Find a parametrization of the following level curves

[evinda]

Hello! (Wave)

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? (Thinking)

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?

 

[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.

 

[evinda]

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 $x \geq 0$.

 

[I like Serena]

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! (Smile)

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

 

[evinda]

Hey evinda! (Smile)

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

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 like Serena]

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 $y=x^2$. (Thinking)

 

[evinda]

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 $y=x^2$. (Thinking)

Ah, I see... And how could we show that it is a parametrization for $x \geq 0$ ? (Thinking)

 

[I like Serena]

Ah, I see... And how could we show that it is a parametrization for $x \geq 0$ ? (Thinking)

For $x \geq 0$ we can pick $t=\sqrt x$, which satisfies the implication. (Mmm)

 

[evinda]

For $x \geq 0$ we can pick $t=\sqrt x$, which satisfies the implication. (Mmm)

Nice... Thank you! (Smile)

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?

 

[I like Serena]

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. (Sweating)

 

[evinda]

It is correct but not sufficient.
We still need to show the reverse implication. (Sweating)

A parametrization of the level curve $y^2-x^2=1$ is $r(t)=(\cosh t, \sinh t)$, because:

• For $x= \sinh t, y= \cosh t$ we have $y^2-x^2=\cosh^2 t- \sinh^2 t=1, t \in \mathbb{R}$.
  
  This is the one direction, right?
  
• Which should be the other direction? I am a little confused right now... (Thinking)

 

[I like Serena]

A parametrization of the level curve $y^2-x^2=1$ is $r(t)=(\cosh t, \sinh t)$, because:

• For $x= \sinh t, y= \cosh t$ we have $y^2-x^2=\cosh^2 t- \sinh^2 t=1, t \in \mathbb{R}$.
  
  This is the one direction, right?
  
• Which should be the other direction? I am a little confused right now... (Thinking)

You have shown that $\{ r(t)=(\cosh t, \sinh t) : t \in \mathbb R \} \subseteq \{ (x,y) : y^2-x^2=1 \}$.
But do we also have that $\{ (x,y) : y^2-x^2=1 \} \subseteq \{ r(t)=(\cosh t, \sinh t) : t \in \mathbb R \}$? (Wondering)

In other words, suppose we pick some point $(x,y)$ that is part of $y^2-x^2=1$, can we always find a corresponding $t$?

 

[evinda]

You have shown that $\{ r(t)=(\cosh t, \sinh t) : t \in \mathbb R \} \subseteq \{ (x,y) : y^2-x^2=1 \}$.
But do we also have that $\{ (x,y) : y^2-x^2=1 \} \subseteq \{ r(t)=(\cosh t, \sinh t) : t \in \mathbb R \}$? (Wondering)

In other words, suppose we pick some point $(x,y)$ that is part of $y^2-x^2=1$, can we always find a corresponding $t$?

So do we have to say the following?

Suppose that we pick some point $(x,y)$ that satisfies $y^2-x^2=1$. We pick $t= arc \cosh y = -arc \sinh x$.

 

[I like Serena]

So do we have to say the following?

Suppose that we pick some point $(x,y)$ that satisfies $y^2-x^2=1$. We pick $t= arc \cosh y = -arc \sinh x$.

For instance yes... do they always work?
What's the domain of $\arcosh$? (Wondering)

 

[evinda]

For instance yes... do they always work?
What's the domain of $\arcosh$? (Wondering)

It is $[1, +\infty)$, right? So we have to mention this restriction, right? (Thinking)

 

[I like Serena]

It is $[1, +\infty)$, right? So we have to mention this restriction, right? (Thinking)

Yep. So we can be sure that the left half of the curve is not covered. (Thinking)

As far as I am concerned, it means that the given parametrization is not a parametrization of the given level curve.
It's only a parametrization of part of the level curve.
We can mention under which restriction it is a parametrization, which would actually be an explanation why it's not. (Nerd)

 

[evinda]

Yep. So we can be sure that the left half of the curve is not covered. (Thinking)

As far as I am concerned, it means that the given parametrization is not a parametrization of the given level curve.
It's only a parametrization of part of the level curve.
We can mention under which restriction it is a parametrization, which would actually be an explanation why it's not. (Nerd)

So to find a right parametrization, do we have to pick a point $(x,y)$ such that $t$ will have $\mathbb{R}$ as its domain? (Thinking)

 

[I like Serena]

So to find a right parametrization, do we have to pick a point $(x,y)$ such that $t$ will have $\mathbb{R}$ as its domain? (Thinking)

Huh?

 

[evinda]

Yep. So we can be sure that the left half of the curve is not covered. (Thinking)

As far as I am concerned, it means that the given parametrization is not a parametrization of the given level curve.
It's only a parametrization of part of the level curve.
We can mention under which restriction it is a parametrization, which would actually be an explanation why it's not. (Nerd)

Huh?

Because of the fact that $t \in [1,+\infty)$ it holds that $(\cosh t, \sinh t)$ is not parametrization of the given level curve.
Or have I understood it wrong? (Thinking)

 

[I like Serena]

Because of the fact that $t \in [1,+\infty)$ it holds that $(\cosh t, \sinh t)$ is not parametrization of the given level curve.
Or have I understood it wrong? (Thinking)

I'm just noticing that $x$ and $y$ and $\cosh t, \sinh t$ have been mixed up.

The equation $y^2-x^2=1$ represents a hyperbola with a top half and a bottom half.
The parametrization $(\sinh t, \cosh t)$ represents the top half of that hyperbola.

It should be that $y=\cosh t \in [1,+\infty)$, which indeed means that only the top half of the hyperbola is covered. (Nerd)

 

[evinda]

I'm just noticing that $x$ and $y$ and $\cosh t, \sinh t$ have been mixed up.

The equation $y^2-x^2=1$ represents a hyperbola with a top half and a bottom half.
The parametrization $(\sinh t, \cosh t)$ represents the top half of that hyperbola.

It should be that $y=\cosh t \in [1,+\infty)$, which indeed means that only the top half of the hyperbola is covered. (Nerd)

So do we have to pick the parametrization $(\pm \cosh x, \sinh x )$ and then pick $t=arc \sinh x= arc \cosh y$ if $t >0$, $t=arc \sinh x=-arc \cosh y$ if $t<0$ ? (Thinking)

 

[I like Serena]

Erm... I think there's a couple of things wrong with your last post...

 

[evinda]

Erm... I think there's a couple of things wrong with your last post...

Should it be maybe as follows?

• For $x= \sinh t, y= \pm \cosh t$ we have $y^2-x^2=(\pm \cosh t)^2- (\sinh t)^2=1, t \in \mathbb{R}$.
• Suppose that it holds $y^2-x^2=1$. We can pick a $t$ such that $x= \sinh t$ since $\sinh t$ is surjective.
  Then we have $y^2- (\sinh t)^2=1 \Rightarrow y^2= 1+ (\sinh t)^2= (\cosh t)^2 \Rightarrow y= \pm \cosh t$.

Thus $r(t)=(\pm \cosh t, \sinh t)$ is a parametrization of the level curve $y^2-x^2=1$.

 

[I like Serena]

Should it be maybe as follows?

• For $x= \sinh t, y= \pm \cosh t$ we have $y^2-x^2=(\pm \cosh t)^2- (\sinh t)^2=1, t \in \mathbb{R}$.
• Suppose that it holds $y^2-x^2=1$. We can pick a $t$ such that $x= \sinh t$ since $\sinh t$ is surjective.
  Then we have $y^2- (\sinh t)^2=1 \Rightarrow y^2= 1+ (\sinh t)^2= (\cosh t)^2 \Rightarrow y= \pm \cosh t$.

Thus $r(t)=(\pm \cosh t, \sinh t)$ is a parametrization of the level curve $y^2-x^2=1$.

Looks fine to me. (Nod)

 

[evinda]

Looks fine to me. (Nod)

Nice! (Happy)

So to show that a parametrization of the level curve $\frac{x^2}{4}+\frac{y^2}{9}=1$ is $(2 \sin t, 3 \cos t)$ do we have to say the following?

• For $x=2 \sin t, y= 3 \cos t$ we have $\frac{x^2}{4}+\frac{y^2}{9}=\frac{4 \sin^2 t}{4}+\frac{9 \cos^2 t}{9}= \sin^2 t+ \cos^2 t=1$
  
• Let $\frac{x^2}{4}+\frac{y^2}{9}=1$. In this case we can't take i.e. a $t$ so that $x=2 \sin t$ since the latter function isn't surjective. Or am I wrong? (Thinking)
  

 

[I like Serena]

[*] Let $\frac{x^2}{4}+\frac{y^2}{9}=1$. In this case we can't take i.e. a $t$ so that $x=2 \sin t$ since the latter function isn't surjective. Or am I wrong? (Thinking)

Surjective from what to what exactly? (Wondering)
That makes quite a difference in this case.

 

[evinda]

Surjective from what to what exactly? (Wondering)
That makes quite a difference in this case.

Do we maybe have to pick the parametrization $( \pm 2 \sin t, 3 \cos t)$ ?
Because I have thought the following:

• For $x= \pm 2 \sin t, y= 3 \cos t$ we have $\frac{x^2}{4}+\frac{y^2}{9}=\frac{4 \sin^2 t}{4}+\frac{9 \cos^2 t}{9}= \sin^2 t+ \cos^2 t=1$
  
• Let $\frac{x^2}{4}+\frac{y^2}{9}=1$. Then $\frac{y^2}{9}=1-\frac{x^2}{4} \leq 1 \Rightarrow -3 \leq y \leq 3$.
  We can pick a $t$ such that $y=3 \cos t $ since the latter is surjective from $\mathbb{R}$ to $[-3,3]$, or isn't it?
  Then we have $\frac{x^2}{4}=1-\frac{9 \cos^2 t}{9}=\sin^2 t \Rightarrow x^2=4 \sin^2 t \Rightarrow x= \pm 2 \sin t$.

 

[I like Serena]

We can pick a $t$ such that $y=3 \cos t $ since the latter is surjective from $\mathbb{R}$ to $[-3,3]$, or isn't it?

Yep. it is. (Nod)

Then we have $\frac{x^2}{4}=1-\frac{9 \cos^2 t}{9}=\sin^2 t \Rightarrow x^2=4 \sin^2 t \Rightarrow x= \pm 2 \sin t$.

Do we need the $\pm$?
Or can we do without? (Wondering)

 

[evinda]

Yep. it is. (Nod)

So does this mean that $3 \cos t$ does never get twice the same value?

Do we need the $\pm$?
Or can we do without? (Wondering)

How can we check this? (Thinking)

 

[I like Serena]

So does this mean that $3 \cos t$ does never get twice the same value?

It does.
That's why it's surjective but not injective. (Mmm)

How can we check this? (Thinking)

As you said, $3 \cos t$ gets each of the values in $(-3,3)$ twice.
Once where $\sin t$ is positive and once where it is negative, which corresponds exactly to the $\pm$. (Thinking)

 

[evinda]

It does.
That's why it's surjective but not injective. (Mmm)

Ah I see... (Nod)

As you said, $3 \cos t$ gets each of the values in $(-3,3)$ twice.
Once where $\sin t$ is positive and once where it is negative, which corresponds exactly to the $\pm$. (Thinking)

Could you explain it further to me? (Thinking)

 

[I like Serena]

Ah I see... (Nod)
Could you explain it further to me? (Thinking)

For each $(x,y)$ in the curve, we can pick $t=\operatorname {atan2}(\frac y 3, \frac x 2)$.
See atan2 on wiki.

As a result we have $x=2\cos t$ and $y=3\sin t$, satisfying the implication.
Therefore every point in $\frac {x^2}{4} + \frac{y^2}{9}=1$ is also in $(2\cos t, 3\sin t)$. (Whew)

 

[evinda]

For each $(x,y)$ in the curve, we can pick $t=\operatorname {atan2}(\frac y 3, \frac x 2)$.
See atan2 on wiki.

As a result we have $x=2\cos t$ and $y=3\sin t$, satisfying the implication.
Therefore every point in $\frac {x^2}{4} + \frac{y^2}{9}=1$ is also in $(2\cos t, 3\sin t)$. (Whew)

How can we explain it without the use of [m] atan2 [/m] ? (Thinking)

 

[I like Serena]

How can we explain it without the use of [m] atan2 [/m] ? (Thinking)

Pick $t$ such that $\cos t = \frac {x/2}{\sqrt{(x/2)^2+(y/3)^2}}$ and $\sin t = \frac {y/3}{\sqrt{(x/2)^2+(y/3)^2}}$.


Or else recognize that $\frac {x^2}{4} + \frac {y^2}{9}=1$ represents the ellipse with semi-axes $2$ and $3$.
And that $(2\cos t, 3\sin t)$ also represents the ellipse with semi-axes $2$ and $3$. (Thinking)