This section covers:

 Introduction to Parametric Equations
 Parametric Equations in the Graphing Calculator
 Converting Parametric Equations to Rectangular: Eliminating the Parameter
 Finding Parametric Equations from a Rectangular Equation
 Simultaneous Solutions
 Applications of Parametric Equations
 Projectile Motion Applications
 Parametric Form of the Equation of a Line in Space
 More Practice
Note that there is information on the parametric form of the equation of a line in space here in the Vectors section.
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 \(\boldsymbol{t}\) (the parameter). Parametric equations are also referred to as plane curves. Note that \(t\) can be negative; this means before motion has started (I won’t get into the Physics part of parameters here!).
Here is a simple set of parametric equations that represent a cubic \(y={{x}^{3}}\) for \(t\) in \([0, 3]\): \(x\left( t \right)=t,\,\,y\left( t \right)={{t}^{3}},\,\,0\le t\le 3\). Many write this simply as \(x=t,\,\,y={{t}^{3}},\,\,0\le t\le 3\).
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, as well as some others. 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.
Note how the domains and ranges aren’t necessarily the same as the order of the points in the parametric tcharts, but they are always from low to high:
Parametric TChart  Parametric Graph  
\(\begin{array}{l}x=t\\y={{t}^{3}}\end{array}\) for \(t\) in \(\left[ {0,3} \right]\)
Domain: \(\left[ {0,3} \right]\) Range: \(\left[ {0,27} \right]\) End Points: \(\left( {0,0} \right)\) and \(\left( {3,27} \right)\) 

\(\begin{array}{l}x=t+2\\y={{t}^{2}}4\end{array}\) for \(t\) in \(\left[ {1,4} \right]\)
Domain: \(\left[ {2,3} \right]\) Range: \(\left[ {4,12} \right]\) End Points: \(\left( {3,3} \right)\) and \(\left( {2,12} \right)\) 

\(\begin{array}{l}x=42t\\y=\sqrt{{{{t}^{2}}+1}}\end{array}\) for \(t\) in \(\left[ {1,2} \right]\)
Domain: \(\left[ {0,6} \right]\) Range: \(\left[ {1,\sqrt{5}} \right]\) End Points: \(\left( {6,\sqrt{2}} \right)\) and \(\left( {0,\sqrt{5}} \right)\) 

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 \(\displaystyle \frac{\pi }{{12}}\), 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.
Note: If you are putting two sets of parametrics equations in the calculator and want to see if the paths collide at the same time, put the calculator in MODE SIMUL for simultaneous. You can put the Tstep in Window low, like 1, and you can the parametrics in “real time”.
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 \(t\) 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 (when trig functions are 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: \(\displaystyle \begin{align}y&=2+4t\\y&=2+4\left( {\frac{{x+2}}{4}} \right)\,\\y&=x+4\,\,\,\text{(line)}\end{align}\) 

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 equation: \(\begin{array}{l}y={{t}^{2}}6t+9\\y={{\left( {x+3} \right)}^{2}}6\left( {x+3} \right)+9\\y={{x}^{2}}+6x+96x18+9\\y={{x}^{2}}\,\,\,\text{(parabola)}\end{array}\) 

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: \(\displaystyle \begin{align}x&=5t\\t&=\frac{x}{5}\end{align}\) Plug this into the second equation: \(\begin{array}{l}y=3{{e}^{t}}\\y=3{{e}^{{\frac{x}{5}}}}\,\,\,\text{(exponential)}\end{array}\) 

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: \(\begin{array}{l}x=\sqrt{{y3}}\\x=\sqrt{{\left( {y2} \right)3}}\\x=\sqrt{{y5}}\\{{x}^{2}}=y5\\y={{x}^{2}}+5\,\,\,\text{(parabola)}\end{array}\) 

Eliminate the parameter and describe the resulting equation:
\(\displaystyle \left\{ \begin{array}{l}x=\sqrt{t}5\\y=\frac{1}{2}\log t\end{array} \right.\)
What is the \(y\)intercept, and what value of \(t\) does this occur? What is the \(x\)intercept, and what value of \(t\) does this occur? 
Solve for \(t\) in the simplest equation and then substitute this in for the \(t\) in the other equation. Note that the first equation is simpler, since it doesn’t have a log.
\(\displaystyle \begin{array}{l}x=\sqrt{t}5\\t={{\left( {x+5} \right)}^{2}}\end{array}\) Plug this into the second equation: \(\begin{array}{l}y=\frac{1}{2}\log {{\left( {x+5} \right)}^{2}}\\y=2\cdot \frac{1}{2}\log \left( {x+5} \right)\\y=\log \left( {x+5} \right);\,\,\,x>5\end{array}\)
The \(y\)intercept is when \(x=0\): \(y=\log \left( {0+5} \right);\,\,y=\log 5\). In this case, \(t={{\left( {0+5} \right)}^{2}}=25\), so the \(y\)intercept is \(\left( {0,25} \right)\). The \(x\)intercept is when \(y=0\): \(0=\log \left( {x+5} \right);\,\,{{10}^{0}}=x+5;\,\,1=x+5;\,\,x=4\). In this case, \(x=\sqrt{t}5;\,\,4=\sqrt{t}5;\,\,\sqrt{t}=1;\,\,t=1\), so the \(x\)intercept is \(\left( {1,0} \right)\). 

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.
\(\displaystyle t=\frac{x}{3},\,\,\,\,y=2{{\left( {\frac{x}{3}} \right)}^{2}}1\) \(\displaystyle y=\frac{{2{{x}^{2}}}}{9}1\,\,\,\,\,\,\text{(parabola)}\) 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 \(\left[ {19,1} \right]\). 
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):
Finding Parametric Equations from a Rectangular Equation
(Note that I showed examples of how to do this via vectors in 3D space here in the Introduction to Vector Section).
Sometimes you may be asked to find a set of parametric equations from a rectangular (cartesian) formula. This seems to be a bit tricky, since technically there are an infinite number of these parametric equations for a single rectangular equation. And remember, you can convert what you get back to rectangular to make sure you did it right!
If no other requirements are given (such as what \(t\) value ranges should be), the easiest way is to always let \(x\left( t \right)=t\) and then \(y\left( t \right)\) is just the function with a \(t\) instead of an \(x\). For example, one way to write equation \(y=4x5\) into a set of parametric equations is \(\left\{ \begin{array}{l}x\left( t \right)=t\\y\left( t \right)=4t5\end{array} \right.\). As another example, to convert \(f\left( x \right)=8{{x}^{2}}+4x2\) into a set of parametric equations, we have \(\left\{ \begin{array}{l}x=t\\y=8{{t}^{2}}+4t2\end{array} \right.\). (I purposely left out the \(t\)’s on the left side of the equations; you see parametrics written both way). Work these the other way (from parametric to rectangular) to see how they work! And remember that this is just one way to write the set of parametric equations; there are many!
You may also be asked come up with parametric equations from a rectangular equation, given an interval or range for \(t\), or write one from a set of points or a point and a slope (these last three would be linear).
Here are some examples:
Problem 
Solution 
Write a set of parametric equations for the line segment starting at \(\left( {1,2} \right)\) and ending at \(\left( {3,8} \right)\).
Graph: (The “regular” rectangular equation for this line is \(y=\frac{3}{2}x+\frac{7}{2}\) .) 
Since we are looking for a line segment, let’s find the set of parametric equations \(\left\{ \begin{array}{l}x\left( t \right)=at+b\\y\left( t \right)=ct+d\end{array} \right.\), where we need to find \(a,\,b,\,c,\) and \(d\).
Since we are starting at \(\left( {1,2} \right)\) and ending at \(\left( {3,8} \right)\), one way to create parametric equations is with \(t=0\) at the first point, and \(t=1\) at the second. Then we have: \(\left\{ \begin{array}{l}1=a\left( 0 \right)+b\\\,\,\,\,2=c\left( 0 \right)+d\end{array} \right.\), or \(b=1\) and \(d=2\). Now, the second point: \(\left\{ \begin{array}{l}3=a\left( 1 \right)1\\8=c\left( 1 \right)+2\end{array} \right.\), or \(a=4\) and \(c=6\).
The parametric equations are \(\displaystyle \left\{ \begin{array}{l}x\left( t \right)=4t1\\y\left( t \right)=6t+2\end{array} \right.\,\,\,\,\,\,0\le t\le 1\).
Easier way using vectors: Start with the point \(\left( {1,2} \right)\), and “add” the vector \(\left\langle {3–1,82} \right\rangle \) to get the next point, so we have \(\left\langle {1,2} \right\rangle +\left\langle {4,6} \right\rangle t=\left\langle {1+4t,\,2+6t} \right\rangle =\left\langle {4t1,\,6t+2} \right\rangle \). 
Write a set of parametric equations for the line segment starting at \(\left( {2,4} \right)\) with slope \(3\).
(The “regular” rectangular equation for this line is \(y=3x+2\) .) 
Since we are looking for a line segment, let’s find the set of parametric equations \(\left\{ \begin{array}{l}x\left( t \right)=at+b\\y\left( t \right)=ct+d\end{array} \right.\), where we need to find \(a,\,b,\,c,\) and \(d\).
Since we are starting at \(\left( {2,4} \right)\) and with slope \(3\), one line could have parametric equations with \(t=0\) at the first point, and \(t=1\) at the point where we go down \(3\) and over \(1\) (thus a slope of \(3\); this point is \(\left( {3,\,7} \right)\)). Then we have: \(\left\{ \begin{array}{l}\,\,\,\,2=a\left( 0 \right)+b\\4=c\left( 0 \right)+d\end{array} \right.\), or \(b=2\) and \(d=4\). Now, the second point: \(\left\{ \begin{array}{l}\,\,\,\,3=a\left( 1 \right)+2\\7=c\left( 1 \right)4\end{array} \right.\), or \(a=1\) and \(c=3\).
The parametric equations are .\(\displaystyle \left\{ \begin{array}{l}x\left( t \right)=t+2\\y\left( t \right)=3t4\end{array} \right.\,\,\,\,\,\,0\le t\le 1\).
Easier way using vectors: Start with the point \(\left( {2,\,4} \right)\), and “add” the vector \(\left\langle {1,\,3} \right\rangle \) (slope –3) to get to the next point, so we have \(\left\langle {2,\,4} \right\rangle +\left\langle {1,\,3} \right\rangle t=\left\langle {2+t,\,\,43t} \right\rangle =\left\langle {t+2,\,\,3t4} \right\rangle \). 
VECTOR METHOD EXAMPLE:
Write a set of parametric equations for the line segment between points \(\left( {2,6} \right)\) and \(\left( {4,–6} \right)\), so that when \(t=0\), we are at \(\left( {2,6} \right)\), and at \(t=2\), we are at \(\left( {4,–6} \right)\). 
Let’s do one more, using the vector method.
Start with the point \(\left( {2,\,6} \right)\), and “add” the vector \(\left\langle {42,66\,} \right\rangle \) to get the next point. Since that point is at \(t=2\), we’ll multiply the vector by \(\displaystyle \frac{t}{2}\) (not sure exactly how this works!), so we have \(\displaystyle \left\langle {2,\,6} \right\rangle +\left\langle {2,\,12} \right\rangle \frac{t}{2}=\left\langle {2+t,\,\,66t} \right\rangle =\left\langle {t+2,6t+6\,} \right\rangle \).
We have \(\displaystyle \left\{ \begin{array}{l}x\left( t \right)=t+2\\y\left( t \right)=6t+6\end{array} \right.\,\,\,\,\,\,0\le t\le 2\). Try it; it works! 
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 \(\boldsymbol{t}\) 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:
Simultaneous Solution Problem  Solution 
Find the \(x\) and \(y\) coordinates of any intersections:
\(\left\{ \begin{array}{l}x={{t}^{2}}+1\\y=5t+6\end{array} \right.\) \(\left\{ \begin{array}{l}x=2t\\y={{t}^{2}}\end{array} \right.\)

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 \(t\).
\(\begin{array}{c}{{t}^{2}}+1=2t\\{{t}^{2}}2t+1=0\\\left( {t1} \right)\left( {t1} \right)=0\\t=1\end{array}\) \(\begin{array}{c}5t+6={{t}^{2}}\\{{t}^{2}}+5t6=0\\\left( {t1} \right)\left( {t+6} \right)=0\\t=1,\,6\end{array}\) Since \(t=1\) works for both sets of equations, we have a solution! Plug in \(t\) for \(x\) and \(y\) in either set of equations to get \((2,1)\). 
Find the \(x\) and \(y\) coordinates of any intersections:
\(\displaystyle \left\{ \begin{array}{l}x=\sin \left( t \right)\\y=\frac{{\sin \left( {2t} \right)}}{2}\end{array} \right.\) \(\displaystyle \left\{ \begin{array}{l}x=\frac{{\sqrt{3}}}{2}\\y=\sqrt{3}{{\cos }^{2}}\left( t \right)\end{array} \right.\)

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 \(t\). Note that we used the identity \(\sin \left( {2t} \right)=2\sin \left( t \right)\cos \left( t \right)\) in the second set:
\(\require {cancel} \displaystyle \begin{array}{c}\sin \left( t \right)=\frac{{\sqrt{3}}}{2}\\t=\frac{\pi }{3},\,\,\frac{{2\pi }}{3}\end{array}\) \(\begin{array}{c}\frac{{\sin \left( {2t} \right)}}{2}=\sqrt{3}{{\cos }^{2}}\left( t \right)\\\frac{{\cancel{2}\sin \left( t \right)\cos \left( t \right)}}{{\cancel{2}}}=\sqrt{3}{{\cos }^{2}}\left( t \right)\\\sqrt{3}{{\cos }^{2}}\left( t \right)+\sin \left( t \right)\cos \left( t \right)=0\\\color{#2E8B57}{{\cos \left( t \right)}}\left( {\color{blue}{{\sqrt{3}\cos \left( t \right)+\sin \left( t \right)}}} \right)=0\end{array}\)
\(\begin{array}{c}\color{#2E8B57}{{\cos \left( t \right)}}=0\\\\\\t=\frac{\pi }{2},\,\,\frac{{3\pi }}{2}\,\end{array}\) \(\begin{array}{c}\color{blue}{{\sqrt{3}\cos \left( t \right)=\sin \left( t \right)}}\\\,\frac{{\sin }}{{\cos }}\left( t \right)=\sqrt{3}\\\,\tan \left( t \right)=\sqrt{3}\\t=\,\frac{{2\pi }}{3},\,\,\frac{{5\pi }}{3}\end{array}\) Since \(\displaystyle t=\frac{{2\pi }}{3}\) works for both sets of equations, we have a solution! Plug in \(t\) for \(x\) and \(y\) in either set of equations to get \(\displaystyle \left( {\frac{{\sqrt{3}}}{2},\,\,\frac{{\sqrt{3}}}{4}} \right)\). 
Find the \(x\) and \(y\) coordinates of any intersections:
\(\left\{ \begin{array}{l}x={{e}^{{2t}}}\\y=\ln t\end{array} \right.\) \(\left\{ \begin{array}{l}x={{e}^{{t4}}}\\y=2\ln \left( {t4} \right)\end{array} \right.\)

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 \(t\). Note that we used the quadratic formula in the second set:
\(\begin{array}{c}{{e}^{{2t}}}={{e}^{{t4}}}\\2t=t4\\t=4\end{array}\) \(\displaystyle \begin{array}{c}\ln t=2\ln \left( {t4} \right)\\\ln t=\ln {{\left( {t4} \right)}^{2}}\\t={{\left( {t4} \right)}^{2}}\\t={{t}^{2}}8t+16\\{{t}^{2}}9t+16=0\\t=\frac{{9\pm \sqrt{{814\left( 1 \right)\left( {16} \right)}}}}{2}=\frac{{9\pm \sqrt{{17}}}}{2}\end{array}\) Since we don’t have the same \(t\) value when we put the \(x\) equations together and the \(y\) equations together, we have no solution.

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, 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) Do 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.
Hiker: \(\displaystyle \begin{align}x&=80.7t\\.7t&=x80\\t&=\frac{{x80}}{{.7}}=\frac{{80x}}{{.7}}\end{align}\) Plug this into the second equation: \(\displaystyle \begin{align}y&=.3t\\y&=.3\left( {\frac{{80x}}{{.7}}} \right)\,\\y&=\frac{3}{7}x+\frac{{240}}{7}\,\,\,\,\,\text{(a line)}\end{align}\)
Bear: \(\displaystyle \begin{align}x&=.2t\\t&=\frac{x}{{.2}}=5x\end{align}\) Plug this into the second equation: \(\displaystyle \begin{array}{l}y=20+.1t\\y=20+.1\left( {5x} \right)\\y=.5x+20\,\,\,\,\,\,\text{(a line)}\end{array}\)
We see that that the two lines are not parallel, so they must intersect! Thus, the 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:
\(\displaystyle \left\{ \begin{array}{l}y=\frac{3}{7}x+\frac{{240}}{7}\\y=.5x+20\end{array} \right.;\,\,\,\,\frac{3}{7}x+\frac{{240}}{7}=.5x+20;\,\,\,\,x=\frac{{200}}{{13}};\,\,y=\frac{{360}}{{13}}\,\,\,\,\,\,\,\,\text{Intersection:}\,\,\left( {\frac{{200}}{{13}},\,\,\frac{{360}}{{13}}} \right)\)
(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 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 \(t\).
\(\begin{array}{c}80.7t=.2t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.3t=20+.1t\\\,\,\,\,\,80=.9t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.2t=20\,\,\,\,\,\,\,\,\\t=88.89\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=100\end{array}\)
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 (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 \(\text{distance}=\text{rate}\times \text{time}\). Let \(t=\) the time they travel (starting at noon), so Julia’s driving time is \(t\), and Marie’s driving time is \(t2\), since she leaves two hours later. Julia’s distance from Austin is \(50t\), and Marie’s distance from Dallas is \(60\left( {t2} \right)\).
We can 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’s distance from Dallas is \(60\left( {t2} \right)\), her distance from Austin is \(20060\left( {t2} \right)\), since both the distances add up to 200 miles.
We’ll 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.
In almost 3 hours from noon (about 3pm), they will pass each other. They will be \(50t\) miles from Austin, or about 145 miles from Austin. This will make them about 200 – 145 = 54 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; don’t confuse these two types of models.)
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 \({{h}_{0}}\); also note that 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 \(\text{distance}=\text{rate}\times \text{time}\), where rate is the initial velocity at a certain angle, and time is \(x\).
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. Then we can solve for \(t\), or time.
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 \(\displaystyle \left( {\frac{b}{{2a}},\,\,f\left( {\frac{b}{{2a}}} \right)} \right)\) to find the vertex of the quadratic \(a{{x}^{2}}+bx+c\).
Here are some examples:
Projectile Motion Problems  Solutions 
The following parametric equations model the path of a soccer ball, where \(t\) is in seconds, and distances are in feet:
\(\left\{ \begin{array}{l}x\left( t \right)=120t\\y\left( t \right)=64t16{{t}^{2}}\end{array} \right.\)
(Notice that \({{h}_{0}}=0\), since the ball starts from the ground).
(a) When will the ball hit the ground? How far does the ball travel? (b) Find the maximum height of the ball. When does this occur? 
(a) The ball hits the ground when the height of the ball is 0; this is when the \(y\) equation equals 0. Notice that it is also at the ground at 0 seconds (before it leaves the ground).
\(\displaystyle \begin{array}{c}64t16{{t}^{2}}=0\\16t\left( {4t} \right)=0\end{array}\) \(\displaystyle \begin{array}{c}16t=0\,\,\,\,\,\,\,\,\,\,4t=0\\t=0, 4\end{array}\) The ball hits the ground after 4 seconds. To get how far the ball travels, we plug this value into the \(x\) equation, which is distance: \(x\left( 4 \right)=120\left( 4 \right)=480\). Therefore, the ball travels 480 feet before it hits the ground again.
(b) The maximum height of the ball occurs at the vertex of the height curve of the ball (\(y\)). Using the standard equation \(y=a{{x}^{2}}+bx+c\), we can use \(\displaystyle \frac{b}{{2a}}\) to find the \(t\) part of the equation (when the maximum height occurs), and then use this value to get the \(y\) (the actual maximum height): \(\displaystyle \frac{b}{{2a}}=\frac{{64}}{{2\left( {16} \right)}}=2\,\,\,\,\,\,\,\,y\left( 2 \right)=64\left( 2 \right)16{{\left( 2 \right)}^{2}}=\,\,64\) The maximum height of the ball is 64 feet, and this happens 2 seconds after it leaves the ground the first time. This makes sense since the ball starts from the ground, and this is half the time for the ball to hit the ground again. 
Jillian, a pro golfer, hits the golf ball with an initial velocity of 50 ft./sec and at an angle of 35° from the horizon.
(a) Find when and where the ball will hit the ground. (b) Find the maximum height of the ball. When does this occur? 
First, let’s set up the parametric equations that models the distance (\(x\)) and height (\(y\)) at a time \(t\): \(\left\{ \begin{array}{l}x\left( t \right)=\left( {{{v}_{0}}\cos \alpha } \right)t\\y\left( t \right)={{h}_{0}}+\left( {{{v}_{0}}\sin \alpha } \right)t16{{t}^{2}}\end{array} \right.\) or \(\left\{ \begin{array}{l}x\left( t \right)=\left( {50\cos 35} \right)t\\y\left( t \right)=0+\left( {50\sin 35} \right)t16{{t}^{2}}\end{array} \right.\) (a) The ball hits the ground when the height of the ball is 0; this is when the \(y\) equation equals 0. Notice that it is also at the ground at 0 seconds (this makes sense). \(\begin{array}{c}\left( {50\sin 35} \right)t16{{t}^{2}}=0\\t\left( {50\sin \left( {35} \right)16t} \right)=0\end{array}\) \(\begin{array}{c}t=0\,\,\,\,\,\,\,\,50\sin \left( {35} \right)16t=0\\t=0, \approx 1.792\end{array}\) The ball hits the ground in 1.792 seconds. To get how far the ball travels, we plug this value into the \(x\) equation, which is distance: \(x\left( {1.792} \right)=\left( {50\cos 35} \right)\left( {1.792} \right)\approx 73.396\). The ball travels about 73.4 feet.
(b) The maximum height of the ball occurs at the vertex of the height curve of the ball (\(y\)). We can use \(\displaystyle \frac{b}{{2a}}\) to find the \(t\) part of the equation (when the maximum height occurs), and then use this value to get the \(y\) (the actual maximum height): \(\displaystyle \frac{b}{{2a}}=\frac{{50\sin 35}}{{2\left( {16} \right)}}\approx .896\) \(\displaystyle y\left( {.896} \right)=50\sin 35\left( {.896} \right)16{{\left( {.896} \right)}^{2}}\approx 12.851\) The maximum height of the ball is 12.851 feet, and this happens .896 seconds after it leaves the ground the first time. This makes sense since the ball starts from the ground, and this is half the time for the ball to hit the ground again. 
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 straightline wind (or if the wind is in a horizontal direction), we can use 0° (or 180°, depending on the direction) 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 \(x\).
Problems  Solutions 
Lisa hits a golf ball off the ground with a velocity of 60 ft./sec at an angle of 45°. The wind is blowing against the path of the ball at 10 ft./sec with an angle of depression of 15°.
(a) Write a set of parametric equations to model this situation.
(b) How long does the ball stay in the air (hang time)?
(c) How far does Lisa hit the ball (in the horizontal direction)? 
(a) Note that wind at an angle of depression affects both the \(x\) and \(y\) equations, and since it is blowing against the path of the ball, we need to subtract the wind from both equations:
\(\displaystyle \left\{ \begin{array}{l}x\left( t \right)=\left( {60\cos 45} \right)t\left( {10\cos 15} \right)t\\y\left( t \right)=\left( {60\sin 45} \right)t16{{t}^{2}}\left( {10\sin 15} \right)t\end{array} \right.\) (If the wind were blowing in the same direction as the ball, we would add to both).
(b) To find out how long the ball stays in the air, we have to set the \(y\) equation to 0 and solve for \(t\). (This is because when the \(y\) equation hits 0, the ball hits the ground again.) To solve, either use quadratic formula, or put in graphing calculator (degree mode): \(\begin{array}{c}\left( {60\sin 45} \right)t16{{t}^{2}}\left( {10\sin 15} \right)t=0\\42.4264t16{{t}^{2}}2.5882t=0\\16{{t}^{2}}+39.8382t=0\\t=0,\,\approx 2.49\,\,\sec \end{array}\) (c) To find how long the ball stays in the air, we’ll use the positive \(t\) that we just got in the \(x\) equation: \(\displaystyle \left( {60\cos 45} \right)\left( {2.49} \right)\left( {10\cos 15} \right)\left( {2.49} \right)\approx 81.59\). Lisa hits the golf ball 81.59 feet in the horizontal direction. Pretty good! 
Jade, a pro softball player, hits the ball when it is 3 feet off the ground with an initial velocity of 100 ft./sec and at an angle of 35° from the horizon. A straightline wind is blowing at 14 ft./sec in the direction the ball is traveling.
She hits the ball towards a 40foot fence that is 120 feet from the plate; if it clears this fence, the ball is a home run.
(a) Will the hit be a home run?
(b) If so, how much does it clear the fence; if not, how much does it miss the fence?

First, let’s set up the parametric equations that models the distance (\(x\)) and height (\(y\)) at a time \(t\):
\(\displaystyle \left\{ \begin{array}{l}x\left( t \right)=\left( {100\cos 35} \right)t+\left( {14\cos 0} \right)t\\y\left( t \right)=3+\left( {100\sin 35} \right)t16{{t}^{2}}+\left( {14\sin 0} \right)t\end{array} \right.\,\) or \(\displaystyle \left\{ \begin{array}{l}x\left( t \right)=\left( {100\cos 35} \right)t+14t\\y\left( t \right)=3+\left( {100\sin 35} \right)t16{{t}^{2}}\end{array} \right.\)
Note we had to add the wind expressions, and use \(\cos \left( 0 \right)\) and \(\sin \left( 0 \right)\), since the wind is blowing in the direction of the ball, and it’s a straightline wind. Also note that \(\cos \left( 0 \right)=1\) and \(\sin \left( 0 \right)=0\) (so we’re not adding any wind to the vertical equation, which makes sense). Note also that we had to add the initial height of 3. (If the wind were blowing in against the ball, we would subtract it).
(a) We need to find the height of the ball when it is 120 feet from the plate; if this height is over 40 feet, then the ball will be home run. Let’s plug in 120 for the \(x\) and solve back for \(t\): . \(\displaystyle \begin{align}120&=\left( {100\cos 35} \right)t+14t\\120&\approx 95.92t\\t&\approx 1.25\end{align}\) At \(t=1.25\) seconds, the ball reaches the fence horizontally. Now let’s see how tall the ball is at this time (height); we’ll use the \(y\left( {1.25} \right)=3+\left( {100\sin 35} \right)\left( {1.25} \right)16{{\left( {1.25} \right)}^{2}}\approx 49.7\). At 120 feet from the home plate, the ball has a height of about 49.7 feet.
(b) Since the ball has a height of 49.7 feet when it is 120 feet from the home plate, it will clear the 40foot fence by 9.7 feet; Jade will make a home run!

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