5.12 Exploring Behaviors of Implicit Relations

Implicit relations like $x^2+y^2=25$ have critical points wherever $\frac{dy}{dx}$ equals zero or does not exist. You find these by using implicit differentiation, then analyzing the first and second derivatives to describe how the curve behaves, just like you would with a regular function. For AP Calculus, keep $x$ and $y$ values attached to each point you classify.

Why This Matters for the AP Calculus Exam

This is the last topic in Unit 5, and it ties together everything you learned about analyzing functions with derivatives. The key skill is extending derivative tools to relations that are not written as $y = f(x)$. On the AP Calculus AB exam, you may need to find $\frac{dy}{dx}$ implicitly, locate horizontal or vertical tangents, set up a second derivative that depends on $x$, $y$, and $\frac{dy}{dx}$, and justify conclusions about the curve's behavior. Clear notation and full justification are important for clear exam work, especially when you refer to $f$, $f'$, and $f''$ instead of vague words like "it."

Key Takeaways

• A critical point of an implicit relation is any point where $\frac{dy}{dx} = 0$ or $\frac{dy}{dx}$ does not exist.
• $\frac{dy}{dx} = 0$ usually signals a horizontal tangent; $\frac{dy}{dx}$ undefined (denominator zero) usually signals a vertical tangent.
• The same increasing, decreasing, and concavity rules from earlier topics apply to implicit relations.
• The second derivative of an implicit relation often contains $x$, $y$, and $\frac{dy}{dx}$, so substitute your first derivative before simplifying.
• To classify a point, check how $\frac{dy}{dx}$ changes sign around it, but remember some points on closed curves can't be labeled max or min.
• Always solve $\frac{dy}{dx}$ in terms of both variables before setting it equal to zero or finding where it is undefined.

Understanding Implicit Relations

An implicit relation is an equation where the variables are mixed together on the same side, instead of being solved as $y = f(x)$. Before this unit you mostly saw explicit functions that pass the vertical line test, like:

$y = f(x)$

An implicit relation can have multiple variables on the same side and sometimes does not pass the vertical line test. Even so, you can still describe its behavior using derivatives.

Here is a classic example:

$x^2 + y^2 = 1$

Implicit differentiation lets you work with these by taking $\frac{d}{dx}$ of both sides to find $\frac{dy}{dx}$, which is usually written in terms of both $x$ and $y$.

First Derivative Applications

Critical Points on Implicit Relations

A point on an implicit relation where the first derivative equals zero or does not exist is a critical point. If you want a refresher on critical points in general, check 5.2 Extreme Value Theorem, Global vs Local Extrema, and Critical Points. These points are potential locations for extrema, horizontal tangents, or vertical tangents.

Steps to Find Critical Points

1. Differentiate both sides of the equation with respect to $x$ by taking $\frac{d}{dx}$.
2. Solve for $\frac{dy}{dx}$. This gives the derivative in implicit form.
3. Set $\frac{dy}{dx} = 0$ to find points with horizontal tangents, and find where $\frac{dy}{dx}$ is undefined (where the denominator is zero) for vertical tangents.
4. Make a conclusion about each critical point by checking how $\frac{dy}{dx}$ changes sign around that point.

Reading Behavior from the First Derivative

The same rules you used for explicit functions still apply:

• If $f'(x)$ is positive, the graph of $f(x)$ is increasing.
• If $f'(x)$ is negative, the graph of $f(x)$ is decreasing.
• If $f'(x)$ is increasing (positive slope), the graph of $f(x)$ is concave up.
• If $f'(x)$ is decreasing (negative slope), the graph of $f(x)$ is concave down.
• If $f'(x)$ changes from positive to negative at $x = k$, then $f(x)$ has a relative maximum at $x = k$.
• If $f'(x)$ changes from negative to positive at $x = k$, then $f(x)$ has a relative minimum at $x = k$.

For implicit relations, relative minimums and maximums can be less intuitive than they are for explicit functions, so analyze the sign changes carefully.

To review these ideas, check out:

• 5.3 Determining Intervals on Which a Function is Increasing or Decreasing
• 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema

Second Derivative Applications

After finding the first derivative implicitly, differentiate again with respect to $x$ to find the second derivative. The second derivative of an implicit relation may involve $x$, $y$, and $\frac{dy}{dx}$, so substitute your expression for $\frac{dy}{dx}$ to simplify.

Once you have the second derivative, you can analyze concavity and look for points of inflection:

1. If $f''(x)$ is positive, the function is concave up.
2. If $f''(x)$ is negative, the function is concave down.
3. A point of inflection is a point where concavity changes and $f''(x) = 0$. Confirm the change by checking the sign of $f''(x)$ just before and just after the candidate point.

To review concavity, check out 5.6 Determining Concavity.

Worked Examples

Example 1: Critical Points on a Circle

Consider the implicit equation $x^2 + y^2 = 25$. Find the critical points and determine their nature.

Differentiate implicitly:

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

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

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

Set $\frac{dy}{dx} = 0$:

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

$x = 0$

From $x^2 + y^2 = 25$, if $x = 0$ then $y^2 = 25$ and $y = \pm 5$. So two candidate points are $(0, 5)$ and $(0, -5)$.

Also set $\frac{dy}{dx}$ undefined. The derivative is undefined when the denominator is zero, so $y = 0$. Plugging back in gives $x = \pm 5$, so $(5, 0)$ and $(-5, 0)$ are also candidates.

First Derivative Test Chart

Why are two of the points not labeled max or min? Look at the graph of the circle. Below $(-5, 0)$ in the third quadrant, the slope is negative; above it in the second quadrant, the slope is positive. The point $(5, 0)$ has a similar sign change.

Despite these sign changes, these points are not conclusively maxima or minima at the level of Calculus AB. To classify a max or min, you need valid points with $x$-values both greater than and less than the critical point. Here there are no points to the left of $(-5, 0)$ and none to the right of $(5, 0)$, so the curve's shape makes these vertical-tangent points instead of clear extrema.

Think of it this way: if you travel the circle clockwise, $(5, 0)$ acts like a minimum and $(-5, 0)$ acts like a maximum. Travel counter-clockwise and they switch. Because the labeling depends on direction, these points are vertical tangents rather than true extrema.

Example 2: Implicit Differentiation with Related Rates

A ladder leans against a wall. The bottom slides away from the wall at a constant rate of 2 ft/s. If the ladder is 10 feet long and its bottom starts 6 feet from the wall, how fast is the top sliding down when the bottom is 8 feet from the wall?

This is a related rates application that uses the same implicit setup. List the given information:

• Horizontal rate of the bottom: $\frac{dx}{dt} = 2$ ft/s
• Length of the ladder: 10 ft
• What to find: vertical rate $\frac{dy}{dt}$ when $x = 8$

Use the Pythagorean Theorem to model the ladder, with $x$ as horizontal distance and $y$ as vertical distance:

$x^2 + y^2 = 10^2$

Differentiate with respect to time:

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

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

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

Find $y$ when $x = 8$:

$8^2 + y^2 = 100$

$y^2 = 100 - 64 = 36$

$y = 6$

Substitute $x = 8$, $y = 6$, and $\frac{dx}{dt} = 2$:

$\frac{dy}{dt} = -\frac{8}{6}\cdot(2) = -\frac{8}{3}\text{ ft/s}$

The top of the ladder slides down at $\frac{8}{3}$ ft/s when the bottom is 8 feet from the wall. The negative sign tells you the height is decreasing, which matches "sliding down."

How to Use This on the AP Calculus Exam

Problem Solving

When you see an implicit relation, follow a consistent process:

1. List the given information and your goal.
2. Differentiate both sides with respect to $x$ (or with respect to $t$ for related rates).
3. Solve for $\frac{dy}{dx}$ (or $\frac{dy}{dt}$).
4. Set the derivative equal to zero for horizontal tangents and find where it is undefined for vertical tangents.
5. For the second derivative, differentiate again and substitute your $\frac{dy}{dx}$ expression.
6. Check your sign changes and confirm any positive or negative results make sense.

Free Response

If a free response question asks you to justify behavior, refer to $f$, $f'$, and $f''$ by name rather than saying "it." For example, state that the curve is concave up on an interval because $\frac{d^2y}{dx^2}$ is positive there. A clear justification names the evidence and connects it to the conclusion.

Common Trap

Watch the denominator. The derivative $\frac{dy}{dx} = -\frac{x}{y}$ is undefined when $y = 0$, and that condition signals a vertical tangent, not a place to ignore. Always check both where the derivative is zero and where it is undefined.

Common Misconceptions

• A critical point is not only where $\frac{dy}{dx} = 0$. It also includes points where $\frac{dy}{dx}$ does not exist, such as vertical tangents.
• A sign change in $\frac{dy}{dx}$ does not always mean a max or min on a closed curve. Points like $(5, 0)$ and $(-5, 0)$ on a circle are vertical tangents, not true extrema.
• The second derivative of an implicit relation is usually not just a function of $x$. It often depends on $x$, $y$, and $\frac{dy}{dx}$, so you must substitute before finishing.
• Forgetting the $\frac{dy}{dx}$ term when differentiating $y$ is a common error. Every time you differentiate a $y$ term, the chain rule adds a $\frac{dy}{dx}$ factor.
• Setting only the numerator to zero misses vertical tangents. Check the denominator for where the derivative is undefined too.

Related AP Calculus Guides

• Unit 5 Overview: Analytical Applications of Differentiation
• 5.1 Using the Mean Value Theorem
• 5.2 Extreme Value Theorem, Global vs Local Extrema, and Critical Points
• 5.3 Determining Intervals on Which a Function is Increasing or Decreasing
• 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema
• 5.11 Solving Optimization Problems