Calculus Review with Matlab

Transcription

1 Calculus Review with Matlab While Matlab is capable of doing symbolic math (i.e. algebra) for us, the real power of Matlab comes out when we use it to implement numerical methods for solving problems, especially problems that cannot be solved symbolically. Let's look at some of the concepts from calculus that will be important in this course through this lens. You should create your own Matlab Live Editor file to do the exercises below. Include explanations for all of your work, and then export to PDF to submit. Differential Calculus In a typical calculus 1 course, we want to answer the question, "How do we find the instantaneous rate of change of a given function at a specified point?" We start by looking at average rate of change. The average rate of change for a function interval is given by the expression over the For example, for the function, the average rate of change over is In addition to interpreting this as average rate of change, we can think of it as the slope of the secant line to f from to. Let's plot the graph of f and its secant line. Here's one way to do so symbolically: syms x m = (1/3)*log(4); f = log(x); secant = m*(x - 1); fplot(f,[0,6]) hold on fplot(secant, [0,6]) title('graph of ln(x) and its secant line') hold off 1

2 Alternatively, we could do so numerically by plotting 100 points and letting Matlab connect the dots: xs = linspace(0,6,100); fx = log(xs); secline = m*(xs-1); plot(xs,fx,'b-') hold on plot(xs,secline,'g-') hold off 2

3 If any of the above commands are unfamiliar, take on of the Matlab tutorials on Mathworks' homepage. To get from secant lines to tangent lines, or from average rate of change to instantaneous rate of change, we just let the length of the interval approach zero using limits. So, the instantaneous rate of change of f at is given by More generally, we define a function's derivative,, to be the function For example, for, Since,, we know the slope of the tangent line to f at is 1. Matlab can find derivatives for us symbolically: f1 = diff(f); tanslope = subs(f1,1); tanline = tanslope*(x-1); 3

4 fplot(f) hold on fplot(tanline) title('tangent line to ln(x) at x=1') hold off Since the derivative is giving us, we can approximate the derivative numerically by taking differences in successive y-values and dividing them by differences in successive x-values over small intervals. We'll make a new vector of x-values here to be a little more precise. dx = 0:0.01:4; fxs = log(dx); dy = diff(fxs); derivf = dy./0.01; plot(dx(1:end-1),derivf) 4

5 Note: The diff function creates an array that's one element shorter than its input. So, in order to plot, I had to take away one element of dx. This graph looks a lot like the graph of! Let's see how close the approximated value of the derivative is at : c = dx==1; approxtanslope = derivf(c) approxtanslope = slopeerror = double(abs(approxtanslope - tanslope)) slopeerror = This is pretty good! If we changed the step-size for dx from 0.01 to 0.001, we'd get an even better answer. Integration In Calculus 2 we ask a very different question whose answer turns out to be very closely related to the work we did in Calc 1: Given a continuous function f, how do we find the net signed area bounded between the graph of f and the x-axis over? 5

6 One way to do this is to split the region up into a bunch of rectangles, whose areas are easy to compute, and then let the number of rectangles increase - i.e. take the limit as the number of rectangles goes to infinity. The fundamental theorem of calculus says that where F is an antiderivative of f. But what if we can't find an antiderivative? For example, the function doesn't have an antiderivative - try doing substitution or integration by parts on it and see what happens. Say we wanted to find the area bounded between f and the x-axis over. That is, we want to know what is. Since we can't take an antiderivative, we have to do so numerically. Let's try it with 100 rectangles: n = 100 ; deltax = 1/n ; xs = 0:deltax:1-deltax ; % I'm only using the left-hand endpoints % If I go up to 1, then I'll have 101 x values fs = sin(xs.^2); plot(xs, fs) 6

7 area = sum(fs*deltax) area = This is an okay approximation. If we up the number of rectangles to 1000 or 10000, we'll get a much better approximation. Exercises You should create a Live Editor file, much like this one, to do the following exercises. Your file should include text, clearly explaining all of your work, in between your code. Export your work to PDF and it to me by Monday, July 2nd. Pre-exercise Get setup with Matlab. Much of what we'll do in this course can be done in Matlab Online. Familiarize your self with Live Editor, specifically, how to use Latex to insert mathematical expressions and how to insert and run code. Play around with this a bit before you start the assignment. Let me know if you have any questions. Exercise 1 This you can do on paper: Take the final exam from Math 122 (posted on our homepage). 7

8 This is intended to be a diagnostic for your benefit, so do it unaided and grade yourself based on the rubric. You'll be graded for completeness, not correctness here (I promise). Submit your original work by scanning it and ing it to me along with your Live Editor PDF. Exercise 2 Pick any function (something not to simple and not too complicated). Take it's derivative by hand. Then, take the derivative symbolically in Matlab. After confirming that your answers are the same, use the derivative to compute the slope of the tangent line to your function at some x-value. Plot the tangent line along with the original graph Repeat these steps using the numerical method described above Exercise 3 Consider the function. Like, this has no antiderivative. Following the method described above in the integration section, approximate the area under f over using 100 rectangles. Repeat for 1000 rectangles and rectangles. See what WolframAlpha gives you when you type in "integral from x=0 to x=2 of e^(x^2) dx." How do your answers stack up? 8