Behaviors of implicit relations — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Implicit differentiation for first and second derivatives, tangent and normal lines to implicit curves, horizontal and vertical tangent identification, critical point classification, and concavity/extrema analysis for non-explicit relations.

You should already know: Chain rule for composite functions of x. Derivative rules for product, quotient, and all elementary functions. Point-slope form for linear equations.

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 BC 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?

Implicit relations are equations relating x and y that cannot be rearranged to write y explicitly as a function of x (or x as a function of y). Unlike explicit functions of the form $y=f(x)$, they often describe closed curves or multi-valued relations, meaning a single x can correspond to multiple y-values. Even though they are not functions, they still have meaningful geometric behavior we can analyze with derivatives.

This topic is formally part of AP Calculus BC CED Unit 5: Analytical Applications of Differentiation, which accounts for 13–18% of the total AP exam score. Questions about behaviors of implicit relations regularly appear on both multiple-choice (MCQ) and free-response (FRQ) sections of the exam, most often as standalone derivative/tangent problems or in combination with related rates, optimization, or curve sketching. The core skill here is using implicit differentiation to extract derivative information, then applying that information to characterize the relation's geometric behavior, just as we do for explicit functions.

2. Implicit Differentiation for First and Second Derivatives

Implicit differentiation is the core technique for analyzing any implicit relation. It relies entirely on the chain rule: because y is a function of x (even if we cannot write it explicitly), any term containing y must be differentiated with respect to y first, then multiplied by $\frac{dy}{dx}$ (or $y^{\prime }$) per the chain rule.

For any implicit equation $F(x,y)=C$ (a constant), we differentiate every term on both sides of the equation with respect to x. For terms that contain both x and y, we first apply the product or quotient rule as needed, then apply the chain rule to any derivative of y. After differentiating, we collect all terms with $\frac{dy}{dx}$ on one side of the equation, factor out $\frac{dy}{dx}$, and solve for the first derivative. This gives $\frac{dy}{dx}$ as an expression in both x and y.

For the second derivative $\frac{d^{2}y}{d{x}^2}$, we differentiate the expression for $\frac{dy}{dx}$ again with respect to x. We still apply the chain rule to any term containing y, so we will get new terms that include $\frac{dy}{dx}$. We substitute the already-solved expression for $\frac{dy}{dx}$ back into the result to get $\frac{d^{2}y}{d{x}^2}$ entirely in terms of x and y.

Worked Example

Find $\frac{dy}{dx}$ and $\frac{d^{2}y}{d{x}^2}$ for the relation $x^2+3xy+y^2=11$ at the point $(2,1)$.

1. Differentiate both sides with respect to x, applying product rule to $3xy$ and chain rule to y-terms: $$2x+3\left(x\frac{dy}{dx}+y\right)+2y\frac{dy}{dx}=0$$
2. Collect terms with $\frac{dy}{dx}$ and solve: $$\frac{dy}{dx}(3x+2y)=-2x-3y⟹\frac{dy}{dx}=\frac{-2x-3y}{3x+2y}$$
3. Evaluate at $(2,1)$: $\frac{dy}{dx}=\frac{-4-3}{6+2}=-\frac{7}{8}$.
4. Differentiate $\frac{dy}{dx}$ with the quotient rule to get the second derivative: $$\frac{d^{2}y}{d{x}^2}=\frac{(3x+2y)(-2-3\frac{dy}{dx})-(-2x-3y)(3+2\frac{dy}{dx})}{(3x+2y{)}^2}$$
5. Substitute $x=2,y=1,\frac{dy}{dx}=-\frac{7}{8}$: Numerator simplifies to $\frac{55}{4}$, denominator is $64$, so $\frac{d^{2}y}{d{x}^2}=\frac{55}{256}\approx 0.215$.

Exam tip: Always evaluate $\frac{dy}{dx}$ at the given point before finding the second derivative, rather than substituting the point at the end. This drastically simplifies arithmetic and reduces the chance of algebra errors on the AP exam.

3. Tangent and Normal Lines, Horizontal and Vertical Tangents

Once you have $\frac{dy}{dx}$ at a point on the implicit curve, $\frac{dy}{dx}$ is exactly the slope of the tangent line to the curve at that point, just like for explicit functions. The normal line is perpendicular to the tangent line at that point, so its slope is the negative reciprocal of the tangent slope (as long as the tangent slope is not zero or undefined). We use point-slope form $y-y_0=m(x-x_0)$ to write the equation of either line, just like with explicit curves.

AP exam questions very frequently ask for all points on an implicit curve that have horizontal or vertical tangents. A horizontal tangent has a slope of zero, so this occurs when $\frac{dy}{dx}=0$. For a rational $\frac{dy}{dx}$, this means the numerator of $\frac{dy}{dx}$ is zero, and the denominator is non-zero (to confirm the slope is actually zero, not undefined). A vertical tangent has an undefined slope, so this occurs when the denominator of $\frac{dy}{dx}$ is zero, and the numerator is non-zero. All candidate points must be checked to confirm they lie on the original implicit curve.

Worked Example

Find the equations of the tangent line and normal line to the ellipse $2x^2+y^2=18$ at the point $(2,\sqrt{10})$.

1. Differentiate implicitly with respect to x: $4x+2y\frac{dy}{dx}=0⟹\frac{dy}{dx}=-\frac{2x}{y}$.
2. Evaluate the tangent slope at $(2,\sqrt{10})$: $m_{\text{tan}}=-\frac{2(2)}{\sqrt{10}}=-\frac{2\sqrt{10}}{5}$.
3. Write the tangent line using point-slope form: $$y-\sqrt{10}=-\frac{2\sqrt{10}}{5}(x-2)$$ Simplified standard form: $2\sqrt{10}x+5y=9\sqrt{10}$.
4. The normal line slope is the negative reciprocal: $m_{\text{norm}}=\frac{\sqrt{10}}{4}$.
5. Write the normal line: $y-\sqrt{10}=\frac{\sqrt{10}}{4}(x-2)$, which simplifies to $\sqrt{10}x-4y+2\sqrt{10}=0$.

Exam tip: If asked to find all points with horizontal/vertical tangents, always check that your candidate points actually lie on the original implicit curve. Solving the slope condition is not enough; the point must satisfy the original relation to be valid.

4. Analyzing Critical Points, Extrema, and Concavity

The same rules we use to analyze extrema and concavity for explicit functions apply directly to implicit relations. A critical point on an implicit curve is a point where $\frac{dy}{dx}=0$ (horizontal tangent, candidate for local maximum or minimum of y with respect to x) or $\frac{dy}{dx}$ is undefined (vertical tangent, candidate for extremum of x with respect to y).

To classify critical points, we use the second derivative test just like for explicit functions: if $\frac{dy}{dx}=0$ at a critical point, then $\frac{d^{2}y}{d{x}^2}>0$ means the point is a local minimum, and $\frac{d^{2}y}{d{x}^2}<0$ means the point is a local maximum. For concavity, if $\frac{d^{2}y}{d{x}^2}>0$ the curve is concave up at that point, and if $\frac{d^{2}y}{d{x}^2}<0$ it is concave down. Inflection points occur where concavity changes, just like for explicit functions.

Worked Example

Find all local maxima and minima of the relation $x^2-xy+y^2=3$.

1. Differentiate implicitly: $2x-\left(x\frac{dy}{dx}+y\right)+2y\frac{dy}{dx}=0$. Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{y-2x}{2y-x}$$
2. Critical points occur when $\frac{dy}{dx}=0$, so numerator $y-2x=0⟹y=2x$. Denominator $2y-x\ne 0$.
3. Substitute $y=2x$ into the original equation to find valid points: $x^2-x(2x)+(2x{)}^2=3⟹3x^2=3⟹x=\pm 1$. The critical points are $(1,2)$ and $(-1,-2)$.
4. Use the quotient rule to find the second derivative, substitute $\frac{dy}{dx}=0$ for both points:

• At $(1,2)$: $\frac{d^{2}y}{d{x}^2}=\frac{(0-2)(4-1)}{3^2}=-\frac{6}{9}=-\frac{2}{3}<0$, so this is a local maximum.
• At $(-1,-2)$: $\frac{d^{2}y}{d{x}^2}=\frac{(0-2)(-4+1)}{(-3{)}^2}=\frac{6}{9}=\frac{2}{3}>0$, so this is a local minimum.

Exam tip: When evaluating the second derivative at a critical point, remember $\frac{dy}{dx}=0$, so this term will drop out of most expressions, drastically simplifying calculation. Always substitute $\frac{dy}{dx}=0$ first before doing any other arithmetic.

5. Common Pitfalls (and how to avoid them)

• Wrong move: After differentiating a product term like $2xy$, you write $2x\frac{dy}{dx}$ and forget the product rule term $2y$. Why: Students remember to apply chain rule to y but forget that terms with both x and y require the product rule before applying the chain rule. Correct move: Always use the product/quotient rule first for any term that has both x and y as factors, then apply the chain rule to any derivative of y.
• Wrong move: When finding $\frac{d^{2}y}{d{x}^2}$, you differentiate $\frac{dy}{dx}$ and leave out the chain rule factor of $\frac{dy}{dx}$ when differentiating terms that include y. Why: Students forget that y is still a function of x when taking the second derivative, so all y terms still require a $\frac{dy}{dx}$ factor. Correct move: Whenever you differentiate a term containing y at any derivative order, multiply by $\frac{dy}{dx}$ (or $\frac{d^{2}y}{d{x}^2}$ as appropriate) per the chain rule.
• Wrong move: When asked for points with horizontal tangents, you set the denominator of $\frac{dy}{dx}$ equal to zero instead of the numerator. Why: Confusion between the conditions for horizontal vs vertical tangents, common when memorizing conditions without context. Correct move: Remember $\frac{dy}{dx}=$ slope: horizontal slope $=0$, so numerator $=0$ (denominator $\ne 0$); vertical slope is undefined, so denominator $=0$ (numerator $\ne 0$). Write this on your exam scratch paper if needed.
• Wrong move: When finding extrema, you keep $\frac{dy}{dx}$ as a variable when evaluating $\frac{d^{2}y}{d{x}^2}$ at a critical point, leading to complicated unsimplified expressions. Why: Students follow a generic formula instead of using the fact that $\frac{dy}{dx}=0$ at all critical points from the first derivative condition. Correct move: Substitute $\frac{dy}{dx}=0$ into the second derivative expression immediately before evaluating, which eliminates most terms and simplifies arithmetic.
• Wrong move: After finding that all horizontal tangents occur at $y=2x$, you just report all points $(x,2x)$ as having horizontal tangents without finding specific x values that lie on the original curve. Why: Students think satisfying the derivative condition is enough, and don't realize only specific points on the line $y=2x$ lie on the original implicit curve. Correct move: Always substitute the derivative condition (e.g. $y=2x$ for $\frac{dy}{dx}=0$) back into the original curve equation to solve for the specific x and y coordinates of the tangent points.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

The curve defined by $x^3+2xy+y^2=5$ contains the point $(1,1)$. What is the slope of the tangent line to the curve at this point? A) $-\frac{5}{4}$ B) $-\frac{5}{2}$ C) $\frac{5}{4}$ D) $-1$

Worked Solution: We use implicit differentiation on both sides with respect to x. Apply the product rule to the $2xy$ term and chain rule to y-terms: $$\frac{d}{dx}\left(x^3+2xy+y^2\right)=\frac{d}{dx}(5)⟹3x^2+2\left(x\frac{dy}{dx}+y\right)+2y\frac{dy}{dx}=0$$ Collect terms with $\frac{dy}{dx}$: $\frac{dy}{dx}(2x+2y)=-3x^2-2y$. Substitute $x=1,y=1$: $\frac{dy}{dx}(2+2)=-3-2=-5$, so $\frac{dy}{dx}=-\frac{5}{4}$. The correct answer is A.

Question 2 (Free Response)

Consider the implicit relation $x^2-2y^2+xy=4$. (a) Find $\frac{dy}{dx}$ in terms of x and y. (b) Find all points on the curve where the tangent line is horizontal. Show your work. (c) For the point $(2,2)$ on the curve, find $\frac{d^{2}y}{d{x}^2}$ and determine if the curve is concave up or concave down at this point.

Worked Solution: (a) Differentiate both sides with respect to x: $$2x-4y\frac{dy}{dx}+\left(x\frac{dy}{dx}+y\right)=0$$ Collect $\frac{dy}{dx}$ terms: $\frac{dy}{dx}(x-4y)=-2x-y⟹\frac{dy}{dx}=\frac{2x+y}{4y-x}$.

(b) Horizontal tangents require $\frac{dy}{dx}=0$, so numerator $2x+y=0⟹y=-2x$. Substitute into the original equation: $$x^2-2(-2x{)}^2+x(-2x)=x^2-8x^2-2x^2=-9x^2=4$$ This has no real solutions for x, so there are no points on the curve with horizontal tangent lines.

(c) At $(2,2)$, first find $\frac{dy}{dx}=\frac{4+2}{8-2}=1$. Use the quotient rule to find the second derivative: $$\frac{d^{2}y}{d{x}^2}=\frac{(2+\frac{dy}{dx})(4y-x)-(2x+y)(4\frac{dy}{dx}-1)}{(4y-x{)}^2}$$ Substitute $x=2,y=2,\frac{dy}{dx}=1$: numerator $=(3)(6)-(6)(3)=0$, denominator $=36$, so $\frac{d^{2}y}{d{x}^2}=0$. Concavity is indeterminate at this point, which is a candidate for an inflection point.

Question 3 (Application / Real-World Style)

A civil engineer is modeling the cross-section of a curved small bridge arch using the implicit relation $x^2+4y^2-0.1x{y}^2=100$, where x and y are measured in meters, and the origin $(0,0)$ is at the center base of the arch. y is the height of the arch above the road (which runs along the x-axis). A horizontal support beam needs to be placed at the highest point of the arch. Find the height of the highest point of the arch, justifying your answer.

Worked Solution: The highest point is a local maximum of y, which occurs at a horizontal tangent where $\frac{dy}{dx}=0$ and $\frac{d^{2}y}{d{x}^2}<0$. Differentiate implicitly: $$2x+8y\frac{dy}{dx}-0.1\left(y^2+2xy\frac{dy}{dx}\right)=0$$ Set $\frac{dy}{dx}=0$, so all $\frac{dy}{dx}$ terms drop out: $2x-0.1y^2=0⟹x=0.05y^2$. Substitute into the original equation: $$(0.05y^2{)}^2+4y^2-0.1(0.05y^2)y^2=100⟹-0.0025y^4+4y^2-100=0$$ Let $u=y^2$, solve the quadratic: $u^2-1600u+40000=0⟹u\approx 25.4$ (the other solution corresponds to the lower edge of the closed arch). This gives $y\approx 5.04$ meters. The second derivative at this point is negative, confirming it is a maximum. In context, the highest point of the bridge arch is approximately 5.04 meters above the road.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the analysis of implicit relations prepares you for core topics across the rest of the AP Calculus BC syllabus. Immediately next, you will apply implicit differentiation to related rates problems, which describe how multiple changing quantities relate to each other over time. Without the ability to correctly differentiate and analyze implicit relations, nearly all non-trivial related rates problems are impossible to solve. This topic also acts as a prerequisite for analyzing parametric and polar curves, where tangent slope calculations rely on implicit differentiation, and for optimization problems where the constraint between variables is given as an implicit relation. It also lays the groundwork for implicit solutions to differential equations later in the course. Follow-on topics to practice next:

Related rates Tangent lines to parametric curves Implicit differential equation solutions Curve sketching for non-explicit relations