# Behaviors of implicit relations

> AP Calculus BC · Unit 5: Analytical Applications of Differentiation
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u5-behaviors-of-implicit-relations/

This guide covers analyzing geometric behavior of implicit relations using implicit differentiation, including finding first/second derivatives, tangent/normal lines, identifying tangents, and classifying critical points, extrema and concavity for non-explicit curves.

**Prerequisites:** Chain rule for composite functions of x; Product and quotient derivative rules; Point-slope form of linear equations

## Learning objectives

- Perform implicit differentiation to find first and second derivatives of implicit relations
- Find equations of tangent and normal lines to implicit curves
- Identify points with horizontal and vertical tangents on implicit curves
- Classify critical points and analyze concavity for implicit relations

## Implicit Differentiation for First and Second Derivatives

Implicit relations are equations relating $x$ and $y$ that cannot be rearranged to write $y$ explicitly as a function of $x$. Even though they are often multi-valued (one $x$ maps to multiple $y$ values), we can still analyze their geometric behavior using derivatives via the technique of implicit differentiation.

**Implicit Differentiation** — A technique to find $\frac{dy}{dx}$ for implicit relations that relies entirely on the chain rule: any term containing $y$ is differentiated with respect to $y$ first, then multiplied by $\frac{dy}{dx}$ to account for $y$ being a function of $x$.

For any implicit equation $F(x,y) = C$ (constant), differentiate every term on both sides with respect to $x$. For terms containing both $x$ and $y$, apply product/quotient rule first, then chain rule to any derivative of $y$. Collect all terms with $\frac{dy}{dx}$ on one side, factor out $\frac{dy}{dx}$, and solve to get $\frac{dy}{dx}$ as an expression in both $x$ and $y$. For the second derivative, differentiate $\frac{dy}{dx}$ again, then substitute the known expression for $\frac{dy}{dx}$ back into the result.

**Worked example:** Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ 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:
2. $$2x + 3\left(x \frac{dy}{dx} + y\right) + 2y \frac{dy}{dx} = 0$$
3. Collect terms with $\frac{dy}{dx}$ and solve:
4. $$\frac{dy}{dx}(3x + 2y) = -2x - 3y \implies \frac{dy}{dx} = \frac{-2x - 3y}{3x + 2y}$$
5. Evaluate at $(2, 1)$:
6. $$\frac{dy}{dx} = \frac{-4 - 3}{6 + 2} = -\frac{7}{8}$$
7. Differentiate $\frac{dy}{dx}$ with the quotient rule to get the second derivative:
8. $$\frac{d^2y}{dx^2} = \frac{(3x + 2y)(-2 - 3\frac{dy}{dx}) - (-2x - 3y)(3 + 2\frac{dy}{dx})}{(3x + 2y)^2}$$
9. Substitute $x=2, y=1, \frac{dy}{dx}=-\frac{7}{8}$: numerator simplifies to $\frac{55}{4}$, denominator is $64$, so final result:
10. $$\frac{d^2y}{dx^2} = \frac{55}{256} \approx 0.215$$

> **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 algebra errors on the AP exam.

## Tangent and Normal Lines, Horizontal and Vertical Tangents

Once you have $\frac{dy}{dx}$ at a point on an implicit curve, $\frac{dy}{dx}$ is exactly the slope of the tangent line at that point, just like for explicit functions. The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope (when tangent slope is non-zero and defined). We use point-slope form $y - y_0 = m(x - x_0)$ to write the equation of either line.

AP exam questions very frequently ask for all points on an implicit curve that have horizontal or vertical tangents. A horizontal tangent has slope zero, so this occurs when $\frac{dy}{dx} = 0$ (numerator of $\frac{dy}{dx}$ is zero, denominator is non-zero). A vertical tangent has undefined slope, so this occurs when the denominator of $\frac{dy}{dx}$ is zero, and the numerator is non-zero.

**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 $x$:
2. $$4x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{2x}{y}$$
3. Evaluate the tangent slope at $(2, \sqrt{10})$:
4. $$m_{\text{tan}} = -\frac{2(2)}{\sqrt{10}} = -\frac{2\sqrt{10}}{5}$$
5. Write the tangent line using point-slope form and simplify:
6. $$y - \sqrt{10} = -\frac{2\sqrt{10}}{5}(x - 2) \implies 2\sqrt{10}x + 5y = 9\sqrt{10}$$
7. Calculate the normal line slope as the negative reciprocal:
8. $$m_{\text{norm}} = \frac{\sqrt{10}}{4}$$
9. Write and simplify the normal line equation:
10. $$y - \sqrt{10} = \frac{\sqrt{10}}{4}(x - 2) \implies \sqrt{10}x - 4y + 2\sqrt{10} = 0$$

> **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 for a valid answer.

## 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 $\frac{d^2y}{dx^2} > 0$ means the point is a local minimum, and $\frac{d^2y}{dx^2} < 0$ means the point is a local maximum. For concavity, $\frac{d^2y}{dx^2} > 0$ means concave up, $\frac{d^2y}{dx^2} < 0$ means concave down, and inflection points occur where concavity changes.

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

1. Differentiate implicitly and solve for $\frac{dy}{dx}$:
2. $$2x - \left(x \frac{dy}{dx} + y\right) + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{y - 2x}{2y - x}$$
3. Critical points for extrema of $y$ occur when $\frac{dy}{dx} = 0$, so numerator $y - 2x = 0 \implies y = 2x$. We confirm denominator $2y - x \neq 0$ for these points.
4. Substitute $y=2x$ into the original equation to find valid points on the curve:
5. $$x^2 - x(2x) + (2x)^2 = 3 \implies 3x^2 = 3 \implies x = \pm 1$$
6. This gives critical points $(1, 2)$ and $(-1, -2)$. We use the second derivative test, substituting $\frac{dy}{dx}=0$ to simplify calculation:
7. At $(1, 2)$:
8. $$\frac{d^2y}{dx^2} = -\frac{2}{3} < 0$$
9. So $(1, 2)$ is a local maximum. At $(-1, -2)$:
10. $$\frac{d^2y}{dx^2} = \frac{2}{3} > 0$$
11. So $(-1, -2)$ 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.

## AP Style Concept Check

**Check your understanding**

Test your understanding with this AP-style multiple choice question:

1. 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?

   - $-\frac{5}{4}$
   - $-\frac{5}{2}$
   - $\frac{5}{4}$
   - $-1$

   *Answer:* $-\frac{5}{4}$

   *Why:* Correct! Implicit differentiation gives $\frac{dy}{dx}(2x+2y) = -3x^2 - 2y$, so substituting $x=1,y=1$ gives $\frac{dy}{dx} = -\frac{5}{4}$.

## Common pitfalls

- **Wrong:** After differentiating a product term like $2xy$, you write $2x \frac{dy}{dx}$ and forget the product rule term $2y$.
  - Why it fails: 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: 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:** When finding $\frac{d^2y}{dx^2}$, you differentiate $\frac{dy}{dx}$ and leave out the chain rule factor of $\frac{dy}{dx}$ when differentiating terms that include y.
  - Why it fails: 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: Whenever you differentiate a term containing y at any derivative order, multiply by the appropriate derivative of y per the chain rule.
- **Wrong:** When asked for points with horizontal tangents, you set the denominator of $\frac{dy}{dx}$ equal to zero instead of the numerator.
  - Why it fails: Confusion between the conditions for horizontal vs vertical tangents, common when memorizing conditions without context.
  - Correct: Remember $\frac{dy}{dx} = $ slope: horizontal slope $= 0$, so numerator $= 0$ (denominator $\neq 0$); vertical slope is undefined, so denominator $= 0$ (numerator $\neq 0$).
- **Wrong:** When finding extrema, you keep $\frac{dy}{dx}$ as a variable when evaluating $\frac{d^2y}{dx^2}$ at a critical point, leading to complicated unsimplified expressions.
  - Why it fails: 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: Substitute $\frac{dy}{dx} = 0$ into the second derivative expression immediately before evaluating, which eliminates most terms and simplifies arithmetic.
- **Wrong:** After finding that all horizontal tangents occur along a line like $y=2x$, you report all points on that line instead of finding specific points on the original curve.
  - Why it fails: Students think satisfying the derivative condition is enough, and don't realize only specific points on the line lie on the original implicit curve.
  - Correct: Always substitute the derivative condition back into the original curve equation to solve for the specific x and y coordinates of valid points.

## Cheatsheet

| Category | Rule/Condition | Notes |
| --- | --- | --- |
| Implicit differentiation of $f(y)$ | $\frac{d}{dx}\left[f(y)\right] = f'(y) \cdot \frac{dy}{dx}$ | Apply chain rule to all y-containing terms; use product/quotient rule first for terms with both x and y |
| First derivative of implicit relation | Solve for $\frac{dy}{dx}$ after differentiating all terms | Result is an expression in both x and y, not just x |
| Second derivative of implicit relation | Differentiate $\frac{dy}{dx}$ again, substitute $\frac{dy}{dx}$ back | Evaluate $\frac{dy}{dx}$ at the point first to simplify calculation |
| Tangent slope at $(x_0,y_0)$ | $m = \frac{dy}{dx}\bigg\|_{(x_0, y_0)}$ | Only valid if $(x_0,y_0)$ lies on the original curve |
| Normal slope at $(x_0,y_0)$ | $m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$ | Undefined if tangent is horizontal; zero if tangent is vertical |
| Horizontal tangent condition | $\frac{dy}{dx} = 0 \implies \text{num}(\frac{dy}{dx}) = 0, \text{den}(\frac{dy}{dx}) \neq 0$ | Candidates for local maxima/minima of y with respect to x |
| Vertical tangent condition | $\frac{dy}{dx}$ undefined $\implies \text{den}(\frac{dy}{dx}) = 0, \text{num}(\frac{dy}{dx}) \neq 0$ | Candidates for local maxima/minima of x with respect to y |
| Second derivative test for extrema | If $\frac{dy}{dx}=0$: $\frac{d^2y}{dx^2} > 0 \to$ local min; $\frac{d^2y}{dx^2} < 0 \to$ local max | Same rule as for explicit functions |
| Concavity for implicit curves | $\frac{d^2y}{dx^2} > 0 \to$ concave up; $\frac{d^2y}{dx^2} < 0 \to$ concave down | Inflection points occur where concavity changes |

## What's next

Mastering the analysis of implicit relations is a foundational skill for many subsequent topics in AP Calculus BC. The core technique of implicit differentiation you learned here is directly applied to related rates problems, where you relate the rates of change of multiple quantities that are implicitly connected to each other. It also underpins the study of parametric and polar curves, which are unique to AP Calculus BC, where derivatives are calculated implicitly for non-cartesian curves. This topic also reinforces your understanding of derivative applications for curve sketching, which is a recurring theme across multiple-choice and free-response questions. Building a strong grasp of implicit behaviors will make more advanced topics in differentiation much easier to master as you progress through your exam preparation.

- [Unit 5: Analytical Applications of Differentiation Overview](https://www.owlsprep.com/study/ap-calculus-bc-u5-overview/)
- [Integration and Accumulation of Change Overview](https://www.owlsprep.com/study/ap-calculus-bc-u6-overview/)
- [Exploring accumulations of change](https://www.owlsprep.com/study/ap-calculus-bc-u6-exploring-accumulations-of-change/)

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