Question

find two different sets of parametric equations for the rectangular equation. y=x^3

Answer 1

Ok, are you familiar with the subject?

Answer 2

no

Answer 3

my teacher gave us this problem without even doing an example in class :/

Answer 4

Ok, parametric equations are set of equations that all are depended on a common 'parameter':
http://en.wikipedia.org/wiki/Parametric_equation

The example they give there is
$$
x = \cot(t) \qquad y = \sin(t)
$$

Answer 5

What's important to understand is that because both \(x\) and \(y\) are depended on \(t\), then it means that you'll get a pair of \(x,y\) values for every value of \(t\).
If you treat this pair as a point in a graph for example, then you can say that you get different points for different values of \(t\).

Answer 6

If you want we can go through an example, but it's up to you =) I don't mind moving on as well

Answer 7

lets try an example

Answer 8

Ok, do you know the equation for a circle?

Answer 9

it (x-h)^2+(y-k)^2=r^2

Answer 10

right. this equation holds true for all the points that are on the circle.
Let's now try instead to make a parametric equations for a circle
To do so we can say that our 'parameter' is some angle \(t\), and we can use the value for the angle to determine the values of \(x,y\) on the circle. makes sense so far?

Answer 11

yeah

Answer 12

Ok, so say we have a circle with radius \(r\), and its center is in the origin (0,0). Can you tell what would be the x,y values for angle \(t\)?

Answer 13

i dont know (0,1)

Answer 14

aha, well we can't say that, because there are many points on the circle, not just one.
So let's try and visualize this.
We have:

Answer 15

So for any possible value of \(t\) we get a different point on the circle. similar to the idea of the unit-circle used in trigonometry we can now form a right angle triangle to find the x,y values of the point

Answer 16

So, for example, we can say that in that \(\frac{y}{r} = \sin(t) \implies y = r \cdot \sin(t)\)
So for any possible \(t\) angle, I can find the \(y\) value of the point on the circle.

Answer 17

Can you find an expression for \(x\) as well?

Answer 18

oh okay and the same is for x/r=cos(t) x=r*cos(t)

Answer 19

exactly. so we can say that our parametric equations are
$$ x = r \cdot cos(t) \qquad y = r \cdot sin(t) $$

Answer 20

Look here https://www.desmos.com/calculator/pbxmlzyghn

Answer 21

i see

Answer 22

This is the graph of our parametric equations. I had to call them a,b instead of x,y.. because it doesn't like it. but then you can see that if we take the points (a,b) for all possible values of t, we get a circle.

Answer 23

you can also see what point you get fora single value for \(t\), using the slider of \(c\).

Answer 24

And you'll notice that there are many values for \(t\) that will give the exact same point. That is perfectly fine.

Answer 25

So, here we did a circle, where we tied x and y some possible value of t. we made sure that no matter what value t has, the values of x,y will be of a point on the circle.

Answer 26

We also made sure that the possible values for \(t\) will create all the possible points on the circle. so we wouldn't left only with half a circle, or a single point on the circle.

Answer 27

We have to do the same in your question. we have to tie x,y into t in some way such that:
A) The x,y values will always match the original equation, no matter what value t has.
B) The possible values of t will generate all the possible points of the original graph, at least once ;)

Answer 28

So let's see what we have. we have:
$$
y = x^3
$$Right?

Answer 29

yes

Answer 30

Now, this tells you everything. In order for the point to be on the graph, y has to be the cube of x.. that simple

Answer 31

I want you to notice that the parametric equations we made for the circle agree with the circle equation. If you'll plug our parametric equations in, you'll get the a true equation:
$$
x^2 + y^2 = r^2 \\ x=r \cdot \cos(t) \qquad y = r \cdot \sin(t) \\
(r \cdot \cos(t))^2 + (r \cdot \sin(t))^2 = r^2 \\
r^2(\cos^2(t) + \sin^2(t)) = r^2 \\
r^2 = r^2
$$

Answer 32

Now, we need a set of equations* that first of all will fit in our original equation. no matter what we set x to be, we have to set y to be the cube of it.

Answer 33

okay so