Critical Points of Implicit Relations (College Board AP® Calculus BC): Revision Note

Consider the curve given by the equation $2y^2-6=y\sin 2x$ for $y>0$.

(a) For $0\le x\le \frac{\pi }{2}$ and $y>0$, find the coordinates of the point where the tangent to the curve is horizontal.

(b) Determine whether the curve has a relative minimum, a relative maximum, or neither at the point found in part (a). Justify your answer.

Answer:

(a)

We need to find where the derivative is equal to zero, as the slope of the tangent is zero at that point

Differentiate both sides of the equation with respect to $x$

Use the product rule for the right-hand side, and don't forget to use the chain rule when differentiating $y$ with respect to $x$

$\begin{array}{rcl}\frac{d}{dx}\left(2y^2-6\right)& =& \frac{d}{dx}\left(y\sin 2x\right)\\ 4y·\frac{dy}{dx}& =& \frac{dy}{dx}·\sin 2x+y·2\cos 2x\end{array}$

Rearrange for $\frac{dy}{dx}$

$\begin{array}{rcl}4y·\frac{dy}{dx}-\frac{dy}{dx}·\sin 2x& =& 2y\cos 2x\\ \frac{dy}{dx}\left(4y-\sin 2x\right)& =& 2y\cos 2x\\ \frac{dy}{dx}& =& \frac{2y\cos 2x}{4y-\sin 2x}\end{array}$

Set this equal to zero to find the critical point

$\frac{2y\cos 2x}{4y-\sin 2x}=0$

$2y\cos 2x=0$

We also need to be careful with the denominator, so that the derivative is not undefined (from dividing by zero)

and $4y-\sin 2x\ne 0$

Solve to find $x$, and use the fact that we know $y>0$, so $2y\cos 2x=0$ will have the same solutions as $\cos 2x=0$

$\begin{array}{rcl}2y\cos 2x& =& 0\\ \cos 2x& =& 0\\ 2x& =& \frac{\pi }{2},\frac{3\pi }{2},...\\ x& =& \frac{\pi }{4},\frac{3\pi }{4},...\end{array}$

We were told in the question that $0\le x\le \frac{\pi }{2}$

$x=\frac{\pi }{4}$

Find the $y$ value by substituting into $2y^2-6=y\sin 2x$

$\begin{array}{rcl}2y^2-6& =& y\sin \left(2·\frac{\pi }{4}\right)\\ 2y^2-6& =& y\sin \left(\frac{\pi }{2}\right)\\ 2y^2-6& =& y\\ 2y^2-y-6& =& 0\\ \left(2y+3\right)\left(y-2\right)& =& 0\end{array}$

$y>0$, so $y=2$

Check this satisfies $4y-\sin 2x\ne 0$

$4\left(2\right)-\sin \left(2·\frac{\pi }{4}\right)=8-1=7\ne 0$

The point where the tangent to the curve is horizontal is $\left(\frac{\pi }{4},2\right)$ because $\frac{dy}{dx}=0$ at this point

(b)

We need to check if the second derivative is positive or negative at $\left(\frac{\pi }{4},2\right)$

Write down the first derivative, and then differentiate both sides with respect to $x$

$\begin{array}{l}\frac{dy}{dx}=\frac{2y\cos 2x}{4y-\sin 2x}\\ \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{2y\cos 2x}{4y-\sin 2x}\right)\end{array}$

Use the quotient rule, ${\left(\frac{u}{v}\right)}^{\text{'}}=\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$ with $u=2y\cos 2x$ and $v=4y-\sin 2x$

Differentiating $v$ is relatively straightforward using the chain rule

$\begin{array}{l}v=4y-\sin 2x\\ v^{\text{'}}=4\frac{dy}{dx}-2\cos 2x\end{array}$

Differentiating $u$ will require the product rule

Use different variables than $u$ and $v$ so you don't get confused, e.g. $p$ and $q$

$u=2y\cos 2x$

$p=2y$ $q=\cos 2x$

$p^{\text{'}}=2\frac{dy}{dx}$ $q^{\text{'}}=-2\sin 2x$

$u^{\text{'}}=p^{\text{'}}q+p{q}^{\text{'}}$

$\begin{array}{l}{u}^{\text{'}}=2\frac{dy}{dx}·\cos 2x+2y·-2\sin 2x\\ u^{\text{'}}=2\cos 2x·\frac{dy}{dx}-4y\sin 2x\end{array}$

Apply the quotient rule, ${\left(\frac{u}{v}\right)}^{\text{'}}=\frac{u\text{'}v-uv\text{'}}{v^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{\left(2\cos 2x·{\displaystyle \frac{dy}{dx}}-4y\sin 2x\right)·\left(4y-\sin 2x\right)-\left(2y\cos 2x\right)·\left(4{\displaystyle \frac{dy}{dx}}-2\cos 2x\right)}{{\left(4y-\sin 2x\right)}^2}$

We know that the point is $\left(\frac{\pi }{4},2\right)$ and we know that at this point, $\frac{dy}{dx}=0$

Substitute

• $x=\frac{\pi }{4}$

• $y=2$

• $\frac{dy}{dx}=0$

$\frac{d^{2}y}{d{x}^2}=\frac{\left(0-4\left(2\right)\sin \left(2·{\displaystyle \frac{\pi }{4}}\right)\right)·\left(4\left(2\right)-\sin \left(2·\frac{\pi }{4}\right)\right)-\left(2\left(2\right)\cos \left(2·\frac{\pi }{4}\right)\right)·\left(0-2\cos \left(2·\frac{\pi }{4}\right)\right)}{{\left(4\left(2\right)-\sin \left(2·\frac{\pi }{4}\right)\right)}^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{\left(-8\right)·\left(8-1\right)-\left(0\right)·\left(0\right)}{{\left(8-1\right)}^2}=-\frac{56}{49}$

Use the sign of the second derivative to classify the nature of the critical point $\left(\frac{\pi }{4},2\right)$

At $\left(\frac{\pi }{4},2\right)$, the first derivative is zero, and $\frac{d^{2}y}{d{x}^2}<0$

Therefore, it is a relative maximum