta,$ where I have included question marks for the limits because, as after most substitutions, they are different. You have a choice: you might go on and put everything back in terms of ${x}$ before you give your numerical answer; or you might find the new limits now. It’s not correct to continue writing down the old limits. The variable has changed, and we really don’t want ${\theta}$ to go from ${3}$ to ${6}$. If you were to find the new limits, then you need to consider: if ${x=3}$ and ${\frac{x}{6} = \sin \theta}$, then we want a ${\theta}$ such that ${\sin \theta = \frac{3}{6}= \frac{1}{2}}$, so we might use ${\theta = \pi/6}$. Similarly, when ${x = 6}$, we want ${\theta}$ such that ${\sin \theta = 1}$, like ${\theta = \pi/2}$. Note that these were two arcsine calculations, which we would have to do even if we waited until after we put everything back in terms of ${x}$ to evaluate. Some people left their answers in terms of these arcsines. As far as mistakes go, this isn’t a very serious one. But this is the sort of simplification that is expected of you on exams, quizzes, and homeworks. In particular, if something can be written in a much simpler way through the unit circle, then you should do it if you have the time. So we could rewrite our integral as $\displaystyle \int_{\pi/6}^{\pi/2} 36 \cos^2 (\theta) \mathrm{d}\theta.$ How do we integrate ${\cos^2 \theta}$? We need to make use of the identity ${\cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}}$. You should know this identity for this midterm. Now we have $\displaystyle 36 \int_{\pi/6}^{\pi/2}\left(\frac{1}{2} + \frac{\cos 2 \theta}{2}\right) \mathrm{d}\theta = 18 \int_{\pi/6}^{\pi/2}\mathrm{d}\theta + 18 \int_{\pi/6}^{\pi/2}\cos 2\theta \mathrm{d}\theta.$ The first integral is extremely simple and yields ${6\pi}$ The second integral has antiderivative ${\dfrac{\sin 2 \theta}{2}}$ (Don’t forget the ${2}$ on bottom!), and we have to evaluate ${\big[9 \sin 2 \theta \big]_{\pi/6}^{\pi/2}}$, which gives ${-\dfrac{9 \sqrt 3}{2}}$. You should know the unit circle sufficiently well to evaluate this for your midterm. And so the final answer is ${6 \pi – \dfrac{9 \sqrt 2}{2} \approx 11.0553}$. (You don’t need to be able to do that approximation). Let’s go back a moment and suppose you didn’t re”{e}valuate the limits once you substituted in ${\theta}$. Then, following the same steps as above, you’d be left with $\displaystyle 18 \int_{?}^{??}\mathrm{d}\theta + 18 \int_{?}^{??}\cos 2\theta \mathrm{d}\theta = \left[ 18 \theta \right]_?^{??} + \left[ 9 \sin 2 \theta \right]_?^{??}.$ Since ${\frac{x}{6} = \sin \theta}$, we know that ${\theta = \arcsin (x/6)}$. This is how we evaluate the left integral, and we are left with ${[18 \arcsin(x/6)]_3^6}$. This means we need to know the arcsine of ${1}$ and ${\frac 12}$. These are exactly the same two arcsine computations that I referenced above! Following them again, we get ${6\pi}$ as the answer. We could do the same for the second part, since ${\sin ( 2 \arcsin (x/6))}$ when ${x = 3}$ is ${\sin (2 \arcsin \frac{1}{2} ) = \sin (2 \cdot \frac{\pi}{6} ) = \frac{\sqrt 3}{2}}$; and when ${x = 6}$ we get ${\sin (2 \arcsin 1) = \sin (2 \cdot \frac{\pi}{2}) = \sin (\pi) = 0}$. Putting these together, we see that the answer is again ${6\pi – \frac{9\sqrt 3}{2}}$. Or, throwing yet another option out there, we could do something else (a little bit wittier, maybe?). We have this ${\sin 2\theta}$ term to deal with. You might recall that ${\sin 2 \theta = 2 \sin \theta \cos \theta}$, the so-called double-angle identity. Then ${9 \sin 2\theta = 18 \sin \theta \cos \theta}$. Going back to our reference triangle, we know that ${\cos \theta = \dfrac{\sqrt{36 – x^2}}{6}}$ and that ${\sin \theta = \dfrac{x}{6}}$. Putting these together, $\displaystyle 9 \sin 2 \theta = \dfrac{ x\sqrt{36 – x^2} }{2}.$ When ${x=6}$, this is ${0}$. When ${x = 3}$, we have ${\dfrac{ 3\sqrt {27}}{2} = \dfrac{9\sqrt 3}{2}}$. And fortunately, we get the same answer again at the end of the day. (phew). 2. The worksheet Finally, here is the worksheet for the day. I’m working on their solutions, and I’ll have that up by late this evening (sorry for the delay). Ending tidbits – when I was last a TA, I tried to see what were the good predictors of final grade. Some things weren’t very surprising – there is a large correlation between exam scores and final grade. Some things were a bit surprising – low homework scores correlated well with low final grade, but high homework scores didn’t really have a strong correlation with final grade at all; attendance also correlated weakly. But one thing that really stuck with me was the first midterm grade vs final grade in class: it was really strong. For a bit more on that, I refer you to my final post from my Math 90 posts. Posted in Brown University, Math 100, Mathematics | | Leave a comment Math 100: Week 3 and pre-midterm This is a post for my Math 100 class of fall 2013. In this post, I give the first three weeks’ worksheets from recitation and the set of solutions to week three’s worksheet, as well as a few administrative details. Firstly, here is the recitation work from the first three weeks: 1. (there was no recitation the first week) 2. A worksheet focusing on review. 3. A worksheet focusing on integration by parts and u-substitution, with solutions. In addition, I’d like to remind you that I have office hours from 2-4pm (right now) in Kassar 018. I’ve had multiple people set up appointments with me outside of these hours, which I’m tempted to interpret as suggesting that I change when my office hours are. If you have a preference, let me know, and I’ll try to incorporate it. Finally, there will be an exam next Tuesday. I’ve been getting a lot of emails about what material will be on the exam. The answer is that everything you have learned up to now and by the end of this week is fair game for exam material. This also means there could be exam questions on material that we have not discussed in recitation. So be prepared. However, I will be setting aside a much larger portion of recitation this Thursday for questions than normal. So come prepared with your questions. Best of luck, and I’ll see you in class on Thursday. Posted in Brown University, Math 100, Mathematics | | Leave a comment An intuitive introduction to calculus This is a post written for my fall 2013 Math 100 class but largely intended for anyone with knowledge of what a function is and a desire to know what calculus is all about. Calculus is made out to be the pinnacle of the high school math curriculum, and correspondingly is thought to be very hard. But the difficulty is bloated, blown out of proportion. In fact, the ideas behind calculus are approachable and even intuitive if thought about in the right way. Many people managed to stumble across the page before I’d finished all the graphics. I’m sorry, but they’re all done now! I was having trouble interpreting how WordPress was going to handle my gif files – it turns out that they automagically resize them if you don’t make them of the correct size, which makes them not display. It took me a bit to realize this. I’d like to mention that this actually started as a 90 minute talk I had with my wife over coffee, so perhaps an alternate title would be “Learning calculus in 2 hours over a cup of coffee.” So read on if you would like to understand what calculus is, or if you’re looking for a refresher of the concepts from a first semester in calculus (like for Math 100 students at Brown), or if you’re looking for a bird’s eye view of AP Calc AB subject material. 1. An intuitive and semicomplete introduction to calculus We will think of a function {f(\cdot)}$as something that takes an input$ {x}$and gives out another number, which we’ll denote by$ {f(x)}$. We know functions like$ {f(x) = x^2 + 1}$, which means that if I give in a number$ {x}$then the function returns the number$ {f(x) = x^2 + 1}$. So I put in$ {1}$, I get$ {1^2 + 1 = 2}$, i.e.$ {f(1) = 2}$. Primary and secondary school overly conditions students to think of functions in terms of a formula or equation. The important thing to remember is that a function is really just something that gives an output when given an input, and if the same input is given later then the function spits the same output out. As an aside, I should mention that the most common problem I’ve seen in my teaching and tutoring is a fundamental misunderstanding of functions and their graphs For a function that takes in and spits out numbers, we can associate a graph. A graph is a two-dimensional representation of our function, where by convention the input is put on the horizontal axis and the output is put on the vertical axis. Each axis is numbered, and in this way we can identify any point in the graph by its coordinates, i.e. its horizontal and vertical position. A graph of a function$ {f(x)}$includes a point$ {(x,y)}$if$ {y = f(x)}$. Thus each point on the graph is really of the form$ {(x, f(x))}$. A large portion of algebra I and II is devoted to being able to draw graphs for a variety of functions. And if you think about it, graphs contain a huge amount of information. Graphing$ {f(x)= x^2 + 1}$involves drawing an upwards-facing parabola, which really represents an infinite number of points. That’s pretty intense, but it’s not what I want to focus on here. 1.1. Generalizing slope – introducing the derivative You might recall the idea of the ‘slope’ of a line. A line has a constant ratio of how much the$ {y}$value changes for a specific change in$ {x}$, which we call the slope (people always seem to remember rise over run). In particular, if a line passes through the points$ {(x_1, y_1)}$and$ {(x_2, y_2)}$, then its slope will be the vertical change$ {y_2 – y_1}$divided by the horizontal change$ {x_2 – x_1}$, or$ {\dfrac{y_2 – y_1}{x_2 – x_1}}$. So if the line is given by an equation$ {f(x) = \text{something}}$, then the slope from two inputs$ {x_1}$and$ {x_2}$is$ {\dfrac{f(x_2) – f(x_1)}{x_2 – x_1}}$. As an aside, for those that remember things like the ‘standard equation’$ {y = mx + b}$or ‘point-slope’$ {(y – y_0) = m(x – x_0)}$but who have never thought or been taught where these come from: the claim that lines are the curves of constant slope is saying that for any choice of$ {(x_1, y_1)}$on the line, we expect$ {\dfrac{y_2 – y_1}{x_2 – x_1} = m}$a constant, which I denote by$ {m}$for no particularly good reason other than the fact that some textbook author long ago did such a thing. Since we’re allowing ourselves to choose any$ {(x_1, y_1)}$, we might drop the subscripts – since they usually mean a constant – and rearrange our equation to give$ {y_2 – y = m(x_2 – x)}$, which is what has been so unkindly drilled into students’ heads as the ‘point-slope form.’ This is why lines have a point-slope form, and a reason that it comes up so much is that it comes so naturally from the defining characteristic of a line, i.e. constant slope. But one cannot speak of the ‘slope’ of a parabola. Intuitively, we look at our parabola$ {x^2 + 1}$and see that the ‘slope,’ or an estimate of how much the function$ {f(x)}$changes with a change in$ {x}$, seems to be changing depending on what$ {x}$values we choose. (This should make sense – if it didn’t change, and had constant slope, then it would be a line). The first major goal of calculus is to come up with an idea of a ‘slope’ for non-linear functions. I should add that we already know a sort of ‘instantaneous rate of change’ of a nonlinear function. When we’re in a car and we’re driving somewhere, we’re usually speeding up or slowing down, and our pace isn’t usually linear. Yet our speedometer still manages to say how fast we’re going, which is an immediate rate of change. So if we had a function$ {p(t)}$that gave us our position at a time$ {t}$, then the slope would give us our velocity (change in position per change in time) at a moment. So without knowing it, we’re familiar with a generalized slope already. Now in our parabola, we don’t expect a constant slope, so we want to associate a ‘slope’ to each input$ {x}$. In other words, we want to be able to understand how rapidly the function$ {f(x)}$is changing at each$ {x}$, analogous to how the slope$ {m}$of a line$ {g(x) = mx + b}$tells us that if we change our input by an amount$ {h}$then our output value will change by$ {mh}$. How does calculus do that? The idea is to get closer and closer approximations. Suppose we want to find the ‘slope’ of our parabola at the point$ {x = 1}$. Let’s get an approximate answer. The slope of the line coming from inputs$ {x = 1}$and$ {x = 2}$is a (poor) approximation. In particular, since we’re working with$ {f(x) = x^2 + 1}$, we have that$ {f(2) = 5}$and$ {f(1) = 2}$, so that the ‘approximate slope’ from$ {x = 1}$and$ {x = 2}$is$ {\frac{5 – 2}{2 – 1} = 3}$. But looking at the graph, we see that it feels like this slope is too large. So let’s get closer. Suppose we use inputs$ {x = 1}$and$ {x = 1.5}$. We get that the approximate slope is$ {\frac{3.25 – 2}{1.5 – 1} = 2.5}$. If we were to graph it, this would also feel too large. So we can keep choosing smaller and smaller changes, like using$ {x = 1}$and$ {x = 1.1}$, or$ {x = 1}$and$ {x = 1.01}$, and so on. This next graphic contains these approximations, with chosen points getting closer and closer to$ {1}$. Let’s look a little closer at the values we’re getting for our slopes when we use$ {1}$and$ {2, 1.5, 1.1, 1.01, 1.001}$as our inputs. We get$ \displaystyle \begin{array}{c|c} \text{second input} & \text{approx. slope} \\ \hline 2 & 3 \\ 1.5 & 2.5 \\ 1.1 & 2.1 \\ 1.01 & 2.01 \\ 1.001 & 2.001 \end{array} $It looks like the approximate slopes are approaching$ {2}$. What if we plot the graph with a line of slope$ {2}$going through the point$ {(1,2)}$? It looks great! Let’s zoom in a whole lot. That looks really close! In fact, what I’ve been allowing as the natural feeling slope, or local rate of change, is really the line tangent to the graph of our function at the point$ {(1, f(1))}$. In a calculus class, you’ll spend a bit of time making sense of what it means for the approximate slopes to ‘approach’$ {2}$. This is called a ‘limit,’ and the details are not important to us right now. The important thing is that this let us get an idea of a ‘slope’ at a point on a parabola. It’s not really a slope, because a parabola isn’t a line. So we’ve given it a different name – we call this ‘the derivative.’ So the derivative of$ {f(x) = x^2 + 1}$at$ {x = 1}$is$ {2}$, i.e. right around$ {x = 1}$we expect a rate of change of$ {2}$, so that we expect$ {f(1 + h) – f(1) \approx 2h}$. If you think about it, we’re saying that we can approximate$ {f(x) = x^2 + 1}$near the point$ {(1, 2)}$by the line shown in the graph above: this line passes through$ {(1,2)}$and it’s slope is$ {2}$, what we’re calling the slope of$ {f(x) = x^2 + 1}$at$ {x = 1}\$.

Let’s generalize. We were able to speak of the derivative at one point, but how about other points? The rest of this post is below the ‘more’ tag below.

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