Question

84. Arc length for polar curves: Prove that the length of the curve r = f(θ) for α ≤ θ ≤ β is L = ∫(α to β) √(f(θ)² + f'(θ)²) dθ.

Accepted Answer

Recall the formula for the arc length of a curve given in Cartesian coordinates: if a curve is parameterized by $$ x(t) $$ and $$ y(t) $$ for $$ t  in$$ $$ [a, b] $$, then the arc length $$ L  is$$ $$ L = \int_a^b \sqrt{\left(\frac{dx}{dt}ight)^2 + \left(\frac{dy}{dt}ight)^2} \, dt . $$Express the polar curve $$ r = f(	heta)  in $$Cartesian coordinates using the relationships $$ x = r \cos 	heta = f(	heta) \cos 	heta $$ and $$ y = r \sin 	heta = f(	heta) \sin 	heta . $$Compute the derivatives $$ \frac{dx}{d	heta} $$ and $$ \frac{dy}{d	heta} $$ using the product rule: $$ \frac{dx}{d	heta} = f'(	heta) \cos 	heta - f(	heta) \sin 	heta $$ and $$ \frac{dy}{d	heta} = f'(	heta) \sin 	heta + f(	heta) \cos 	heta . $$Substitute these derivatives into the arc length formula: $$ L = \int_\alpha^\beta \sqrt{\left(f'(	heta) \cos 	heta - f(	heta) \sin 	hetaight)^2 + \left(f'(	heta) \sin 	heta + f(	heta) \cos 	hetaight)^2} \, d	heta . $$Simplify the expression inside the square root by expanding and combining like terms, using the Pythagorean identity $$ \sin^2 	heta + \cos^2 	heta = 1 $$, to show that it reduces to $$ \sqrt{f'(	heta)^2 + f(	heta)^2} $$, thus proving the formula $$ L = \int_\alpha^\beta \sqrt{f(	heta)^2 + f'(	heta)^2} \, d	heta .$$

84. Arc length for polar curves: Prove that the length of the curve r = f(θ) for α ≤ θ ≤ β is
L = ∫(α to β) √(f(θ)² + f'(θ)²) dθ.

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Key Concepts

Polar Coordinates and Curves

Arc Length Formula for Parametric Curves

Differentiation and Chain Rule in Polar Coordinates