1. Hello!!!

Is $r(t)=(t^2,t^4)$ a parametrization of the parabola $y=x^2$?

I have written the following:

For $x(t)=t^2$ and $y(t)=t^4$ we have that $y(t)=t^4=(t^2)^2=x^2(t)$, so $r(t)=(t^2,t^4)$ a parametrization of the parabola $y=x^2$.

Is it right? How could we say it more formally?

Also I want to find a parametrization of the following level curves :

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

I have tried the following:

• A parametrization of the level curve $y^2-x^2=1$ is $r(t)=(\cosh t, \sinh t)$ since $\cosh^2 t- \sinh^2 t=1$.
• A parametrization of the level curve $\frac{x^2}{4}+\frac{y^2}{9}=1$ is $(2 \sin t, 3 \cos t)$ since $\frac{4 \sin^2 t}{4}+\frac{9 \cos^2 t}{9}=1$

Is it right?

2.

3. Hi,

What is the value of $t$ for parametrizing the point $(-1,1)$ that is in the parabola?

It seems that you are thinking about parametrization just in one direction, you need any point given by the parametrization map being in the object you want to parametrize, but you also need that for any point in the object there exist some parmeter that gives you that point, and that is what is missing in your arguments.

Originally Posted by Fallen Angel
Hi,

What is the value of $t$ for parametrizing the point $(-1,1)$ that is in the parabola?

It seems that you are thinking about parametrization just in one direction, you need any point given by the parametrization map being in the object you want to parametrize, but you also need that for any point in the object there exist some parmeter that gives you that point, and that is what is missing in your arguments.
So is it complete as follows?

For $x(t)=t^2$ and $y(t)=t^4$ we have that $y(t)=t^4=(t^2)^2=x^2(t)$, so $r(t)=(t^2,t^4)$ a parametrization of the parabola $y=x^2$ where $x \geq 0$.

5. Originally Posted by evinda
So is it complete as follows?

For $x(t)=t^2$ and $y(t)=t^4$ we have that $y(t)=t^4=(t^2)^2=x^2(t)$, so $r(t)=(t^2,t^4)$ a parametrization of the parabola $y=x^2$ where $x \geq 0$.
Hey evinda!

It is correct.
But I think we're supposed to verify that each of the implications hold in both directions.

Originally Posted by I like Serena
Hey evinda!

It is correct.
But I think we're supposed to verify that each of the implications hold in both directions.
So should it be as follows?

Since $t^4=(t^2)^2$ for each $t$ the coordinates $x=t^2$ , $y=t^4$ of $r(t)$ satisfy the relation $y=x^2$ for $x \geq 0$.

Conversely, we want to parametrize the parabola $y=x^2$. We set $x=t^2$ and then we have $y=t^4$ so we get the parametrization $r(t)=(t^2, t^4)$.

Therefore, $r(t)=(t^2,t^4)$ is a parametrization of the parabola $y=x^2$ with $x \geq 0$.

7. Originally Posted by evinda
So should it be as follows?

Since $t^4=(t^2)^2$ for each $t$ the coordinates $x=t^2$ , $y=t^4$ of $r(t)$ satisfy the relation $y=x^2$ for $x \geq 0$.

Conversely, we want to parametrize the parabola $y=x^2$. We set $x=t^2$ and then we have $y=t^4$ so we get the parametrization $r(t)=(t^2, t^4)$.

Therefore, $r(t)=(t^2,t^4)$ is a parametrization of the parabola $y=x^2$ with $x \geq 0$.
I think it should be like:
$$r(t)=(x(t), y(t)) = (t^2,t^4) \Rightarrow y=x^2$$
But for $x=-1$ the following does not hold:
$$y=x^2 \Rightarrow r(t)=(x(t), y(t)) = (t^2,t^4)$$

Therefore $r(t)=(t^2,t^4)$ is not a parametrization of $y=x^2$.

Originally Posted by I like Serena
I think it should be like:
$$r(t)=(x(t), y(t)) = (t^2,t^4) \Rightarrow y=x^2$$
But for $x=-1$ the following does not hold:
$$y=x^2 \Rightarrow r(t)=(x(t), y(t)) = (t^2,t^4)$$

Therefore $r(t)=(t^2,t^4)$ is not a parametrization of $y=x^2$.
Ah, I see... And how could we show that it is a parametrization for $x \geq 0$ ?

9. Originally Posted by evinda
Ah, I see... And how could we show that it is a parametrization for $x \geq 0$ ?
For $x \geq 0$ we can pick $t=\sqrt x$, which satisfies the implication.

Originally Posted by I like Serena
For $x \geq 0$ we can pick $t=\sqrt x$, which satisfies the implication.
Nice... Thank you!!!

Also I want to find a parametrization of the following level curves :

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

I have tried the following:

• A parametrization of the level curve $y^2-x^2=1$ is $r(t)=(\cosh t, \sinh t)$ since $\cosh^2 t- \sinh^2 t=1$.
• A parametrization of the level curve $\frac{x^2}{4}+\frac{y^2}{9}=1$ is $(2 \sin t, 3 \cos t)$ since $\frac{4 \sin^2 t}{4}+\frac{9 \cos^2 t}{9}=1$

Is it right?

11. Originally Posted by evinda
Also I want to find a parametrization of the following level curves :

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

I have tried the following:

• A parametrization of the level curve $y^2-x^2=1$ is $r(t)=(\cosh t, \sinh t)$ since $\cosh^2 t- \sinh^2 t=1$.
• A parametrization of the level curve $\frac{x^2}{4}+\frac{y^2}{9}=1$ is $(2 \sin t, 3 \cos t)$ since $\frac{4 \sin^2 t}{4}+\frac{9 \cos^2 t}{9}=1$

Is it right?
It is correct but not sufficient.
We still need to show the reverse implication.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•