Graded Assignment 2 Maple plots

Transcription

1 Graded Assignment 2 Maple plots The Maple part of the assignment is to plot the graphs corresponding to the following problems. I ll note some syntax here to get you started see tutorials for more. Problem 5: Basic equations of lines and planes. Find the equation as specified below: (a) Of the line passing through the point (3, 1,4) and parallel to the line with symmetric equation x 2 y 3 z Put your final answer in parametric form (you ll need that for plotting). Plot (on the same graph) (a) The given line (you ll need to turn that symmetric form into parametric first) (b) The given point (c) Your line (the answer) I ll give you parts (a) and (b) as freebies, to show you the syntax for plotting lines and points. First, we do x 2 y 3 z need to turn into parametric form. You have to recall that the direction vector can be read off the denominator, and the point from the numerator this is the line passing through the point (2, 3,0) with direction vector v 4, 1,5. Its equation can be expressed as Which is in parametric form xyz,, 2, 3,0 t 4, 1,5 x 2 4t y 3 t z 0 5t That is the first key takeaway: if you re plotting a line, it has to be in parametric form (whether the directions explicitly say that or not!) Here is the Maple that plots that line:

2 Notice how I m setting up functions of t. The syntax for plotting lines (and curves) in space is the spacecurve command as shown. Next, we need the syntax for plotting a point. That looks like this: Now, if I wanted to display both of those things together, I would use the display command: Try entering all of the above, and see what happens. However, this problem isn t done you need to plot the line that is the answer. That means you ll be defining a second set of parametric equations: x2:=t -> [your answer] : y2:=t -> [your answer] : z2:=t -> [your answer] : and defining a third plot: > p3:=spacecurve([x2(t), y2(t), z2(t)], t = , color = green, thickness = 2): and adding it to the display: That is how you are checking the problem what you see had better match the conditions set forth (passes through point, and is parallel to given line).

3 (b) Of a plane with normal vector passing through the point (1,2,3) and perpendicular to the line x, y, z 2, 1, 1 t 5,1,2 Put your answer in general form first, then solve it out further to z f( x, y) form for plotting. Problem 5 (b): Plot (a) The given line (b) The given point (c) Your plane (the answer) Now you know the syntax for plotting a line. You can define that line in parametric form and plot it. If you keep working in the same document, make sure to give it a new symbol name at this point, we d be up to You also know the syntax for plotting the point: What you don t know is the syntax for plotting a plane, and the problem is, I don t want to work out the answer for you to show you. Let me show you a general example: Suppose you have the equation for a plane in general form Ax By Cz D 0. The first thing you have to do is solve for z : defining that function is Ax By D z. That is a function of x and y, and the syntax for C z:=(x,y)-> [your expression for the plane]: Plotting that function is a 3d surface plot, and the syntax for that is: p5:=plot3d(z(x,y), x = -5..5, y = -5..5): With any of this stuff, you may need to adjust the ranges on your t s, x s, and y s to get a good picture of the thing. After getting that entered, you can

4 Notice I ve got one new thing in there too the view parameter. What I am doing is squaring off the view (it s like setting the window in your calculator). The reason for this is, if I use Maple s default, the different axes will have different scales, and I won t be able to tell if my line and plane are perpendicular. And that basically covers it. There s only three types of things to plot here lines, points, and planes, and you ve now got the syntax for all three of those things. The last tip is don t try to plot everything in the same document when Maple gets crowded, Maple gets confused. I used separate documents for P5, P6, and P7. The remaining questions will just show the short form of what I want you to plot. (c) Of the plane containing the points (2,0,1), ( 3,7,2), and (4, 2,5). Put your answer in general form first, then solve it out further to z f( x, y) form for plotting. Problem 5 (c): Plot a) The three points b) Your plane (the answer) The points probably won t show up that well if done correctly, because they ll be in the plane. If a point is floating outside of the plane, you ve got a problem. Problem 6: Two lines are given: l : x, y, z 1,9,8 t 1,5,3 1 2 l : x, y, z 3, 7, 1 t 2,2,1 (a) Find the point of intersection of the two lines. (b) Write the equation of third line that also passes through that point, and is perpendicular to both the given lines. Problem 6: Plot (a) The two given lines (b) The point you found to be the point of intersection (for a quick check before proceeding, display those things first and make sure your intersection point really is the intersection point) (c) Your line (the final answer)

5 Your line should pass through the point of intersection, and appear to be perpendicular to both lines (again, don t forget to square off the view) Problem 7: For the planes 2x 3y 2z 4and x 4y 7z 1 (a) Find the equation of the line of intersection of the planes. (b) Find the angle between the planes. (c) Find the distance from the point (2,1, 4) to the plane x 4y 7z 1. Problem 7: Plot (a) The two given planes (b) The solution for the line of intersection (c) There s nothing to plot for angle or distance all you re checking is part (a) above Your line of intersection should be lying right in the crease where the planes do in fact intersect.