**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 ** t-chart** 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 **t**-charts, but they are always from **low to high**:

## 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 **Tmin** and **Tmax** values for $ t$, and also the **Xmin** and **Xmax** 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:

You can also put **trigonometric** parametric equations in the calculator. Make sure the calculator is in **radians**, and I like to use a **Tstep** of $ \displaystyle \frac{\pi }{{12}}$, with $ t$ from **0** to **2 π. **You might also 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 parametric equations** in the calculator and want to see if the paths collide at the same time, put the calculator in **mode SIMUL** (instead of SEQUENTIAL) for simultaneous. You can make the **Tstep** in **window** low, like **1**, and you can see the the parametric graphed “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$ 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 **trig** **parametric equations **(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 (Pythagorean) **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?

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=4x-5$ into a set of parametric equations is $ \left\{ \begin{array}{l}x\left( t \right)=t\\y\left( t \right)=4t-5\end{array} \right.$. As another example, to convert $ f\left( x \right)=8{{x}^{2}}+4x-2$ into a set of parametric equations, we have $ \left\{ \begin{array}{l}x=t\\y=8{{t}^{2}}+4t-2\end{array} \right.$. (I purposely left out the $ t$’s on the left side of the equations; you see parametric equations 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 (which would be linear).

## 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:

## 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 are some examples of the type of **Parametric** **Simultaneous Solution** problems you may 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, **eliminate the parameter** in both cases, by solving for $ t$ in one of the equations and then substituting 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! 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 \begin{array}{l}\,\,\,\,\,\,\,\left\{ \begin{array}{l}y=\displaystyle -\frac{3}{7}x+\displaystyle \frac{{240}}{7}\\y=.5x+20\end{array} \right.;\,\,\,\,\,\displaystyle -\frac{3}{7}x+\displaystyle \frac{{240}}{7}=.5x+20;\,\,\,\,\\\\\,\,x=\displaystyle \frac{{200}}{{13}};\,\,\,\,\,y=\displaystyle \frac{{360}}{{13}}\,\,\,\,\,\,\,\,\text{Intersection:}\,\,\left( {\displaystyle \frac{{200}}{{13}},\,\,\displaystyle \frac{{360}}{{13}}} \right)\end{array}$

(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. The easiest way is 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; 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\approx 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 **2**pm), 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:** 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 $ t-2$, since she leaves two hours later.

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. Julia’s distance from Austin is $ 50t$, and Marie’s distance from **Dallas** is $ 60\left( {t-2} \right)$. Thus, her distance from **Austin** is $ 200-60\left( {t-2} \right)$. (Put real numbers in: if Marie is **50** miles from Dallas, she is $ 200-50=150$ miles from Austin.

To see when (and where) they meet, 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 about **2.9 hours** from noon (about **3**pm), they will pass each other. They will be about $ 50t=50\left( {2.9} \right)\approx 145$ miles from Austin. This will make them about $ 200-45=155$ miles from Dallas.

## Projectile Motion Applications

Parametric equations are also 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, $ -16$ is the **gravity** (in feet per seconds per seconds), $ {{v}_{0}}$ is the **initial velocity** (in feet per seconds) and $ {{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$).

Remember these rules:

- 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 (the non-zero) $ t$, or time. - When the problem asks
**how far the object travels**, we typically want to find when the ball hits the ground (setting the $ y$ equation to**0**) 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:

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 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$.

**Learn these rules, and practice, practice, practice!**

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