Unit 9.1 – Defining and Differentiating Parametric Equations BC ONLY

AP® Calculus BC | Parametric Curves and Calculus

Why This Matters: Parametric equations open a new world in calculus! Instead of \(y = f(x)\), we express both \(x\) and \(y\) as functions of a third variable \(t\) (parameter). This allows us to describe curves that can't be expressed as single functions, like circles, loops, and particle motion. Mastering parametric calculus is essential for BC students and appears on virtually every BC exam!

📚 What Are Parametric Equations?

PARAMETRIC EQUATIONS DEFINED

A parametric curve is defined by a pair of equations:

\[ x = f(t), \quad y = g(t) \]

where \(t\) is the parameter (often representing time)

Key Characteristics:

Both \(x\) and \(y\) depend on the parameter \(t\)
As \(t\) varies over an interval, the point \((x(t), y(t))\) traces out a curve
The parameter often represents time, but not always
One curve can have multiple parametric representations

📝 Why Use Parametric Equations? They can represent curves that fail the vertical line test (like circles), curves that double back on themselves, and motion along a path where direction and speed matter.

🔄 The First Derivative: dy/dx

First Derivative of Parametric Equations

THE FORMULA:

For parametric equations \(x = f(t)\) and \(y = g(t)\):

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)} \]

provided \(\frac{dx}{dt} \neq 0\)

Why This Works:

Using the chain rule:

\[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy/dt}{dx/dt} \]

💡 Key Insight: This is NOT \(\frac{g(t)}{f(t)}\)! You must find the derivatives \(g'(t)\) and \(f'(t)\) first, then divide.

🔁 The Second Derivative: d²y/dx²

Second Derivative of Parametric Equations

THE FORMULA:

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d/dt(dy/dx)}{dx/dt} \]

Step-by-Step Process:

Find \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)
Differentiate \(\frac{dy}{dx}\) with respect to \(t\) (use quotient rule!)
Divide by \(\frac{dx}{dt}\)

The Formula Expanded:

\[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy/dt}{dx/dt}\right)}{dx/dt} = \frac{(dx/dt)(d^2y/dt^2) - (dy/dt)(d^2x/dt^2)}{(dx/dt)^3} \]

(using quotient rule)

📐 Horizontal and Vertical Tangent Lines

Special Tangent Lines

Horizontal Tangent Lines:

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

\[ \frac{dy}{dt} = 0 \quad \text{AND} \quad \frac{dx}{dt} \neq 0 \]

Solve \(g'(t) = 0\) where \(f'(t) \neq 0\)

Vertical Tangent Lines:

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

\[ \frac{dx}{dt} = 0 \quad \text{AND} \quad \frac{dy}{dt} \neq 0 \]

Solve \(f'(t) = 0\) where \(g'(t) \neq 0\)

⚠️ WARNING: If both \(\frac{dx}{dt} = 0\) AND \(\frac{dy}{dt} = 0\) at the same value of \(t\), the tangent is indeterminate. This is called a cusp or singular point.

📖 Comprehensive Worked Examples

Example 1: Finding dy/dx

Problem: For \(x = t^2 + 1\) and \(y = t^3 - 3t\), find \(\frac{dy}{dx}\) at \(t = 2\).

Solution:

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

\[ \frac{dx}{dt} = 2t \]

\[ \frac{dy}{dt} = 3t^2 - 3 \]

Step 2: Apply the formula

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2 - 3}{2t} \]

Step 3: Evaluate at \(t = 2\)

\[ \frac{dy}{dx}\Big|_{t=2} = \frac{3(4) - 3}{2(2)} = \frac{9}{4} \]

ANSWER: \(\frac{dy}{dx} = \frac{9}{4}\) at \(t = 2\)

Example 2: Finding d²y/dx²

Problem: For \(x = t^2\) and \(y = t^3\), find \(\frac{d^2y}{dx^2}\).

Step 1: Find first derivatives

\[ \frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2 \]

Step 2: Find \(\frac{dy}{dx}\)

\[ \frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2} \]

Step 3: Differentiate \(\frac{dy}{dx}\) with respect to \(t\)

\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2} \]

Step 4: Divide by \(\frac{dx}{dt}\)

\[ \frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t} \]

Example 3: Horizontal Tangents

Problem: Find all points where \(x = t^2 - 4\), \(y = t^3 - 3t\) has a horizontal tangent.

Condition: \(\frac{dy}{dt} = 0\) and \(\frac{dx}{dt} \neq 0\)

\[ \frac{dy}{dt} = 3t^2 - 3 = 0 \]

\[ 3(t^2 - 1) = 0 \quad \Rightarrow \quad t = \pm 1 \]

Check \(\frac{dx}{dt} \neq 0\)

\[ \frac{dx}{dt} = 2t \]

At \(t = 1\): \(\frac{dx}{dt} = 2 \neq 0\) ✓

At \(t = -1\): \(\frac{dx}{dt} = -2 \neq 0\) ✓

Find the points:

At \(t = 1\): \((x, y) = (-3, -2)\)

At \(t = -1\): \((x, y) = (-3, 2)\)

Example 4: Vertical Tangents

Problem: For \(x = \cos t\), \(y = \sin t\), find vertical tangent lines on \([0, 2\pi]\).

Condition: \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} \neq 0\)

\[ \frac{dx}{dt} = -\sin t = 0 \]

On \([0, 2\pi]\): \(t = 0, \pi, 2\pi\)

Check \(\frac{dy}{dt} \neq 0\)

\[ \frac{dy}{dt} = \cos t \]

At \(t = 0\): \(\cos 0 = 1 \neq 0\) ✓

At \(t = \pi\): \(\cos \pi = -1 \neq 0\) ✓

At \(t = 2\pi\): \(\cos 2\pi = 1 \neq 0\) ✓

Points with vertical tangents:

\((1, 0)\) and \((-1, 0)\)

📊 Complete Formula Reference

Parametric Derivatives Summary
Quantity	Formula	Notes
First Derivative	\(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)	Slope of tangent line
Second Derivative	\(\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}\)	Concavity
Horizontal Tangent	\(dy/dt = 0, dx/dt \neq 0\)	Slope = 0
Vertical Tangent	\(dx/dt = 0, dy/dt \neq 0\)	Undefined slope

💡 Essential Tips & Strategies

✅ Success Strategies:

Always find derivatives with respect to \(t\) first: Don't confuse \(\frac{dy}{dx}\) with \(\frac{g(t)}{f(t)}\)
For second derivative: Use quotient rule on \(\frac{dy}{dx}\) with respect to \(t\)
Check both conditions: For horizontal/vertical tangents
Simplify \(\frac{dy}{dx}\) before differentiating: Makes second derivative easier
Keep track of what you're differentiating: With respect to \(t\) or \(x\)?
Sketch if possible: Helps understand the curve

🔥 Common Parametric Curves:

Circle: \(x = r\cos t\), \(y = r\sin t\)
Ellipse: \(x = a\cos t\), \(y = b\sin t\)
Line: \(x = x_0 + at\), \(y = y_0 + bt\)
Cycloid: \(x = r(t - \sin t)\), \(y = r(1 - \cos t)\)

❌ Common Mistakes to Avoid

Mistake 1: Using \(\frac{g(t)}{f(t)}\) instead of \(\frac{g'(t)}{f'(t)}\) for \(\frac{dy}{dx}\)
Mistake 2: Forgetting to differentiate \(\frac{dy}{dx}\) with respect to \(t\) for second derivative
Mistake 3: Not checking both conditions for horizontal/vertical tangents
Mistake 4: Confusing when to use chain rule vs quotient rule
Mistake 5: Evaluating at wrong value of \(t\) vs \((x, y)\)
Mistake 6: Sign errors when differentiating trig functions
Mistake 7: Not simplifying before taking second derivative
Mistake 8: Missing the division by \(\frac{dx}{dt}\) in second derivative
Mistake 9: Assuming \((dx/dt)^2\) in denominator (it's cubed!)
Mistake 10: Not checking domain/interval for parameter \(t\)

📝 Practice Problems

Solve these:

Find \(\frac{dy}{dx}\) for \(x = t^3\), \(y = t^2\) at \(t = 1\).
Find \(\frac{d^2y}{dx^2}\) for \(x = e^t\), \(y = e^{-t}\).
Find horizontal tangents for \(x = t^2 - 1\), \(y = t^3 - t\).
Find vertical tangents for \(x = \sin^2 t\), \(y = \cos t\) on \([0, 2\pi]\).

Answers:

\(\frac{dy}{dx} = \frac{2}{3}\) at \(t = 1\)
\(\frac{d^2y}{dx^2} = \frac{2}{e^{3t}}\)
\(t = \pm\frac{1}{\sqrt{3}}\) giving points \((-\frac{2}{3}, \pm\frac{2\sqrt{3}}{9})\)
\(t = 0, \pi, 2\pi\) (but check: only \(t = \pi\) gives vertical tangent)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Write the formula: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) explicitly
Show both derivatives: Calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
Show division: Don't skip steps
For second derivative: Show differentiation of \(\frac{dy}{dx}\) w.r.t. \(t\)
For tangents: State conditions being checked
Evaluate at specific \(t\): Substitute correctly
Find \((x, y)\) coordinates: When asked for points
Simplify answers: Leave in exact form unless specified

💯 Exam Strategy:

Identify it's a parametric problem
Write down \(x(t)\) and \(y(t)\) clearly
Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) separately
Apply appropriate formula
Simplify before further operations
Check domain/conditions
Evaluate at specific values if asked
State final answer clearly

⚡ Quick Reference Guide

PARAMETRIC CALCULUS ESSENTIALS

First Derivative:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)} \]

Second Derivative:

\[ \frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt} \]

Tangent Lines:

Horizontal: \(dy/dt = 0\), \(dx/dt \neq 0\)
Vertical: \(dx/dt = 0\), \(dy/dt \neq 0\)

Remember:

Differentiate with respect to \(t\) first!
Then divide for \(dy/dx\)
Check both conditions for tangents

Master Parametric Derivatives! Parametric equations: \(x = f(t)\), \(y = g(t)\) express both coordinates as functions of parameter \(t\). First derivative: \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}\)—find derivatives with respect to \(t\), then divide. Second derivative: \(\frac{d^2y}{dx^2} = \frac{d/dt(dy/dx)}{dx/dt}\)—differentiate \(dy/dx\) with respect to \(t\), then divide by \(dx/dt\). Using quotient rule: \(\frac{(dx/dt)(d^2y/dt^2) - (dy/dt)(d^2x/dt^2)}{(dx/dt)^3}\). Horizontal tangents: \(dy/dt = 0\) AND \(dx/dt \neq 0\). Vertical tangents: \(dx/dt = 0\) AND \(dy/dt \neq 0\). If both zero: cusp/singular point. Common error: using \(g(t)/f(t)\) instead of \(g'(t)/f'(t)\). This is BC-ONLY and fundamental for Unit 9—appears on every BC exam! Practice until automatic! 🎯✨