Finding the Paramertric Coordinates of the Maximum and Minimum points of a Certain Curve

Question

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?

Answer 1

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}

Answer 2

First of all, a comment: since no interval is given for $\theta$, the convention then is to choose the maximal one, meaning $[0, 2\pi)$. Notice that this a closed curve (i.e. it gets back at its starting point), and as such it makes no rigorous sense to speak about its extrema. On the other hand, I believe that the problem is informally asking you about the vertices of the curve, which I am going to find next.

Almost surely the problem does not expect you to use differential calculus to solve it, because the computations can get a bit ugly (and time-consuming during an exam).

You have correctly noticed that the given curve can be presented in Cartesian coordinates as

$$\frac {x^2} 9 + \frac {y^2} {16} = 1$$

which I guess that you are supposed to recognize as an ellipse centered in $(0,0)$ and having the coordinate axes as its own axes of symmetry. You are next supposed to know how to sketch such an ellipse, from which it becomes apparent that the vertices are $(\pm 3, 0)$ and $(0, \pm 4)$.

Finally, you are required to find the parametric coordinates of thse points. Keep in mind that when $\theta = 0$ you are on the $x$ axis to the right, when $\theta = \frac \pi 2$ you are upward on the $y$ axis, when $\theta = \pi$ you are on the $x$ axis to the left, and when $\theta = \frac {3\pi} 2$ you are downward on the $y$ axis. This means that $(3,0) = (3 \cos 0, \sin 0)$, $(0, 4) = (\cos \frac \pi 2, 4 \sin \frac \pi 2)$, $(-3,0) = (3 \cos \pi, \sin \pi)$ and $(0, -4) = (\cos \frac {3\pi} 2, 4 \sin \frac {3\pi} 2)$.