Answer to Question #265033 in Calculus for Hema

Question #265033

The graph of the parametric equations is called a cycloid.

x=θ-sinθ and y=1-cosθ for 0≤θ≤2π

(a) find dy/dx

(b) find an equation of the tangent to the cycloid at the point where θ=π/3

(c) at what point is the tangent horizontal?

(d) graph the cycloid and the tangent lines in parts (b) and (c)



1
Expert's answer
2021-11-16T10:39:32-0500

a)

dy/dx=dydθdθdxdy/dx=\frac{dy}{d\theta}\frac{d\theta}{dx}


dy/dθ=sinθdy/d\theta=sin\theta

dx/dθ=1cosθdx/d\theta=1-cos\theta


dy/dx=sinθ1cosθdy/dx=\frac{sin\theta}{1-cos\theta}


b)

equation of the tangent:

yy0=f(x0)(xx0)y-y_0=f'(x_0)(x-x_0)


then, θ=π/3:

1cosθ(1cos(π/3))=sin(π/3)1cos(π/3)(θsinθ(π/3sin(π/3)))1-cosθ-(1-cos(\pi/3))=\frac{sin(\pi/3)}{1-cos(\pi/3)}(θ-sinθ-(\pi/3-sin(\pi/3)))

1/2cosθ=3(θsinθπ/3+3/2)1/2-cosθ=\sqrt3(θ-sinθ-\pi/3+\sqrt3/2)

cosθ+3(θsinθ)=π/31cosθ+\sqrt3(θ-sinθ)=\pi/\sqrt3-1


c)

the tangent horizontal if f(x)=0f'(x)=0

so,

sinθ1cosθ=0\frac{sin\theta}{1-cos\theta}=0


sinθ=0sin\theta=0

cosθ1cos\theta\neq 1

θ=π\theta=\pi

y(π)=2,x(π)=πy(\pi)=2,x(\pi)=\pi


d)






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