
MHB Master
#1
September 29th, 2015,
12:42
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 :
 $$y^2x^2=1$$
 $$\frac{x^2}{4}+\frac{y^2}{9}=1$$
I have tried the following:
 A parametrization of the level curve $y^2x^2=1$ 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?
Last edited by evinda; September 29th, 2015 at 14:14.

September 29th, 2015 12:42
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MHB Craftsman
#2
September 29th, 2015,
13:07
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.

MHB Master
#3
September 29th, 2015,
13:11
Thread Author
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 $x \geq 0$.

MHB Seeker
#4
September 29th, 2015,
15:08
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.

MHB Master
#5
September 29th, 2015,
15:22
Thread Author
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 $x \geq 0$.

MHB Seeker
#6
September 29th, 2015,
15:40
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 $y=x^2$.

MHB Master
#7
September 29th, 2015,
15:45
Thread Author
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 $x \geq 0$ ?

MHB Seeker
#8
September 29th, 2015,
16:02
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.

MHB Master
#9
September 30th, 2015,
14:36
Thread Author
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 :
 $$y^2x^2=1$$
 $$\frac{x^2}{4}+\frac{y^2}{9}=1$$
I have tried the following:
 A parametrization of the level curve $y^2x^2=1$ 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?

MHB Seeker
#10
September 30th, 2015,
15:26
Originally Posted by
evinda
Also I want to find a parametrization of the following level curves :
 $y^2x^2=1$
 $\frac{x^2}{4}+\frac{y^2}{9}=1$
I have tried the following:
 A parametrization of the level curve $y^2x^2=1$ 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.