# Behaviors of implicit relations

> AP Calculus AB · Analytical Applications of Differentiation
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u5-behaviors-of-implicit-relations/

This module covers analyzing slope, tangent location, increasing/decreasing behavior, and concavity for equations not solved explicitly for $y$. You will learn to use implicit differentiation to find first and second derivatives for these relations.

**Prerequisites:** Basic derivative rules (power, product, quotient); Chain rule for composite functions; Derivative sign interpretation for increasing/decreasing; Second derivative interpretation for concavity

## Learning objectives

- Find first derivatives of implicit relations via implicit differentiation
- Identify points with horizontal and vertical tangents on implicit curves
- Calculate second derivatives of implicit relations to determine concavity
- Interpret increasing/decreasing behavior for implicit relations

## Core Concepts of Implicit Relation Analysis

Behaviors of implicit relations refers to the analysis of slope, tangent location, increasing/decreasing tendency, and concavity for equations that are not solved explicitly for $y$ as a function of $x$. Unlike explicit functions of the form $y = f(x)$, implicit relations can have multiple $y$-values for a single $x$, so we cannot simply differentiate an explicit expression for $y$ to get derivative information. Instead, we use implicit differentiation to extract behavior information directly from the original implicit equation. This topic accounts for ~2-4% of the total AP Calculus AB exam score, appearing in both multiple-choice and free-response sections.

**Implicit Relation** — An equation relating $x$ and $y$ that is not written in the explicit form $y = f(x)$, which may have multiple $y$ values for a given $x$.

*Example:* $x^2 + y^2 = 4$ is an implicit relation representing a circle.

## Tangent Slopes and Special Tangents

To find the slope of a tangent line at a point on an implicit relation, we use implicit differentiation, which leverages the chain rule to differentiate terms containing $y$ (since $y$ is a function of $x$).

1. Differentiate every term on both sides of the equation with respect to $x$
2. Apply the chain rule to any term with a $y$, multiplying by $\frac{dy}{dx}$ to account for the derivative of the inner function $y(x)$
3. Collect all terms containing $\frac{dy}{dx}$ on one side of the equation and all other terms on the opposite side
4. Factor out $\frac{dy}{dx}$ and solve for it in terms of $x$ and $y$, then substitute the coordinates of the point of interest to get the slope

**Worked example:** Find the slope of the tangent line to the implicit relation $x^3 + 2xy + y^2 = 5$ at the point $(1, 1)$.

1. Differentiate both sides of the equation with respect to $x$:
2. $$frac{d}{dx}left(x^3right) + frac{d}{dx}left(2xyright) + frac{d}{dx}left(y^2right) = frac{d}{dx}left(5right)$$
3. Apply power rule, product rule, and chain rule:
4. $$3x^2 + 2left(y + xfrac{dy}{dx}right) + 2yfrac{dy}{dx} = 0$$
5. Expand and collect terms with $\frac{dy}{dx}$, then solve for $\frac{dy}{dx}$:
6. $$3x^2 + 2y + frac{dy}{dx}left(2x + 2yright) = 0 implies frac{dy}{dx} = frac{-3x^2 - 2y}{2x + 2y}$$
7. Substitute $x=1, y=1$ to get the slope:
8. $$frac{dy}{dx} = frac{-3(1)^2 - 2(1)}{2(1) + 2(1)} = -frac{5}{4}$$
9. The slope of the tangent line at $(1,1)$ is $-frac{5}{4}$.

A common AP exam question asks to find all points on an implicit relation where the tangent line is horizontal or vertical. For $\frac{dy}{dx}$ written as a fraction $\frac{N(x,y)}{D(x,y)}$:

- **Horizontal tangents**: Occur when $\frac{dy}{dx} = 0$, which requires $N(x,y) = 0$, as long as $D(x,y) \neq 0$ at that point.
- **Vertical tangents**: Occur when slope is undefined, which means $D(x,y) = 0$, as long as $N(x,y) \neq 0$ at that point.
- If both $N(x,y) = 0$ and $D(x,y) = 0$, the point is singular, and AP rarely asks for tangents at these points.

> **tip**
>
> Always find the full coordinates of points by substituting your solution for $x$ or $y$ back into the original implicit relation, not just the derivative.

**Worked example:** For the relation $x^2 + 4y^2 - 2x = 3$, find all points where the tangent line is horizontal.

1. Differentiate to find $\frac{dy}{dx}$:
2. $$2x + 8yfrac{dy}{dx} - 2 = 0 implies frac{dy}{dx} = frac{1 - x}{4y}$$
3. Set numerator equal to zero for horizontal tangents:
4. $$1 - x = 0 implies x = 1$$
5. Substitute $x=1$ back into the original relation to find corresponding $y$-values:
6. $$(1)^2 + 4y^2 - 2(1) = 3 implies 4y^2 = 4 implies y = 1 text{ or } y = -1$$
7. Check denominator at both points: $D=4y \neq 0$ for both, so both are valid. The points with horizontal tangents are $(1, 1)$ and $(1, -1)$.

> **tip**
>
> If you are asked for the equations of tangent lines, remember: horizontal tangents have the form $y = k$, vertical tangents have the form $x = h$, where $(h,k)$ is the point of tangency. Many students mix these forms up and lose points.

> **Exam tip:** Do not waste time simplifying the general expression for $\frac{dy}{dx}$ if you only need the slope at a specific point; substitute the point's coordinates immediately after solving for $\frac{dy}{dx}$ to reduce algebra errors.

## Second Derivatives and Concavity

To determine concavity for an implicit relation at a point, we need the second derivative $\frac{d^2y}{dx^2}$, which follows the same sign interpretation as for explicit functions: $\frac{d^2y}{dx^2} > 0$ means concave up, $\frac{d^2y}{dx^2} < 0$ means concave down.

1. First find $\frac{dy}{dx}$ via implicit differentiation, as before.
2. Differentiate the entire expression for $\frac{dy}{dx}$ with respect to $x$, using the product/quotient/chain rule as needed.
3. Substitute the expression you already found for $\frac{dy}{dx}$ into the second derivative to eliminate $\frac{dy}{dx}$ from the right-hand side.
4. Substitute the coordinates of the point to get a numerical value for $\frac{d^2y}{dx^2}$.

> **warning**
>
> The most common mistake here is substituting the point into $\frac{dy}{dx}$ before differentiating, which gives an incorrect constant second derivative of zero.

**Worked example:** Find the concavity of the relation $x^2 + xy + y^2 = 3$ at the point $(1, 1)$.

1. Find $\frac{dy}{dx}$ via implicit differentiation:
2. $$2x + y + xfrac{dy}{dx} + 2yfrac{dy}{dx} = 0 implies frac{dy}{dx} = frac{-2x - y}{x + 2y}$$
3. Calculate $\frac{dy}{dx}$ at $(1, 1)$:
4. $$frac{dy}{dx} = frac{-2 -1}{1 + 2} = -1$$
5. Differentiate $\frac{dy}{dx}$ using the quotient rule to get $\frac{d^2y}{dx^2}$:
6. $$frac{d^2y}{dx^2} = frac{(-2 - frac{dy}{dx})(x + 2y) - (-2x - y)(1 + 2frac{dy}{dx})}{(x + 2y)^2}$$
7. Substitute $x=1, y=1, \frac{dy}{dx} = -1$:
8. $$frac{d^2y}{dx^2} = frac{(-2 - (-1))(3) - (-3)(1 - 2)}{9} = frac{-3 - 3}{9} = -frac{2}{3}$$
9. Since $\frac{d^2y}{dx^2} = -frac{2}{3} < 0$, the relation is concave down at $(1,1)$.

> **tip**
>
> Always substitute $\frac{dy}{dx}$ into the second derivative expression before evaluating at the point; leaving $\frac{dy}{dx}$ unsubstituted will lead to incorrect numerical values for concavity.

## AP-Style Worked Practice Problems

**Worked example:** Multiple Choice: The relation $2x^2 + y^2 = 12$ has a horizontal tangent at which of the following points?<br>A) $(0, 2\sqrt{3})$<br>B) $(2, 2)$<br>C) $(\sqrt{3}, -\sqrt{6})$<br>D) $(-2, 2)$

1. To find points with horizontal tangents, first compute $\frac{dy}{dx}$ via implicit differentiation:
2. $$4x + 2y frac{dy}{dx} = 0 implies frac{dy}{dx} = frac{-2x}{y}$$
3. Horizontal tangents require $\frac{dy}{dx} = 0$, so numerator $-2x = 0 \implies x=0$. Only option A has $x=0$, and $(0, 2\sqrt{3})$ satisfies the original relation with non-zero denominator.
4. Correct answer: **A**

**Worked example:** Free Response: Consider the implicit relation $x^2 + xy - y^2 = 5$.<br>(a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$.<br>(b) Find all points on the curve where the tangent line is vertical, and write the equation of each vertical tangent.<br>(c) Find $\frac{d^2y}{dx^2}$ at the point $(1, 1)$ and state whether the curve is concave up or concave down at this point.

1. Part (a): Differentiate both sides with respect to $x$:
2. $$2x + y + x frac{dy}{dx} - 2y frac{dy}{dx} = 0$$
3. Collect terms with $\frac{dy}{dx}$ and solve:
4. $$frac{dy}{dx} = frac{-(2x + y)}{x - 2y}$$
5. Part (b): Vertical tangents occur when denominator of $\frac{dy}{dx}$ is zero and numerator is non-zero. Set denominator equal to zero:
6. $$x - 2y = 0 implies x = 2y$$
7. Substitute $x=2y$ into the original relation:
8. $$(2y)^2 + (2y)y - y^2 = 5 implies 5y^2 = 5 implies y = 1, y = -1$$
9. Corresponding $x$-values are $x=2$ and $x=-2$, and both have non-zero numerator. Valid points are $(2,1)$ and $(-2,-1)$, so vertical tangent equations are $x=2$ and $x=-2$.
10. Part (c): First find $\frac{dy}{dx}$ at $(1,1)$:
11. $$frac{dy}{dx} = frac{-(2 + 1)}{1 - 2(1)} = 3$$
12. Use quotient rule to find $\frac{d^2y}{dx^2}$:
13. $$frac{d^2y}{dx^2} = frac{ - (2 + frac{dy}{dx})(x - 2y) + (2x + y)(1 - 2frac{dy}{dx}) }{(x - 2y)^2}$$
14. Substitute $x=1, y=1, \frac{dy}{dx}=3$:
15. $$frac{d^2y}{dx^2} = frac{ - (5)(-1) + (3)(-5) }{(-1)^2} = 5 - 15 = -10$$
16. Since $\frac{d^2y}{dx^2} = -10 < 0$, the curve is concave down at $(1,1)$.

## Common pitfalls

- **Wrong:** Differentiating $y^n$ as $n y^{n-1}$ instead of $n y^{n-1} \frac{dy}{dx}$
  - Why it fails: Students treat $y$ as an independent variable like $x$, and forget $y$ is a function of $x$, so chain rule applies.
  - Correct: Always multiply any derivative of a $y$-term by $\frac{dy}{dx}$ when doing implicit differentiation.
- **Wrong:** Claiming a vertical tangent exists at all points that make the denominator of $\frac{dy}{dx}$ zero, even when the numerator is also zero.
  - Why it fails: Students memorize 'denominator zero = vertical tangent' without checking for the 0/0 indeterminate case.
  - Correct: After finding points that make the denominator zero, plug into the numerator of $\frac{dy}{dx}$ to confirm it is non-zero before confirming a vertical tangent.
- **Wrong:** Calculating the second derivative by differentiating $\frac{dy}{dx}$ evaluated at the point, instead of differentiating the general $\frac{dy}{dx}$ first.
  - Why it fails: Students try to save work by plugging in the point early, which eliminates all variables and gives an incorrect second derivative of zero.
  - Correct: Always differentiate the general expression for $\frac{dy}{dx}$ to get $\frac{d^2y}{dx^2}$, then substitute values last.
- **Wrong:** When finding horizontal tangents, stopping after solving for $x$ and not finding the corresponding $y$-coordinate from the original equation.
  - Why it fails: Questions ask for points, so $x$ alone is not a complete answer, and students forget that points must satisfy the original relation.
  - Correct: After solving for the $x$ (or $y$) that makes $\frac{dy}{dx}$ zero, substitute back into the original implicit relation to get full coordinates for all valid points.
- **Wrong:** When finding $\frac{d^2y}{dx^2}$, leaving $\frac{dy}{dx}$ unsubstituted in the final expression before plugging in the point.
  - Why it fails: Students forget $\frac{dy}{dx}$ is already known from the first step, so leaving it in leads to wrong numerical values for $\frac{d^2y}{dx^2}$.
  - Correct: Always substitute the expression for $\frac{dy}{dx}$ into the second derivative before evaluating at a point.

## Cheatsheet

| Category | Rule/Formula | Notes |
| --- | --- | --- |
| Implicit Differentiation Step | 1. Differentiate both sides w.r.t $x$<br>2. Multiply all $y$-terms by $\frac{dy}{dx}$<br>3. Collect $\frac{dy}{dx}$ terms<br>4. Solve for $\frac{dy}{dx}$ | Works for any implicit relation, for general or point-specific slope |
| Tangent Slope | Slope = $\frac{dy}{dx}$ evaluated at $(x_0, y_0)$ | Confirm $(x_0, y_0)$ is on the original relation first |
| Horizontal Tangent Condition | $N(x,y) = 0$, $D(x,y) \neq 0$ | $\frac{dy}{dx} = \frac{N}{D}$; find full points via original relation |
| Vertical Tangent Condition | $D(x,y) = 0$, $N(x,y) \neq 0$ | Ignore points where both are zero (singular points) |
| Increasing/Decreasing | $\frac{dy}{dx} > 0 \implies$ increasing<br>$\frac{dy}{dx} < 0 \implies$ decreasing | Same as explicit functions; applies to local curve branch |
| Second Derivative Process | Differentiate general $\frac{dy}{dx}$, substitute $\frac{dy}{dx}$, evaluate | Differentiate first, substitute values last to avoid errors |
| Concavity | $\frac{d^2y}{dx^2} > 0 \implies$ concave up<br>$\frac{d^2y}{dx^2} < 0 \implies$ concave down | Same interpretation as explicit functions, applies at the point |

## What's next

This topic is the foundation for analyzing non-explicit relations, which come up repeatedly in applied problems across the rest of Unit 5. You will apply the skills from this chapter to related rates and optimization problems, where you almost always start with an implicit relation between changing variables before solving for the required quantity. Without mastering how to find derivatives, slope behavior, and concavity for implicit relations, you will struggle to set up and solve these applied problems correctly, which make up ~8-11% of the total AP Calculus AB exam score. This topic also builds directly into full curve sketching for all relation types, connecting derivative behavior to visual graph shape.

- [Integration and Accumulation of Change Overview](https://www.owlsprep.com/study/ap-calculus-ab-u6-overview/)
- [Exploring accumulations of change](https://www.owlsprep.com/study/ap-calculus-ab-u6-exploring-accumulations-of-change/)
- [Approximating Areas with Riemann Sums](https://www.owlsprep.com/study/ap-calculus-ab-u6-approximating-areas-with-riemann-sums/)

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