Behaviors of implicit relations — AP Calculus BC
1. Implicit Differentiation for First and Second Derivatives ★★☆☆☆ ⏱ 4 min
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.
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.
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.
2. Tangent and Normal Lines, Horizontal and Vertical Tangents ★★☆☆☆ ⏱ 4 min
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.
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.
3. Analyzing Critical Points, Extrema, and Concavity ★★★☆☆ ⏱ 4 min
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.
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.
4. AP Style Concept Check ★★★☆☆ ⏱ 2 min
Common Pitfalls
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.
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.
Why: Confusion between the conditions for horizontal vs vertical tangents, common when memorizing conditions without context.
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.
Why: Students think satisfying the derivative condition is enough, and don't realize only specific points on the line lie on the original implicit curve.