## 17Calculus - Parametric Equations

##### 17Calculus

Parametrics is a way to plot a graph by specifying the x, y (and z for three dimensions) values separately. We do so using a parameter, another variable used to link equations together. It is simple but deceptively powerful.
If you want a full-length lecture on this topic, we recommend this video.

### Prof Leonard - Introduction to Parametric Equations [1hr-38min-25secs]

video by Prof Leonard

For starters we will use the parameter t. Here is an example.

 $$x = \cos(t)$$ $$y = \sin(t)$$

Notice that, although we have two separate equations, the x and y are linked by the value of t. In this example, we take a value for the parameter t and plug that value into both equations, to get a corresponding point $$(x,y)$$. When $$t=0$$, we get $$x=\cos(0)=1$$ and $$y=\sin(0)=0$$. Writing this as an ordered pair, we have $$(x,y)=(1,0)$$. So, the point $$(1,0)$$ corresponds to the value $$t=0$$.
Before we go on, let's watch a good introductory video.

### Michel vanBiezen - What is a Parametric Equation?

video by Michel vanBiezen

One of the advantages of using parametric equations is that we can describe many more graphs than we could when we had only x and y. Also, we are not going to require the graphs to be functions now, i.e. they may not pass the vertical line test.

Another piece of information we get from parametric equation is direction. The use of the variable t as a parameter is not random. Often, we assign a meaning to the parameter and sometimes that meaning is time. When you graph a set of parametric equations, the graph is swept out in a certain direction. This is an inherent feature of the parametric equations. We will often start at $$t=0$$ and increase t, giving the idea that time is passing. By adjusting the parametric equations, we can reverse the direction that the graph is swept.

Graphing Parametric Equations

As you learned when graphing functions, you need some experience with the equations and what the graphs look like in order to be able to know what a graph looks like just from the equation. As you are learning, I suggest using winplot to graph parametric equations. It is easy to use, has a very short learning curve and, best of all, it's free. You can find out more information on the Tools page.
Here is a video showing how to plot parametric equations.

### MIP4U - Introduction to Parametric Equations

video by MIP4U

Eliminating The Parameter

Sometimes we are given the set of parametric equations and we are asked to write the equation without the parameter by eliminating the parameter. This not always possible but with some equations there are ways to do it.

The easiest technique to try is to solve one of the equations for the parameter and then substitute the result in the other equation. Here is an example.
$$x = 2t$$ and $$y = t^2$$
Solve the first equation for t giving $$t=x/2$$ and substitute into the second equation.
$$y = (x/2)^2 = x^2/4$$

A second technique involves the use of trig identities. For example, given the parametric equations $$x=\cos(t)$$ and $$y=\sin(t)$$, we know that $$\cos^2(x) + \sin^2(x) = 1$$. So we can write $$x^2 + y^2 = 1$$ which eliminates the parameter.

A third way is by inspection. Sometimes it is obvious what a substitution might be. For example, if we have $$x=e^t$$ and $$y=e^{3t}+1$$ then we can rewrite $$y=(e^t)^3+1$$. Then we can replace $$e^t$$ with $$x$$ in the last equation (because our first parametric equation was $$x=e^t$$) to get $$y=x^3+1$$. Of course, the first technique would have worked too by solving $$x=e^t$$ for $$t$$ and then substituting and simplifying but by standing back and looking at the equations more carefully, the solution was much easier.
Okay, time for some practice problems.

Practice

Basic

Eliminate the parameter $$t$$ from $$x=3t$$ and $$y=t^2$$.

Problem Statement

Eliminate the parameter $$t$$ from $$x=3t$$ and $$y=t^2$$.

Final Answer

$$y = x^2/9$$

Problem Statement

Eliminate the parameter $$t$$ from $$x=3t$$ and $$y=t^2$$.

Solution

Solve the first equation for t giving $$t=x/3$$ and substitute into the second equation.
$$y = (x/3)^2 = x^2/9$$

Final Answer

$$y = x^2/9$$

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Eliminate the parameter $$t$$ from $$x=3\cos(t)$$ and $$y=2\sin(t)$$.

Problem Statement

Eliminate the parameter $$t$$ from $$x=3\cos(t)$$ and $$y=2\sin(t)$$.

Final Answer

$$\displaystyle{ \frac{x^2}{9} + \frac{y^2}{4} = 1 }$$

Problem Statement

Eliminate the parameter $$t$$ from $$x=3\cos(t)$$ and $$y=2\sin(t)$$.

Solution

We will use the trig identity $$\cos^2(t) + \sin^2(t) = 1$$.

$$x = 3\cos(t) \to x/3 = \cos(t)$$

$$y = 2\sin(t) \to y/2 = \sin(t)$$

$$(x/3)^2 + (y/2)^2 = 1$$

Final Answer

$$\displaystyle{ \frac{x^2}{9} + \frac{y^2}{4} = 1 }$$

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Eliminate the parameter t from $$x=5\cos(t)$$ and $$y=3\sin(t)$$.

Problem Statement

Eliminate the parameter t from $$x=5\cos(t)$$ and $$y=3\sin(t)$$.

Final Answer

$$9x^2 + 25y^2 = 225$$

Problem Statement

Eliminate the parameter t from $$x=5\cos(t)$$ and $$y=3\sin(t)$$.

Solution

In the video, she shows a really difficult way to solve this problem. Her solution is correct but it is overly tedious and unnecessarily complicated. Here is an easier and more common solution.
In this problem, you are given that $$x=5\cos(t)$$ and $$y=3\sin(t)$$. We solve each of these for the sine and cosine terms since we use $$\cos^2(t)+\sin^2(t)=1$$. So we have $$x/5=\cos(t) \to (x/5)^2=\cos^2(t)$$ and $$y/3=\sin(t) \to (y/3)^2=\sin^2(t)$$.
$$\begin{array}{rcl} 1 & = & \cos^2(t)+\sin^2(t) \\ 1 & = & (x/5)^2+(y/3)^2 \\ 1 & = & x^2/25+y^2/9 \\ 225 & = & 9x^2 + 25y^2 \end{array}$$

### Krista King Math - 719 video solution

video by Krista King Math

Final Answer

$$9x^2 + 25y^2 = 225$$

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Eliminate the parameter from the parametric equations $$x=t-1$$, $$y=4t+1$$.

Problem Statement

Eliminate the parameter from the parametric equations $$x=t-1$$, $$y=4t+1$$.

Solution

### PatrickJMT - 48 video solution

video by PatrickJMT

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Eliminate the parameter from the parametric equations $$x=\sqrt{t+1}$$, $$y=3t+2$$.

Problem Statement

Eliminate the parameter from the parametric equations $$x=\sqrt{t+1}$$, $$y=3t+2$$.

Solution

### PatrickJMT - 49 video solution

video by PatrickJMT

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Graph the parametric equations $$x=1+\sqrt{t}$$, $$y=t^2-4t$$ on $$0\leq t\leq4$$.

Problem Statement

Graph the parametric equations $$x=1+\sqrt{t}$$, $$y=t^2-4t$$ on $$0\leq t\leq4$$.

Solution

### PatrickJMT - 51 video solution

video by PatrickJMT

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Graph the parametric equations $$x=\sqrt{t}$$, $$y=1-t$$ and eliminate the parameter.

Problem Statement

Graph the parametric equations $$x=\sqrt{t}$$, $$y=1-t$$ and eliminate the parameter.

Solution

### PatrickJMT - 52 video solution

video by PatrickJMT

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Eliminate the parameter t from $$x=\ln t$$ and $$y=\sqrt{t}$$.

Problem Statement

Eliminate the parameter t from $$x=\ln t$$ and $$y=\sqrt{t}$$.

Solution

### Michel vanBiezen - 2466 video solution

video by Michel vanBiezen

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Intermediate

Eliminate the parameter $$t$$ from $$x=e^t-1$$ and $$y=e^{2t}$$.

Problem Statement

Eliminate the parameter $$t$$ from $$x=e^t-1$$ and $$y=e^{2t}$$.

Final Answer

$$y=x^2+2x+1$$

Problem Statement

Eliminate the parameter $$t$$ from $$x=e^t-1$$ and $$y=e^{2t}$$.

Solution

### Michel vanBiezen - 2464 video solution

video by Michel vanBiezen

Final Answer

$$y=x^2+2x+1$$

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For the parametric equations $$x=1+\sin(t)$$, $$y=-2+\cos(t)$$, find the point that corresponds to $$t=\pi/2$$, graph the equations and eliminate the parameter.

Problem Statement

For the parametric equations $$x=1+\sin(t)$$, $$y=-2+\cos(t)$$, find the point that corresponds to $$t=\pi/2$$, graph the equations and eliminate the parameter.

Solution

### PatrickJMT - 47 video solution

video by PatrickJMT

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Eliminate the parameter from the parametric equations $$x=4e^{t/4}$$, $$y=3e^t$$.

Problem Statement

Eliminate the parameter from the parametric equations $$x=4e^{t/4}$$, $$y=3e^t$$.

Solution

### PatrickJMT - 50 video solution

video by PatrickJMT

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Find all points of intersection of the parametric curves $$C_1: x=t+1; y=t^2$$ and $$C_2: x=3t+1; y=t^2+1$$.

Problem Statement

Find all points of intersection of the parametric curves $$C_1: x=t+1; y=t^2$$ and $$C_2: x=3t+1; y=t^2+1$$.

Solution

### PatrickJMT - 53 video solution

video by PatrickJMT

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### parametrics 17calculus youtube playlist

Here is a playlist of the videos on this page.

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