<s>[INST] Finding the Paramertric Coordinates of the Maximum and Minimum points of a Certain Curve I was studying for some quizzes when a wild question appears. It goes like this:

Find the paramertric coordinates of the maximum and minimum points of 
  the following curves. The curve is $x = 3 \cos \theta$ ;$\space$ $ y = 4 \sin \theta$

My work:
I need to get the rectangular equation described by the parametric equation.
I remember the relation $(\sin \theta)^2 + (\cos \theta)^2 = 1$
Getting the $\cos \theta:$ $\cos \theta = \frac{x}{3}$
Getting the $\sin \theta:$ $\sin \theta = \frac{y}{4}$
Substituting the newly-found variables to $(\sin \theta)^2 + (\cos \theta)^2 = 1$,
it becomes $$\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1$$ 
By using the implicit differentiation, we got:
$$\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1$$ 
$$\left( \frac{1}{9} \right)\left( 2x \right) + \left( \frac{1}{16} \right)\left( 2 y y' \right) = 0 $$
$$\left( \frac{2}{9} \right)\left( 2x \right) + \left( \frac{1}{8} \right)\left(y y' \right) = 0$$
$$\left( \frac{1}{8} \right)\left(y y' \right) = \left( \frac{-2}{9} \right)\left( 2x \right) $$
$$y' = \frac{-16x}{9y}$$
We got to make the derivative expressed as a single variable, so we need to get the $y$ in terms of $x$ in the equation 
$\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1.$
getting the $y$ from $\left( \frac{x}{3} \right)^2 + \left( \frac{y}{4} \right)^2 = 1,$ it is:
$$y = \frac{4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
Then ,substituting it to the equation $y' = \frac{-16x}{9y},$ it becomes:
$$y' = \frac{-16x}{9\left( \frac{4}{3}\left((9 - x^2)^{\frac{1}{2}}\right) \right)}$$
$$ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
To get the minimum and maximum points of the rectangular equation, we set $\frac{dy}{dx} = y'$ to zero.
$$ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
$$ 0 = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right)$$
$$ 0 = (9 - x^2)^{\frac{1}{2}}$$
$$ 0 = (9 - x^2)$$
$$x^2 = 9$$
We get $x = \pm 3$
To determine whether $x = -3$ is a minimum or maximum point, we use the first derivative test.
Letting $x = -3$ as the critical/base point, we let $x = -99$ and $x = +99.$ If $x = -99$ is plugged into 
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary. 
If $x = +99$ is plugged into 
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary too. 
To determine whether $x =  +3$ is a minimum or maximum point, we use the first derivative test.
Letting $x = -3$ as the critical/base point, we let $x = -99$ and $x = +99.$ If $x = -99$ is plugged into 
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary. 
If $x = +99$ is plugged into 
the equation $ y' = \frac{-4}{3}\left((9 - x^2)^{\frac{1}{2}}\right),$ the value of $y$ is imaginary too. 
At this point, I'm stuck, because I can't determine if $x = \pm 3$ is a minimum or maximum point. I used large numbers
 just to make sure I see the change of signs as I pass through the derivative (+, 0, -) or (-, 0, +)
How do you get the minimum and maximum points of the curve shown above?
 [/INST] The max/min values occur at the values of $\theta$ for which
$$ \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=0 $$
which occurs when
$$ \frac{dy}{d\theta}=4\cos\theta=0 $$
So the solution is
$$\theta=\pm\frac{\pi}{2}$$
(or, more generally)
$$\theta=2\pi n\pm\frac{\pi}{2}$$
But $\theta=\pm\dfrac{\pi}{2}$ will be sufficient to find the max/min.
\begin{eqnarray}
\left(3\cos\frac{\pi}{2},4\sin\frac{\pi}{2}\right)&=&(0,4)\text{ max}\\
\left(3\cos\frac{-\pi}{2},4\sin\frac{-\pi}{2}\right)&=&(0,-4)\text{ min}
\end{eqnarray}
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