Mathematica Proficiency and motivation for 6.1 and 6.2

Transcription

1 Calculus Page 1 Master Lab Friday, August 19, :13 PM Mathematica Proficiency and motivation for 6.1 and 6.2 This lab is designed to test some of your Mathematica background, to be useful as a reference, and to provide some hands-on examples to motivate the material in 6.1 and 6.2. It is not designed to stand-alone. I have included some more advanced material that will be useful to some of you but is absolute over-kill for most of you. This material has a gray background and can be safely ignored. However if you are going to continue using Mathematica in your field of study you may find the backgrounded material very helpful. You will NOT be tested over any of that material. I also use a yellow background in order to hide Mathematica commands that are at an intermediate level of difficulty. I want you to try to come up with the command yourself first, but if you can not then highlighting the yellow box should reveal the answer (if it doesn't then copy and paste into another program). A gentle Introduction to Lists: Lists are extremely useful in Mathematica and I expect you to be able to do some simple list manipulations. Lists are collections of stuff (very technical non?). The simplest way to build a list is to use the 'curly brackets' (also known as 'curly braces'). Those look like this: { and }. For example: {2,3, 2,x,10-a} is a list with five elements. It contains numbers and expressions (Mathematica is flexible in this respect). Create a list with 6 elements (anything you care to include) and assign it to the variable m. (I have included an example just a bit further down below to show you how this is done) This can be just about anything You can extract individual elements using [[ and ]]. Here's an example where I create a list of both numbers and expressions and then extract the 2nd element: m = {1, 20, a, 2+x, 3, 6} m[[2]] Here's how to extract the second-to-last element: m[[-2]] And finally here is how to extract the 3rd through 5th elements: m[[3;;5]] Now you should do the same thing with your list: Pretty self-explanatory A useful way to generate a list of numbers is the Range[] command. The command can take one, two, or three arguments. Play around a bit with Range[] and see what it does. Just try various things Unfortunately, it is a bit misleading to use the curly braces for lists because in the rest of the mathematical world we use those curly braces for sets. Why is this unfortunate? Because in a list ORDER matters. A list has a first element, and a second element, and so on and so forth. Lists can also have repeated elements. Part of the useful properties of a set is that there is NO replication and no order.

2 Calculus Page 2 Use Range[] to generate a list of numbers starting at 6, ending at 10, and incrementing by 0.5 each time. Store the list in m. You could do this by hand like this: m={6,6.5,7,7.5,8,8.5,9,9.5,10} m = Range[6,10,0.5] Evaluate m + 2 and 2 m and explain what you observe: m m Adding or multiplying a number with a list adds (or multiplies) that number to EACH element in the list. The list m contains 9 elements. Generate another list of 9 elements containing numbers and assign it to the variable n. Then add m and n. Explain what happened: n = Range[10, 2, -1] (my example) n + m The corresponding elements were added element by element. Motivation for 6.1: Plot the graph of y = x 3-1 and y = -4x simultaneously from x = -10 to x = 10 using the Plot[] command: Plot[{x^3-1,-4 x^2+2},{x,-10,10}] What's misleading about this image? Functions only appear to cross in one place Find the x-values at which the two functions cross as exact values (Solve[] is useful here): Solve[x^3-1 == -4 x^2 + 2, x] Now express them as decimals (you can either redo the last step with 'Nsolve', or use 'N': Solve[x^3-1 == -4 x^2 + 2, x] // N NSolve[x^3-1 == -4 x^2 + 2, x] Save these three x-values (as exact values) as the variables min, mid, and max. At first this might seem problematic, but try highlighting the expression like this:

3 Calculus Page 3 Now you can copy the expression and paste it wherever you would like for example: It will help to know which of the three expressions is smallest, middle, and largest, but that information can be gleaned from the decimal form. There is also a fancy way to do this: {mid,min,max}=x /. Solve[x^3-1 == -4 x^2 + 2, x] Notice the order of the variables on the left hand side of the expression. They match the order in which the solutions occur. If I was using numerical approximations I could do this: {min,mid, max}=sort[x /. NSolve[x^3-1 == -4 x^2 + 2, x]] Exact expressions can be ordered differently so be careful! Double check that min is the smallest, max the largest and mid the one in between (if this is not correct the rest of the assignment is going to be very wonky): You can check the decimal form of a variable like this: N[min] Now find 10% of the distance between the largest and smallest x-values and save it as a variable called d. Plot the two functions again from min - d to max + d: d=.1*(max-min) Plot[{x^3-1, -4 x^2 + 2}, {x, min - d, max + d}] Verify the values of min, mid, and max using the 'Get Coordinates' option (right-click on the graph) Repeat using mid and max (be sure to recalculate d): d=.1*(max-mid) Plot[{x^3-1, -4 x^2 + 2}, {x, mid - d, max + d}]

4 Calculus Page Why does this image look so different from the last? Because Mathematica chooses the scale for the y-axis based on the function values to be displayed. The previous graph is effectively zoomed out so far that it is hard to see the details visible in this graph. A Brief Digression into Mathematica Concerns: Now create two functions f and g. The function f will represent y = x 3-1, and g will represent y = -4x 2 + 2: f[x_] := x^3-1;g[x_]:=-4 x^2 + 2 Experiment a bit with plugging in values for x in f and g (for example f[1]): There are lots of different ways to do this. One final check that min,mid, and max are correct: f[mid] - g[mid] Now you do the same for mid, and max. f[mid] - g[mid] f[min] - g[min] f[max] - g[max] These numbers should all be zero, but on my computer they are not. I get a complicated expression for mid, and max. Explain how to convert the expression into decimal form. Is the decimal non-zero? If not why? I use the postfix operator // to pipe the results into N: f[max] - g[max] // N When Mathematica evaluates the expressions it does some rounding, this results in a number that is NEAR zero, but not exactly zero. Explain why this does NOT mean that Mathematica is broken. The SYMBOLIC answer is still correct Back to 6.1: We want to try to estimate the area between these two curves using rectangles. That last statement does not make any sense unless we includes some boundaries, so in deference to convention we will call the leftmost x-value a, and we will call the right-most x-value b. We can, of course choose a and b to be whatever we like (thought typically a is less than b). I have a bit of an agenda so I want you to set a to the value of mid

5 Calculus Page 5 and b to the value of max. While you are at it undefine mid and max. a=mid; b=max; mid=. max=. With those preliminaries out of the way, we will start our approximation using 5 rectangles. We want the left-most rectangle to line up with x = a and the rightmost to line up with x = b. That means the width of each rectangle is going to be 1/5 the distance between b and a. Typically we would call this x (pronounce that as delta x). It is common to use 'delta' to stand for 'change'. Think about those rectangles: If we start at a, the first rectangle should have one vertical side at a, and the next at a + x. The second rectangle should have vertical sides at a + x and a + 2 x, and so on. It is a bit of a pain to represent x in Mathematica so we will use w as a stand-in. Type this: w=(b-a)/5 Now we are going to need to figure out where the LEFT and RIGHT x-values for the sides of each rectangle are going to live: lefts=a + w*range[0,4] rights=a + w*range[1,5] Understanding those last two commands is a bit tricky. It might help to remember that a and w are both NUMBERS, and Range[0,4] and Range[1,5] are both lists with 5 elements. Go back to the Gentle Introduction to Lists if you are still unclear over what is happening (or ask me). At this point we have stored the LEFT sides of the rectangles. They are in a list that we have assigned to the variable lefts. We have done a similar thing for the RIGHT sides of the rectangles So now we have found the left and right sides but what about the tops and the bottoms? We have, many, many places where we can put the tops and the bottoms of the rectangles. The fundamental ideas is that the bottom of each approximating rectangle should cross the lower graph somewhere and the top should cross the upper graph somewhere. It does not really matter where we do this as long as our left and right sides are in the proper places, but it is nice to do it systematically and since it tends to be easier to see if we choose the middle of the rectangle, lets do that: middles=(lefts+rights)/2 That command is also a bit tricky to understand. Explain what it does: The variables 'lefts' and 'rights' are both lists with 5 elements. When we add them together we generate another list of 5 elements where the addition is performed element by element. Dividing by the *number* 2 divides every entry in that new list by 2. This has the effect of finding the mid-point between the left and right sides. I want to draw the boxes so you can see what is happening, but the commands needed to draw those boxes goes quite a bit beyond what you will probably ever need so don't think about it too much and just type it out. The end result will be a variable called funcs that contains the images of the functions and another variable called boxes that contains the images of the approximating rectangles (if you want to know more, please ask): funcs=plot[{x^3-1, -4 x^2 + 2}, {x, a - d, b + d}]; boxes = Array[ Graphics[ { EdgeForm[Black], White, Rectangle[ { lefts[[#]], g[ middles[[#]] ] }, {rights[[#]], f[middles[[#]] ]} ] } ] &, 5];

6 Calculus Page 6 Show[{funcs,boxes,funcs}] Now calculate the area of those rectangles. Use Total[] to add all the elements in a list together. You should get Try to figure this bit out yourself first, but if you can not highlight the box below to reveal the answer: Total[(g[middles] - f[middles])*w] // N or N[Total[(g[middles] - f[middles])*w] ] Now go back through the process but with double the number of boxes. You should get an answer around and a graph that looks something like this: We can keep increasing and increasing and increasing the number of rectangles-- getting closer and closer and closer to the real answer. As we learned in class, the exact answer is calculated using an integral. The command below will *almost* give you the correct answer. What goes wrong and why? NIntegrate[f[x]-g[x],{x,a,b}] Do it correctly (give the command and the right answer to 3 decimal places): NIntegrate[g[x]-f[x],{x,a,b}] Now suppose that we went from a = -2 to b = 0.5. Generate a picture with 10 rectangles with these new values of a and b. Pretty much everything you typed to generate the graph with the 10 rectangles above can be reused after you change the values of the a and b. Don't forget to recalculate d and re-run the command funcs=plot[{x^3-1, -4 x^2 + 2}, {x, a - d, b + d}]; If you forget that step the Show[] command is going to have incompatible graphs and rectangles. The graph should look like this:

7 Calculus Page Now type this: Integrate[g[x]-f[x],{x,a,b}] The answer you get is LESS than that from before but how can that be? Sure we're not going quite as far to the right but that bit from x = -2 to x = -1 should more than compensate right? So what's the deal? We are integrating across TWO regions. In one region the f[x] is ABOVE g[x] and in the other region g[x] is above f[x]. There are TWO ways to correct this. One involves using TWO integrals and the other involves using Abs[]. Use both ways to answer this question and ensure you get the same answer both ways: What is the area between the curves y = x 3-1 and y = -4x from a = -2 and b = -0.5? Integrate[f[x] - g[x], {x, -2, -1}] + Integrate[g[x] - f[x], {x, -1, 0.5}] Integrate[Abs[f[x] - g[x]], {x, -2, 0.5}] Both answers should be around Volumes (6.2): I don't want to get too bogged down in the auxiliary details, like the approximating rectangles we used in the last section. They're useful but I don't want to introduce too many new ideas at once. However, the main point of section 6.2 is very similar to what was done in 6.1, but instead of using approximating rectangles to figure out how to calculate an area we use approximating slices to figure out how to find a volume. I trust at this point you can use Mathematica to evaluate definite integrals. Consider a cone centered on the x-axis, that can be formed by rotating the line y = x around the x-axis.

8 Calculus Page 8 What is the area of the cross-section at x = 3? The cross-section is a circle. Since the defining line of the cone has slope 1, the radius of this circle is the same as the x-value. So in the case the cross-section is a circle of radius 3. That has area. Use Mathematica to calculate the volume of the cone between x = 2 and x = 4. You should get an answer somewhere near Integrate[Pi x^2,{x,2,4}] // N This is the general technique, but we focus on a particular subset of solids-- those that can be formed by rotating the graph of a function around the x-axis or the y-axis. The "skin" of these solids are called surfaces of revolution. Unfortunately, in order to show you how to draw these surfaces of revolution we are going to have to cheat a little bit and use some material from Chapter 10. If the details of the commands don't make sense right now, don't worry about it-- view this section instead as a how-to. At the end of this lab you should be able to graph a surface of revolution that has been formed by rotating a graph of the form y = f(x) or x = g(y) around an axis. This won't be enough to generate graphs for all of your homework but it will hopefully be enough to get you started. So, onto the gory details: Let's start with that perennial classic y = command: Graph it from x = -0.2 to x = 2, using the normal Plot[]

9 Calculus Page The book (page 433) asks you to find the volume of the solid of revolution (from x = 0 to x = 1) obtained by rotating this graph around the x-axis. They thoughtfully show the graph of that surface. The key command to duplicating this graph is ParametricPlot3D[]. It's easier to reuse the commands if we store the function as f[x]. Then we can reuse the command after defining f[x] to reflect the function that we are rotating. I'll use f[x] for y as a function of x and g[y] for x as a function of y. This is an arbitrary convention but I'm hoping it will help reduce some confusion. So Define f[x] as the sqrt of x: f[x_]:= Sqrt[x] Go ahead and reproduce the plot you did above, but this time use f[x] (I'm not going to reproduce the graph because it's the same:) Plot[f[x],{x,-.2,2}] The graph is only defined for x >= 0, so lets graph it again without the undefined bits off to the left: Plot[f[x],{x,0,2}] We don't want to include those undefined bits in the surface of revolution. Here is the command for rotating f[x] around the x-axis: ParametricPlot3D[{t,f[t]*Cos[theta],f[t]*Sin[theta]},{t,0,2},{theta,0,2Pi}] Your graph should look something like this:

10 Calculus Page 10 You can grab it with the mouse and move it around. It's up to you to keep track of which axis is which though so be careful! Draw the graph again, but this time let x range from 1 to 4. Your graph should look like this:

11 Calculus Page 11 Now let's get ready to rotate the same graph around the y-axis. To do this we are going to think of x as a function of y. Since the relationship is y = (as long as x >= 0) The inverse relationship is x = y 2 and here we require that y >= 0. Consider the path that was traced out by the point (2, ). I'll show the point in red, and it's traced out path in red (poorly): When we rotate this graph around the y-axis that point is going to form a different circle. Notice that the resulting graph is going to range from y=0 to y =. So let's define g[y] and make the graph: g[y_] := y^2 ParametricPlot3D[{t*Cos[theta],t*Sin[theta],g[t]},{t,0,Sqrt[2]},{theta,0,2Pi}] You should get something like this: You should sketch both of these surfaces of revolution by hand and convince yourself that you understand the differences between the two images. The gridwork of lines can be annoying that can be turned off by adding Mesh->None. I highlighted it to make it easier to see. I also rotated the image a bit before pasting it into this handout: ParametricPlot3D[{t*Cos[theta],t*Sin[theta],g[t]},{t,0,Sqrt[2]},{theta,0,2Pi},Mesh->None]

12 Calculus Page 12 You can also add a bit of transparency to the image using PlotStyle->Opacity[number]. The number can vary from 0 to 1 (0 is completely see-through, 1 is complete opaque): ParametricPlot3D[{t*Cos[theta],t*Sin[theta],g[t]},{t,0,Sqrt[2]},{theta,0,2Pi}, Mesh->None, PlotStyle->Opacity[0.4]] Try changing the value in Opacity and rotate the graph to get a good idea of what happens. Where this is particularly helpful is when you are plotting TWO surfaces at the same time. Pay extra close attention to the curly braces in the example below: g1[y_]:=y g2[y_]:=sqrt[y] ParametricPlot3D[{{t*Cos[theta], t*sin[theta], g1[t]}, {t*cos[theta], t*sin[theta], g2[t]}}, {t, 0, Sqrt[2]}, {theta, 0, 2 Pi}, Mesh -> None, PlotStyle -> Opacity[0.4]] What did I do differently up above? I put two sets of triplets inside a set of curly braces. Here's a picture of the graph I produced above: Now you define two functions f1 and f2 and show the two surfaces of rotation rotated around the x-axis (I did it around the y-axis). Play with the opacity and mesh setting. Lots of possible answers here