Behaviors of implicit relations — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: This chapter covers finding first and second derivatives via implicit differentiation, identifying horizontal/vertical tangents, classifying increasing/decreasing behavior, and determining concavity for implicit relations.

You should already know: Basic derivative rules including chain rule and quotient rule, derivative sign interpretation for increasing/decreasing behavior, second derivative sign interpretation for concavity.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Behaviors of implicit relations?

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 the same behavior information we use for explicit functions, directly from the original implicit equation. According to the AP Calculus AB Course and Exam Description (CED), this topic is part of Unit 5: Analytical Applications of Differentiation, accounting for ~2-4% of total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam. It is often combined with other topics like tangent line equations or curve sketching in multi-part questions.

2. Finding Slopes of Tangent Lines for Implicit Relations

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 $x$). The process follows four core steps: (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.

This method works because we only need the slope at a specific point, so we do not need to solve for $y$ explicitly first. For most AP problems, you only need the slope at a given point, so you can substitute values early to reduce algebraic complexity.

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 $x$: $$\frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(2xy\right)+\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}\left(5\right)$$
2. Apply power rule, product rule, and chain rule: $3x^2+2\left(y+x\frac{dy}{dx}\right)+2y\frac{dy}{dx}=0$
3. Expand and collect terms with $\frac{dy}{dx}$: $$3x^2+2y+\frac{dy}{dx}\left(2x+2y\right)=0⟹\frac{dy}{dx}=\frac{-3x^2-2y}{2x+2y}$$
4. Substitute $x=1,y=1$: $\frac{dy}{dx}=\frac{-3(1{)}^2-2(1)}{2(1)+2(1)}=\frac{-5}{4}$

The slope of the tangent line at $(1,1)$ is $-\frac{5}{4}$.

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.

3. Finding Horizontal and Vertical Tangents

A key AP exam question type asks to find all points on an implicit relation where the tangent line is horizontal or vertical. We use the properties of $\frac{dy}{dx}$ (written as a fraction $\frac{N(x,y)}{D(x,y)}$, where $N$ is numerator and $D$ is denominator) to find these points:

• Horizontal tangents occur when $\frac{dy}{dx}=0$, which requires $N(x,y)=0$, as long as $D(x,y)\ne 0$ at that point. This matches what we know for explicit functions: slope zero means horizontal tangent.
• Vertical tangents occur when the slope is undefined, which means $D(x,y)=0$, as long as $N(x,y)\ne 0$ at that point. This is equivalent to $\frac{dx}{dy}=0$, since a vertical line has an undefined slope.

If both $N(x,y)=0$ and $D(x,y)=0$, the point is singular, and AP rarely asks for the tangent at these points. 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 $\frac{dy}{dx}$: $2x+8y\frac{dy}{dx}-2=0⟹\frac{dy}{dx}=\frac{2-2x}{8y}=\frac{1-x}{4y}$
2. Set numerator equal to zero for horizontal tangents: $1-x=0⟹x=1$
3. Substitute $x=1$ back into the original relation to find corresponding $y$-values: $$(1{)}^2+4y^2-2(1)=3⟹4y^2=4⟹y^2=1⟹y=1\text{ or }y=-1$$
4. Check the denominator at both points: at $(1,1)$, $D=4(1)=4\ne 0$; at $(1,-1)$, $D=4(-1)=-4\ne 0$, so both points are valid.

The points with horizontal tangents are $(1,1)$ and $(1,-1)$.

Exam 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.

4. Finding Second Derivatives and Concavity

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

The process for finding the second derivative is: (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 $\frac{d^{2}y}{d{x}^2}$.

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: $2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0⟹\frac{dy}{dx}=\frac{-2x-y}{x+2y}$
2. At $(1,1)$, $\frac{dy}{dx}=\frac{-2-1}{1+2}=-1$
3. Differentiate $\frac{dy}{dx}$ using the quotient rule to get $\frac{d^{2}y}{d{x}^2}$: $$\frac{d^{2}y}{d{x}^2}=\frac{(-2-\frac{dy}{dx})(x+2y)-(-2x-y)(1+2\frac{dy}{dx})}{(x+2y{)}^2}$$
4. Substitute $x=1,y=1,\frac{dy}{dx}=-1$: $$\frac{d^{2}y}{d{x}^2}=\frac{(-2-(-1))(3)-(-3)(1-2)}{9}=\frac{(-1)(3)-(-3)(-1)}{9}=\frac{-6}{9}=-\frac{2}{3}$$

Since $\frac{d^{2}y}{d{x}^2}=-\frac{2}{3}<0$, the relation is concave down at $(1,1)$.

Exam 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.

5. Common Pitfalls (and how to avoid them)

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

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

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

Worked Solution: To find points with horizontal tangents, first compute $\frac{dy}{dx}$ via implicit differentiation. Differentiating both sides of $2x^2+y^2=12$ with respect to $x$ gives $4x+2y\frac{dy}{dx}=0$, so solving for $\frac{dy}{dx}$ gives $\frac{dy}{dx}=\frac{-2x}{y}$. Horizontal tangents occur when $\frac{dy}{dx}=0$, which requires the numerator to be zero (and non-zero denominator), so $-2x=0⟹x=0$. Only option A has $x=0$, and substituting back into the original relation confirms $(0,2\sqrt{3})$ lies on the curve with a non-zero denominator. Correct answer: A

Question 2 (Free Response)

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

Worked Solution: (a) Differentiate both sides with respect to $x$: $$2x+y+x\frac{dy}{dx}-2y\frac{dy}{dx}=0$$ Collect terms with $\frac{dy}{dx}$: $$\frac{dy}{dx}(x-2y)=-2x-y⟹\frac{dy}{dx}=\frac{-(2x+y)}{x-2y}$$

(b) Vertical tangents occur when the denominator of $\frac{dy}{dx}$ is zero and the numerator is non-zero. Set denominator equal to zero: $x-2y=0⟹x=2y$. Substitute $x=2y$ into the original relation: $$(2y{)}^2+(2y)y-y^2=5⟹5y^2=5⟹y^2=1⟹y=1,y=-1$$ Corresponding $x$-values are $x=2(1)=2$ and $x=2(-1)=-2$. Checking the numerator: both points have non-zero numerator, so valid points are $(2,1)$ and $(-2,-1)$. Vertical tangent equations are $\boxed{x=2}$ and $\boxed{x=-2}$.

(c) First, find $\frac{dy}{dx}$ at $(1,1)$: $\frac{dy}{dx}=\frac{-(2+1)}{1-2(1)}=3$. Use quotient rule to find $\frac{d^{2}y}{d{x}^2}$: $$\frac{d^{2}y}{d{x}^2}=\frac{-(2+\frac{dy}{dx})(x-2y)+(2x+y)(1-2\frac{dy}{dx})}{(x-2y{)}^2}$$ Substitute $x=1,y=1,\frac{dy}{dx}=3$: $$\frac{d^{2}y}{d{x}^2}=\frac{-(5)(-1)+(3)(-5)}{(-1{)}^2}=5-15=-10$$ Since $\frac{d^{2}y}{d{x}^2}=-10<0$, the curve is concave down at $(1,1)$.

Question 3 (Application / Real-World Style)

A civil engineer models the cross-section of a curved overpass arch using the implicit relation $x^2+20y-100y^2=100$, where $x$ is horizontal distance in meters from the center of the arch, and $y$ is vertical height in meters above ground level. At the point $(10,1)$, find the slope of the arch, then classify the arch as increasing/decreasing and concave up/concave down at this point.

Worked Solution: First, differentiate both sides with respect to $x$: $2x+20\frac{dy}{dx}-200y\frac{dy}{dx}=0$. Collect terms and solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{-2x}{20-200y}=\frac{-x}{10-100y}$$ Substitute $x=10,y=1$: $\frac{dy}{dx}=\frac{-10}{10-100}=\frac{1}{9}>0$, so the arch is increasing at this point. Next, use quotient rule to find $\frac{d^{2}y}{d{x}^2}$: $$\frac{d^{2}y}{d{x}^2}=\frac{-(10-100y)+x(100\frac{dy}{dx})}{(10-100y{)}^2}$$ Substitute values: $\frac{d^{2}y}{d{x}^2}=\frac{-(-90)+10(100\cdot \frac{1}{9})}{8100}=\frac{90-\frac{1000}{9}}{8100}=\frac{-190}{72900}<0$, so the arch is concave down.

Interpretation: 10 meters right of the arch center, the overpass road surface rises as you move right and curves downward, matching the expected shape of the right side of a symmetric arch.

7. Quick Reference Cheatsheet

8. 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. Next, 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. Follow-on topics: Related rates Optimization of functions Curve sketching Tangent and normal lines