Unit 5.12 – Exploring Behaviors of Implicit Relations

AP® Calculus AB & BC | Analyzing Curves Defined Implicitly

Why This Matters: Most functions we've studied are explicit (\(y = f(x)\)), but many important curves can't be written this way! Implicit relations like \(x^2 + y^2 = 25\) (circle) or \(x^2 + xy + y^2 = 3\) (ellipse) define \(y\) implicitly in terms of \(x\). To analyze these curves—find slopes, tangent lines, extrema, concavity—we use implicit differentiation. This technique is essential for understanding circles, ellipses, lemniscates, and many other curves that appear in physics, engineering, and economics. It's a powerful tool that extends all our derivative techniques to a much broader class of curves!

🔄 Review: Implicit Differentiation

KEY CONCEPT

Implicit differentiation allows us to find \(\frac{dy}{dx}\) when \(y\) is defined implicitly by an equation \(F(x, y) = 0\), even if we can't solve for \(y\) explicitly.

The Process:

Differentiate both sides with respect to \(x\)
Treat \(y\) as a function of \(x\) (use Chain Rule: \(\frac{d}{dx}[y^n] = ny^{n-1}\frac{dy}{dx}\))
Collect all terms with \(\frac{dy}{dx}\) on one side
Solve algebraically for \(\frac{dy}{dx}\)

Essential Rules for Implicit Differentiation

Differentiation Formulas (with respect to \(x\))
Expression	Derivative	Explanation
\(y\)	\(\frac{dy}{dx}\)	\(y\) is function of \(x\)
\(y^n\)	\(ny^{n-1}\frac{dy}{dx}\)	Power Rule + Chain Rule
\(xy\)	\(x\frac{dy}{dx} + y\)	Product Rule
\(\frac{y}{x}\)	\(\frac{x\frac{dy}{dx} - y}{x^2}\)	Quotient Rule
\(\sin(y)\)	\(\cos(y)\frac{dy}{dx}\)	Chain Rule
\(e^y\)	\(e^y\frac{dy}{dx}\)	Chain Rule
\(\ln(y)\)	\(\frac{1}{y}\frac{dy}{dx}\)	Chain Rule

📏 Finding Tangent Lines to Implicit Curves

Standard Procedure:

Find \(\frac{dy}{dx}\) using implicit differentiation
Evaluate \(\frac{dy}{dx}\) at the given point \((x_0, y_0)\)
Use point-slope form: \(y - y_0 = m(x - x_0)\)
Simplify to desired form

📝 Important: The answer \(\frac{dy}{dx}\) from implicit differentiation usually contains both \(x\) and \(y\). You must substitute the specific point to get a numerical slope!

📖 Comprehensive Worked Examples

Example 1: Circle - Basic Implicit Differentiation

Problem: For the circle \(x^2 + y^2 = 25\), find:
(a) \(\frac{dy}{dx}\)
(b) The equation of the tangent line at \((3, 4)\)

Solution:

Part (a): Find \(\frac{dy}{dx}\)

Differentiate both sides with respect to \(x\):

\[ \frac{d}{dx}[x^2 + y^2] = \frac{d}{dx}[25] \]

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

Solve for \(\frac{dy}{dx}\):

\[ 2y\frac{dy}{dx} = -2x \]

\[ \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y} \]

Part (b): Tangent line at (3, 4)

Evaluate slope at \((3, 4)\):

\[ m = -\frac{3}{4} \]

Use point-slope form:

\[ y - 4 = -\frac{3}{4}(x - 3) \]

\[ y - 4 = -\frac{3}{4}x + \frac{9}{4} \]

\[ y = -\frac{3}{4}x + \frac{9}{4} + 4 = -\frac{3}{4}x + \frac{25}{4} \]

Answer: (a) \(\frac{dy}{dx} = -\frac{x}{y}\) | (b) \(y = -\frac{3}{4}x + \frac{25}{4}\)

Example 2: Ellipse with Product Term

Problem: Find \(\frac{dy}{dx}\) for the curve \(x^2 + xy + y^2 = 3\).

Solution:

Step 1: Differentiate both sides

\[ \frac{d}{dx}[x^2 + xy + y^2] = \frac{d}{dx}[3] \]

Step 2: Apply rules term by term

\(\frac{d}{dx}[x^2] = 2x\)
\(\frac{d}{dx}[xy] = x\frac{dy}{dx} + y\) (Product Rule)
\(\frac{d}{dx}[y^2] = 2y\frac{dy}{dx}\) (Chain Rule)
\(\frac{d}{dx}[3] = 0\)

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

Step 3: Collect terms with \(\frac{dy}{dx}\)

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

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

Step 4: Solve for \(\frac{dy}{dx}\)

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

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

Example 3: Horizontal and Vertical Tangent Lines

Problem: For the curve \(x^3 + y^3 = 6xy\), find all points where:
(a) The tangent line is horizontal
(b) The tangent line is vertical

Solution:

Step 1: Find \(\frac{dy}{dx}\)

Differentiate:

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

Collect terms:

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

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

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

Part (a): Horizontal tangent (\(\frac{dy}{dx} = 0\))

Numerator = 0:

\[ 2y - x^2 = 0 \quad \Rightarrow \quad y = \frac{x^2}{2} \]

Substitute into original equation \(x^3 + y^3 = 6xy\):

\[ x^3 + \left(\frac{x^2}{2}\right)^3 = 6x \cdot \frac{x^2}{2} \]

\[ x^3 + \frac{x^6}{8} = 3x^3 \]

\[ \frac{x^6}{8} = 2x^3 \]

\[ x^6 = 16x^3 \quad \Rightarrow \quad x^3(x^3 - 16) = 0 \]

Solutions: \(x = 0\) or \(x = \sqrt[3]{16} = 2\sqrt[3]{2}\)

For \(x = 0\): \(y = 0\), but \((0,0)\) doesn't satisfy original equation

For \(x = 2\sqrt[3]{2}\): \(y = \frac{(2\sqrt[3]{2})^2}{2} = 2\sqrt[3]{4}\)

Part (b): Vertical tangent (\(\frac{dy}{dx}\) undefined)

Denominator = 0:

\[ y^2 - 2x = 0 \quad \Rightarrow \quad x = \frac{y^2}{2} \]

Substitute and solve similarly...

Answer: (a) Horizontal tangent at \(\left(2\sqrt[3]{2}, 2\sqrt[3]{4}\right)\)

Example 4: Second Derivative (Concavity)

Problem: For \(x^2 + y^2 = 25\), find \(\frac{d^2y}{dx^2}\) at the point \((3, 4)\).

Solution:

Step 1: Find first derivative (from Example 1)

\[ \frac{dy}{dx} = -\frac{x}{y} \]

Step 2: Differentiate again (use Quotient Rule)

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

\[ = -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} \]

\[ = -\frac{y - x\frac{dy}{dx}}{y^2} \]

Step 3: Substitute \(\frac{dy}{dx} = -\frac{x}{y}\)

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

\[ = -\frac{y^2 + x^2}{y^3} \]

Since \(x^2 + y^2 = 25\):

\[ \frac{d^2y}{dx^2} = -\frac{25}{y^3} \]

Step 4: Evaluate at (3, 4)

\[ \frac{d^2y}{dx^2}\bigg|_{(3,4)} = -\frac{25}{4^3} = -\frac{25}{64} \]

Answer: \(\frac{d^2y}{dx^2} = -\frac{25}{64}\) at \((3, 4)\). Since negative, curve is concave down.

⚠️ Special Cases and Important Points

Key Scenarios:

Horizontal Tangent Lines

Occur when \(\frac{dy}{dx} = 0\)

Numerator of \(\frac{dy}{dx}\) equals zero
Denominator ≠ 0
These are typically extrema of the curve

Vertical Tangent Lines

Occur when \(\frac{dy}{dx}\) is undefined

Denominator of \(\frac{dy}{dx}\) equals zero
Numerator ≠ 0
Graph has vertical slope

Singular Points

When both numerator AND denominator = 0

Point where curve has a cusp or crosses itself
\(\frac{dy}{dx}\) is indeterminate (0/0)
Requires special analysis

📊 Finding Local Extrema on Implicit Curves

Procedure for Finding Extrema:

Find \(\frac{dy}{dx}\) using implicit differentiation
Find critical points:
- Set numerator = 0 (horizontal tangents)
- Substitute back into original equation
- Solve for actual points
Verify points are on the curve
Test using Second Derivative Test or sign analysis
Classify as max, min, or neither

🎯 Applications of Implicit Differentiation

Common Applications:

Where Implicit Relations Appear
Application	Implicit Equation	Why Implicit?
Circles	\(x^2 + y^2 = r^2\)	Can't write as single function \(y=f(x)\)
Ellipses	\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)	Two \(y\) values for each \(x\)
Hyperbolas	\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)	Multiple branches
Folium of Descartes	\(x^3 + y^3 = 3xy\)	Cannot solve for \(y\)
Lemniscate	\((x^2 + y^2)^2 = a^2(x^2 - y^2)\)	Figure-eight shape
Related Rates	Various physical constraints	Multiple related variables

💡 Tips, Tricks & Strategies

✅ Essential Tips for Implicit Differentiation:

Use Chain Rule automatically: Every \(y\) term needs \(\frac{dy}{dx}\)
Product Rule for \(xy\) terms: \(\frac{d}{dx}[xy] = x\frac{dy}{dx} + y\)
Don't solve for \(y\) first: Differentiate the equation as given
Collect all \(\frac{dy}{dx}\) terms: Factor them out
Simplify algebraically: Factor numerator and denominator
Substitute points AFTER differentiating: Get general formula first
Check your answer: Verify point is on curve
For second derivative: Differentiate the first derivative (it has both \(x\) and \(y\)!)

🔥 Common Patterns to Remember:

Quick Reference
If you see...	Derivative is...
\(y^n\)	\(ny^{n-1}\frac{dy}{dx}\)
\(xy\)	\(x\frac{dy}{dx} + y\)
\(x^2y\)	\(x^2\frac{dy}{dx} + 2xy\)
\(xy^2\)	\(x \cdot 2y\frac{dy}{dx} + y^2\)
\(\sqrt{y}\)	\(\frac{1}{2\sqrt{y}}\frac{dy}{dx}\)
\(\sin(xy)\)	\(\cos(xy)\left(x\frac{dy}{dx} + y\right)\)

❌ Common Mistakes to Avoid

Mistake 1: Forgetting \(\frac{dy}{dx}\) when differentiating \(y\) terms
Mistake 2: Not using Product Rule for \(xy\) terms
Mistake 3: Treating \(y\) as a constant (it's a function of \(x\)!)
Mistake 4: Substituting the point too early (find general formula first)
Mistake 5: Forgetting to solve for \(\frac{dy}{dx}\) (leaving it implicit)
Mistake 6: Algebraic errors when collecting \(\frac{dy}{dx}\) terms
Mistake 7: Not simplifying the final answer
Mistake 8: For second derivative: forgetting that \(\frac{dy}{dx}\) contains both \(x\) and \(y\)
Mistake 9: Not verifying that point is actually on the curve
Mistake 10: Confusing when tangent is horizontal (numerator = 0) vs vertical (denominator = 0)

📝 Practice Problems

Set A: Basic Implicit Differentiation

Find \(\frac{dy}{dx}\) for \(x^2 + y^2 = 16\)
Find \(\frac{dy}{dx}\) for \(x^3 + y^3 = 6xy\)
Find the slope of tangent to \(x^2 - xy + y^2 = 3\) at \((1, 2)\)

Answers:

\(\frac{dy}{dx} = -\frac{x}{y}\)
\(\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x}\)
\(m = 0\) (horizontal tangent)

Set B: Tangent Lines

Find equation of tangent line to \(x^2 + y^2 = 25\) at \((4, 3)\)
Find all points on \(x^2 + y^2 = 25\) where tangent is vertical

Answers:

\(y = -\frac{4}{3}x + \frac{25}{3}\)
\((5, 0)\) and \((-5, 0)\)

Set C: Second Derivative

For \(x^2 + y^2 = 9\), find \(\frac{d^2y}{dx^2}\) at \((0, 3)\)
Determine concavity of \(xy = 1\) at \((1, 1)\)

Answers:

\(\frac{d^2y}{dx^2} = -\frac{1}{3}\) (concave down)
\(\frac{d^2y}{dx^2} = \frac{2}{1} = 2\) (concave up)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Clear differentiation: Show derivative of each term
\(\frac{dy}{dx}\) notation: Write it explicitly when differentiating \(y\) terms
Algebraic work: Show collection and solving for \(\frac{dy}{dx}\)
Substitution shown: When evaluating at a point
Simplification: Final answer in simplified form
Point-slope form: For tangent line equations
Verification: Check point is on curve if requested
Complete answers: Include all requested information

💯 Earning Full Credit:

1 point: Correct implicit differentiation
1 point: Solving for \(\frac{dy}{dx}\)
1 point: Evaluating at specific point
1 point: Correct tangent line equation (if asked)
Key: Show ALL steps—partial credit for correct method!

⚡ Quick Reference Card

Implicit Differentiation Quick Reference
Concept	Formula/Method	Key Point
Basic Process	Differentiate both sides w.r.t. \(x\)	Treat \(y\) as function of \(x\)
Chain Rule	\(\frac{d}{dx}[y^n] = ny^{n-1}\frac{dy}{dx}\)	Always include \(\frac{dy}{dx}\)
Product \(xy\)	\(\frac{d}{dx}[xy] = x\frac{dy}{dx} + y\)	Use Product Rule
Horizontal Tangent	Numerator of \(\frac{dy}{dx} = 0\)	Slope = 0
Vertical Tangent	Denominator of \(\frac{dy}{dx} = 0\)	Slope undefined
Second Derivative	Differentiate \(\frac{dy}{dx}\) again	Contains both \(x\) and \(y\)
Tangent Line	\(y - y_0 = m(x - x_0)\)	Evaluate \(m\) at point

Master Implicit Relations! Implicit differentiation extends derivative techniques to curves that can't be written as \(y = f(x)\), like circles, ellipses, and complex curves. The key principle: differentiate both sides with respect to \(x\), treating \(y\) as an implicit function of \(x\). Always use the Chain Rule—every \(y\) term needs \(\frac{dy}{dx}\). For products like \(xy\), use the Product Rule: \(\frac{d}{dx}[xy] = x\frac{dy}{dx} + y\). After differentiating, collect all \(\frac{dy}{dx}\) terms, factor out \(\frac{dy}{dx}\), and solve algebraically. For tangent lines, evaluate \(\frac{dy}{dx}\) at the specific point. Horizontal tangents occur when numerator = 0; vertical tangents when denominator = 0. For second derivatives and concavity, differentiate \(\frac{dy}{dx}\) again (remember it contains both \(x\) and \(y\)!). Common applications include circles, ellipses, and exotic curves like the folium of Descartes. Always verify points are on the curve, show all work clearly, and simplify final answers. This technique is essential for analyzing complex geometric relationships! 🎯✨