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Exploring Behaviors of
Implicit Relations

Discover how curves defined by equations like \( x^2 + y^2 = 1 \) reveal tangent lines, critical points, and concavity through implicit differentiation.

AP Calculus BC

What Is an Implicit Relation?

An implicit relation is an equation involving \( x \) and \( y \) that is not solved for one variable in terms of the other. Instead of writing \( y = f(x) \), we write:

Many important curves: ellipses, hyperbolas, folium of Descartes, and lemniscates are most naturally described implicitly.

Why Study Implicit Curves?

Multi-valued
A circle at \( x = 3 \) has two \( y \)-values. Implicit form captures both branches at once.
Natural Form
Physics and engineering equations often arise implicitly: pressure-volume curves, constraint surfaces.
Calculus Tools
We can find \( \frac{dy}{dx} \), tangent lines, and concavity without ever solving for \( y \).

Visualizing an Implicit Curve

x y -1 1 1 -1 tangent \(\left(\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right)\) \( x^2 + y^2 = 1 \)

The Core Rule: Implicit Differentiation

Given \( F(x,y) = 0 \), differentiate both sides with respect to \( x \), treating \( y \) as a function of \( x \). Every time you differentiate a \( y \)-term, use the Chain Rule and multiply by \( \frac{dy}{dx} \).

Finding \( \frac{d^2y}{dx^2} \) for Concavity

To determine concavity, we need the second derivative. Differentiate \( \frac{dy}{dx} \) with respect to \( x \) again, using the quotient rule and substituting back for \( \frac{dy}{dx} \).

\( \frac{d^2y}{dx^2} > 0 \)
Curve is concave up at that point.
\( \frac{d^2y}{dx^2} < 0 \)
Curve is concave down at that point.

Explicit vs. Implicit Differentiation

Explicit Implicit
Form \( y = f(x) \) \( F(x,y) = 0 \)
Solve for \( y \)? Yes, required Not necessary
Chain rule on \( y \) N/A: \( y \) already isolated Multiply by \( \frac{dy}{dx} \)
Multi-valued curves Must split into branches Handles all branches at once
Example \( y = \sqrt{25 - x^2} \) \( x^2 + y^2 = 25 \)

Key Behaviors to Explore

1
Horizontal tangents: Where the numerator of \( \frac{dy}{dx} \) is 0, but the denominator is not.
2
Vertical tangents: Where the denominator of \( \frac{dy}{dx} \) is 0, but the numerator is not.
3
Concavity: Sign of \( \frac{d^2y}{dx^2} \) determines whether the curve bends up or down.
4
Singular points: Where both the numerator and denominator of \( \frac{dy}{dx} \) are 0: cusps, self-intersections, or isolated points.

Step-by-Step Process

1
Write the equation in the form \( F(x,y) = 0 \).
2
Differentiate implicitly: apply \( \frac{d}{dx} \) to every term. Tag each \( y \)-derivative with \( \frac{dy}{dx} \).
3
Solve algebraically for \( \frac{dy}{dx} \).
4
Substitute the point \( (x_0, y_0) \) to evaluate slope, tangent line, or classify behavior.
5
For concavity: differentiate \( \frac{dy}{dx} \) again implicitly to get \( \frac{d^2y}{dx^2} \).

Example 1: Circle

Find \( \frac{dy}{dx} \) for \( x^2 + y^2 = 25 \) at the point \( (3, 4) \).

1
Differentiate: \( 2x + 2y\,\frac{dy}{dx} = 0 \)
2
Solve: \( \frac{dy}{dx} = -\frac{x}{y} \)
3
At \( (3,4) \): \( \frac{dy}{dx} = -\frac{3}{4} \)

Example 1: Visual

x y 5 5 (3, 4) slope = \(-\frac{3}{4}\)

Example 2: Folium of Descartes

Find \( \frac{dy}{dx} \) for \( x^3 + y^3 = 6xy \).

1
Differentiate: \( 3x^2 + 3y^2\,\frac{dy}{dx} = 6y + 6x\,\frac{dy}{dx} \)
2
Group: \( (3y^2 - 6x)\,\frac{dy}{dx} = 6y - 3x^2 \)
3
Solve: \( \frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x} \)

Example 3: Concavity of \( x^2 + y^2 = 25 \)

We already found \( \frac{dy}{dx} = -\frac{x}{y} \). Now find \( \frac{d^2y}{dx^2} \).

1
Apply quotient rule: \( \frac{d^2y}{dx^2} = -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} \)
2
Substitute \( \frac{dy}{dx} = -\frac{x}{y} \): \( \frac{d^2y}{dx^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3} \)
3
Since \( x^2 + y^2 = 25 \): \( \frac{d^2y}{dx^2} = -\frac{25}{y^3} \)

Detecting Horizontal and Vertical Tangents

horizontal vertical \( \frac{x^2}{4} + y^2 = 1 \)
Horizontal
Set numerator of \( \frac{dy}{dx} \) to zero (and check denominator \( \neq 0 \)).
Vertical
Set denominator of \( \frac{dy}{dx} \) to zero (and check numerator \( \neq 0 \)).

Common Mistakes to Avoid

✕ Forgetting the Chain Rule
When differentiating \( y^2 \), writing \( 2y \) instead of \( 2y\,\frac{dy}{dx} \). Every \( y \)-term needs the chain rule factor.
✕ Product Rule Omission
For terms like \( 6xy \), forgetting to use the product rule: \( 6\bigl(y + x\,\frac{dy}{dx}\bigr) \), not just \( 6\,\frac{dy}{dx} \).
✕ Wrong Point Substitution
Substituting the point before differentiating. Always differentiate first, then plug in \( (x_0, y_0) \).

Behavior Summary

Behavior Condition Interpretation
Horizontal tangent Numerator of \( \frac{dy}{dx} \) = 0 Local max/min candidates
Vertical tangent Denominator of \( \frac{dy}{dx} \) = 0 Curve turns back on itself
Singular point Numerator & Denom. of \( \frac{dy}{dx} \) = 0 Cusp, node, or isolated point
Concave up \( \frac{d^2y}{dx^2} > 0 \) Curve opens upward
Concave down \( \frac{d^2y}{dx^2} < 0 \) Curve opens downward

Question 1 of 5

Given \( x^2 + y^2 = 16 \), what is \( \frac{dy}{dx} \)?

Question 2 of 5

For \( x^3 + y^3 = 6xy \), the curve has a horizontal tangent when:

Question 3 of 5

For \( x^2 + y^2 = 25 \), the second derivative is \( \frac{d^2y}{dx^2} = -\frac{25}{y^3} \). At the point \( (3, -4) \), the curve is:

Question 4 of 5

When differentiating \( \sin(xy) = x \) implicitly, which expression correctly handles the left side?

Question 5 of 5

For the ellipse \( \frac{x^2}{4} + y^2 = 1 \), at which point(s) does a vertical tangent occur?