Practice with Parameterizing Curves and Surfaces. (Part 1)

Transcription

1 M:8 Spring 7 J. Simon Practice with Parameterizing Curves and Surfaces (Part ) A "parameter" is a number used to measure something. In particular, if we want to describe and analyzie some curve or surface or other set, we usually begin by assigning numbers to act as coordinates on the set. A"parameterization" of some set S is a function from R^ or R^ (or R^n for appropriate n) onto the set S. Each number t (or n-tuple [t, t,..., tn]) describes a point in the set, and the function p: R^n --> S can be used to measure geometric properties of the set such as length, area, curvature, and integrals of functions whose domain is the set S. The domain of parameterization is "oriented". For R^, i.e there is a notion of positive and negative, and a sense of movement from lower to higher; for R^ we have a sense of "left-to-right" in the x-coordinate and "low-to-high" in the y-coordinate. We talked in class about how to generalize this idea to "equivalence classes of ordered bases" for the vector space R^n, but right now, we can be informal. The point is that when we parameterize a set, we are imposing on the set a sense of orientation. For example, when we parameterize a curve, we are tacitly stating a sense of direction along the curve. Usually, when we tell a computer to draw a picture of some set, we are actually parameterizing the set. Here first are examples of parameterizations of one-dimensional sets. A one-dimensional set is parameterized by one number (use "t" and think of "time"), regardless of where the set lives. For all the examples, I will use only one orientation; you can create more practice exercises for yourself by changing the desired orientations and/or the desired starting points. EXAMPLE. S = line segment in the xy-plane between [,] and [,], and oriented from [,] towards [,]. Solution a.

2 The point-slope form of equation for the line through [,] and [,] is y- = ( (-)/(-) ) (x-) > eq:=y- = ( (-)/(-) ) *(x-); eq := y - = x - > y=solve(eq,y); y = 7 + x Since the line can be described by an equation that gives the y-coordinate in terms of the x-coordinate, we can use the x-coordinate as a parameter to describe the line segment. The x coordinates start at and end at. So we can parameterize the line segment via the function p(x) = [x, 7/ + (/) x ], x =.. Using the parameterization, we can tell the computer to draw the set. > plot([x, 7/ + (/)*x, x=..], color=blue, thickness=, scaling=constrained, axes=normal, view=[-..6, -..]);

3 - 6 - Solution b. Use the vector form of the line x = [,] + t [-, -] with t=.. x = [+t, +t] > plot([+*t, +t, t=..], color=green, thickness=, scaling=constrained, axes=normal, view=[-..6, -..]);

4 - 6 - Exercises: Write a parameterization for this line segment that reverses the direction. ########### EXAMPLE. S = circle of radius centered at the point [,], oriented counter-clockwise. Solution a. The usual equation for such a circle is (x-)^ + (y-)^ = ^ > EqCirc:=(x-)^+(y-)^ = ^; > with(plots): EqCirc := ( x - ) + ( y - ) = 9 Warning, the name changecoords has been redefined

5 > implicitplot(eqcirc, x=-..6, y=-.., thickness=, color=blue, axes=normal, scaling=constrained); y - - x - The circle cannot be described by a single equation of the form y = f(x), so if we want to use x as the parameter to describe the set, then we will need to parameterize the set in two pieces. So we will parameterize the upper and lower half-circles separately. > solve(eqcirc,y); We will have two parameterizations: + - x + x, - - x + x x --> [x, +sqrt(-x^+x)] and x --> [x, -sqrt(-x^+x)]. What are the ranges of x-values we should use?? We can reason geometrically (circle of radius, centered at [,], so x is beween - and +) or we can reason from the equation for the circle ( (x-)^ + (y-)^ = ^; when y=, (x-)^ = 9, so x- = or - ==> x= or -. The function x --> [x, +sqrt(-x^+x)], x =..(-) is a parameterization of the upper half-circle, oriented counter-clockwise, and starting from the point [,] > plot([x, +sqrt(-x^+*x), x=..-], color=red, thickness=, scaling=constrained);

6 ... - And the function x --> [x, -sqrt(...)], x = -.., is a parameterization of the lower half-circle, oriented counter-clockwise. > plot([x, -sqrt(-x^+*x), x=-..], color=blue, thickness=, scaling=constrained);

7 Solution b. Whenever we see a circle, we should think about using polar coordinates. The usual polar equation of a circle of radius centered at the origin is (in terms of parameter t = angle) x = cos(t) y = sin(t) But in this problem, the circle is not centered at the origin. Think about obtaining the desired circle by translating the circle centered at the origin. We add to each point on the circle at [,], the translation vector [,]. That moves the origin to the new center, and moves the whole circle to where we want it. > plot([*cos(t) +, *sin(t) +, t=..*pi], color=green, thickness=, scaling=constrained);

8 - - - Exercises: Using polar coordinates, find parameterizations of this circle that **orient the circle clockwise **have various starting points **include only half or a quarter of the circle. Notice how much harder it would be to do these exercises if you use x or y as the parameter; but it's not impossible ######################## EXAMPLE. S is a circular helix in R^ having radius and height, having uniform pitch and making full turns. The axis of the helix is the z-axis. > display(p);

9 Think about tracing out a circle of radius centered at the origin in the xy-plane, but as we go around the circle, we also climb ("the axis of the helix is the z-axis") at a constant rate ("uniform pitch"). This says that the height parameter and the angle parameter should be the same, or perhaps one a constant multiple of the other. We want the height to go from to during the "time" that the angle goes around times, i.e. from to Pi to Pi to 6Pi. We could write height as a function of angle, or vice versa, but it may be simpler to write each of them as a multiple of "time" t. As t goes from to, we want height z to go from to at a constant rate, so use z = *t. As t goes from to, we want the angle to go from to 6*Pi. So use theta = 6*Pi *t. To describe the [x,y]-movement as going around the circle of radius centered at the origin, we want x = *cos(theta) = *cos(6*pi*t), and y = *sin(theta) = *sin(6*pi*t). > tubeplot([*cos(6*pi*t), *sin(6*pi*t), *t], t=.., radius =., scaling=constrained, axes=boxed);

10 Exercises: **Write a parameterization of this helix that uses z as the parameter, i.e. find a parameterization of the form [x(z), y(z), z], z=.., where x(z) and y(z) are appropriate functions of z. **Write a parameterization that starts at the top point and winds down. **Explain how the parameterization [*cos(6*pi*s^), *sin(6*pi*s^), *s^], s=.., differs from the one shown above.