Behavior of Implicit Relations — AP® Calculus AB

Part 1 · Hook

Why this matters

A race track is not the graph of a function: it loops back, and at some points it runs straight up the page. Designers still need to know where the track runs exactly east-west or north-south. Implicit differentiation gives the slope at every point of such a curve, and the slope formula shows where it is flat or vertical.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Differentiating $y^2$ with respect to $x$ gives

$2y\frac{dy}{dx}$
$2y$
$2x$
$y\frac{dy}{dx}$

Show the answer

Chain rule: $y$ depends on $x$ .

Correct: $2y\frac{dy}{dx}$ :
$2y$ :
$2x$ :
$y\frac{dy}{dx}$ :

2. A critical point of a function $f$ is an interior point where

$f'(x) = 0$ or $f'(x)$ does not exist
$f(x) = 0$
$f''(x) = 0$
$f$ is largest

Show the answer

The same idea, for curves, uses $\frac{dy}{dx}$ .

Correct: $f'(x) = 0$ or $f'(x)$ does not exist:
$f(x) = 0$ :
$f''(x) = 0$ :
$f$ is largest:

3. If $f'(c) = 0$ and $f''(c) < 0$ , $f$ has

a relative maximum at $c$
a relative minimum at $c$
a point of inflection at $c$
no conclusion

Show the answer

The Second Derivative Test.

Correct: a relative maximum at $c$ :
a relative minimum at $c$ :
a point of inflection at $c$ :
no conclusion:

Part 4 · See it

See it first

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. — On $x^2 + xy + y^2 = 3$ , the numerator of $\frac{dy}{dx}$ vanishes at the blue points (horizontal tangent lines) and the denominator at the green ones (vertical tangent lines). LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Differentiate both sides of the curve's equation with respect to $x$ , then solve $\frac{dy}{dx}$ $= \frac{N(x, y)}{D(x, y)}$ , a slope at each point $(x, y)$ of the curve
$N = 0$ , $D \ne 0$ , and the point is on the curvethe tangent line is horizontal there
$D = 0$ , $N \ne 0$ , and the point is on the curvethe tangent line is vertical there
$N = 0$ and $D = 0$ at oncethe formula decides nothing at that point
$\frac{dy}{dx} = 0$ and the sign of $\frac{d^2y}{dx^2}$ is knownthe curve has a local high point ( $< 0$ ) or low point ( $> 0$ )

Part 6 · Key ideas

Key ideas

Horizontal tangent: numerator of $\frac{dy}{dx}$ is 0, denominator is not, point is on the curve. Vertical tangent: denominator 0, numerator not 0, point on the curve.
To find the points: set the numerator (or denominator) to 0, solve for one variable, substitute into the curve's equation, then check the other part is not 0.
$\frac{d^2y}{dx^2}$ : differentiate $\frac{dy}{dx}$ (quotient rule), then substitute the point and the value of $\frac{dy}{dx}$ there.
Write " $= 0$ " after differentiating, and apply $\frac{d}{dx}$ to both sides.

Part 7 · Misconception

A common mistake

The wrong idea: The tangent line is vertical where $\frac{dy}{dx} = 0$ .

What actually happens: $\frac{dy}{dx} = 0$ means a horizontal tangent. A vertical tangent is where the slope is undefined because the denominator is 0 and the numerator is not. And every candidate point must satisfy the curve's equation; a point that only makes the numerator zero may not be on the curve at all.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The curve $x^2 - xy + y^2 = 12$

The graph shows the tilted oval

x^2 - xy + y^2 = 12

. Marked points:

(2, 4)

,

(-2, -4)

,

(4, 2)

,

(-4, -2)

and

(-2, 2)

.

1. Differentiate the oval's equation implicitly and solve: find $\frac{dy}{dx}$ in terms of $x$ and $y$ .

Type an expression for $\frac{dy}{dx} =$ .

Show the answer

$\frac{d}{dx}$ of both sides: $2x - \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx}$ $= 0$ . So $\frac{dy}{dx}(2y - x) = y - 2x$ and $\frac{dy}{dx}$ $= \frac{y - 2x}{2y - x}$ .

Answer: $\frac{dy}{dx} =$ $\frac{y - 2x}{2y - x}$

2. At which points is the line tangent to the curve horizontal?

$(2, 4)$ and $(-2, -4)$
$(4, 2)$ and $(-4, -2)$
$(0, 2\sqrt{3})$ and $(0, -2\sqrt{3})$
$(-2, 2)$ and $(2, -2)$

Show the answer

Horizontal: numerator $y - 2x = 0$ with the point on the curve and the denominator not 0. Substituting $y = 2x$ : $3x^2 = 12$ , so $(2, 4)$ and $(-2, -4)$ ; the denominator there is $6$ and $-6$ .

Correct: $(2, 4)$ and $(-2, -4)$ : $y = 2x$ makes the numerator 0; on the curve, $x^2 - 2x^2 + 4x^2 = 12$ gives $x = \pm 2$ .
$(4, 2)$ and $(-4, -2)$ : There the denominator $2y - x$ is 0: vertical tangent lines.
$(0, 2\sqrt{3})$ and $(0, -2\sqrt{3})$ : At $(0, 2\sqrt{3})$ , $\frac{dy}{dx}$ $= \frac{2\sqrt{3}}{4\sqrt{3}}$ $= \frac{1}{2}$ .
$(-2, 2)$ and $(2, -2)$ : At $(-2, 2)$ , $\frac{dy}{dx} = 1$ .

3. At which points is the line tangent to the curve vertical?

$(2, 4)$ and $(-2, -4)$
$(2\sqrt{3}, 0)$ and $(-2\sqrt{3}, 0)$
$(0, 0)$ , where both parts are 0
$(4, 2)$ and $(-4, -2)$

Show the answer

Vertical: denominator $2y - x = 0$ , numerator not 0, point on the curve. $x = 2y$ gives $4y^2 - 2y^2 + y^2 = 12$ , so $y = \pm 2$ : $(4, 2)$ and $(-4, -2)$ .

$(2, 4)$ and $(-2, -4)$ : There the numerator is 0: horizontal tangent lines.
$(2\sqrt{3}, 0)$ and $(-2\sqrt{3}, 0)$ : At $(2\sqrt{3}, 0)$ , $\frac{dy}{dx}$ $= \frac{-4\sqrt{3}}{-2\sqrt{3}}$ $= 2$ .
$(0, 0)$ , where both parts are 0: $(0, 0)$ is not on the curve: $0 \ne 12$ .
Correct: $(4, 2)$ and $(-4, -2)$ : $x = 2y$ makes the denominator 0; on the curve, $3y^2 = 12$ gives $y = \pm 2$ ; the numerator is $-6$ and $6$ , not 0.

4. Write an equation for the line tangent to the curve at $(-2, 2)$ . Enter the right side.

Type an expression for $y =$ .

Show the answer

$\frac{dy}{dx}$ at $(-2, 2)$ : $\frac{2 - 2(-2)}{2(2) - (-2)}$ $= \frac{6}{6}$ $= 1$ . Tangent line: $y - 2 = 1(x + 2)$ , so $y = x + 4$ .

Answer: $y =$ $x + 4$

5. At the highest point $(2, 4)$ of the oval, find $\frac{d^2y}{dx^2}$ , which tells how the curve bends there.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Quotient rule: $\frac{d^2y}{dx^2}$ $= \frac{\left(\frac{dy}{dx} - 2\right)(2y - x) - (y - 2x)\left(2\frac{dy}{dx} - 1\right)}{(2y - x)^2}$ . At $(2, 4)$ , $\frac{dy}{dx} = 0$ : $\frac{(-2)(6) - 0}{36} = -\frac{1}{3}$ . Concave down at a horizontal tangent: $(2, 4)$ is the highest point.

Answer: $\frac{-1}{3}$ ≈ -0.333

6. For a curve with $\frac{dy}{dx}$ $= \frac{N(x, y)}{D(x, y)}$ , which statement earns credit for "the tangent line is vertical at $P$ "?

" $\frac{dy}{dx} = 0$ at $P$ ."
" $D = 0$ and $N \ne 0$ at $P$ , and $P$ is on the curve."
" $D = 0$ at $P$ ."
" $N = 0$ and $D \ne 0$ at $P$ , and the point $P$ is on the curve."

Show the answer

Vertical tangent: the denominator is 0, the numerator is not, and the point lies on the curve. Chief readers see " $\frac{dy}{dx} = 0$ " used for vertical tangents every year; it is the condition for horizontal ones.

" $\frac{dy}{dx} = 0$ at $P$ .": That is a horizontal tangent.
Correct: " $D = 0$ and $N \ne 0$ at $P$ , and $P$ is on the curve.": All three conditions.
" $D = 0$ at $P$ .": If $N = 0$ too, nothing is decided; and $P$ must be on the curve.
" $N = 0$ and $D \ne 0$ at $P$ , and the point $P$ is on the curve.": That is a horizontal tangent.

Part 9 · Summary

Summary

For an implicit curve,

\frac{dy}{dx} = \frac{N}{D}

. Horizontal tangents:

N = 0

,

D \ne 0

, on the curve. Vertical tangents:

D = 0

,

N \ne 0

, on the curve. Both zero: no conclusion. At a horizontal tangent, the sign of

\frac{d^2y}{dx^2}

tells whether the curve has a local high or low point.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The curve x2−xy+y2=12x^2 - xy + y^2 = 12

Summary

What comes next

Connections

The curve $x^2 - xy + y^2 = 12$