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Unit 8.13 – Arc Length of Smooth, Planar Curves & Distance Traveled BC ONLY

AP® Calculus BC | Measuring Curve Length

Why This Matters: Arc length measures the actual distance along a curve—not just the straight-line distance! This BC-only topic extends integration to find the length of curves defined by functions or parametric equations. From the path of a particle to the length of a suspension cable, arc length has countless applications. Master this and you've added a powerful tool to your calculus arsenal!

🎯 The Arc Length Concept

FROM PYTHAGOREAN THEOREM TO ARC LENGTH

Imagine breaking a curve into tiny segments. Each segment is approximately a straight line with length given by the Pythagorean theorem:

\[ \Delta s \approx \sqrt{(\Delta x)^2 + (\Delta y)^2} \]

The total length is the sum of all these segments. Taking the limit as segments become infinitesimally small:

\[ L = \int ds = \int \sqrt{(dx)^2 + (dy)^2} \]

📐 Arc Length for y = f(x)

Arc Length Formula: Functions of x

THE FORMULA:

For a smooth curve \(y = f(x)\) from \(x = a\) to \(x = b\):

\[ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]
\[ L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx \]
Derivation:

Starting from \(ds = \sqrt{(dx)^2 + (dy)^2}\):

\[ ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]

Factor out \(dx\) from the square root, integrate from \(a\) to \(b\).

📝 Key Point: The curve must be smooth (continuous derivative) on \([a, b]\).

📐 Arc Length for x = g(y)

Arc Length Formula: Functions of y

THE FORMULA:

For a smooth curve \(x = g(y)\) from \(y = c\) to \(y = d\):

\[ L = \int_c^d \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy \]
\[ L = \int_c^d \sqrt{1 + [g'(y)]^2} \, dy \]

💡 When to Use: Use this form when the curve is more naturally expressed as \(x = g(y)\) or when \(dy\) is easier to work with.

🔄 Arc Length for Parametric Curves

Arc Length: Parametric Equations

THE FORMULA:

For parametric curve \(x = x(t)\), \(y = y(t)\) 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 \(ds = \sqrt{(dx)^2 + (dy)^2}\), divide and multiply by \(dt\):

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

🚗 Distance Traveled vs Arc Length

DISTANCE TRAVELED FOR PARAMETRIC CURVES

The Formula:

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

This is the same as arc length for parametric curves!

📝 Important Distinction: Distance traveled is the total path length, even if the particle backtracks. Arc length is the length of the curve itself.

📖 Comprehensive Worked Examples

Example 1: Arc Length for y = f(x)

Problem: Find the arc length of \(y = \frac{2}{3}x^{3/2}\) from \(x = 0\) to \(x = 3\).

Solution:

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

\[ \frac{dy}{dx} = \frac{2}{3} \cdot \frac{3}{2}x^{1/2} = x^{1/2} = \sqrt{x} \]

Step 2: Square the derivative

\[ \left(\frac{dy}{dx}\right)^2 = x \]

Step 3: Set up arc length integral

\[ L = \int_0^3 \sqrt{1 + x} \, dx \]

Step 4: Evaluate

Let \(u = 1 + x\), \(du = dx\)

When \(x = 0\): \(u = 1\); when \(x = 3\): \(u = 4\)

\[ L = \int_1^4 \sqrt{u} \, du = \left[\frac{2u^{3/2}}{3}\right]_1^4 = \frac{2}{3}(8 - 1) = \frac{14}{3} \]

ANSWER: \(L = \frac{14}{3}\) units

Example 2: Arc Length for x = g(y)

Problem: Find the arc length of \(x = \frac{1}{3}(y^2 + 2)^{3/2}\) from \(y = 0\) to \(y = 1\).

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

\[ \frac{dx}{dy} = \frac{1}{3} \cdot \frac{3}{2}(y^2+2)^{1/2} \cdot 2y = y\sqrt{y^2+2} \]

Step 2: Square and simplify

\[ \left(\frac{dx}{dy}\right)^2 = y^2(y^2+2) = y^4 + 2y^2 \]
\[ 1 + \left(\frac{dx}{dy}\right)^2 = 1 + y^4 + 2y^2 = (y^2+1)^2 \]

Step 3: Evaluate

\[ L = \int_0^1 \sqrt{(y^2+1)^2} \, dy = \int_0^1 (y^2+1) \, dy \]
\[ = \left[\frac{y^3}{3} + y\right]_0^1 = \frac{1}{3} + 1 = \frac{4}{3} \]

Example 3: Parametric Arc Length

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

Step 1: Find derivatives

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

Step 2: Set up formula

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

Step 3: Evaluate using substitution

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

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

Example 4: Distance Traveled

Problem: A particle moves along the path \(x = \cos t\), \(y = \sin t\) from \(t = 0\) to \(t = 2\pi\). Find the distance traveled.

Solution:

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

This is the circumference of a unit circle!

📊 Complete Formula Reference

Arc Length Formulas
Curve Type Formula Variable
\(y = f(x)\) \(\int_a^b \sqrt{1 + [f'(x)]^2}\,dx\) Integrate w.r.t. \(x\)
\(x = g(y)\) \(\int_c^d \sqrt{1 + [g'(y)]^2}\,dy\) Integrate w.r.t. \(y\)
Parametric \(\int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt\) Integrate w.r.t. \(t\)

💡 Essential Tips & Strategies

✅ Success Strategies:

  • Find derivative carefully: Chain rule, product rule as needed
  • Square the derivative: Don't forget this step!
  • Simplify under square root: Before integrating if possible
  • Look for perfect squares: \(1 + [f']^2 = (\text{something})^2\)
  • For parametric: Find both \(x'(t)\) and \(y'(t)\)
  • Use substitution: When integrand is complicated
  • Calculator okay: These integrals are often calculator problems
  • Units are length units: Not squared

🔥 Special Cases:

  • Straight lines: Arc length = distance formula
  • Circles: Arc length = circumference or portion thereof
  • Perfect squares under radical: Simplify before integrating
  • Trigonometric identities: Use \(\sin^2 + \cos^2 = 1\)

❌ Common Mistakes to Avoid

  • Mistake 1: Forgetting to square the derivative
  • Mistake 2: Forgetting the "+1" in \(\sqrt{1 + [f']^2}\)
  • Mistake 3: Using distance formula instead of arc length
  • Mistake 4: Wrong derivative (chain rule errors)
  • Mistake 5: Not simplifying under square root before integrating
  • Mistake 6: For parametric: forgetting to square BOTH derivatives
  • Mistake 7: Integration errors (especially with substitution)
  • Mistake 8: Wrong bounds (using x-values instead of t-values for parametric)
  • Mistake 9: Saying square units instead of just units
  • Mistake 10: Calculator mode errors (radians vs degrees)

📝 Practice Problems

Find the arc length:

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

Answers:

  1. \(\frac{1}{27}(80\sqrt{10} - 8)\) units
  2. \(\ln(\sqrt{2} + 1)\) units
  3. \(\sqrt{2}(e^\pi - 1)\) units
  4. \(\frac{1}{27}(10\sqrt{10} - 1)\) units

✏️ AP® Exam Success Tips

What AP® Graders Look For:

  • Identify formula type: State which arc length formula you're using
  • Show derivative: Calculate \(f'(x)\) or \(x'(t)\), \(y'(t)\)
  • Show squaring: Explicitly square the derivative(s)
  • Set up integral: \(L = \int \sqrt{1 + [f']^2}\) or parametric version
  • Simplify integrand: If possible before evaluating
  • Show integration work: Or clearly state using calculator
  • Evaluate: Find numerical or exact answer
  • Include units: Length units (not squared)

💯 Exam Strategy:

  1. Identify curve type: \(y = f(x)\), \(x = g(y)\), or parametric?
  2. Write appropriate arc length formula
  3. Find required derivative(s)
  4. Square derivative(s)
  5. Substitute into formula
  6. Simplify under square root if possible
  7. Evaluate integral (calculator allowed for these)
  8. State answer with proper units

⚡ Quick Reference Guide

ARC LENGTH ESSENTIALS

For \(y = f(x)\):

\[ L = \int_a^b \sqrt{1 + [f'(x)]^2} \, dx \]

For \(x = g(y)\):

\[ L = \int_c^d \sqrt{1 + [g'(y)]^2} \, dy \]

For Parametric \(x(t), y(t)\):

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

Remember:

  • Square the derivative(s)!
  • Don't forget +1 for function forms
  • Simplify before integrating
  • Units: length, not area!

Master Arc Length! The fundamental concept: arc length measures actual distance along a curve. For \(y = f(x)\): \(L = \int_a^b\sqrt{1+[f'(x)]^2}\,dx\). For \(x = g(y)\): \(L = \int_c^d\sqrt{1+[g'(y)]^2}\,dy\). For parametric \(x(t), y(t)\): \(L = \int_\alpha^\beta\sqrt{[x'(t)]^2+[y'(t)]^2}\,dt\). The derivation: from Pythagorean theorem on tiny segments: \(ds = \sqrt{(dx)^2+(dy)^2}\). Critical steps: (1) find derivative(s), (2) SQUARE derivative(s), (3) add 1 (for function form) or add squares (parametric), (4) take square root, (5) integrate. Common errors: forgetting to square, missing +1, wrong derivative. Distance traveled for parametric = arc length formula. These are typically calculator problems on AP® exams. This is BC-ONLY content—appears regularly on BC exams! Practice all three forms until mastered! 🎯✨