Dave Posted June 28, 2004 Share Posted June 28, 2004 Okay, since a few people seem quite keen on this idea, I'm going to start it all off and begin with an introductory online calculus tutorial - for free Lessons Syllabus I'll attempt to cover the following topics over a 10-lesson period (notice I'm not constricted by the number of lessons, they're pretty flexible): Background to differentiation, first principles and an introduction to limits (2 lessons) Basic rules of differentiation, essential notation and examples (1 lesson) Maxima and Minima (1 lesson) Applied differentiation (2 lesson) An introduction to integration (2 lessons) The chain rule (1 lesson - maybe 2) Rounding it all off (however many lessons ) Some Information Here's some basic information for you all to read: Each lesson will be posted mid-afternoon every 3-4 days, depending on how much time I have. I'll post an introductory post at the beginning of the thread with the core knowledge required, and then some questions for you to attempt. If you have problems with anything to do with a particular day's topic, then please post it in the same thread rather than creating a new one. Please read other people's replies to save duplication of questions Please try and use the LaTeX facilities to typeset your mathematics properly if you have any questions. Please rate each thread so I can see what the response is. If you're really confident of the answer and get it within 2 seconds or you're quite advanced already, please give people a hand and don't just post the full solution to a problem. Finally, the most important thing: spread the word! These things are only so successful if we have a limited number of people. Please not that this course is not intended to be a replacement for real teaching - it is only intended as a supplement. Prerequisites There's only so much I can do: I'm assuming that you've got a solid basis in algebra, and I will start from about the level of maths GCSE. I assume that you will understand the concept of a function (e.g. f(x) = x^{2}) and understand various concepts such as graphing techniques. For the later stages, I assume some knowledge in the area of trigonometry, mainly the sine and cosine functions. For the more advanced calculus, I will be working in radians instead of degrees for the measurement of angles. There is one other thing: GRADIENTS - know that the definition of a gradient of a straight line between two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is [math]\frac{y_2 - y_1}{x_2 - x_1}[/math]. Having said all of this, let's get stuck in Lesson 1 - A background to differentiation Now, a lot of people have asked me the question: "Just what the hell is calculus anyway?" And my response is: "Probably the most important area of mathematics you're likely to come across, and the basis of most modern physics." And this is not an understatement. I won't go into much more detail historywise, but needless to say the original inventor of calculus was Sir Isaac Newton, who used it to solve the classic two body problem back in the 1600's. Since then, a new notation has been commonly accepted (created by Leibniz) and many, many advances have been made. And now onto the mathematics. Calculus encompasses three major topics: the study of limits, differentation of functions and integration. We will be starting to look at limits and differentation over the next couple of weeks, and integration will come towards the end of the course. So what actually is a limit? It's a very hard concept to define in layman's terms (although relatively easy from a strictly analyitical point of view). I think the best way to think of it is in terms of sequences. Imagine you have a sequence of numbers that goes like this: 1, 1/2, 1/3, 1/4, 1/5, ... and so on. If we call the n^{th} number a_{n}, then it's fairly clear to see that a_{1} is 1, a_{2} is 1/2, and so on. The mathematical definition for the n^{th} number is obviously a_{n} = 1/n. Now we look at what happens as we get bigger and bigger values of n. We can notice that each term in the sequence gets progessively smaller as we increase the value of n, and it doesn't take a genius to work out that as we get really big values of n, we get excruciatingly small values for a_{n}. So we can say that the "limit" of a_{n} is 0 as n gets really big (i.e. as n tends to infinity). Don't start crying just yet over how complex this all is; it's an abstract concept to understand, and it'll take some time just to understand the idea, let alone how it all works. A quick remark on this: we won't be using limits that tend to infinity much in calculus at all, I just used it as an example. A very important idea to understand is the fact that we're not actually saying that the sequence will ever hold a value of 0 - what we are saying is that if you were to go on and extend the sequence forever, then you'd be continually getting closer and closer to zero. Quickly, some notation. You won't be using this every day, but it's handy to know. Instead of writing "the limit of f(x) as x tends to infinity is 971283.2", we can write: [math]\lim_{x\to\infty} f(x) = 971283.2[/math] - so remember this, it's a very useful shorthand! So now onto first principles of differentiation: this is where I tell you how to actually go about differentiating something and what we actually mean by the term 'differentiation'. A classic problem in a GCSE maths paper is to sketch a curve out (like the classic y = x^{2}) and then they say to you: "draw a tangent to the curve at the point x = 2, and hence find the gradient at that point". And you grudgingly scrawl out a quick graph between this and that, shove a quick tangent on and get an approximate value for the gradient. We all know it's dead easy to find the gradient between two points on a straight line, but on a curve? Bah, impossible. But this is not so. Let's draw ourselves a graph of y= x^{2}, and have a look at a better way of doing things. Have a look at the graph at the bottom of the page, and observe the points I've drawn on it. We have a point P, and a position (1,1) and then a point Q at position (1+h, (1+h)^{2}). I initially got confused here: basically, we're looking at a point where x = 1, and then a point a little bit further down the x-axis at x = 1+h, where h is some value (we don't really care all that much). Now we're going to consider what happens to the gradient of the line PQ when we decrease the value of h - i.e. we move the point Q closer to the point P. If you look closely, you'll notice that as you draw the two points closer and closer to one another, you're going to get a better and better approximation for the value of the gradient at the point x = 2 on the curve. And now you scream "so what?!?!?" - and then I say "Well look - we have a method for finding the precise value of a gradient at a point on a curve!" And then you probably say "so?!?!" again, and I don't really care by this point What we really want to know is how we find this value. Let's look at the gradient of the line PQ. This is equal to: [math]\frac{(1+h)^2 - 1}{(1+h) - 1} = \frac{(1+h)^2 - 1}{h}[/math] Now just a second. We want to find the limit of this as we decrease h (i.e. as h tends to zero). But we've got a silly h lying around all by itself on the bottom, meaning that if we just stick h=0 into here, we get something divided by zero - we can't do that. So now we have to play around a bit with the fraction, and this is the key operation of this lesson. Make sure you watch very, very carefully and understand each step in the minutest of detail. First of all, notice that (1+h)^{2} = 1 + 2h + h^{2}. So now we have the gradient equal to: [math]\frac{1 + 2h + h^2 - 1}{h} = \frac{2h + h^2}{h} = 2 + h[/math] Hurrah! Now we have something that we can work with. Notice now that as h gets very, very small we have that the gradient of PQ will tend to 2+0 = 2. So we have a method for finding the exact value of the gradient at a certain point. All those hours of drawing tangents to curve wasted whilst your teacher can work out the answer in their heads Next time, I'll extend the method to work for all values (i.e. find the gradient at any point) and introduce some very important notation that you're likely to carry around for a very long time. Questions 1. Find the gradient of the function y= x^{3} at the point x = 5. 2. Find the gradient of the function [math]y = \frac{1}{x^2}[/math] at the point x = 5. I may post further questions; I just can't think of any yet. Hope you enjoyed this and found it useful, Dave 2 Link to comment Share on other sites More sharing options...

Dave Posted June 28, 2004 Author Share Posted June 28, 2004 Sorry, forgot to attach my graph: Link to comment Share on other sites More sharing options...

YT2095 Posted June 28, 2004 Share Posted June 28, 2004 1. y= 125 2. y= 0.04 if I`ve got those right it`ll be a miricle as most all of that went right over my head Link to comment Share on other sites More sharing options...

Dave Posted June 28, 2004 Author Share Posted June 28, 2004 You're slightly out on the second one; first one is quite a way off. Link to comment Share on other sites More sharing options...

Sayonara Posted June 28, 2004 Share Posted June 28, 2004 First one looks like a solution for y= x^{3}where x=5... I'm going ot read that again and then have a crack. Not done academic maths* for a while so this will be good. * Academic maths being the opposite of Excel maths Link to comment Share on other sites More sharing options...

Dave Posted June 28, 2004 Author Share Posted June 28, 2004 Yeah, you need to apply first principles before you can put the value in. Link to comment Share on other sites More sharing options...

YT2095 Posted June 28, 2004 Share Posted June 28, 2004 First one looks like a solution for y= x^{3}where x=5... that`s exactly what it was. 5 cubed = y but I was posting before the graph was in place, so that`s probably why (said the liar LOL) the before the graph part`s true, but after seeing it, it would have been of no use to me anyway, Maths is NOT my strong point, and the fact I was close on the second one was simply that 5 squared is 25 and the reciprocal of it is 0.04 Link to comment Share on other sites More sharing options...

jordan Posted June 28, 2004 Share Posted June 28, 2004 I am a little lost myself. Just two quick questions should clear things up: 1) Will the denominator always be (1+h)-1? 2) How does the value of x play into this? In the example, you solved for h and then plugged 0 in for h. Is that what you do for everything? Link to comment Share on other sites More sharing options...

Sayonara Posted June 28, 2004 Share Posted June 28, 2004 Right. I'll be using d (difference) instead of h so it doesn't look like I just copied Dave. Part One: Point A on y=x^{3} is (1, 1) Point B on y=x^{3} is (1+d, (1+d)^{3}) Gradient between A and B is [math]\frac{(1+d)^3 - 1}{(1+d) - 1}[/math] [math]\frac{(1+d)^3 - 1}{(1+d) - 1}[/math] = [math]\frac{(1+d)^3 - 1}{d}[/math] Apply binomial series thingy: (a + b)^{3} = a^{3} + 3a^{2}b + 3b^{2}a + b^{3} [math]\frac{(1+d)^3 - 1}{d}[/math] = [math]\frac{1 + 3d + 3(d^2) + d^3 - 1}{d}[/math] = [math]\frac{3d + 3(d^2) + d^3}{d}[/math] = [math]3 + 3d + d^2[/math] As d gets teeny weeny gradient AB tends to [math]3 + 0 + 0 = 3[/math] For [math]x = 5[/math], [math]d = (5 - 1) = 4[/math] Gradient AB = [math]\frac{1 + 12 + 3(16) + 64 - 1}{4}[/math] = [math]\frac{124}{4}[/math] = [math]31[/math] That doesn't look right Link to comment Share on other sites More sharing options...

JaKiri Posted June 28, 2004 Share Posted June 28, 2004 Mainly because that doesn't look like the method at all. The method is from setting a point (in the eg. (1,1)) and then finding the gradient of a point 'h' away from it on the x axis, and finding the gradient as h -> 0. You don't set h as a value and solve it, you set the initial value to (5, 125) and solve it, rather than (1, 1). The limit as h -> 0 is the answer, not some crazy subbing with h. SPOILER! (spoiler tags please sayo) ( ((d+5)^3 - 125)/((d+5) - 5) = d^2 + 15d + 75, as d -> 0, approximation of dy/dx -> 75, as required) Link to comment Share on other sites More sharing options...

Sayonara Posted June 28, 2004 Share Posted June 28, 2004 Gah. Everything after "for x=5" is rubbish. Link to comment Share on other sites More sharing options...

jordan Posted June 28, 2004 Share Posted June 28, 2004 For [math]x = 5[/math]' date=' [math']d = (5 - 1) = 4[/math] Gradient AB = [math]\frac{1 + 12 + 3(16) + 64 - 1}{4}[/math] = [math]\frac{124}{4}[/math] = [math]31[/math] I did exactly what you did up to this point and then was a little lost. MrL's post made sense, though. So that would make number 2...I'm getting -4 but that can't be right. As soon as I learn how to make the cool math text stuff I'll post my work so you can all help me find the glaringly obvious mistake. Link to comment Share on other sites More sharing options...

Sayonara Posted June 28, 2004 Share Posted June 28, 2004 Ignore everything after "for x=5"... p.s. MrL - spoiler tags made as requested: [hide]hahaha[/hide] Link to comment Share on other sites More sharing options...

Dave Posted June 28, 2004 Author Share Posted June 28, 2004 I am a little lost myself. Just two quick questions should clear things up: 1) Will the denominator always be (1+h)-1? Using this method' date=' yes. 2) How does the value of x play into this? In the example, you solved for h and then plugged 0 in for h. Is that what you do for everything? Yeah; you need to get it in a form which makes sense for h = 0. Link to comment Share on other sites More sharing options...

bloodhound Posted June 28, 2004 Share Posted June 28, 2004 great post. Link to comment Share on other sites More sharing options...

Dave Posted June 28, 2004 Author Share Posted June 28, 2004 Sorry if i chucked everyone in at the deep end a bit; if you'd like, I'll do the first question as another example. Link to comment Share on other sites More sharing options...

Dave Posted June 28, 2004 Author Share Posted June 28, 2004 great post. Cheers Link to comment Share on other sites More sharing options...

Sayonara Posted June 28, 2004 Share Posted June 28, 2004 Sorry if i chucked everyone in at the deep end a bit; if you'd like, I'll do the first question as another example. That would be good. This is what comes of stopping maths after GCSE Link to comment Share on other sites More sharing options...

Dave Posted June 28, 2004 Author Share Posted June 28, 2004 rofl In retrospect, I shouldn't've posted the second question because it does require quite a lot of tricky algebra to get the limit out. Nevermind. Okay, question number 1. We want to find the gradient of y = x^{3} at x = 5. Let's do exactly the same as what we did before, and pick a point P. When we pick this point, we want it to be the point at which we want the gradient. So in our case, P is going to be the point (5, 5^{3}) = (5, 125). Now we want to pick Q, the point which we're going to move towards P. We basically want the x-coordinate to be slightly bigger than the one for P, so let's assign some value h to be the distance between P and Q. Then the co-ordinate representing Q will be the same as P's, plus some value - we'll call it h. So we have our two points: P at (5, 125) and Q at (5+h, (5+h)^{3}). Now we're going to find the gradient of the line PQ, because we know that as we decrease the distance between the points, we're going to get a better and better approximation for the gradient of x^{3} at x = 5. So our gradient is: [math]\frac{(5+h)^3 - 125}{h} = \frac{h^3 + 15h^2 + 75h}{h} = h^2 + 15h + 75[/math]. Now the bit to where people are getting confused. Basically, we want to work out the limit of this function as h gets very small. The easiest way to do this is just to put h=0 into the equation and we're going to see that the first two terms disappear, leaving us with the correct answer of 75. Don't worry too much if you're quite confused by this; it's hard to get your head around at first. Once we get into the differentiation a bit more, it'll become easier to understand. Link to comment Share on other sites More sharing options...

Sayonara Posted June 28, 2004 Share Posted June 28, 2004 Don't worry too much if you're quite confused by this; it's hard to get your head around at first. Once we get into the differentiation a bit more, it'll become easier to understand. I get what you mean now. This bit was confusing: Now we have something that we can work with. Notice now that as h gets very, very small we have that the gradient of PQ will tend to 2+0 = 2. So we have a method for finding the exact value of the gradient at a certain point. I didn't see how you got from one thing to the other, so tried to reverse engineer a solution. lollz. Link to comment Share on other sites More sharing options...

Dave Posted June 28, 2004 Author Share Posted June 28, 2004 Yes, that's really the worst bit; whilst I was writing it I thought it might confuse people a bit. That's the most abstract part of the entire thing. When you do differentiation from day to day, you'll find that you more or less never use a limit. It's only in very rare circumstances (or if you're doing a maths degree) that you will. But I do think it's important to realise and understand what's going on when you differentiate a function. Next lesson will probably be published either on Wednesday or Thursday; doesn't take me too long to write, it's just thinking of examples. I'll be extending the method and starting to talk about some real applications to different functions. Link to comment Share on other sites More sharing options...

Dave Posted June 28, 2004 Author Share Posted June 28, 2004 And please: * If you know anyone that's interested in this stuff, tell them about the thread. * Rate the thread according to how good you think it is Link to comment Share on other sites More sharing options...

jordan Posted June 28, 2004 Share Posted June 28, 2004 [math](1+\frac{1}{h^2}-125)[/math] for the numerator. This simplifies to : [math](\frac{h^2+1+125h^2}{h^2})[/math] For the denominator I used: [math](1+h)-5[/math] which obviously simplifies to [math]h-4[/math] The result is [math]\frac{(\frac{h^2+1+125h^2}{h^2})}{h-4}[/math] Taking the [math]h^2[/math] in the middle down to the denominator causes problems though. I can't get it out and subsequently can't susbtitute 0 to get anything reasonable. Wow, this math text stuff works great. Any help would be apreciated. Thanks. Link to comment Share on other sites More sharing options...

JaKiri Posted June 28, 2004 Share Posted June 28, 2004 Using this method, yes. You've explained yourself really badly here. The denominator will always be (where n is the initial x value) (x+n) - n Not (x+1) - 1. This may partially explain people always having 1 as the 'base' value, despite the questions being based around 5. Link to comment Share on other sites More sharing options...

Dave Posted June 28, 2004 Author Share Posted June 28, 2004 Fair enough. However, it still stands that there will only ever be a h on the bottom of the fraction when you consider this particular limit. There is only so much I can explain over an internet forum; as I've said, it's only a supplement to proper teaching. Trying to convey an abstract idea like this over the net is not an easy thing to accomplish, and I'm sorry for the confusion. If it would help, I will post the application of this method for differentiating x^{2} for any value of x, rather than just a specific value. Link to comment Share on other sites More sharing options...

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