Calculus 2 Quiz: Polar Coordinates And Differentiation
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Polar Coordinates And DifferentiationQuestion 1 of 4

A polar curve is given by r=f(θ)r = f(\theta) where ff is differentiable. At a point where r=3r = 3, drdθ=−2\frac{dr}{d\theta} = -2, and θ=π4\theta = \frac{\pi}{4}, what is the value of dxdθ\frac{dx}{d\theta}?

−522-\frac{5\sqrt{2}}{2}
22\frac{\sqrt{2}}{2}
−22-\frac{\sqrt{2}}{2}
522\frac{5\sqrt{2}}{2}
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Calculus 2 Quiz

Calculus 2 Quiz: Polar Coordinates And Differentiation

Practice Polar Coordinates And Differentiation 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 Polar Coordinates And Differentiation, 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.

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Question 1

A polar curve is given by r=f(θ)r = f(\theta) where ff is differentiable. At a point where r=3r = 3, drdθ=−2\frac{dr}{d\theta} = -2, and θ=π4\theta = \frac{\pi}{4}, what is the value of dxdθ\frac{dx}{d\theta}?

  1. −522-\frac{5\sqrt{2}}{2} (correct answer)
  2. 22\frac{\sqrt{2}}{2}
  3. −22-\frac{\sqrt{2}}{2}
  4. 522\frac{5\sqrt{2}}{2}
Explanation: We use dxdθ=drdθcos⁡θ−rsin⁡θ\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta. Given r=3r = 3, drdθ=−2\frac{dr}{d\theta} = -2, and θ=π4\theta = \frac{\pi}{4}, we have cos⁡(π4)=sin⁡(π4)=22\cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}. Therefore, dxdθ=(−2)(22)−(3)(22)=−2−322=−22+322=−522\frac{dx}{d\theta} = (-2)\left(\frac{\sqrt{2}}{2}\right) - (3)\left(\frac{\sqrt{2}}{2}\right) = -\sqrt{2} - \frac{3\sqrt{2}}{2} = -\frac{2\sqrt{2} + 3\sqrt{2}}{2} = -\frac{5\sqrt{2}}{2}. Choice B omits the negative signs. Choice C only includes the first term. Choice D omits the negative sign from the final result.

Question 2

Consider the polar curve r=sin⁡(θ)+cos⁡(θ)r = \sin(\theta) + \cos(\theta). At the point where this curve intersects the positive xx-axis, what is the value of dydx\frac{dy}{dx}?

  1. dydx=1\frac{dy}{dx} = 1 (correct answer)
  2. dydx=−1\frac{dy}{dx} = -1
  3. dydx=0\frac{dy}{dx} = 0
  4. dydx\frac{dy}{dx} is undefined
Explanation: The curve intersects the positive xx-axis when y=0y = 0 and x>0x > 0. Since y=rsin⁡θy = r\sin\theta, we need rsin⁡θ=0r\sin\theta = 0. Either r=0r = 0 or sin⁡θ=0\sin\theta = 0. If sin⁡θ=0\sin\theta = 0, then θ=0\theta = 0 or π\pi. At θ=0\theta = 0: r=sin⁡(0)+cos⁡(0)=1>0r = \sin(0) + \cos(0) = 1 > 0, giving point (1,0)(1,0). At θ=π\theta = \pi: r=sin⁡(π)+cos⁡(π)=−1<0r = \sin(\pi) + \cos(\pi) = -1 < 0, giving point (1,0)(1,0) as well (since negative rr reverses direction). Using dydx=drdθsin⁡θ+rcos⁡θdrdθcos⁡θ−rsin⁡θ\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta} with drdθ=cos⁡θ−sin⁡θ\frac{dr}{d\theta} = \cos\theta - \sin\theta, at θ=0\theta = 0: dydx=(1)(0)+(1)(1)(1)(1)−(1)(0)=1\frac{dy}{dx} = \frac{(1)(0) + (1)(1)}{(1)(1) - (1)(0)} = 1. The other choices result from computational errors or using the wrong intersection point.

Question 3

A polar curve is defined implicitly by the equation r2=cos⁡(2θ)r^2 = \cos(2\theta). At the point where θ=π6\theta = \frac{\pi}{6} and r>0r > 0, what is the slope of the tangent line?

  1. 33\frac{\sqrt{3}}{3}
  2. −33-\frac{\sqrt{3}}{3}
  3. 13\frac{1}{\sqrt{3}}
  4. −13-\frac{1}{\sqrt{3}} (correct answer)
Explanation: From r2=cos⁡(2θ)r^2 = \cos(2\theta), we get r=cos⁡(2θ)r = \sqrt{\cos(2\theta)} (taking positive square root). At θ=π6\theta = \frac{\pi}{6}: r2=cos⁡(π3)=12r^2 = \cos(\frac{\pi}{3}) = \frac{1}{2}, so r=12r = \frac{1}{\sqrt{2}}. To find drdθ\frac{dr}{d\theta}, we differentiate implicitly: 2rdrdθ=−2sin⁡(2θ)2r\frac{dr}{d\theta} = -2\sin(2\theta), giving drdθ=−sin⁡(2θ)r\frac{dr}{d\theta} = -\frac{\sin(2\theta)}{r}. At θ=π6\theta = \frac{\pi}{6}: drdθ=−sin⁡(π3)12=−3212=−322=−62\frac{dr}{d\theta} = -\frac{\sin(\frac{\pi}{3})}{\frac{1}{\sqrt{2}}} = -\frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}} = -\frac{\sqrt{3}\sqrt{2}}{2} = -\frac{\sqrt{6}}{2}. Using dydx=drdθsin⁡θ+rcos⁡θdrdθcos⁡θ−rsin⁡θ\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta} with sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2} and cos⁡(π6)=32\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}, we get dydx=−13\frac{dy}{dx} = -\frac{1}{\sqrt{3}}. The other choices result from sign errors or incorrect trigonometric calculations.

Question 4

A curve is defined by the polar equation r=3+2cos⁡(θ)r = 3 + 2\cos(\theta). At the point where θ=π3\theta = \frac{\pi}{3}, what is the slope of the tangent line to the curve?

  1. −33-\frac{\sqrt{3}}{3}
  2. −237-\frac{2\sqrt{3}}{7} (correct answer)
  3. 37\frac{\sqrt{3}}{7}
  4. −437-\frac{4\sqrt{3}}{7}
Explanation: To find the slope, we use dydx=drdθsin⁡θ+rcos⁡θdrdθcos⁡θ−rsin⁡θ\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}. Here r=3+2cos⁡θr = 3 + 2\cos\theta and drdθ=−2sin⁡θ\frac{dr}{d\theta} = -2\sin\theta. At θ=π3\theta = \frac{\pi}{3}: r=3+2(12)=4r = 3 + 2(\frac{1}{2}) = 4, drdθ=−2(32)=−3\frac{dr}{d\theta} = -2(\frac{\sqrt{3}}{2}) = -\sqrt{3}. So dydx=−3⋅32+4⋅12−3⋅12−4⋅32=−32+2−32−23=12−732=−237\frac{dy}{dx} = \frac{-\sqrt{3} \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2}}{-\sqrt{3} \cdot \frac{1}{2} - 4 \cdot \frac{\sqrt{3}}{2}} = \frac{-\frac{3}{2} + 2}{-\frac{\sqrt{3}}{2} - 2\sqrt{3}} = \frac{\frac{1}{2}}{-\frac{7\sqrt{3}}{2}} = -\frac{2\sqrt{3}}{7}. Choice A is the negative reciprocal error. Choice C omits the negative sign. Choice D doubles the numerator incorrectly.