This section covers:
 Introduction to Parametric Equations
 Parametric Equations in the Graphing Calculator
 Converting Parametric Equations to Rectangular: Eliminating the Parameter
 Simultaneous Solutions
 Applications of Parametric Equations
 Projectile Motion Applications
 Parametric Form of the Equation of a Line in Space
 More Practice
Introduction to Parametric Equations
So far, we’ve dealt with Rectangular Equations, which are equations that can be graphed on a regular coordinate system, or Cartesian Plane.
Parametric Equations are a little weird, since they take a perfectly fine, easy equation and make it more complicated. But sometimes we need to know what both x and y are, for example, at a certain time, so we need to introduce another variable, say t. The t is called the parameter. Parametric equations are also referred to as plane curves.
Here is a simple set of parametric equations that represent a cubic \(y={{x}^{3}}\) for t in [0, 3]: \(x\left( t \right)=t,\,\,\,\,y\left( t \right)={{t}^{3}},\,\,\,\,0\le t\le 3\).
Do you see how when we introduce the parametric variable t, we can see how the curve is being drawn for certain values of t? For example, for t = 0, we are at the point (0, 0), for t = 1, we are at the point (1, 1), for t = 2, we are at the point (2, 8), and so on. We can even put arrows on a graph to show the direction, or orientation of the set of parametric equations.
Here is a tchart and graph for this parametric equation. Note that the domain is the lowest x value to the highest x value, regardless what the value for t is. The range is the lowest y value to the highest y value, again regardless what the value for t is. The end points are the points with the lowest t value and the highest t value.
Here’s another that’s a little more complicated. Note how the range isn’t the same as the order of the y points:
Parametric Equations in the Graphing Calculator
We can graph the set of parametric equations above by using a graphing calculator:
First change the MODE from FUNCTION to PARAMETRIC, and enter the equations for X and Y in “Y =”.
For the WINDOW, you can put in the min and max values for t, and also the min and max values for x and y if you want to. Tstep will determine how many points are graphed; the smaller the Tstep, the more points will be graphed (smoother curve); you can play around with this. Then hit GRAPH to see the graph:
Let’s put Trigonometry parametric equations in the calculator. Make sure the calculator is in radians. I like to use a Tstep of \(\frac{\pi }{{12}}\), with t from 0 to 2π, and you might want to use ZOOM Zsquare to make the screen square. Note that when the coefficients of cos(t)and sin(t) are the same, we get a circle; we will show this below algebraically.
Converting Parametric Equations to Rectangular: Eliminating the Parameter
Sometimes we want to get a set of parametric equations back to its simplest form – without the parameter (usually if we don’t care about extra variable, which in many cases is time). There’s a trick to do this; you have to solve for t in one of the equations (typically the simplest one), and then plug what you get into the other equation, so you only are left with x’s and y’s. This new equation is called a rectangular equation.
When dealing with parametric equations with trig functions, and you have trig functions in both equations, you typically don’t want to solve for t, but solve for the trig functions with argument t. Then you can plug this expression in the other parametric equation and many times a Trigonometric Identity can be used to simplify. In these cases, we sometimes get equations for a circle, ellipse, or hyperbola (found in the Conics section). But if we don’t have the trig functions in both parametric equations, we’ll want to get the t by itself by taking the inverse of the trig function.
Here are some examples; let’s do problems without trig first. Do you see how our goal is to not have t in our equation at all?
Parametric Equations  Rectangular Equations 
Eliminate the parameter and describe the resulting equation: \(\left\{ \begin{array}{l}x=4t2\\y=2+4t\end{array} \right.\)  Solve for t in one of the equations and then substitute this in for the t in the other equation: \(\displaystyle \begin{align}x&=4t2\\x+2&=4t\\t&=\frac{{x+2}}{4}\end{align}\) Plug this into the second equation: \(\displaystyle \begin{array}{l}y=2+4t\\y=2+4\left( {\frac{{x+2}}{4}} \right)\,\\y=x+4\,\,\,\text{(}line\text{)}\end{array}\) 
Eliminate the parameter and describe the resulting equation: \(\left\{ \begin{array}{l}x=t3\\y={{t}^{2}}6t+9\end{array} \right.\)  Solve for t in the simplest equation and then substitute this in for the t in the other equation: \(\begin{array}{l}x=t3\\t=x+3\end{array}\) Plug this into the second equation: \(\begin{array}{l}y={{t}^{2}}6t+9\\y={{\left( {x+3} \right)}^{2}}6\left( {x+3} \right)+9\\y={{x}^{2}}+6x+96x18+9\\y={{x}^{2}}\,\,\,\text{(}parabola\text{)}\end{array}\) 
Eliminate the parameter and describe the resulting equation: \(\left\{ \begin{array}{l}x=5t\\y=3{{e}^{t}}\end{array} \right.\)  Solve for t in the simplest equation and then substitute this in for the t in the other equation: \(\begin{array}{l}x=5t\\t=\frac{x}{5}\end{array}\) Plug this into the second equation: \(\begin{array}{l}y=3{{e}^{t}}\\y=3{{e}^{{\frac{x}{5}}}}\,\,\,\text{(}exponential\text{)}\end{array}\) 
Eliminate the parameter and describe the resulting equation:
\(\left\{ \begin{array}{l}x=\sqrt{{t3}}\\y=2+t\end{array} \right.\)  Solve for t in the simplest equation and then substitute this in for the t in the other equation. Note that the second equation is simpler, since it doesn’t have the square root. \(\begin{array}{l}y=2+t\\t=y2\end{array}\) Plug this into the second equation: \(\begin{array}{l}x=\sqrt{{y3}}\\x=\sqrt{{\left( {y2} \right)3}}\\x=\sqrt{{y5}}\\{{x}^{2}}=y5\\y={{x}^{2}}+5\,\,\,\text{(}parabola\text{)}\end{array}\) 
Eliminate the parameter and describe the resulting equation:
\(\left\{ \begin{array}{l}x=3t\\y=2{{t}^{2}}1\end{array} \right.\)
\(t\,\,\text{on}\,\,\left[ {1,3} \right]\)
 Solve for t in the simplest equation and then substitute this in for the t in the other equation. Note that we are given an interval for t, so we are expected to find the domain and range for the rectangular equations. To find the domain and range, make a tchart:
Domain is\(\left[ {3,\,9} \right]\) and Range is \(\left[ {19,\,1} \right]\).
\(\begin{array}{l}t=\frac{x}{3},\,\,\,\,y=2{{\left( {\frac{x}{3}} \right)}^{2}}1\\y=\frac{{2{{x}^{2}}}}{9}1\,\,\,\,\,\,\text{(}parabola\text{)}\end{array}\) 
Here are more problems where you have to eliminate the parameter with trig. Notice that when we have trig arguments in both equations, we can sometimes use a Pythagorean Trig Identity to eliminate the parameter (and we end up with a Conic):
Parametric Equations  Rectangular Equation 
Eliminate the parameter and describe the resulting equation:
\(\left\{ \begin{array}{l}x=3\sin t+2\\y=3\cos t1\end{array} \right.\)
\(\displaystyle t\,\,\text{on}\,\,\left[ {\frac{\pi }{2},\frac{\pi }{2}\,} \right]\)
 Solve for \(\sin t\) in the first equation and \(\cos t\) in the second (since it’s too complicated to solve for t). Then use the Pythagorean identity \({{\sin }^{2}}t+{{\cos }^{2}}t=1\) and substitute:
\(\begin{align}x&=3\sin t+2\\3\sin t&=x2\\\sin t&=\frac{{x2}}{3}\end{align}\) \(\begin{align}y&=3\cos t1\\3\cos t&=y+1\\\cos t&=\frac{{y+1}}{3}\end{align}\) \(\begin{align}{{\sin }^{2}}t+{{\cos }^{2}}t&=1\\{{\left( {\frac{{x2}}{3}} \right)}^{2}}+{{\left( {\frac{{y+1}}{3}} \right)}^{2}}&=1\\{{\left( {x2} \right)}^{2}}+{{\left( {y+1} \right)}^{2}}&=9\,\text{ }\left( {circle} \right)\end{align}\)
This looks like a circle, but since the interval for t is \(\left[ {\frac{\pi }{2},\frac{\pi }{2}\,} \right]\), we have a semicircle. 
Eliminate the parameter and describe the resulting equation: \(\left\{ \begin{array}{l}x=4\sec t+1\\y=3\tan t\end{array} \right.\)  Solve for \(\sec t\) in the first equation \(\tan t\) and in the second. Then use the Pythagorean identity \({{\sec }^{2}}t{{\tan }^{2}}t=1\) and substitute: \(\begin{align}x&=4\sec t+1\\4\sec t&=x1\\\sec t&=\frac{{x1}}{4}\end{align}\) \(\begin{align}y&=3\tan t\\\tan t&=\frac{y}{3}\end{align}\) \(\begin{align}{{\sec }^{2}}t{{\tan }^{2}}t&=1\\{{\left( {\frac{{x1}}{4}} \right)}^{2}}{{\left( {\frac{y}{3}} \right)}^{2}}&=1\\\frac{{{{{\left( {x1} \right)}}^{2}}}}{{16}}\frac{{{{y}^{2}}}}{9}&=1\,\,\,\,\,\text{(}hyperbola\text{)}\end{align}\) 
Eliminate the parameter and describe the resulting equation: \(\left\{ \begin{array}{l}x=5\cot t4\\y=4\csc t\end{array} \right.\)  Solve for \(\cot t\) in the first equation and \(\csc t\) in the second. Then use the Pythagorean identity \({{\csc }^{2}}t{{\cot }^{2}}t=1\) and substitute: \(\begin{align}x&=5\cot t4\\5\cot t&=x+4\\\cot t&=\frac{{x+4}}{5}\end{align}\) \(\begin{align}y&=4\csc t\\\csc t&=\frac{y}{4}\end{align}\) \(\begin{align}{{\csc }^{2}}t{{\cot }^{2}}t&=1\\{{\left( {\frac{y}{4}} \right)}^{2}}{{\left( {\frac{{x+4}}{5}} \right)}^{2}}&=1\\\frac{{{{y}^{2}}}}{{16}}\frac{{{{{\left( {x+4} \right)}}^{2}}}}{{25}}&=1\,\,\,\,\,\text{(}hyperbola\text{)}\end{align}\) 
Eliminate the parameter: \(\left\{ \begin{array}{l}x=3t\\y=\sin t\end{array} \right.\)  Solve for t in the simplest equation and then substitute in the t in the other equation. Since we don’t have a trig function in both equations, we can’t use the Pythagorean identity like we did above.
\(\begin{align}x&=3t\\t&=\frac{x}{3}\end{align}\) Plug this into the second equation: \(\begin{align}y&=\sin t\\y&=\sin \left( {\frac{x}{3}} \right)\,\,\,\,\,\left( {sin\,graph} \right)\end{align}\)

Simultaneous Solutions
Sometimes we need to find the x and y coordinates of any intersections of parametric sets of equations. To do this, we want to set the x’s together and solve for t, and then set the y’s together and also solve for t. Where we have the same t when setting both the x and y equations together, we have an intersection. Then we have to put the t back in either x equation and either y equation to get the intersection.
Here are some examples; find the x and y coordinates of any intersections:
Applications of Parametric Equations
Parametric Equations are very useful applications, including Projectile Motion, where objects are traveling on a certain path at a certain time. Let’s first talk about Simultaneous Solution examples, though, where we might find out whether or not certain objects collide (are at the same place at the same time).
Simultaneous Solution Examples
Here is an example of type of Parametric Simultaneous Solution problem you might see:
Problem:
A hiker in the woods travels along the path described by the parametric equations \(\left\{ \begin{array}{l}x=80.7t\\y=.3t\end{array} \right.\). A bear leaves another area of the woods to the west and travels along the path described by the parametric equations \(\left\{ \begin{array}{l}x=.2t\\y=20+.1t\end{array} \right.\).
(a) Do the pathways of the hiker and the bear intersect?
(b) Does the hiker and bear collide?
Solution:
(a) The reason we might want to have the paths of the hiker and the bear represented by parametric equations is because we are interested in where they are at a certain time.
It appears that each of the set of parametric equations form a line, but we need to make sure the two lines cross, or have an intersection, to see if the paths of the hiker and the bear intersect. To do this, we’ll want to eliminate the parameter in both cases, by solving for t in one of the equations and then substituting this in for the t in the other equation. We can eliminate the parameter in this case, since we don’t care about the time.
We see that that the two lines are not parallel, so they must intersect! So that answer to (a) above is yes, the pathways of the hiker and bear intersect.
We can see where the two lines intersect by solving the system of equations (We could also find these intersections by putting the two equations in the graphing calculator without using parametrics).
Here’s what it looks like in a graphing calculator using parametric equations (I had to play around with the WINDOW to get the graph to display properly):
(b) We can’t really tell from the graph whether or not the hiker and and bear collide, although we might be able to by looking at the TABLE in the graphing calculator. But the easiest way would be to set the two x equations together, and then set the two y equations together, and see if we have the same t (like we did above in the Simultaneous Solutions section):
Solve for t by setting the “x” equations together, and do the same for the “y” equations. If any t values are the same in both, we have a solution; we then solve for x and y in either equation for that t.
Since the t values aren’t the same for the x and y, the hiker and bear won’t be at the same place at the same time. Whew!
Problem:
At noon, Julia starts out from Austin and starts driving towards Dallas; she drives at a rate of 50 mph. Marie starts out in Dallas and starts driving towards Austin; she leaves two hours later Julia (leave at 2pm), and drives at a rate of 60 mph. The cities are roughly 200 miles apart. When will Julia and Marie pass each other? How far will they be from Dallas when they pass each other?
Solution:
Let’s first draw this situation and then try to come up with a pair of parametric equations. Remember that distance = rate * time! Let’s let t = the time after noon, so Marie’s driving time is t, and Marie’s driving time is t – 2, since she leaves two hours later.
What we can do is make the x equations the distance from Austin for each of the girls, and the y equations the paths of the two girls, so we can randomly assign y = 0 to Julia, and y = 1 to Marie.
Note that since Marie starts out 200 miles away from Julia, and she’s is going “backwards” towards Austin, her distance is –60(t – 2) + 200, with relationship to Austin. This is because she starts out 200 miles from Austin, and every passing hour she’s 60 miles closer to Austin, so we subtract (put in real numbers to see how this works – the distance is getting smaller as time passes).
We’ll just set the two distances from Austin together (the x part of the equations) and solve for t to get the time that they meet, measured in the time from Julia leaving Austin (noon). We really don’t need to use the y equations, but it’s important to see how we can model a situation with them.
So in almost 3 hours (about 3pm), they will pass each other. They will be 50t miles from Austin, or about 145.5 miles from Austin. This will make them 200 – 145.5 = 54.5 miles from Dallas.
Projectile Motion Applications
Again, parametric equations are very useful for projectile motion applications.
With parametric equations and projectile motion, think of x as the distance along the ground from the starting point, y as the distance from the ground up to the sky, and t as the time for a certain x value and y value. This is called the trajectory, or path of the object.
If we remember from the Quadratic Applications section here (Quadratic Projectile Problem), we can define a parabolic curve of an object going up into the sky and back down as \(h\left( t \right)=16{{t}^{2}}+{{v}_{0}}t+{{h}_{0}}\), where, in simplistic terms, the \(16\) is the gravity (in feet per seconds per seconds), the \({{v}_{0}}\) is the initial velocity (in feet per seconds) and the \({{h}_{0}}\) is the initial height (in feet).
With a quadratic equation, we could also model the height of an object, given a certain distance from where it started.
Now we can model both distance and time of this object using parametric equations to get the trajectory of an object. Note that we’re using trigonometry again:
(Note that the y equation includes an initial height \({{h}_{0}}\); we assume the object starts at x = 0; if not, we have to add an initial value \({{x}_{0}}\) to the x equation).
These equations make sense since the horizontal and vertical distances use the “famous” equation distance = rate x time, where rate is the initial velocity at a certain angle, and time is t.
To solve these problems, we’ll typically want to use one equation first to get the time t, depending on what we know about either the distance from the starting point (x) or how high up the object is (y). We then want to see either how far away the object is from the starting point (x), or how high up it is (y).
When the problems ask how long the object is in the air, we typically want to set the y equation to 0, since this is when the ball is on the ground.
When the problem asks how far the object travels, we typically want to find when the ball hits the ground (from the y equation) and then plug that t into the x equation to see how far it traveled. The x part of the equation is typically linear.
When the problem asks the maximum height of the object, and when it hits that height, we typically want to find the vertex of the y equation, since this is the height curve (parabola) for the object. Remember that we can use \(\left( {\frac{b}{{2a}},\,\,f\left( {\frac{b}{{2a}}} \right)} \right)\) to find the vertex of the quadratic \(a{{x}^{2}}+bx+c\).
Here are some examples:
Here are some projectile motion problems with wind; we’ll have to add or subtract vectors. Remember that the wind against the object will have to subtracted from the x equation, and the wind in the same direction of the object will have to be added.
For any straight line wind (or if the wind is in a horizontal direction), we’ll use 0° for the trig arguments, since it comes straight across. When the wind is straight line, it turns out that we’re not adding or subtracting anything from the y equation, only the x.
There is information on the parametric form of the equation of a line in space here in the Vectors section.
Learn these rules, and practice, practice, practice!
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