MHB Cartesian Equation for Parametrized Curve: Find Solution

[evinda]

Hello! (Wave)

I want to find the cartesian equation of the following parametrized curve:

$$r(t)=(\cos^2 t, \sin^2 t)$$

I have tried the following:

Since $\cos^2 t+ \sin^2 t=1, \forall t$, the coordinates $x= \cos^2 t, y= \sin^2 t$ of $r(t)$ satisfy $x+y=1$.

Is the above sufficient or is a reverse implication needed? (Thinking)

 

[I like Serena]

Hello! (Wave)

I want to find the cartesian equation of the following parametrized curve:

$$r(t)=(\cos^2 t, \sin^2 t)$$

I have tried the following:

Since $\cos^2 t+ \sin^2 t=1, \forall t$, the coordinates $x= \cos^2 t, y= \sin^2 t$ of $r(t)$ satisfy $x+y=1$.

Is the above sufficient or is a reverse implication needed? (Thinking)

Hey evinda! (Smile)

Indeed, something more is needed.
Let's just consider what the ranges of $\cos^2 t$ and $\sin^2 t$ are...
Are they surjective on $\mathbb R$? (Thinking)

 

[evinda]

Hey evinda! (Smile)

Indeed, something more is needed.
Let's just consider what the ranges of $\cos^2 t$ and $\sin^2 t$ are...
Are they surjective on $\mathbb R$? (Thinking)

They are surjective from $\mathbb{R}$ to $[0,1]$, right?

 

[I like Serena]

They are surjective from $\mathbb{R}$ to $[0,1]$, right?

Yep.
So perhaps we should set some bounds on the curve defined by $x+y=1$. (Thinking)

 

[evinda]

Yep.
So perhaps we should set some bounds on the curve defined by $x+y=1$. (Thinking)

Don't we have to pick these restrictions? (Thinking)

$$0 \leq x \leq 1 \\ 0 \leq y \leq 1$$

 

[I like Serena]

Don't we have to pick these restrictions? (Thinking)

$$0 \leq x \leq 1 \\ 0 \leq y \leq 1$$

Yep. (Happy)

 

[evinda]

Yep. (Happy)

So can we say the following?

Since $\cos^2 t+ \sin^2 t=1, \forall t$, the coordinates $x= \cos^2 t, y= \sin^2 t$ of $r(t)$ satisfy $x+y=1$ with $0 \leq x \leq 1, 0 \leq y \leq 1$.

- - - Updated - - -

So if we want to find the cartesian equation of $r(t)=(e^t, t^2)$ can we say the following?

Since $t^2= (\ln e^t)^2 , \forall t$, the coordinates $x=e^t, y=t^2$ satisfy $y=(\ln x)^2$ for $x >0 , y \geq 0$.

 

[I like Serena]

So can we say the following?

Since $\cos^2 t+ \sin^2 t=1, \forall t$, the coordinates $x= \cos^2 t, y= \sin^2 t$ of $r(t)$ satisfy $x+y=1$ with $0 \leq x \leq 1, 0 \leq y \leq 1$.

- - - Updated - - -

So if we want to find the cartesian equation of $r(t)=(e^t, t^2)$ can we say the following?

Since $t^2= (\ln e^t)^2 , \forall t$, the coordinates $x=e^t, y=t^2$ satisfy $y=(\ln x)^2$ for $x >0 , y \geq 0$.

Seems fine to me. (Smile)

 

[evinda]

Seems fine to me. (Smile)

Nice... Thanks a lot! (Smirk)