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)
(a) "\\frac{dy}{dx}=\\frac{\\frac{dy}{d\\theta}}{\\frac{dx}{d\\theta}}"
"y=1-cos\\theta ,\\ \\frac{dy}{d\\theta}=sin\\theta"
"x=\\theta-sin\\theta,\\" "\\frac{dx}{d\\theta}=1-cos\\theta"
"therefore \\ \\frac{dy}{dx}=\\frac{sin\\theta}{1-cos\\theta}"
(b)Equation of tangent line to the cycloid
"when \\ \\theta=\\frac{\\pi}{3}\\ we \\ have"
"x=r(\\frac{\\pi}{3}-sin\\frac{\\pi}{3})=r(\\frac{\\pi}{3}-\\frac{\\sqrt{3}}{2})"
"y=r(1-cos\\frac{\\pi}{3})=\\frac{r}{2}"
and"\\frac{dy}{dx}=\\frac{sin(\\frac{\\pi}{3})}{1-cos\\frac{\\pi}{3}}=\\frac{\\frac{\\sqrt{3}}{2}}{1-\\frac{1}{2}}=\\sqrt{3}"
The slope of the tangent is "\\sqrt{3}"
its equation is "y-\\frac{r}{2}=\\sqrt{3}(x-\\frac{r\\pi}{3}+\\frac{r\\sqrt{3}}{3})\\ or \\sqrt{3}x-y=r(\\frac{\\pi}{\\sqrt{3}}-2)"
(c)The tangent is horizontal when "\\frac{dy}{dx}=0"
This occurs when sin θ = 0 and 1 – cos θ ≠ 0,
that is, θ = (2n – 1)π, n an integer.
The corresponding point on the cycloid is
((2n – 1)π, 2).
When θ = 2nπ, both dx/dθ and dy/dθ are 0.
(d)Graph
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