Exploring Behaviors of Implicit Relations - AP Calculus BC Study Guide

The Core Idea: Exploring Behaviors of Implicit Relations

Implicit relations define curves that are not necessarily functions and cannot always be written in the form $y=f(x)$. While first-order implicit differentiation allows us to determine the slope of a tangent line at a point on such a curve, it does not provide a complete picture of the curve's local behavior. This topic extends the concept to the second derivative, $\frac{d^{2}y}{d{x}^2}$.

By finding the second derivative of an implicitly defined function, we can analyze its concavity. The process involves differentiating the first derivative, $\frac{dy}{dx}$, again with respect to $x$. The resulting expression for $\frac{d^{2}y}{d{x}^2}$ allows us to determine whether the curve is concave up or concave down at a specific point, providing a deeper understanding of the shape and behavior of the implicit relation.

Key Processes

The primary focus of this topic is the process of finding the second derivative of an implicitly defined relation and using it to determine concavity.

1. Finding the First Derivative ($\frac{dy}{dx}$)

   This is the foundational step. Differentiate the entire implicit equation with respect to $x$, remembering to apply the chain rule for any term involving $y$. Then, algebraically solve for $\frac{dy}{dx}$. The resulting expression will typically involve both $x$ and $y$.

2. Finding the Second Derivative ($\frac{d^{2}y}{d{x}^2}$)

   This is achieved by differentiating the expression for $\frac{dy}{dx}$ with respect to $x$. This step often requires the quotient rule or product rule. During this differentiation, whenever the variable $y$ is differentiated, it produces a $\frac{dy}{dx}$ term due to the chain rule.

3. Substitution and Simplification

   The initial expression for $\frac{d^{2}y}{d{x}^2}$ will contain $x$, $y$, and $\frac{dy}{dx}$. To express $\frac{d^{2}y}{d{x}^2}$ in terms of only $x$ and $y$, substitute the expression for $\frac{dy}{dx}$ (found in the first step) back into the equation for $\frac{d^{2}y}{d{x}^2}$.

Understanding Concavity with Implicit Relations

The essential link between the second derivative and the behavior of an implicit relation is concavity. Just as with explicitly defined functions, the sign of the second derivative at a point reveals the shape of the curve at that point.

• If $\frac{d^{2}y}{d{x}^2}>0$ at a point $(a,b)$, the graph of the relation is concave up at that point.

• If $\frac{d^{2}y}{d{x}^2}<0$ at a point $(a,b)$, the graph of the relation is concave down at that point.

A critical nuance in the process is the evaluation of $\frac{d^{2}y}{d{x}^2}$ at a specific point. To find the value of the second derivative at $(a,b)$, you must first find the value of the first derivative, $\frac{dy}{dx}$, at that same point. Then, substitute the numerical values of $x$, $y$, and $\frac{dy}{dx}$ into the expression for $\frac{d^{2}y}{d{x}^2}$. It is often more efficient to substitute the numerical values into the un-simplified second derivative expression rather than performing a full algebraic substitution and simplification first.

Core Concepts & Rules

• The first derivative, $\frac{dy}{dx}$, of an implicit relation represents the slope of the tangent line to the curve at any point $(x,y)$.

• The second derivative, $\frac{d^{2}y}{d{x}^2}$, is found by implicitly differentiating the expression for the first derivative, $\frac{dy}{dx}$.

• The value of the second derivative at a point determines the concavity of the implicit relation's graph at that point.

• A positive value for $\frac{d^{2}y}{d{x}^2}$ at a point indicates the curve is concave up at that point.

• A negative value for $\frac{d^{2}y}{d{x}^2}$ at a point indicates the curve is concave down at that point.

• To evaluate $\frac{d^{2}y}{d{x}^2}$ at a point, you must use the coordinates of the point ($x$ and $y$) as well as the value of $\frac{dy}{dx}$ at that same point.

Step-by-Step Example 1: Finding the Second Derivative

Problem: For the curve defined by the equation $x^3+y^3=9$, find an expression for $\frac{d^{2}y}{d{x}^2}$ in terms of $x$ and $y$.

Step 1: Find the first derivative, $\frac{dy}{dx}$, using implicit differentiation.

Differentiate both sides of the equation with respect to $x$:

$$\frac{d}{dx}(x^3+y^3)=\frac{d}{dx}(9)$$

$$3x^2+3y^2\frac{dy}{dx}=0$$

Now, solve for $\frac{dy}{dx}$:

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

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

Step 2: Differentiate the expression for $\frac{dy}{dx}$ to find $\frac{d^{2}y}{d{x}^2}$.

We will use the quotient rule on $\frac{dy}{dx}=-\frac{x^2}{y^2}$.

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

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

$$\frac{d^{2}y}{d{x}^2}=-\frac{2x{y}^2-2x^{2}y\frac{dy}{dx}}{y^4}$$

Step 3: Substitute the expression for $\frac{dy}{dx}$ back into the equation.

Substitute $\frac{dy}{dx}=-\frac{x^2}{y^2}$ into the expression for $\frac{d^{2}y}{d{x}^2}$:

$$\frac{d^{2}y}{d{x}^2}=-\frac{2x{y}^2-2x^{2}y\left(-\frac{x^2}{y^2}\right)}{y^4}$$

$$\frac{d^{2}y}{d{x}^2}=-\frac{2x{y}^2+\frac{2x^4}{y}}{y^4}$$

Step 4: Simplify the expression.

To eliminate the complex fraction, multiply the numerator and denominator by $y$:

$$\frac{d^{2}y}{d{x}^2}=-\frac{y(2x{y}^2)+y(\frac{2x^4}{y})}{y(y^4)}$$

$$\frac{d^{2}y}{d{x}^2}=-\frac{2x{y}^3+2x^4}{y^5}$$

Factor out $2x$ from the numerator:

$$\frac{d^{2}y}{d{x}^2}=-\frac{2x(y^3+x^3)}{y^5}$$

From the original equation, we know $x^3+y^3=9$. Substitute this in for a final simplification:

$$\frac{d^{2}y}{d{x}^2}=-\frac{2x(9)}{y^5}=-\frac{18x}{y^5}$$

Step-by-Step Example 2: Determining Concavity at a Point

Problem: Consider the curve given by $y^2-2x=1+xy$. Determine if the curve is concave up or concave down at the point $(0,-1)$.

Step 1: Find the first derivative, $\frac{dy}{dx}$.

Differentiate the equation implicitly with respect to $x$. Note the product rule is needed for the $xy$ term.

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

$$2y\frac{dy}{dx}-2=0+\left((1)(y)+(x)\frac{dy}{dx}\right)$$

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

Group the $\frac{dy}{dx}$ terms and solve:

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

$$\frac{dy}{dx}(2y-x)=y+2$$

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

Step 2: Evaluate $\frac{dy}{dx}$ at the point $(0,-1)$.

Substitute $x=0$ and $y=-1$ into the expression for $\frac{dy}{dx}$:

$${\frac{dy}{dx}|}_{(0,-1)}=\frac{(-1)+2}{2(-1)-0}=\frac{1}{-2}=-\frac{1}{2}$$

Step 3: Find the second derivative, $\frac{d^{2}y}{d{x}^2}$.

Apply the quotient rule to the expression for $\frac{dy}{dx}$:

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

Step 4: Evaluate $\frac{d^{2}y}{d{x}^2}$ at the point $(0,-1)$.

Substitute the known values $x=0$, $y=-1$, and $\frac{dy}{dx}=-\frac{1}{2}$ into the expression for $\frac{d^{2}y}{d{x}^2}$.

$${\frac{d^{2}y}{d{x}^2}|}_{(0,-1)}=\frac{\left(-\frac{1}{2}\right)(2(-1)-0)-((-1)+2)\left(2(-\frac{1}{2})-1\right)}{(2(-1)-0{)}^2}$$

$${\frac{d^{2}y}{d{x}^2}|}_{(0,-1)}=\frac{\left(-\frac{1}{2}\right)(-2)-(1)(-1-1)}{(-2{)}^2}$$

$${\frac{d^{2}y}{d{x}^2}|}_{(0,-1)}=\frac{1-(1)(-2)}{4}$$

$${\frac{d^{2}y}{d{x}^2}|}_{(0,-1)}=\frac{1+2}{4}=\frac{3}{4}$$

Step 5: Interpret the result.

Since $\frac{d^{2}y}{d{x}^2}=\frac{3}{4}>0$ at the point $(0,-1)$, the curve is concave up at that point.

Using Your Calculator

This topic is primarily analytical, and a calculator cannot be used to find a symbolic second derivative of an implicit relation. The entire process of differentiation and substitution must be done by hand.

A calculator is useful for performing the arithmetic calculations required when evaluating $\frac{d^{2}y}{d{x}^2}$ at a specific point, especially if the coordinates or the value of $\frac{dy}{dx}$ are fractions or decimals.

For example, in Step 4 of the second example, you would use your calculator to compute the final numerical value after substituting $x$, $y$, and $\frac{dy}{dx}$ into the expression. The calculator serves to check arithmetic, not to perform the calculus steps.

AP Exam Quick Hit

Common Question Types

• Finding the expression for $\frac{d^{2}y}{d{x}^2}$: A multiple-choice or free-response part might ask you to find the second derivative in terms of $x$ and $y$, requiring full simplification.

  • Example: "For the curve $y^2-x^2=4$, which of the following is $\frac{d^{2}y}{d{x}^2}$?"

• Determining concavity at a point: Given an implicit relation and a point on the curve, you will be asked to find the concavity.

  • Example: "Show that the curve $x^2+2y^2=6$ is concave down at the point $(2,1)$."

• Applying the Second Derivative Test: You may be asked to find points where the tangent line is horizontal ($\frac{dy}{dx}=0$) and then use the sign of $\frac{d^{2}y}{d{x}^2}$ at those points to classify them as local minima or maxima.

  • Example: "Find the coordinates of the points on the curve $x^2+xy+y^2=3$ where the tangent line is horizontal, and determine if each point is a local maximum or a local minimum."

Common Mistakes

• Forgetting the Chain Rule: When differentiating a term with $y$ (e.g., $y^3$), a common mistake is to write $3y^2$ instead of the correct $3y^2\frac{dy}{dx}$.

• Incorrectly Applying Product/Quotient Rules: The expressions for $\frac{dy}{dx}$ can be complex, and errors often occur when applying differentiation rules to find $\frac{d^{2}y}{d{x}^2}$.

• Forgetting to Substitute for $\frac{dy}{dx}$: When evaluating $\frac{d^{2}y}{d{x}^2}$ at a point, students sometimes forget to first calculate the numerical value of $\frac{dy}{dx}$ at that point and substitute it into the second derivative expression.

• Algebraic Errors During Simplification: Simplifying the complex fraction that often results when finding $\frac{d^{2}y}{d{x}^2}$ is a frequent source of error.

• Substituting Before Differentiating: Students may mistakenly plug in the numerical values of $x$ and $y$ into the first derivative expression before differentiating it a second time. This is incorrect, as the derivative of a constant is zero. You must find the general expression for $\frac{d^{2}y}{d{x}^2}$ first.