Introduction

This topic extends derivative applications to implicitly defined relations. The essential knowledge tells us that a point on an implicit relation where the first derivative equals zero or does not exist is a critical point of the function, applications of derivatives can be extended to implicitly defined functions, and second derivatives involving implicit differentiation may be relations of $x$, $y$, and $\frac{dy}{dx}$. Understanding these concepts allows us to analyze curves that cannot be expressed as explicit functions.

Review of Implicit Differentiation

Basic Implicit Differentiation

For a relation like $x^2+y^2=25$, we differentiate both sides with respect to $x$, treating $y$ as a function of $x$.

$\frac{d}{dx}(x^2+y^2)=\frac{d}{dx}(25)$

$2x+2y\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{x}{y}$

This derivative represents the slope of the tangent line at any point $(x,y)$ on the circle.

Critical Points of Implicit Relations

Definition and Finding Critical Points

A critical point of an implicit relation occurs where:

Example 1:

Find all critical points on the circle $x^2+y^2=25$.

From above: $\frac{dy}{dx}=-\frac{x}{y}$

Horizontal tangents ($\frac{dy}{dx}=0$):

$-\frac{x}{y}=0$

$x=0$

Substitute into original equation:

$0+y^2=25$

$y=\pm 5$

Critical points: $(0,5)$ and $(0,-5)$

Vertical tangents ($\frac{dy}{dx}$ undefined):

$y=0$

Substitute:

$x^2+0=25$

$x=\pm 5$

Points with vertical tangents: $(5,0)$ and $(-5,0)$

More Complex Implicit Relations

Example 2:

Find all points on the ellipse $4x^2+9y^2=36$ where there are horizontal or vertical tangents.

Differentiate implicitly:

$8x+18y\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{8x}{18y}=-\frac{4x}{9y}$

Horizontal tangents:

$-\frac{4x}{9y}=0$

$x=0$

$9y^2=36$

$y=\pm 2$

Points: $(0,2)$ and $(0,-2)$

Vertical tangents:

$y=0$

$4x^2=36$

$x=\pm 3$

Points: $(3,0)$ and $(-3,0)$

Determining Local Extrema

Using the First Derivative Test

For implicit relations, local extrema occur at critical points where the curve changes direction.

Example 3:

Classify the critical points of $x^2+y^2=25$ as local maxima or minima.

From Example 1, critical points with horizontal tangents: $(0,5)$ and $(0,-5)$

At $(0,5)$:

$y$ changes from increasing to decreasing → local maximum

At $(0,-5)$:

$y$ changes from decreasing to increasing → local minimum

Analyzing More Complex Relations

Example 4:

For the relation $x^3+y^3=6xy$, find where horizontal tangents occur and interpret.

Differentiate implicitly:

$3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}$

Solve for $\frac{dy}{dx}$:

$3y^2\frac{dy}{dx}-6x\frac{dy}{dx}=6y-3x^2$

$(3y^2-6x)\frac{dy}{dx}=6y-3x^2$

$\frac{dy}{dx}=\frac{6y-3x^2}{3y^2-6x}=\frac{2y-x^2}{y^2-2x}$

Horizontal tangents:

Let $f(x)=0$

$2y-x^2=0$

$y=\frac{x^2}{2}$

This must also satisfy $x^3+y^3=6xy$. The specific points require solving a system (often done with technology).

Second Derivatives of Implicit Relations

Finding the Second Derivative

The second derivative $\frac{d^{2}y}{d{x}^2}$ tells us about concavity, even for implicit relations.

Key Point: The second derivative often involves $x$, $y$, AND $\frac{dy}{dx}$.

Example 5:

Find $\frac{d^{2}y}{d{x}^2}$ for the circle $x^2+y^2=25$.

From earlier: $\frac{dy}{dx}=-\frac{x}{y}$

Differentiate again with respect to $x$:

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(-\frac{x}{y}\right)$

Use quotient rule:

$\frac{d^{2}y}{d{x}^2}=-\frac{y(1)-x\frac{dy}{dx}}{y^2}$

Substitute $\frac{dy}{dx}=-\frac{x}{y}$:

$\frac{d^{2}y}{d{x}^2}=-\frac{y-x\left(-\frac{x}{y}\right)}{y^2}=-\frac{y+\frac{x^2}{y}}{y^2}$

$\frac{d^{2}y}{d{x}^2}=-\frac{y^2+x^2}{y^3}$

Since $x^2+y^2=25$: $\frac{d^{2}y}{d{x}^2}=-\frac{25}{y^3}$

Determining Concavity

Example 6:

Determine the concavity of $x^2+y^2=25$ at the point $(3,4)$.

From Example 5: $\frac{d^{2}y}{d{x}^2}=-\frac{25}{y^3}$

At $(3,4)$: $\frac{d^{2}y}{d{x}^2}=-\frac{25}{64}<0$

Since $\frac{d^{2}y}{d{x}^2}<0$, the curve is concave down at $(3,4)$.

At $(3,-4)$: $\frac{d^{2}y}{d{x}^2}=-\frac{25}{-64}>0$

The curve is concave up at $(3,-4)$.

Applications to Curve Analysis

Complete Behavior Analysis

Example 7:

Analyze the behavior of the relation $x^2-y^2=1$ (hyperbola).

Find $\frac{dy}{dx}$:

$2x-2y\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{x}{y}$

Critical points:

Horizontal tangents: $x=0$, but $x^2-y^2=1$ gives $-y^2=1$ (no real solution)

Vertical tangents: $y=0$, which gives $x=\pm 1$

Points with vertical tangents: $(1,0)$ and $(-1,0)$

Find $\frac{d^{2}y}{d{x}^2}$: $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{x}{y}\right)=\frac{y-x\frac{dy}{dx}}{y^2}$

Substitute $\frac{dy}{dx}=\frac{x}{y}$: $\frac{d^{2}y}{d{x}^2}=\frac{y-x\cdot \frac{x}{y}}{y^2}=\frac{y^2-x^2}{y^3}$

Since $x^2-y^2=1$, we have $y^2-x^2=-1$: $\frac{d^{2}y}{d{x}^2}=-\frac{1}{y^3}$

Concavity:

Justifying Conclusions

Using Derivative Information

Example 8:

Given the implicit relation $xy=1$, justify that there are no horizontal tangents.

Find $\frac{dy}{dx}$:

$y+x\frac{dy}{dx}=0$

$\frac{dy}{dx}=-\frac{y}{x}$

For a horizontal tangent, we need $\frac{dy}{dx}=0$:

$-\frac{y}{x}=0$

$y=0$

But if $y=0$, then $xy=0\ne 1$, so $y=0$ is not on the curve.

Conclusion: There are no horizontal tangents because setting $\frac{dy}{dx}=0$ leads to a contradiction with the original relation.

Practice Section