Find a parametrization of the following level curves

[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)