Source: Paper Homework Assignment 8, Texas A&M University
Tags: arc length, polar arc length, polar curve length, dr/dθ, Math 2414, calculus II, Texas A&M
The length of a polar curve r = f(θ) from θ = α to θ = β is computed with an integral involving both r and its derivative dr/dθ. The formula is L = ∫ √(r² + (dr/dθ)²) dθ. The Pythagorean identity sin²x + cos²x = 1 often simplifies the integrand dramatically.
Arc length (polar)
The total length along a polar curve between two angles. Computed via L = ∫_α^β √(r² + (dr/dθ)²) dθ.
dr/dθ (derivative of r with respect to θ)
The rate of change of the radial distance as the angle changes. Required for the arc length formula.
For a curve r = f(θ), the arc length from θ = α to θ = β is:
L = ∫_α^β √(r² + (dr/dθ)²) dθ
The derivation comes from the parametric arc length formula with x = r cos θ and y = r sin θ, where dx/dθ and dy/dθ are expanded using the product rule and then simplified via the Pythagorean identity.
The typical workflow:
Compute dr/dθ
Form r² + (dr/dθ)²
Simplify (often using sin² + cos² = 1)
Integrate
Step 1: Compute dr/dθ
r = sin²(θ/2)
By the chain rule:
dr/dθ = 2 sin(θ/2) · cos(θ/2) · (1/2) = sin(θ/2) cos(θ/2)
This can also be written as (1/2) sin θ using the double-angle identity, but leaving it in the product form is more useful here.
Step 2: Form the expression under the square root
r² + (dr/dθ)² = sin⁴(θ/2) + sin²(θ/2) cos²(θ/2)
Factor out sin²(θ/2):
= sin²(θ/2) [sin²(θ/2) + cos²(θ/2)]
= sin²(θ/2) · 1
= sin²(θ/2)
Step 3: Take the square root and integrate
√(sin²(θ/2)) = |sin(θ/2)| = sin(θ/2) (since sin(θ/2) ≥ 0 on [0, π])
L = ∫₀^π sin(θ/2) dθ
Let u = θ/2, du = dθ/2, so dθ = 2 du. When θ = 0, u = 0; when θ = π, u = π/2.
L = 2 ∫₀^(π/2) sin u du = 2 [−cos u] from 0 to π/2
= 2 [−cos(π/2) + cos(0)] = 2 [0 + 1]
L = 2 units
The key move was the Pythagorean identity collapsing sin²(θ/2) + cos²(θ/2) = 1, which turned a nasty-looking integrand into something simple.
Step 1: Compute dr/dθ
r = 4 sin θ − cos θ
dr/dθ = 4 cos θ + sin θ
Step 2: Form r² + (dr/dθ)²
r² = (4 sin θ − cos θ)² = 16 sin²θ − 8 sin θ cos θ + cos²θ
(dr/dθ)² = (4 cos θ + sin θ)² = 16 cos²θ + 8 sin θ cos θ + sin²θ
Add them:
r² + (dr/dθ)² = 16 sin²θ − 8 sin θ cos θ + cos²θ + 16 cos²θ + 8 sin θ cos θ + sin²θ
The −8 sin θ cos θ and +8 sin θ cos θ cancel:
= 16 sin²θ + cos²θ + 16 cos²θ + sin²θ
= 16(sin²θ + cos²θ) + (cos²θ + sin²θ)
= 16 · 1 + 1
= 17
Step 3: Integrate
L = ∫₀^π √17 dθ = √17 · π
L = π√17 units
Again, the cross terms cancelled and the Pythagorean identity reduced everything to a constant. This pattern is very common in textbook problems and is worth watching for.
When you expand r² + (dr/dθ)², look for:
Cross terms that cancel (this happens whenever r = a sin θ + b cos θ or similar linear combinations)
sin² + cos² = 1 consolidation
Common factors to pull out of the square root
If the integrand does not simplify to something elementary, the problem likely requires numerical methods or there is an algebraic error. At the MATH 2414 level, the integrand almost always simplifies cleanly.
Polar arc length: L = ∫_α^β √(r² + (dr/dθ)²) dθ
Useful identities for simplification: sin²x + cos²x = 1 2 sin x cos x = sin 2x sin²(x/2) = (1 − cos x)/2
⚠️ The most common mistake is forgetting to compute dr/dθ correctly, especially with chain rule applications (as in r = sin²(θ/2)).
⚠️ Cross terms in r² and (dr/dθ)² frequently cancel. If your final integrand looks messy, recheck your algebra for cancellation opportunities.
⚠️ On [0, π], sin(θ/2) ≥ 0, so √(sin²(θ/2)) = sin(θ/2) without absolute values. On a different interval, you might need to handle the sign more carefully.
⚠️ When the integrand simplifies to a constant (like √17), the integral is trivial: just multiply by the interval length. This is a strong signal you've done the algebra correctly.
⚠️ Don't confuse the polar arc length formula with the polar area formula. Area uses (1/2)∫ r² dθ. Arc length uses ∫ √(r² + (r')²) dθ.
Q: What is the arc length formula for a polar curve r = f(θ) from α to β?
A: L = ∫_α^β √(r² + (dr/dθ)²) dθ.
Q: For r = 4 sin θ − cos θ, what is dr/dθ?
A: dr/dθ = 4 cos θ + sin θ.
Q: Why did the integrand for r = 4 sin θ − cos θ simplify to √17?
A: When you expand r² + (dr/dθ)², the cross terms (±8 sin θ cos θ) cancel out, and the remaining sin² + cos² groups each collapse to 1 via the Pythagorean identity, leaving 16 + 1 = 17.
Q: If r = 3 cos θ, what is the arc length from 0 to π?
A: dr/dθ = −3 sin θ. r² + (dr/dθ)² = 9 cos²θ + 9 sin²θ = 9. So L = ∫₀^π √9 dθ = 3π.
Q: What's the difference between the polar area integral and the polar arc length integral?
A: Area = (1/2) ∫ r² dθ. Arc length = ∫ √(r² + (dr/dθ)²) dθ. The area formula does not involve the derivative of r.
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