Cycloids.....Does the name ring a bell ???

Calculus Level 4

Find the length of the complete cycloid given by

x = a × θ + a × sin ⁡ θ x=a \times \theta + a \times \sin \theta ; y = a − a × cos ⁡ θ y=a - a \times \cos \theta ;

The answer is of the form m × a m \times \ a , Find m m where m m is a positive constant .

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The answer is 8.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

X = a ( θ − S i n θ ) , Y = a ( 1 − C o s θ ) , d x d θ = a ( 1 − C o s θ ) , d y d θ = a S i n θ . A r c l e n g t h S = ∫ a b ( d x d θ ) 2 + d y d θ ) 2 d θ . d x d θ ) 2 + ( d x d θ ) 2 = a 2 ( ( 1 − c o s θ ) 2 + s i n 2 θ ) = a 2 ( 1 − 2 c o s θ + c o s 2 θ + s i n 2 θ ) = 2 a 2 ( 1 − c o s θ ) = 4 a 2 s i n 2 ( θ 2 ) S = ∫ 0 2 π 2 a s i n ( θ 2 ) d θ = ∣ − 4 a c o s ( θ / 2 ) ∣ 0 2 π = 8 a X=a(\theta - Sin\theta),\ \ Y=a(1- Cos\theta),\ \ \ \ \\ \dfrac{dx}{d\theta}=a(1- Cos\theta),\ \ \dfrac{dy}{d\theta}=aSin\theta.\\ \displaystyle Arc\ length\ S=\int_a^b\sqrt{(\dfrac{dx}{d\theta})^2 + \dfrac{dy}{d\theta})^2}d\theta.\\ \dfrac{dx}{dθ})^2+(\dfrac{dx}{dθ})^2 = a^2(\ (1-cosθ)^2+sin^2θ) \\ = a^2(1-2cosθ+cos^2θ+sin^2θ) \\ = 2a^2(1-cosθ) \\ = 4a^2sin^2(\dfrac θ 2) \\ \displaystyle S = \int_0^{2\pi}2asin(\dfrac θ 2)dθ \\ ={\Large |}-4acos(θ/2) {\Large |}_0^{2\pi}\\ = 8a

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Jun Arro Estrella
Jul 11, 2015

By integration one the length of a cycloid is 8a

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