Differentiation 1. The METRIC Project, Imperial College. Imperial College of Science Technology and Medicine, 1996.

Transcription

1 Differentiation 1 The METRIC Project, Imperial College. Imperial College of Science Technology and Medicine,

2 Launch Mathematica. Type <<Mathetic`diffpack` Instructions for Getting Started hold down the shift key, and press the return key. Wait for Mathematica s response. (Note: be sure to use the ` symbol rather than the '. You may need to hunt for it on your keyboard: on most, it s in the top left corner.) This essential first step sets up Mathematica for this module. If you omit this bit, the special commands (see below) will not work. Mathematica Commands The following Mathematica commands may be useful to you in this module. Commands that come with Mathematica: Plot, Limit, D, Derivative, Show, InverseFunction Special commands for this module: PlotChord, GiveQuestion, LastAnswer For information on, say, the Limit command, type?limit hold down shift, and press return. 2

3 Experiment 1: Gradients of straight lines Preparatory reading The gradient of a straight line is a measure of its steepness. It s equal to the number of units the line rises for each unit step to the right. The following graph, then, has gradient 2: it goes up 2 units for every 1 unit along. If the graph falls instead of rises, then the gradient is negative: here s a line with gradient 3. To calculate the gradient of a straight line, fix two points on it, (x 0, y 0 ) and (x 1, y 1 ). The line s gradient is then 3

4 y 1 y 0 x 1 x 0, change in y or, more simply, change in x. You ll have spotted, probably, that there s a correspondence between a function s equation and its graph. The function with equation y = 2x 4 has a graph with gradient 2, and the function with equation y = 6 3x has a graph with gradient 3. In general, the function with gradient y = mx + c has a graph with gradient m. 1) Type in the following Mathematica input, hold down the shift key, and press the return key (called enter on some machines). Plot[{x-5, 2-x, 2x+1}, {x, -6, 6}, AspectRatio->Automatic] The AspectRatio->Automatic part sets an option. In this case, all you re doing is ensuring that the scales are the same on both axes. If you leave out this bit, the plot will still work but the scales might be different. 2) Experiment with plotting different functions, keeping the following questions in mind: what kinds of equations have straight line graphs? what kinds don t? for a straight line graph, how can we predict its slope from its equation? for a straight line graph, how can we predict, from its equation, where it crosses the y-axis? 3) Type the following Mathematica code: StraightLinesPicture This should generate a graph of three straight lines. Using the Plot command, reproduce this graph. Note: this section uses this module's special functions. If they fail to work, try going back to the Instructions for Getting Started at the beginning. Experiment 2: Chords and tangents Preparatory reading It's often important for us, as users of mathematics, to know about gradients of graphs. If the horizontal axis represents time, for instance, then the gradient of the graph represents how rapidly the quantity is changing in time (velocity = gradient of distance time graph, acceleration = gradient of velocity time graph, inflation = gradient of price time graph, etc.) The problem is, few graphs we study are straight lines. We need, then, some idea of gradient for curves. 4

5 Suppose we're interested in the gradient of a curve at a given point. Now: we already know about gradients of straight lines. So we can draw a straight line through our point, exactly as steep as the curve and then ask: what's the gradient of this line? This line, which just touches the curve, is called a tangent. The bad news is that, in general, it's hard to calculate gradients of tangents. Far harder, for example, than it is to calculate the gradient of a line that crosses the curve at two given points. Lines like this are called chords or secants. Because they pass through two points whose coordinates we know, we can use the formula to find their gradients. change in y change in x BIG IDEA: The closer together the two points, the more the chord looks like a tangent 5

6 The first part of the experiment involves generating a certain diagram. To produce this particular diagram type PlotChord[x^2, {x, -5, 5}, 2, 4, PlotRange->{-5, 25}] then shift-return in the normal way. 1) Generate a diagram showing the graph of the function y = x 2, together with a chord passing through the points corresponding to x = 2 and x = 4. Note down the gradient of this chord. 2) Generate a fresh diagram. This time, keep the first point at x = 2, but move the second point a little nearer: say to x = 3. Again, note down the gradient of the chord. Repeat, moving the second point nearer and nearer to x = 2. Note down the gradient each time. 3) Write a short report stating what you believe to be the gradient of the tangent at the point x = 2. Explain why you believe this to be the case, based on the results of your experiment. 4) Use the same technique to find the gradients of the tangents at the points x = 1, x = 3 and x = 4. 5) What do your findings from part 4 suggest? Write down your answer. Note: this section uses this module s special functions. If they fail to work, try going back to the Instructions for Getting Started at the beginning. Experiment 3: Limits and derivatives Preparatory reading We can express the ideas you met in Experiment 2 like this: the gradient of the tangent at the point P is the limit, as Q approaches P, of the gradient of the chord PQ. A limit is a number that we can get as close to as we like, even though we may not be able to quite reach it. 6

7 Let's think about the equation y = x 2. If P has coordinates (2, 4), and Q has coordinates (2 + h, [2 + h] 2 ), then the gradient of the chord PQ is given by change in y change in x = ( 2+h)2 4 h As h gets progressively smaller (we say it tends to zero) Q approaches P and the gradient of the chord PQ gets closer to that of the tangent at P. The gradient of this tangent, then, is the limit as h tends to zero of the above quantity. We write this as Notice, though, that and therefore that ( 2 + h) 2 4 lim. h (2 + h) 2 4 = 4 + 4h + h 2 4 lim ( 2 + h ) 2 4 h = 4h + h 2, 4h + h2 = lim h = lim 4 + h This shows us, then, that the gradient of the tangent at x = 2 is 4. Does this fit in with what you found? More generally, the gradient of the graph y = f(x) at the point (x, f(x)) is given by lim = 4. f (x + h) f (x) h This is called the derivative of f, and is written f '(x) or sometimes just f '.. Mathematica is good at handling symbolic quantities. In this experiment, you get to use some of these symbolic capabilities as you explore an algebraic approach to the numerical and graphical work you did in Experiment 2. 1) Begin by defining, in Mathematica, a function f such that f(x) = x 2. You do this by typing f[x_] = x^2 then holding down shift and pressing return. Be sure you type it exactly as it appears: notice that Mathematica uses square brackets where functions are concerned, and that all variables on the left-hand side of a function definition need to be followed with an underscore, _. 2) Use Mathematica to find the gradient of the tangent to the curve y = f(x) at x = 2. You can do this through the following Mathematica commands: 7

8 chordgradient = (f[2+h] - f[2])/h tangentgradient = Limit[chordGradient, h->0] 3) Repeat for x = 1, 3 and 4. Compare your answers to those from Experiment 2. 4) You can streamline this process considerably. To find the gradient of the tangent to the curve y = f(x) at x = 2, simply type f'[2] then hold down shift and press return. Repeat this for x = 1, 3 and 4, and for any other values you care to choose. 5) Based on your results, and on those of Experiment 2, what would you predict would be the output from the following Mathematica input command: f'[x] Try it: were you right? Experiment 4: Differentiation Preparatory reading In the background reading for Experiment 3, we saw a mathematical argument proving that the gradient of the graph y = x 2 at the point (2, 4) was 4. We can use a very similar argument to prove that gradient of the graph y = x 2 at the point (x, x 2 ) is 2x, whatever the value of x may be: lim The derivative of x 2, then, is 2x. We write or ( x + h ) 2 x 2 h = 2 xh + h2 lim h lim 2 x + h = = 2x. if f(x) = x 2 then f '(x) = 2x if y = x 2 then dy dx = 2x. Finding a function's derivative is called differentiating the function. We could repeat these kinds of calculation every time we need to differentiate (what we call differentiation from first principles), but that would be time-consuming. Alternatively, we could rely on Mathematica to do the job for us every time; that, though, would be a kind of cheating. 8

9 It would be ideal to have sets of rules that allow us to differentiate relatively quickly without relying on technology and without getting bogged down in calculations involving limits. In this experiment, you'll use Mathematica to explore some rules of that type. We've already met one Mathematica approach to differentiation: the f'[x] idea you saw in Experiment 3. Another uses something called the D operator. Try the following: D[x^2, x] 1) Using either the D operator or the f'[x] technique, find the derivatives of x 3, x 4 and x. What do you notice? Explore further. 2) Find the derivatives of x, 1/x, and 1/x 2. Do these fit in with your findings from part 1? 3) Write a short report summarising your findings about derivatives of functions of the form f(x) = x n. 4) Try the following: 5x 3 + 2x 2 + 1/x; 11x x; 2x 3-3 x+ 7. Write down what you observe. Try some other examples of this type. Post-experiment reading The observations etc. can be summed up in the single rule if f(x) = x 2 then f '(x) = 2x, if f(x) = x 3 then f '(x) = 3x 2, if f(x) = x 4 then f '(x) = 4x 3, if f(x) = x n then f '(x) = nx n 1. This rule isn't hard to prove from first principles. It applies to any function of the form f(x) = x n, including cases where n is fractional or negative. Thus, for example, suppose f(x) = x. Then and thus f (x) = x 1/2 9

10 f ' ( x) = 1 2 x 1/2 = 1 2 x. You'll notice, too, that the derivative of, say, 5x 3 is 5 3x 2 (or 15x 2 ), and that the derivative of, say, 5x 3 + 2x 2 is 5 3x x (or 15x 2 + 4x). This works quite generally: the derivative of a f(x) + b g(x) is a f '(x) + b g'(x). Practice Questions We've included a feature which allows you to get Mathematica to generate practice questions and their answers. There are three sets of questions on differentiation of x n. To generate a question of the simplest kind, type GiveQuestion["diff, simple x^n"] not forgetting to shift-enter. To generate the answer for checking, type LastAnswer["diff, simple x^n"] You can do this as often as you want: the questions are randomly generated, and repetitions should be rare. The other two sets of questions are accessed by the commands GiveQuestion["diff, harder x^n"] and GiveQuestion["diff, several terms"] Note: this section uses this module's special functions. If they fail to work, try going back to the Instructions for Getting Started at the beginning. Experiment 5: Trigonometric functions In this short experiment, we look at the derivatives of the principal trigonometric functions, sin x and cos x. Throughout, we'll be working in radians: if you've forgotten this way of measuring angle, or if you've never met it, consult the Trigonometry module. 1) Generate a plot of the graph of y = sin x, for x between and, such that the scales are the same on both axes. If you need a reminder about how to force the scales to be the same, consult the instructions for Experiment 1. (Note: Mathematica recognises the word Pi as meaning.) 2) What would you estimate to be the gradient of this graph at x = 0? At x = /2? At x =? On a piece of paper, sketch a graph of this gradient against x. 3) On the basis of your sketch, what would you conjecture to be the derivative of sin x? Use Mathematica to check your answer. 10

11 4) Repeat this process for y = cos x. Post-experiment reading It isn t hard to show, from first principles, that the derivative of sin x is, in fact, cos x. You need to know two facts: that, for small values of h, sin h is very close to h and cos h is very close to 1 h 2 /2; that sin (A + B) = sin A cos B + cos A sin B. The argument then goes like this. Suppose f(x) = sin x. Then f ' ( x) = = lim sin x + h h lim = lim ( ) sin x sin xcosh + cosx sinh sin x h sin x 1 h2 2 + hcos x sin x h = lim cosx h 2 sin x = cos x. A similar method can be used to show that the derivative of cos x is sin x. Note, though, that all this only works if you measure all angles in radians. Experiment 6: Exponential functions and e Preparatory reading Exponential functions are functions such as y = 2 x, y = 4 x, y = 5 x, etc. Note that these functions are not the same as things like y = x 2 or y = x 3 : we can t apply the same rules when we differentiate them. The general shape of the graph of y = a x is more or less the same for all positive values of a: they all look roughly like this: 11

12 The graph begins very flat, so for x negative the gradient is close to zero. Then, as x gets larger, the graph gets very quickly steeper: the gradient rises rapidly. If we were to plot a graph of gradient against x, then, we d get something that begins close to zero, then rises very rapidly: something very like the graph of the function itself! This suggests that the derivative of an exponential function might be another exponential function, or something very like one. In this experiment, we take a graphical look at the problem of differentiating exponential functions. You re asked early on to plot, for 3 < x < 3, a graph of the function f(x) = 2 x and its derivative. You re asked to do this in such a way that the graph of the function appears in red, and that of the derivative in blue. The quickest way to do this is to type f[x_] = 2^x Plot[{f[x], f'[x]}, {x, -3, 3}, PlotStyle->{RGBColor[1,0,0],RGBColor[0,0,1]}] remembering to shift-return after each command. 1) Generate a diagram showing a plot, on the same axes, of the function f(x) = 2 x and its derivative for x between 3 and 3. Show the function in red and the derivative in blue. Describe the apparent relationship between the two curves. 2) Repeat for the function f(x) = 3 x. 3) Find a value of a for which the graph of f(x) = a x is indistinguishable from that of its derivative. Post-experiment reading The results of this experiment strongly suggest that there is a close relationship between exponential functions and their derivatives. This is indeed the case, as the following argument makes clear. If f(x) = a x, then 12

13 f' ( x) = lim a x + h a x h = a x lim ah 1 h = L a x, where L = lim ah 1. h This means that the derivative of a x is just some multiple of a x itself. If we choose the value of a carefully, we can make L equal to 1. In this case, the derivative of a x will simply be a x. The value of a for which L = 1 is known as e, and is about The function f(x) = e x is so important that it is known as the exponential function, sometimes also written exp(x). Remember the derivative of e x is e x. Check using Mathematica if you like! Experiment 7: Derivatives of inverses Preparatory reading Reminder: the inverse of a function is that function in reverse: if f 1 is the inverse of f then f ( a) = b f 1 ( b) = a. The inverse of the function f(x) = a x is known as the logarithmic function f 1 (x) = log a x. The inverse of the exponential function f(x) = e x is known as the natural logarithm, and is written f 1 (x) = ln x,or sometimes f 1 (x) = log x, though the latter notation may be more familiar to you as standing for log 10 x. In Mathematica, the natural logarithm is written Log[x] (and the base-10 logarithm as Log[10,x]). In this experiment, we examine the problem of differentiating inverse functions. You re asked early on to plot, for 3 < x < 3, a graph of the function f(x) = e x and its inverse, Log[x]. You re asked to do this in such a way that the graph of the function appears in red, and that of the inverse in blue. One way to do this is to type f = Exp plot1 = Plot[f[x],{x,-3,3}, PlotStyle -> RGBColor[1,0,0]] plot2 = Plot[InverseFunction[f][x],{x,0,3}, PlotStyle -> RGBColor[0,0,1]] Show[{plot1, plot2}, PlotRange -> {{-3,3},{-3,3}}, AspectRatio -> Automatic] remembering to shift-return after each command. When you move on to part 2, you should begin instead by typing f = Sin and then adjust the x- and y-ranges in the Plot and Show commands. 13

14 1) Generate a diagram showing a plot, on the same axes, of the function f(x) = e x and its inverse. Define a horizontal range of 3 < x < 3 for f and one of 0 < x < 3 for the inverse. Show the function in red and the inverse in blue. Describe the apparent relationship between the two curves. 2) Repeat for the function f(x) = sin x, with a horizontal range of /2 < x < /2 for f and 1 < x < 1 for the inverse. 3) Repeat for any other functions you care to choose. 4) Define the Mathematica function g as the exponential function, using the command g = Exp or g[x_] = Exp[x] Calculate the gradient of the graph of this function at each of the points (0, 1), (0.693, 2) and (1.098, 3). [Hint: the first one is just g'[0].] Define the Mathematica function ginv as the natural logarithm, and calculate its gradient at each of the points (1, 0), (2, 0.693) and (3, 1.098). Compare your results with those for g. Do they fit in with what you expected? Could they have been predicted from the first graph you generated? 5) Explore further if you wish, perhaps using other functions and their inverses. Write a short report summarising your findings and explaining them. Post-experiment reading We can conjecture: if the gradient of the graph of f at the point (a, b) is m 1, and the gradient of f 1 at the point (b, a) is m 2, then m 2 = 1 m 1. This is indeed the case, as a little bit of thought about the graphs of f and f 1 makes clear. One consequence of all this is that if we know how to differentiate a function, we can often differentiate its inverse. For example, if f(x) = e x, then f '(x) = e x. It follows that the gradient of the graph of f at the point (a, b) is e a. Now, if g(x) = ln x, we can deduce from the above that the gradient of the graph of g at the point (b, a) is 1/e a = 1/b. It follows that g'(x) = 1/x. We can use much the same method to show that the derivative of arcsin x is derivative of arctan x is x 2. 1, and that the 1 x2 14

15 Summary Function x n sin x cos x e x Derivative nx n 1 cos x sin x e x ln x 1 x arcsin x 1 1 x 2 arctan x x 2 Practice Questions There is one more set of questions in this module. To generate a question from it, type GiveQuestion["diff, miscellaneous"] not forgetting to shift-enter. To generate the answer for checking, type LastAnswer["diff, miscellaneous"] 15