## Assignment 10

Exploring Parametric Equations and Curves## Nicolina Scarpelli

A curve in the plane is said to be parameterized if the set of coordinates on the curve (x , y) are represented as functions of a variable t, namely x = f (t) and y = g (t). These two equations are usually called the parametric equations of a curve. The variable t is called a parameter. In many applications, parametric equations are extremely useful in describing the movement of a point along a curve, so in many cases we think of x and y "varying with time t".

Let's begin our investigation by exploring the parametric equations x = a cos(t) and y = b sin(t) for 0 ≤ t ≤ 2π while varying the a and b values. We will begin with a = b = 1.

Now, let's switch x and y to see what happens to the graph. So let x = a sin(t) and y = b cos(t). How do you think the graph will be different?

Observe that the two graphs above are the same graphs because it is essentially a reflection of each other over the line y = x.

Now, let's observe what happens when a = b = 2. Observe the graph below.

Notice from the graphs above when a = b, the paramaterization of x and y is a circle centered at the origin (0,0) with radius equal to a. When a = b = 1, we have a graph of the unit circle. Since the domain of t is between 0 and 2π (approximately 6.28) or 360 degrees, the graph starts at t = 0, the point (a,0) or in this case (1,0), and moves in the counterclockwise direction around the origin to create a full circle that ends at t =2π, which is also the point (a,0) or in this case (1,0). The same holds for when a = b = 2 except the circle begins at t = 0, the point (2,0), and ends at t = 2π, also the point (2,0).

To further investigate, hypothesize what will happen if your t - values range from 0 to π instead of 0 to 2π. How will your graph be different than the graph above? What if t = (0, π/2) or t = (0, 3π /2)? Observe the graphs below.

The graph to the far left is the graph of the parametrics equations with t ranging from 0 to π /2 or 90 degrees, the middle graph is the graph of the parametric equations with t ranging from 0 to π or 180 degrees, and the far right graph is the graph of the parametric equations with t ranging from 0 to 3π/2 or 270 degrees. Focusing on these properties, we can extract an important trigonometric identity from the unit circle above. Observe the unit circle below that I created using Geometer's Sketchpad.

Using the right triangle CPE in the above diagram, CP= cos(θ), EP = sin(θ) and CE = 1 (the radius of the circle. Thus, by the Pythagorean Theorem CP

^{2}+ EP^{2}= CE^{2}. Substituting for CP, EP, and CE we get, cos^{2}(θ) + sin^{2}(θ) = 1^{2}. We will use this formula later on in our investigation.

Now, let's examine what happens when a < b. Let's begin by setting a = 2, and b = 5.

Observe the graph above. When a < b the graph no longer takes the shape of a circle. We now have a graph that draws an ellipse parametrically centered around the origin. Our t - value still ranges from 0 to 2π. As you can see from the graph, this ellipse has a vertical major axis along the y - axis, x- intercepts at (-a,0) and (a,0) or in this case (-2,0) and (2,0), and y - intercepts at (0,-b) and (0,b) or in this case (0,-5) and (0,5). The equation of an ellipse in standard form is x

^{2}/a^{2}+ y^{2}/b^{2}= 1. Thus, to find the equation for this ellipse in standard form, we must solve the equation x = a cos(t) for cos(t) and solve the equation y = bsin(t) for sin(t) in order to use our trigonometric identity we established above. We find that cos(t) = x/a and sin(t) = y / b. Substituting those values into the trigonometric identity cos^{2}(t) + sin^{2}(t) = 1, we obtain the equation (x/a)^{2}+ (y/b)^{2}= 1. This is in fact the equation of an ellipse in standard form centered at the origin (0,0) with a vertical major axis and vertices at (0,-b) and (0,b). This ellipse has a vertical major axis because the b^{2}value is greater than the a^{2}value and b^{2}is the denominator under the y. In our case above, we have a = 2 and b = 5, thus the equation of this ellipse in standard form is (x/2)^{2}+ (y/5)^{2}= 1 which simplifies to (x^{2}/4) + (y^{2}/25) = 1. 25 is obviously greater than 4 and since 25 is under the y, the ellipse has the y- axis as its major axis.Now, let's examine what happens to the graph when a > b. Let's set a = 3 and b = 1. Observe the graph below.

As you can see from the graph above, when a > b we also have parameterization of an ellipse except this time the ellipse has a horizontal major axis instead of a vertical major axis. We can use the pythagorean trigonometric identity once again to find the equation of this ellipse in standard form. Solve for cos (t) in the equation x = acos(t) and solve for y in the equation y = bsin(t) to get cos(t) = x/a and sin(t) = y/b. Substitute the values into the trigonometric identity cos

^{2}(t) + sin^{2}(t) = 1 to obtain the equation (x/a)^{2}+ (y/b)^{2}= 1. Thus, in standard form, the equation of this ellipse is (x^{2}/a^{2}) + (y^{2}/b^{2}) = 1. Above, we have a = 3, and b = 1. Substituting these values in for a and b into the equation gives us a new equation x^{2}/3^{2}+ y^{2}/1^{2}= 1 which simplifies to x^{2}/9 + y^{2}/1 = 1. Since a^{2}> b^{2}and a^{2}is the denominator of x^{2}this determines that the ellipse has the x-axis as its major axis. Since the major axis is horizontal, the vertices of the ellipse will be located on the x-axis at the points (-a,0) and (a,0) or (-3,0) and (3,0). The y-intercepts are (0,-b) and (0,b) or (0,-1) and (0,1).From the two investigations above, you may be able to tell that the equations for the ellipses formed are very alike. However, when a > b, meaning a

^{2}> b^{2}, and the larger value a^{2}is the denonminator of the x^{2}term, the major axis of the ellipse will be horizontal along the x-axis.When a < b, meaning a^{2}< b^{2}, and the larger value b^{2}is the denominator of the y^{2}term, the major axis of the ellipse will be vertical along the y - axis. Therefore, the a and b values determine the major axis of an ellipse.Observe the graph below that shows many graphs on the same set of axes and observe the changes when a < b and when a > b.

For further investigation, let's explore the parametric equations x = cos(at) and y = sin(at) for various values of a and b. Can you predict what kind of graphs will be formed when a = b, when a >b, and when a < b? Observe the graphs and animations below.

When a = b, the graph is a circle.

View the animation below. The t-values range from 0 to 2π (6.28)

Now, let's observe what happens when a > b.

Observe when a = 2 and b = 1, a sideways parabola is formed. Why is this?

If you take the function y = sin(t) you can see that it equals the function y = sin(1

^{2}t). Now, solve for t in the function x = cos (2t) you get t = (cos^{-1}(x))/2. Then if you substitute this value of t into the y function y = sin(t) you get a new function y = sin^{2}((cos^{-1}(x))/2). To solve for y, you must take the square root of this function which leaves you with two values y = sin(cos^{-1}(x))/2) and y = - sin(cos^{-1}(x))/2). These two graphs make up the parabola shown above. Observe the graph below, "acos" stands for arc(cos) which is the inverse of the cosine function.Now, let's observe more graphs where a > b.

From these graphs, we can see that the period is increasing while the value of a increases. In addition, we can see that the cosine function is symmetric about the y - axis. This leads to a very important concept of the cosine function being an even function. cos (x) is an even function because cos (-x) = cos (x). Observe the animation below to see how the period of the graphs of the cosine function change when the parameter n gets smaller and larger. Why does the graph seem to be moving around the x-axis?

Next, let's observe what happens when a < b. Any hypotheses on what the graph will look like?

As you can see from the graphs, when a < b , it seems that the period of the sine function is getting larger. The sine function is an odd function because sin(-x) = -sin(x), and it's graph is symmetric to the origin. Observe the animation below when a is held constant and equal to 1, and the parameter n ranges from -10 to 10. Why does the graph seem to be moving around the origin in such a way?

For further investigation, try exploring the tangents of parametric curves, finding the area under parametric curves, and examine different parametric curves such as x = t

^{2}+ t and y = 2t - 1. Also, you can have your students explore the path of a particle given by a set of parametric equations, and have them describe the path of the particle, and give the range of t-values for which the path will be traced out exactly once.