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Unit 9.3 – Arc Lengths of Parametric Curves BC ONLY

AP® Calculus BC | Measuring Parametric Curve Length

Why This Matters: Arc length for parametric equations extends the arc length concept from Unit 8 to curves defined parametrically. Instead of \(y = f(x)\), we work with \(x(t)\) and \(y(t)\), measuring the actual distance traveled along the curve as the parameter varies. This is essential for physics applications (distance traveled) and appears regularly on BC exams!

📐 The Arc Length Formula

Arc Length for Parametric Curves

THE FORMULA:

For parametric curve \(x = f(t)\), \(y = g(t)\) from \(t = \alpha\) to \(t = \beta\):

\[ L = \int_\alpha^\beta \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
\[ L = \int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2} \, dt \]

Why This Formula?

From the Pythagorean theorem on infinitesimal segments:

\[ ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

Integrating gives the total arc length!

🚗 Speed and Distance Traveled

SPEED FORMULA

The speed at time \(t\) is the magnitude of velocity:

\[ \text{Speed} = \left|\vec{v}(t)\right| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \]
Distance Traveled:

The total distance traveled from \(t = \alpha\) to \(t = \beta\):

\[ \text{Distance} = \int_\alpha^\beta \text{Speed} \, dt = \int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2} \, dt \]

This is exactly the arc length formula!

📝 Key Insight: Arc length and distance traveled are the same thing for parametric curves—both measure the path length.

🔄 Relationship to Regular Arc Length

Connecting to y = f(x) Form

Recall: Arc Length for y = f(x)
\[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]
Parametric Form Reduces to This:

If we use parameter \(t = x\), then \(x = t\), \(y = f(t)\):

\[ \frac{dx}{dt} = 1, \quad \frac{dy}{dt} = f'(t) \]
\[ L = \int_a^b \sqrt{1 + [f'(t)]^2} \, dt = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx \]

The parametric formula is more general!

📖 Comprehensive Worked Examples

Example 1: Circle Arc Length

Problem: Find the arc length of the circle \(x = \cos t\), \(y = \sin t\) from \(t = 0\) to \(t = 2\pi\).

Solution:

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

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

Step 2: Square and add

\[ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \sin^2 t + \cos^2 t = 1 \]

Step 3: Set up and evaluate integral

\[ L = \int_0^{2\pi} \sqrt{1} \, dt = \int_0^{2\pi} 1 \, dt = 2\pi \]

ANSWER: \(L = 2\pi\) (the circumference of a unit circle!)

Example 2: Parabolic Path

Problem: Find the arc length of \(x = t^2\), \(y = t^3\) from \(t = 0\) to \(t = 1\).

Step 1: Find derivatives

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

Step 2: Set up integral

\[ L = \int_0^1 \sqrt{(2t)^2 + (3t^2)^2} \, dt = \int_0^1 \sqrt{4t^2 + 9t^4} \, dt \]
\[ = \int_0^1 \sqrt{t^2(4 + 9t^2)} \, dt = \int_0^1 t\sqrt{4 + 9t^2} \, dt \]

Step 3: Use substitution

Let \(u = 4 + 9t^2\), \(du = 18t \, dt\), so \(t \, dt = \frac{du}{18}\)

When \(t = 0\): \(u = 4\); when \(t = 1\): \(u = 13\)

\[ L = \int_4^{13} \sqrt{u} \cdot \frac{du}{18} = \frac{1}{18} \cdot \frac{2u^{3/2}}{3}\Big|_4^{13} = \frac{1}{27}(13^{3/2} - 8) \]

Example 3: Cycloid

Problem: Find one arch length of the cycloid \(x = t - \sin t\), \(y = 1 - \cos t\) from \(t = 0\) to \(t = 2\pi\).

Find derivatives:

\[ \frac{dx}{dt} = 1 - \cos t, \quad \frac{dy}{dt} = \sin t \]

Set up integral:

\[ L = \int_0^{2\pi} \sqrt{(1-\cos t)^2 + \sin^2 t} \, dt \]
\[ = \int_0^{2\pi} \sqrt{1 - 2\cos t + \cos^2 t + \sin^2 t} \, dt \]
\[ = \int_0^{2\pi} \sqrt{2 - 2\cos t} \, dt = \int_0^{2\pi} \sqrt{2(1-\cos t)} \, dt \]

Use identity: \(1 - \cos t = 2\sin^2(t/2)\)

\[ L = \int_0^{2\pi} \sqrt{4\sin^2(t/2)} \, dt = \int_0^{2\pi} 2\left|\sin(t/2)\right| \, dt = 2\int_0^{2\pi} \sin(t/2) \, dt \]
\[ = 2 \cdot \left[-2\cos(t/2)\right]_0^{2\pi} = -4[\cos\pi - \cos 0] = -4[-1-1] = 8 \]

Example 4: Distance Traveled

Problem: A particle moves with position \(x = e^t\cos t\), \(y = e^t\sin t\). Find distance traveled from \(t = 0\) to \(t = \pi\).

Find derivatives (using product rule):

\[ \frac{dx}{dt} = e^t\cos t - e^t\sin t = e^t(\cos t - \sin t) \]
\[ \frac{dy}{dt} = e^t\sin t + e^t\cos t = e^t(\sin t + \cos t) \]

Set up:

\[ L = \int_0^\pi \sqrt{e^{2t}(\cos t - \sin t)^2 + e^{2t}(\sin t + \cos t)^2} \, dt \]
\[ = \int_0^\pi e^t\sqrt{(\cos t - \sin t)^2 + (\sin t + \cos t)^2} \, dt \]
\[ = \int_0^\pi e^t\sqrt{2} \, dt = \sqrt{2}e^t\Big|_0^\pi = \sqrt{2}(e^\pi - 1) \]

📊 Arc Length Formula Comparison

Arc Length Formulas
Curve Type Formula Notes
\(y = f(x)\) \(\int_a^b \sqrt{1 + [f'(x)]^2}\,dx\) Standard form
\(x = g(y)\) \(\int_c^d \sqrt{1 + [g'(y)]^2}\,dy\) Inverse form
Parametric \(\int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt\) Most general

💡 Essential Tips & Strategies

✅ Success Strategies:

  • Find both derivatives first: Calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
  • Square each separately: Before adding under square root
  • Look for simplifications: Under the square root before integrating
  • Trig identities help: Especially \(\sin^2 + \cos^2 = 1\)
  • Factor when possible: Can often pull terms out of square root
  • Use substitution: For complicated integrals
  • Calculator allowed: Many problems are calculator problems
  • Units are length: Not squared

🔥 Special Cases & Shortcuts:

  • Circle: \(x = r\cos t\), \(y = r\sin t\) → Arc length = \(2\pi r\)
  • Line segment: Arc length = straight-line distance
  • Perfect square under radical: Simplifies nicely
  • Constant speed: If \(\sqrt{(dx/dt)^2 + (dy/dt)^2}\) is constant

❌ Common Mistakes to Avoid

  • Mistake 1: Forgetting to square both derivatives before adding
  • Mistake 2: Using \(|x'(t)| + |y'(t)|\) instead of \(\sqrt{[x'(t)]^2 + [y'(t)]^2}\)
  • Mistake 3: Not simplifying under square root before integrating
  • Mistake 4: Product rule errors when finding derivatives
  • Mistake 5: Wrong bounds (using x or y values instead of t values)
  • Mistake 6: Forgetting absolute value when simplifying square root
  • Mistake 7: Integration errors (especially with substitution)
  • Mistake 8: Calculator mode issues (radians vs degrees)
  • Mistake 9: Saying square units instead of just units
  • Mistake 10: Not recognizing special forms (circles, lines)

📝 Practice Problems

Find the arc length:

  1. \(x = 3t\), \(y = 4t\) from \(t = 0\) to \(t = 1\)
  2. \(x = \cos^3 t\), \(y = \sin^3 t\) from \(t = 0\) to \(t = \pi/2\)
  3. \(x = e^t\sin t\), \(y = e^t\cos t\) from \(t = 0\) to \(t = \pi\)
  4. \(x = t^2\), \(y = \frac{2}{3}t^3\) from \(t = 0\) to \(t = 2\)

Answers:

  1. \(L = 5\) (straight line)
  2. \(L = \frac{3}{2}\)
  3. \(L = \sqrt{2}(e^\pi - 1)\)
  4. \(L = \frac{2}{3}(5\sqrt{5} - 1)\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

  • Write the formula: \(L = \int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt\)
  • Show both derivatives: Calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
  • Show squaring: Square each derivative
  • Set up integral: Substitute into formula with correct bounds
  • Show simplification: If any under the square root
  • Evaluate integral: Show work or clearly state using calculator
  • Include units: Length units (not area)
  • Exact or decimal: Follow problem instructions

💯 Exam Strategy:

  1. Identify it's an arc length problem
  2. Write the arc length formula
  3. Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)
  4. Square both derivatives
  5. Add under square root
  6. Simplify if possible
  7. Set up integral with correct bounds (t values!)
  8. Evaluate (calculator often allowed)
  9. State answer with units

⚡ Quick Reference Guide

PARAMETRIC ARC LENGTH ESSENTIALS

The Formula:

\[ L = \int_\alpha^\beta \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

Speed Formula:

\[ \text{Speed} = \sqrt{[x'(t)]^2 + [y'(t)]^2} \]

Distance = Arc Length:

For parametric curves, they're the same!

Remember:

  • Square BOTH derivatives
  • Add under ONE square root
  • Bounds are t-values
  • Units: length (not area!)

Master Parametric Arc Length! The fundamental formula: \(L = \int_\alpha^\beta\sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt\). Derived from Pythagorean theorem on infinitesimal segments: \(ds = \sqrt{(dx)^2+(dy)^2}\). Process: (1) find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), (2) square each derivative, (3) add under square root, (4) simplify if possible, (5) integrate from \(\alpha\) to \(\beta\) (t-bounds!). This equals speed = \(\sqrt{[x'(t)]^2+[y'(t)]^2}\) integrated over time = distance traveled. Special cases: circles give \(2\pi r\), straight lines give actual distance. Common errors: forgetting to square, using wrong bounds, not simplifying under radical. Calculator problems common on AP® exams. Connects to regular arc length: parametric form is more general. This is guaranteed BC content—appears regularly! Practice until automatic! 🎯✨