find the points on the given curve where the tangent line is horizontal or vertical. (order your answers…

find the points on the given curve where the tangent line is horizontal or vertical. (order your answers from smallest to largest r, then from smallest to largest θ.)\n\nr = 1 + cos(θ), 0 ≤ θ < 2π\n\nhorizontal tangent\n\n(r, θ) = ( )\n\n(r, θ) = ( )\n\n(r, θ) = ( )\n\nvertical tangent\n\n(r, θ) = ( )\n\n(r, θ) = ( )\n\n(r, θ) = ( )

find the points on the given curve where the tangent line is horizontal or vertical. (order your answers from smallest to largest r, then from smallest to largest θ.)\n\nr = 1 + cos(θ), 0 ≤ θ < 2π\n\nhorizontal tangent\n\n(r, θ) = ( )\n\n(r, θ) = ( )\n\n(r, θ) = ( )\n\nvertical tangent\n\n(r, θ) = ( )\n\n(r, θ) = ( )\n\n(r, θ) = ( )

Answer

Explanation:

Step1: Convert polar to parametric

For a polar curve (r = f(\theta)), the parametric equations are (x=r\cos\theta=(1 + \cos\theta)\cos\theta=\cos\theta+\cos^{2}\theta) and (y = r\sin\theta=(1+\cos\theta)\sin\theta=\sin\theta+\sin\theta\cos\theta).

Step2: Differentiate (x) and (y) with respect to (\theta)

Using the sum - rule and product - rule:

  • (\frac{dx}{d\theta}=-\sin\theta - 2\cos\theta\sin\theta=-\sin\theta(1 + 2\cos\theta))
  • (\frac{dy}{d\theta}=\cos\theta+\cos^{2}\theta-\sin^{2}\theta=\cos\theta+\cos^{2}\theta-(1 - \cos^{2}\theta)=2\cos^{2}\theta+\cos\theta - 1=(2\cos\theta - 1)(\cos\theta+1))

Step3: Find horizontal tangents ((\frac{dy}{d\theta}=0) and (\frac{dx}{d\theta}\neq0))

Set (\frac{dy}{d\theta}=(2\cos\theta - 1)(\cos\theta + 1)=0)

  • Case 1: (\cos\theta=\frac{1}{2}), then (\theta=\frac{\pi}{3}) or (\theta=\frac{5\pi}{3})
    • When (\theta=\frac{\pi}{3}), (r = 1+\cos\frac{\pi}{3}=1+\frac{1}{2}=\frac{3}{2})
    • When (\theta=\frac{5\pi}{3}), (r = 1+\cos\frac{5\pi}{3}=1+\frac{1}{2}=\frac{3}{2})
  • Case 2: (\cos\theta=-1), then (\theta=\pi), and (\frac{dx}{d\theta}=-\sin\pi(1 + 2\cos\pi)=0) (rejected as (\frac{dx}{d\theta}) and (\frac{dy}{d\theta}) cannot both be (0) for a non - singular tangent)

Step4: Find vertical tangents ((\frac{dx}{d\theta}=0) and (\frac{dy}{d\theta}\neq0))

Set (\frac{dx}{d\theta}=-\sin\theta(1 + 2\cos\theta)=0)

  • Case 1: (\sin\theta = 0), then (\theta = 0) or (\theta = 2\pi) (but (\theta\in[0,2\pi))). When (\theta = 0), (r=1+\cos0 = 2), (\frac{dy}{d\theta}=2\cos^{2}0+\cos0 - 1=2 + 1-1 = 2\neq0)
  • Case 2: (\cos\theta=-\frac{1}{2}), then (\theta=\frac{2\pi}{3}) or (\theta=\frac{4\pi}{3})
    • When (\theta=\frac{2\pi}{3}), (r = 1+\cos\frac{2\pi}{3}=1-\frac{1}{2}=\frac{1}{2})
    • When (\theta=\frac{4\pi}{3}), (r = 1+\cos\frac{4\pi}{3}=1-\frac{1}{2}=\frac{1}{2})

Answer:

Horizontal tangent: ((\frac{3}{2},\frac{\pi}{3})), ((\frac{3}{2},\frac{5\pi}{3})) Vertical tangent: ((2,0)), ((\frac{1}{2},\frac{2\pi}{3})), ((\frac{1}{2},\frac{4\pi}{3}))