How to derive a Parametric Equation?

In summary, the author provides a summary of the content and states that there is no unique parametrization for a given graph and that all that is necessary is to replace x with t in the equation for y.
  • #1
DrummingAtom
659
2
I've been looking everywhere for a tutorial or lesson on Parametric equations and can't find one that shows how to derive the actual equations. Either they show how to eliminate the parameter or sketch the graph. Let's say you had a function like y = x^2 + x - 3

What would you do to find the Parametric equation? Thanks.
 
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  • #2
How about these parametric functions.

y = t^2 + t - 3
x = t
 
  • #3
Yeah, I've seen a lot of that. That's pretty clear cut but there was one that given as an example that was so confusing. The original function was y = -x^2/72 + x

Then they said from this we obtain the parametric equations:

x = 24(sqrt2)(t)
y = -16t^2 + 24(sqrt2)(t)

I'm pretty lost on to how they got those ones.
 
  • #4
When you're working with functions, I'm fairly certain that you can always just replace the x's with t's in the parameter for y and then use the parameter x = t.
 
  • #5
There is no unique parametrization of a given graph. For example, note that mapping t to 4t does not change the resulting points (x, y) that make up the graph. If the graph is the path of a particle, some physics texts will say that you have simply changed the speed at which the point is moving along the graph. Indeed, changing t to f(t) for any bijective function f (over the domain required for the parametrization to cover the graph) gives exactly the same graph. For example, consider the unit semicircle above the x-axis centered at the origin. The parametrizations (x, y) = (cos(t), sin(t)) for t in [0, Pi] and (x, y) = (t, Sqrt[1 - t^2]) for t in [-1, 1] both cover it.
 
  • #6
DrummingAtom said:
Yeah, I've seen a lot of that. That's pretty clear cut but there was one that given as an example that was so confusing. The original function was y = -x^2/72 + x

Then they said from this we obtain the parametric equations:

x = 24(sqrt2)(t)
y = -16t^2 + 24(sqrt2)(t)

I'm pretty lost on to how they got those ones.
All they did for y was replace x in the equation with 24(sqrt2)(t). You can do whatever you want to x, as long as you then make sure y and x satisfy the equation.
 
  • #7
slider142 said:
All they did for y was replace x in the equation with 24(sqrt2)(t). You can do whatever you want to x, as long as you then make sure y and x satisfy the equation.

Yeah, I understand that part. Maybe I'm not being clear. The whole example is about a projectile that is shot at an angle of 45 degrees and a initial velocity of 48 ft/sec. Then they say it follows the path given by y = -x^2/72 + x

They go on and say how this equation does not give all the information possible. Then how we need to introduce a third variable t for time. Then they by writing x and y as functions of t we get:

x = 24(sqrt2)(t)
y = -16t^2 + 24(sqrt2)(t)

That's where I'm confused, they don't explain how they got to that point (Mathematically).
 
  • #8
DrummingAtom said:
Yeah, I understand that part. Maybe I'm not being clear. The whole example is about a projectile that is shot at an angle of 45 degrees and a initial velocity of 48 ft/sec. Then they say it follows the path given by y = -x^2/72 + x

They go on and say how this equation does not give all the information possible. Then how we need to introduce a third variable t for time. Then they by writing x and y as functions of t we get:

x = 24(sqrt2)(t)
y = -16t^2 + 24(sqrt2)(t)

That's where I'm confused, they don't explain how they got to that point (Mathematically).
They are using the additional physical information to get the horizontal speed of the particle, so that they can replace x with x(t). Being fired at a 45 degree angle, the horizontal and vertical components of the velocity are the same, as vx = 48*cos(45) = 48*(Sqrt(2)/2) = 24sqrt(2), etc. Splitting a vector into horizontal and vertical components should have been covered in a previous chapter.
Using the kinematic equation for constant velocity (since there is no horizontal acceleration), we get x = vx*t = 24sqrt(2)*t.
 
  • #9
Yay, thanks.
 

1. What is a parametric equation?

A parametric equation is a set of equations that describe the relationship between two or more variables in terms of one or more parameters. These equations are often used to represent curves, surfaces, or other mathematical objects.

2. How do you derive a parametric equation?

To derive a parametric equation, you must first determine the variables and parameters involved in the equation. Then, you can use techniques such as substitution or elimination to solve for the parameters. Finally, you can express the variables in terms of the parameters to obtain the parametric equation.

3. What is the purpose of using a parametric equation?

A parametric equation allows for a more efficient and concise way of representing mathematical objects. It also allows for more flexibility in manipulating and graphing these objects, making it easier to analyze and understand them.

4. Are there any limitations to using parametric equations?

One limitation of parametric equations is that they may not always provide an accurate representation of a mathematical object. In some cases, a parametric equation may not be able to fully capture all the properties of the object, leading to potential errors or inaccuracies.

5. Can parametric equations be used in different fields of science?

Yes, parametric equations are commonly used in various fields of science, including physics, engineering, and computer science. They are particularly useful in modeling and analyzing complex systems, such as motion and curves, and can also be applied to real-life problems in these fields.

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