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),  y = g(t)  tD. where D is a set of real numbers. The variable t is called a parameter and the relations between x, y and t are called parametric equations. The set D is called the domain of f and g and it is the set of values t takes.

Conversely, given a pair of parametric equations with parameter t, the set of points (f(t), g(t)) form a curve in the plane.

As an example, the graph of any function can be parameterized. For, if y = f(x) then let t = x so that

Example

The parametric equations x = t,   y = t2 are an example of how to parameterize the graph of the function y=x2. We arrived at this pair of parametric equations as described above.

In the last example there were no restrictions on the parameter. The curve defined parametrically by the equations was identical to the graph of the function. Note that by restricting the values of the parameter, we can parameterize part of the graph of the function.

Example

The parametric equations x = t,   y = t2; t [-1,2] are an example of how to parameterize part of the graph of the function y=x2. The part of the graph we get is from x=-1 to x=2.

There are several techniques we use to sketch a curve generated by a pair of parametric equations. The simplest is to evaluate f(t) and g(t) for several values of t. We then plot the points (f(t), g(t)) in the plane and through them draw a smooth curve (assuming this is valid!!!). This idea of plotting points is identical to the elementary graphing techniqus of graphing functions and is illustrated in the following two diagrams.

In this example the parametric equations are x = 2t and y = t² and we have evaluated t at -2, -1.5, -1, -0.5, -0.25, 0, 0.25, 0.5, 1.5 and 2. We have determined the corresponding values of x and y and plotted these points. The result is shown in the first diagram. Through these points we have drawn a smooth curve and the result is shown in the second diagram.

The orientation of a parameterized curve is the direction determined by increasing values of the parameter. Sometimes arrows are drawn on the curve to denote the orientation. The diagram shows the same parametric curve we have just studied where we have included some arrows to illustrate the orientation. In this case, the direction of t increasing is from left to right.

Plotting a parametric curve

You should try this technique. In this exercise, you are given a pair of parametric equations. Select 6 valid values of t and substitute each into both equations to get the coordinates of a point in the plane. Plot all six points in the plane. The "Delete" button deletes the points in reverse order. The "Delete all" button will remove all the points you have plotted. Once you are satisfied, press the "Check" button. If you have at least two points correct you can either correct the points or try sketching the graph. To sketch the graph, hold down the mouse button and drag the mouse. Finally, press "Graph" and you will be shown the correct graph.
You are also asked to provide the orientation of the curve. Make the best selection of the four choices. You should only choose clockwise and anticlockwise for circles and ellipses. All other curves have orientation which can be described as left to right or right to left.

A second technique to identifying the curve of the parametric equations is to try to eliminate the parameter from the equations. This will result in an equation involving only x and y which we may recognize. For example, let's look again at the previous example. Since x = 2t then solving for t gives t = x/2. Substitute for t into the equation for y to get y = (x/2)² = x²/4. This we recognize as the graph of a parabola and we can sketch its graph usinfromg function graphing techniques.
Be careful, however, to take into account any restrictions on the value of the parameter. If, in this example we add the condition t > 0, then the curve defined by the parametric equations would be the graph of y=x²/4 on the positive x axis.

Eliminating the parameter

In this set of exercises you are given parametric equations. You are to eliminate the parameter and find an expression between y and x. All answers in this set can be written in the form y=f(x).

Once you have entered the expression, press CHECK to see if you have obtained the correct answer.

A parametric representation of a curve is not unique. That is, a curve C may be represented by two (or more) different pairs of parametric equations.

Example

We saw earlier that the parametric equations x = t,   y = t2; t [-1,2] parameterize part of the graph of the function y=x2. from x=-1 to x=2. However, the equations x = 1-2u,   y = (1-2u)2; u [-1/3,1] also parameterize the same part of the graph of the function y=x2. from x=-1 to x=2.

Example

Consider x = cos(t),  y = sin(t); t [0, 2]. To see what curve these equations define let's square and add x and y. x2 + y2 = cos2(t) + sin2(t) = 1. This is the equation of the unit circle and so we have found a parameterization of the unit circle. Now, consider x = sin(t),  y = cos(t); t [0, 2]. We do the same trick to eliminate the parameter, namely square and add x and y. x2 + y2 = sin2(t) + cos2(t) = 1. This parametric curve is also the unit circle and we have found two different parameterizations of the unit circle.

A thing to note in this previous example was how we obtained an equation in terms of x and y. We eliminated the parameter but not in a direct way by solving one of the equations. This technique is useful in many parametric equations involving sine and cosine. That is, solve the equations for sin(t) and cos(t) then square and add the equations.

One nice interpretation of parametric equations is to think of the parameter as time (measured in seconds, say) and the functions f and g as functions that describe the x and y position of an object moving in a plane. We give four examples of parametric equations that describe the motion of an object around the unit circle.

Example

The parametric equations x = sin(t),  y = cos(t); t [0, 2] for example, describe an object moving around the unit circle. If t=0, we think of the object as starting its journey at the point (0,1) in the plane and if t=2 we think of the object ending its journey at (0,1) in the plane. The orientation of the curve represents the direction of travel of the object. In this case, the object is moving around the circle in a clockwise direction. It takes the object 2 seconds to travel around the circle.

Example

The parametric equations x = sin(t),  y = cos(t); t > 0 could describe an object moving around the unit circle indefinitely (i.e. making infinitely many revolutions). It would take the object 2 seconds to complete a revolution.

Example

The parametric equations x = sin(2t),  y = cos(2t); t [0, 2] describe an object moving twice around the unit circle. At t=0, the object starts its journey, at t= the object has made one revolution, and at t=2 the object ends its journey. It takes the object seconds to travel around the circle.

Example

The parametric equations x = cos(t),  y = sin(t); t [0, 2] describe an object moving around the unit circle in an anticlockwise direction. The object starts and ends at (1, 0) in the plane. It takes the object 2 seconds to travel around the circle.

We illustrate these four examples with the applets below. Click the start on each one to see the motion we just described.

A projectile is an object moving only under the force of gravity such as when we throw a basketball, or fire a missile or water the garden. In modeling the motion of a projectile we assume no air resistance. In this case, the projectile follows a parabolic path.

In order to obtain parametric equations that describe the motion of a projectile it is useful to set up a rectangular coordinate system with the positive x-axis along the (horizontal) ground and in the direction we will send the projectile and the y-axis perpendicular to the ground. With that done, if an object is projected at an angle to the ground with an initial speed V, then the path of the projectile is given by the parametric equations

x = Vcos()t,   y = Vsin()t-gt²/2

x = Vcos()t,   y = Vsin()t-4.9t².

x = Vcos()t,   y = Vsin()t-16t².

From these equations we can find how long the object is in the air (time of flight), how far horizontally the object will travel (range of the projectile) and the maximum height reached by the object.

Time of flight

The projectile reaches the ground at y=0. Solving Vsin()t-gt2/2=0 gives t= 2Vsin()/g (note we have ignored the solution t=0). That is, the time of flight is 2Vsin()/g.

Range

The range is the value of x when the projectile reaches the ground. Since we have just found the time of flight, we substitute this time of flight into the equation connecting x and t to get Range = Vcos()2Vsin()/g = 2V2sin()cos()/g = V2[2sin()cos()]/g = V2sin(2)/g.

Maximum height

By the symmetry of the path of the object, you can intuitively reason that the projectile reaches its greatest height at one-half of the time of flight. That is when t=Vsin()/g. If we substitute this value of t into the equation for y we obtain yMax = Vsin()Vsin()/g - gVsin()Vsin()/2g2 =V2sin2()/2g.

We summarize these formulas for the general case and for the typical system of units.

	Time of flight	Range	Max Height
General	2Vsin()/g.	V2sin(2)/g.	V2sin2()/2g
Metric	Vsin()/4.9	V2sin(2)/9.8	V2sin2()/19.6
Feet	Vsin()/16	V2sin(2)/32	V2sin2()/64

Example

A projectile is fired at an angle of /4 to the horizontal with an initial speed of 100 m/s. The range is 10,000 sin(/2)/9.8 which is approximately 1,020 meters. The time of flight is 100sin(/4)/4.9 which is approximately 14.4 seconds. The maximum height is 10,000sin2 (/4)/19.6 which is approximately 255 meters.

Projectiles

In this exercise you are required to find the range or the time of flight of a projectile. You will be given the initial speed and the angle at which the projectile is fired relative to the horizontal.

Your answer should be rounded to two decimal places.