Topic 5.12 Notes – Exploring Behaviors of Implicit Relations

What an Implicit Relation Is and How We Differentiate It

An implicit relation looks like
$$F(x,y)=0$$ Example: $x^2 + xy + y^2 = 7$.

We don’t solve for $y$. Instead, we differentiate both sides with respect to $x$.

Key idea: treat $y$ as a function of $x$. So whenever you differentiate a $y$-term, multiply by $dy/dx$.

Quick example:

Differentiate $x^2 + xy + y^2 = 7$.

• $d/dx(x^2) = 2x$
• $d/dx(xy)$ → product rule
  $= x \frac{dy}{dx} + y$
• $d/dx(y^2) = 2y \frac{dy}{dx}$

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

Now solve algebraically for $\frac{dy}{dx}$.

Every time you find $\frac{dy}{dx}$, that expression gives the slope of the tangent line to the curve at a point.

Critical Points of an Implicit Relation

A critical point is any point on the curve where:

• $\frac{dy}{dx} = 0$, or
• $\frac{dy}{dx}$ does not exist.

Same definition as with explicit functions.

How to Find Them

1. Compute $\frac{dy}{dx}$.
2. Set it equal to 0.
   • Usually this means set the numerator = 0, while denominator ≠ 0.
3. Find where it’s undefined.
   • Usually where the denominator = 0, while numerator ≠ 0.
4. Plug back into the original equation to find actual points $(x,y)$.

You must report points, not just x-values.

Why Undefined Slopes Matter

Undefined slopes often mean vertical tangents.

For example, consider the circle $x^2 + y^2 = 9$.

Circle $x^2 + y^2 = 9$ with vertical tangents

• At $(3,0)$ and $(-3,0)$, the slope is undefined.
• These are critical points.
• They are not automatically maxima or minima.

That’s where sign analysis comes in.

Classifying Critical Points Using the First Derivative

Use the First Derivative Test exactly like before.

• If $\frac{dy}{dx}$ changes from positive to negative → relative maximum.
• If it changes from negative to positive → relative minimum.
• If there is no sign change → neither.

Subtle but important:
For implicit curves, a vertical tangent might not correspond to a local max or min because the curve could approach that x-value from only one side or behave differently along different branches.

On FRQs, you must justify your conclusion using sign changes, not just say “it’s a maximum.”

Concavity and the Second Derivative

To analyze concavity, differentiate again.

You start with your expression for $\frac{dy}{dx}$, then take $d/dx$ again.

Important: the second derivative may involve:

• $x$
• $y$
• $\frac{dy}{dx}$

That’s normal.

Interpreting $\frac{d^2y}{dx^2}$

• $>0$ → concave up
• $<0$ → concave down

A possible inflection point occurs where:

• $\frac{d^2y}{dx^2} = 0$ or undefined
• AND concavity changes sign.

You must show a sign change in concavity to justify an inflection point.

AP graders look for explicit reasoning like:
“Since $d^2y/dx^2$ changes from positive to negative at (a,b), the curve changes concavity, so there is an inflection point.”

Extending Derivative Applications to Implicit Functions

Everything you learned about derivatives still works:

• Increasing/decreasing behavior
• Relative extrema
• Concavity
• Inflection points

It also connects to related rates.

If both $x$ and $y$ depend on time $t$:

$$\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$$

That chain rule connection is huge. Many related rates setups start with an implicit equation and require this exact idea.

Common Mistakes That Cost Points

• Forgetting to multiply by $dy/dx$ when differentiating $y$-terms.
• Setting both numerator and denominator equal to zero.
• Calling every vertical tangent a max or min.
• Forgetting to verify points satisfy the original equation.
• Claiming an inflection point without proving a concavity change.
• Stopping at $dy/dx$ when the question asks for $dy/dt$.