Critical Points of Implicit Relations (College Board AP® Calculus AB)

Revision Note

Author

Jamie Wood

Expertise

Maths

Critical points of implicit relations

Do implicit equations have critical points?

Equations defined implicitly can have critical points
Critical points are defined in the same way as they are for any other function
- A critical point occurs where the derivative is equal to zero
- or where the derivative does not exist
The applications of first and second derivatives to classify the nature of points on a graph can be extended to implicit functions
- These properties are summarized in the table below

Type of point	First derivative	Second derivative
Local minimum	Zero	Positive or zero
Local maximum	Zero	Negative or zero
Point of inflection (critical)	Zero	Zero
Point of inflection (non-critical)	Non-zero	Zero

Remember that second derivative equal to zero is not enough for a point to be a point of inflection
- The second derivative must change sign at the point as well

How can I find points on an implicitly-defined curve where the tangent is horizontal or vertical?

The tangent line to an implicitly-defined curve will be horizontal at a point on the curve where $\frac{d y}{d x} = 0$
The tangent line to an implicitly-defined curve will be vertical at a point on the curve where $\frac{d x}{d y} = 0$
- This is the same as a point at which $\frac{d y}{d x}$ has a denominator equal to zero
  - and a numerator not equal to zero
- Recall $\frac{d x}{d y} = \frac{1}{(\frac{d y}{d x})}$

Worked Example

Consider the function $y = f (x)$ whose curve is given by the equation $2 y^{2} - 6 = y \sin 2 x$ for $y > 0$ .

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

Answer:

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 $x$

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

Rearrange for $\frac{d y}{d x}$

$\begin{array}{rcl} 4 y \cdot \frac{d y}{d x} - \frac{d y}{d x} \cdot \sin 2 x & = & 2 y \cos 2 x \\ \frac{d y}{d x} (4 y - \sin 2 x) & = & 2 y \cos 2 x \\ \frac{d y}{d x} & = & \frac{2 y \cos 2 x}{4 y - \sin 2 x} \end{array}$

Set this equal to zero to find the critical point

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

$2 y \cos 2 x = 0$

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

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

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

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

We were told in the question that $0 \leq x \leq \frac{π}{2}$

$x = \frac{π}{4}$

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

$\begin{array}{rcl} 2 y^{2} - 6 & = & y \sin (2 \cdot \frac{π}{4}) \\ 2 y^{2} - 6 & = & y \sin (\frac{π}{2}) \\ 2 y^{2} - 6 & = & y \\ 2 y^{2} - y - 6 & = & 0 \\ (2 y + 3) (y - 2) & = & 0 \end{array}$

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

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

$4 (2) - \sin (2 \cdot \frac{π}{4}) = 8 - 1 = 7 \neq 0$

The point where the tangent to the curve is horizontal is $(\frac{π}{4}, 2)$

(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:

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

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

$\frac{d y}{d x} = \frac{2 y \cos 2 x}{4 y - \sin 2 x} \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{2 y \cos 2 x}{4 y - \sin 2 x})$

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

Differentiating $v$ is relatively straightforward using the chain rule

$v = 4 y - \sin 2 x v^{'} = 4 \frac{d y}{d x} - 2 \cos 2 x$

Differentiating $u$ will require the product rule

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

$u = 2 y \cos 2 x$

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

$p^{'} = 2 \frac{d y}{d x}$ $q^{'} = - 2 \sin 2 x$

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

$u^{'} = 2 \frac{d y}{d x} \cdot \cos 2 x + 2 y \cdot - 2 \sin 2 x u^{'} = 2 \cos 2 x \cdot \frac{d y}{d x} - 4 y \sin 2 x$

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

$\frac{d^{2} y}{d x^{2}} = \frac{(2 \cos 2 x \cdot \frac{d y}{d x} - 4 y \sin 2 x) \cdot (4 y - \sin 2 x) - (2 y \cos 2 x) \cdot (4 \frac{d y}{d x} - 2 \cos 2 x)}{{(4 y - \sin 2 x)}^{2}}$

This is quite a tricky expression, but thankfully we do not need to rearrange it, or even substitute in the expression $\frac{d y}{d x}$

We know that the point is $(\frac{π}{4}, 2)$ and we know that at this point, $\frac{d y}{d x} = 0$

So we can substitute these values of $x$ , $y$ , and $\frac{d y}{d x}$ into the expression for $\frac{d^{2} y}{d x^{2}}$

$\frac{d^{2} y}{d x^{2}} = \frac{(0 - 4 (2) \sin (2 \cdot \frac{π}{4})) \cdot (4 (2) - \sin (2 \cdot \frac{π}{4})) - (2 (2) \cos (2 \cdot \frac{π}{4})) \cdot (0 - 2 \cos (2 \cdot \frac{π}{4}))}{{(4 (2) - \sin (2 \cdot \frac{π}{4}))}^{2}}$

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

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

At $(\frac{π}{4}, 2)$ , the first derivative is zero, and the second derivative is negative.

Therefore it is a relative maximum.

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