Behavior of Implicit Relations — AP® Calculus AB Notes

A circle, an oval or a looping curve fails the vertical line test, so it is not the graph of a function $y = f(x)$ . You already know how to find its slope with implicit differentiation. This topic asks what that slope formula tells you about the shape: where the curve is flat, where it stands straight up, and where it has a local high or low point.

The slope formula as a fraction

Differentiate both sides of the curve's equation with respect to $x$ , write " $= 0$ " (or the derivative of the right side), and solve. The result usually looks like a fraction in both $x$ and $y$ :

$\frac{dy}{dx} = \frac{N(x, y)}{D(x, y)}$

Worked example 1. Find $\frac{dy}{dx}$ for $x^2 - xy + y^2 = 12$ .

$\frac{d}{dx}$ of both sides: $2x - \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx}$ $= 0$ . (The product rule on $xy$ gives two terms.)

Collect the $\frac{dy}{dx}$ terms: $\frac{dy}{dx}(2y - x) = y - 2x$ , so $\frac{dy}{dx}$ $= \frac{y - 2x}{2y - x}$ .

Horizontal and vertical tangent lines

A horizontal tangent needs slope 0, so the numerator must be 0 while the denominator is not. A vertical tangent line needs the slope to be undefined because the denominator is 0 while the numerator is not. In both cases the point must actually be on the curve.

Tangent line	Numerator $N$	Denominator $D$	Also
horizontal	$0$	not $0$	point on the curve
vertical	not $0$	$0$	point on the curve
no conclusion	$0$	$0$	the formula gives $\frac{0}{0}$

The tilted oval x squared plus xy plus y squared = 3, with dy/dx = -(2x + y)/(x + 2y). Horizontal tangent lines touch it at (-1, 2) and (1, -2), where the numerator 2x + y is zero; vertical tangent lines touch it at (-2, 1) and (2, -1), where the denominator x + 2y is zero and the numerator is not. — **Figure 1.** On $x^2 + xy + y^2 = 3$ , each condition picks out a line through the origin; where that line meets the curve, the tangent line is horizontal (blue) or vertical (green). LevlPrep original diagram.

Worked example 2. Find the points of $x^2 - xy + y^2 = 12$ with horizontal and with vertical tangent lines.

Horizontal: $y - 2x = 0$ , so $y = 2x$ . Substitute into the curve: $x^2 - 2x^2 + 4x^2 = 12$ , so $3x^2 = 12$ and $x = \pm 2$ . Points $(2, 4)$ and $(-2, -4)$ ; the denominator there is $6$ and $-6$ , not 0.

Vertical: $2y - x = 0$ , so $x = 2y$ . Substitute: $4y^2 - 2y^2 + y^2 = 12$ , so $y = \pm 2$ . Points $(4, 2)$ and $(-4, -2)$ ; the numerator there is $-6$ and $6$ , not 0.

The substitution step is the one students skip. The point $(1, 2)$ makes the numerator $y - 2x$ zero, but $1 - 2 + 4 = 3 \ne 12$ , so it is not on the curve, and its "slope" means nothing.

When both parts are zero

For the curve $x^3 + y^3 = 6xy$ , implicit differentiation gives $\frac{dy}{dx}$ $= \frac{2y - x^2}{y^2 - 2x}$ . At the origin, which is on the curve, both the numerator and the denominator are 0. The formula is undefined and decides nothing; this curve actually crosses itself there. Never write " $\frac{0}{0} = 0$ " or " $\frac{0}{0} = 1$ ": that is a notation error.

High and low points of a curve

Near a point with a horizontal tangent, the curve is the graph of a function of $x$ , so the Second Derivative Test still works. Find $\frac{d^2y}{dx^2}$ by differentiating the fraction for $\frac{dy}{dx}$ (quotient rule, treating $y$ as a function of $x$ ), then substitute the point and the value $\frac{dy}{dx} = 0$ .

Worked example 3. For $x^2 - xy + y^2 = 12$ , find $\frac{d^2y}{dx^2}$ at $(2, 4)$ .

$\frac{d^2y}{dx^2}$ $= \frac{\left(\frac{dy}{dx} - 2\right)(2y - x) - (y - 2x)\left(2\frac{dy}{dx} - 1\right)}{(2y - x)^2}$ .

At $(2, 4)$ , $\frac{dy}{dx} = 0$ and $y - 2x = 0$ : $\frac{(-2)(6) - 0}{36} = -\frac{1}{3}$ .

A horizontal tangent where the curve is concave down: $(2, 4)$ is a local high point of the curve, and in fact its highest point.

What earns the points

Apply $\frac{d}{dx}$ to both sides and keep " $= 0$ " on the right.
Use the product rule on every $xy$ -type term and the chain rule on every power of $y$ .
For a vertical tangent, say "the denominator is 0 and the numerator is not," never " $\frac{dy}{dx} = 0$ ."
Substitute candidate points back into the curve's equation.