5.12 Exploring Behaviors of Implicit Relations

Learning objectives

—FUN-4.D—Determine critical points of implicit relations• FUN-4.D Determine critical points of implicit relations.• FUN-4.D.1 A point on an implicit relation where the first derivative equals zero or does not exist is a critical point of the function.• Enduring understanding FUN-4: A function’s derivative can be used to understand some behaviors of the function.—FUN-4.E—Justify conclusions about the behavior of an implicitly defined function based on evidence from its derivatives• FUN-4.E Justify conclusions about the behavior of an implicitly defined function based on evidence from its derivatives.• FUN-4.E.1 Applications of derivatives can be extended to implicitly defined functions.• FUN-4.E.2 Second derivatives involving implicit differentiation may be relations of x, y, and dy/dx.• Enduring understanding FUN-4: A function’s derivative can be used to understand some behaviors of the function.

For a branch of an implicit relation, a critical point is a point $(x,y)$ on the relation where $dy/dx=0$ or where $dy/dx$ does not exist. The answer is a coordinate pair, so a derivative condition alone is not enough.

Differentiate implicitly and solve two cases: set the numerator of $dy/dx$ equal to zero while the denominator is nonzero; then set the denominator equal to zero and examine where the derivative is undefined. In both cases, solve together with the original relation and verify that the relevant branch exists near the point.

For $x^2+4y^2=4$ , implicit differentiation gives $dy/dx=-x/(4y)$ . The derivative is zero when $x=0$ ; the original relation then gives $(0,1)$ and $(0,-1)$ . It is undefined when $y=0$ ; the relation gives $(2,0)$ and $(-2,0)$ . Thus all four are critical points of the relation; the first pair has horizontal tangents and the second pair has vertical tangents.

Do not list only $x$ -values or cancel a factor before checking where it is zero. If both numerator and denominator vanish, the simplified derivative may hide a singular point; inspect the original relation and its local branches before classifying it.

An implicitly defined branch can be analyzed without solving explicitly for $y$ . Use the sign of $dy/dx$ to justify where the branch increases or decreases, and the sign of $d^2y/dx^2$ to justify where it is concave up or concave down.

First find $dy/dx$ from the relation. Differentiate that equation again with respect to $x$ , treating $y$ as a function of $x$ ; the result may contain $x$ , $y$ , and $dy/dx$ . Substitute the first-derivative relation when useful, then determine signs using the coordinates and branch conditions supplied by the original relation.

x^2+y^2=25 \qquad \frac{dy}{dx}=-\frac{x}{y} \qquad \frac{d^2y}{dx^2}=-\frac{25}{y^3}

On the upper semicircle, $y>0$ , so $d^2y/dx^2<0$ : the entire upper branch is concave down. There, $dy/dx=-x/y$ is positive when $x<0$ and negative when $x>0$ , so the branch increases to $(0,5)$ and then decreases. On the lower branch, $y<0$ , so the concavity conclusion reverses.

A sign conclusion belongs only to the branch and interval on which its conditions hold. Do not assign one increasing, decreasing, or concavity statement to an entire relation when different branches give different signs, and do not cross points where the required derivative is undefined.

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—AP Calculus BCRead

5.12 Exploring Behaviors of Implicit Relations

Learning objectives

Find Critical Points on an Implicit Relation

Use Derivatives to Describe an Implicit Branch

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