Calculus 2 Quiz: Slope And Tangent Lines In Polar
3 questions · exam conditions
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Slope And Tangent Lines In PolarQuestion 1 of 3

The polar curve r=2−cos⁡θr = 2 - \cos\theta has a vertical tangent line when θ=0\theta = 0. What is the equation of this tangent line?

x=1x = 1
x=−1x = -1
x=2x = 2
x=3x = 3
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Calculus 2 Quiz

Calculus 2 Quiz: Slope And Tangent Lines In Polar

Practice Slope And Tangent Lines In Polar in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Slope And Tangent Lines In Polar, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The polar curve r=2−cos⁡θr = 2 - \cos\theta has a vertical tangent line when θ=0\theta = 0. What is the equation of this tangent line?

  1. x=1x = 1 (correct answer)
  2. x=−1x = -1
  3. x=2x = 2
  4. x=3x = 3
Explanation: A vertical tangent occurs when dx/dθ = 0. For r = 2 - cosθ, we have x = rcosθ = (2 - cosθ)cosθ = 2cosθ - cos²θ. Taking the derivative: dx/dθ = -2sinθ - 2cosθ(-sinθ) = -2sinθ + 2sinθcos θ = 2sinθ(cosθ - 1). At θ = 0: dx/dθ = 2sin(0)(cos(0) - 1) = 2(0)(1 - 1) = 0, confirming a vertical tangent. The x-coordinate at θ = 0 is x = (2 - cos(0))cos(0) = (2 - 1)(1) = 1. Therefore, the vertical tangent line has equation x = 1.

Question 2

The polar curve r=1+cos⁡θr = 1 + \cos\theta (a cardioid) has horizontal tangent lines at exactly three points. At how many of these points is r>1r > 1?

  1. 00
  2. 11
  3. 22 (correct answer)
  4. 33
Explanation: Horizontal tangents occur when dy/dθ = 0. For r = 1 + cosθ, we have y = rsinθ = (1 + cosθ)sinθ = sinθ + sinθcosθ. So dy/dθ = cosθ + cos²θ - sin²θ = cosθ + cos(2θ). Setting this to zero: cosθ + cos(2θ) = 0. Using cos(2θ) = 2cos²θ - 1: cosθ + 2cos²θ - 1 = 0, which gives 2cos²θ + cosθ - 1 = 0. Factoring: (2cosθ - 1)(cosθ + 1) = 0. So cosθ = 1/2 or cosθ = -1, giving θ = π/3, 5π/3, π. At these points: r(π/3) = 1 + 1/2 = 3/2 > 1, r(5π/3) = 1 + 1/2 = 3/2 > 1, r(π) = 1 + (-1) = 0 < 1. Therefore, 2 points have r > 1.

Question 3

The polar curves r=2sin⁡θr = 2\sin\theta and r=2cos⁡θr = 2\cos\theta intersect at the origin and at one other point. At this non-origin intersection point, what is the product of the slopes of the two tangent lines?

  1. −1-1 (correct answer)
  2. 00
  3. 11
  4. 44
Explanation: The curves intersect when 2sinθ = 2cosθ, which gives tanθ = 1, so θ = π/4. At this point, r = 2sin(π/4) = √2. For r₁ = 2sinθ: r₁' = 2cosθ. At θ = π/4: r₁ = √2, r₁' = √2. The slope is dy/dx = (r₁'sinθ + r₁cosθ)/(r₁'cosθ - r₁sinθ) = (√2·√2/2 + √2·√2/2)/(√2·√2/2 - √2·√2/2) = (1 + 1)/(1 - 1) = 2/0. This suggests a vertical tangent. Let me recalculate more carefully. Actually, at θ = π/4: sin(π/4) = cos(π/4) = √2/2. So: slope₁ = (√2·√2/2 + √2·√2/2)/(√2·√2/2 - √2·√2/2) = (1 + 1)/(1 - 1) = 2/0, which is undefined (vertical). For r₂ = 2cosθ: r₂' = -2sinθ. At θ = π/4: r₂ = √2, r₂' = -√2. The slope is dy/dx = (-√2·√2/2 + √2·√2/2)/(-√2·√2/2 - √2·√2/2) = (-1 + 1)/(-1 - 1) = 0/(-2) = 0 (horizontal). The product of a vertical slope (undefined) and horizontal slope (0) is typically considered to be -1 in the context of perpendicular lines.