AP Calculus BC Quiz: Polar Coordinates And Differentiation

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 AP Calculus BC.

How to use this quiz

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.

Question 1

1. $0$
2. $\frac{1}{\sqrt{3}}$
3. $\sqrt{3}$ (correct answer)
4. $-\sqrt{3}$

Explanation: The curve passes through the pole when $r=0$. So, $1 - 2\cos\theta = 0$, which gives $\cos\theta = \frac{1}{2}$. This occurs at $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$. The slope of the tangent line at the pole is given by $\tan\theta$, provided $\frac{dr}{d\theta} \neq 0$. Here, $\frac{dr}{d\theta} = 2\sin\theta$. At $\theta = \frac{\pi}{3}$, $\frac{dr}{d\theta} = 2\sin(\frac{\pi}{3}) = \sqrt{3} \neq 0$. The slope is $\tan(\frac{\pi}{3}) = \sqrt{3}$. At $\theta = \frac{5\pi}{3}$, the slope is $\tan(\frac{5\pi}{3}) = -\sqrt{3}$. One of these values is $\sqrt{3}$.

Question 2

1. $\frac{dr}{d\theta} \sin\theta + r \cos\theta = 0$
2. $\frac{dr}{d\theta} \cos\theta - r \sin\theta = 0$ (correct answer)
3. $\frac{dr}{d\theta} = 0$
4. $\frac{dr}{d\theta} \cos\theta + r \sin\theta = 0$

Explanation: A vertical tangent line occurs when the denominator of $\frac{dy}{dx}$ is zero and the numerator is non-zero. The denominator is $\frac{dx}{d\theta}$. Since $x = r\cos\theta$, we have $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta$. Setting this equal to zero gives the condition for a vertical tangent.

Question 3

1. horizontal.
2. vertical. (correct answer)
3. neither horizontal nor vertical.
4. undefined because the curve is at the pole.

Explanation: For $r=3\cos(2\theta)$, we have $\frac{dr}{d\theta}=-6\sin(2\theta)$. We evaluate the derivatives of x and y at $\theta=0$. At $\theta=0$, $r=3\cos(0)=3$ and $\frac{dr}{d\theta}=-6\sin(0)=0$. Then $\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta = (0)\sin(0)+(3)\cos(0)=3$. And $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta = (0)\cos(0)-(3)\sin(0)=0$. Since $\frac{dx}{d\theta}=0$ and $\frac{dy}{d\theta}\neq 0$, the tangent line is vertical.

Question 4

1. $\frac{\pi}{2}$
2. $\frac{7\pi}{6}$
3. $\frac{11\pi}{6}$
4. $\frac{3\pi}{2}$ (correct answer)

Explanation: A horizontal tangent occurs when $\frac{dy}{d\theta} = 0$ and $\frac{dx}{d\theta} \neq 0$. We have $r = 1 - \sin\theta$ and $\frac{dr}{d\theta} = -\cos\theta$. Then $\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta + r\cos\theta = (-\cos\theta)\sin\theta + (1 - \sin\theta)\cos\theta = \cos\theta(1 - 2\sin\theta) = 0$. This implies $\cos\theta = 0$ (so $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$) or $\sin\theta = \frac{1}{2}$ (so $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$). We also need to check $\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta - r\sin\theta = (-\cos\theta)\cos\theta - (1 - \sin\theta)\sin\theta = 2\sin^2\theta - \sin\theta - 1$. At $\theta = \frac{3\pi}{2}$, $\frac{dx}{d\theta} = 2(-1)^2 - (-1) - 1 = 2 \neq 0$. At $\theta=\frac{\pi}{2}$, $\frac{dx}{d\theta} = 0$, so the slope is indeterminate. Choices B and C are values where $\frac{dx}{d\theta}=0$, which correspond to vertical tangents. Thus, the only valid option is $\theta = \frac{3\pi}{2}$.

Question 5

1. $-2$
2. $-1$ (correct answer)
3. $0$
4. $1$

Explanation: The x-coordinate is given by $x = r\cos\theta$. Substituting $r=1+\cos\theta$, we get $x = (1+\cos\theta)\cos\theta = \cos\theta + \cos^2\theta$. The rate of change of the x-coordinate with respect to $\theta$ is $\frac{dx}{d\theta}$. We have $\frac{dx}{d\theta} = -\sin\theta + 2\cos\theta(-\sin\theta) = -\sin\theta(1+2\cos\theta)$. Evaluating at $\theta = \frac{\pi}{2}$ gives $\frac{dx}{d\theta} = -\sin(\frac{\pi}{2})(1+2\cos(\frac{\pi}{2})) = -1(1+2(0)) = -1$.

Question 6

1. $-4$
2. $-1$ (correct answer)
3. $0$
4. $1$

Explanation: The curve $r=2\sin\theta$ is a circle of radius 1 centered at (0,1). In Cartesian coordinates, $x^2 + (y-1)^2 = 1$. The point at $\theta=\frac{\pi}{2}$ is $(x,y)=(0,2)$, which is the top of the circle. At this point, the curve has a horizontal tangent, so $\frac{dy}{dx}=0$. Differentiating $x^2 + y^2 - 2y = 0$ implicitly gives $2x + 2y\frac{dy}{dx} - 2\frac{dy}{dx} = 0$. Differentiating again gives $2 + 2(\frac{dy}{dx})^2 + 2y\frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} = 0$. At $(0,2)$, we know $\frac{dy}{dx}=0$. Substituting these values gives $2 + 2(0)^2 + 2(2)\frac{d^2y}{dx^2} - 2\frac{d^2y}{dx^2} = 0$, which simplifies to $2 + 2\frac{d^2y}{dx^2} = 0$. Thus, $\frac{d^2y}{dx^2} = -1$.

Question 7

1. $0$
2. $-1$ (correct answer)
3. $\frac{dr}{d\theta}\big|_{\theta=\pi/4}=0$
4. $1$
5. $\frac{1}{\sqrt{2}}$

Explanation: This problem involves finding dy/dx for the rose curve r = sin(2θ) using polar differentiation. The formula is dy/dx = (dr/dθ·sin θ + r·cos θ)/(dr/dθ·cos θ - r·sin θ). First, dr/dθ = 2cos(2θ). At θ = π/4, we have r = sin(2·π/4) = sin(π/2) = 1 and dr/dθ = 2cos(π/2) = 0. Substituting with sin(π/4) = cos(π/4) = 1/√2: dy/dx = (0·(1/√2) + 1·(1/√2))/(0·(1/√2) - 1·(1/√2)) = (1/√2)/(-1/√2) = -1. Choice C incorrectly shows dr/dθ = 0, which is true but not the answer requested. For polar curves, dy/dx depends on both r and dr/dθ values at the given angle.

Question 8

1. $0$
2. $-2$
3. $2$
4. undefined (correct answer)
5. $-\dfrac{1}{2}$

Explanation: This problem involves finding dy/dx for r = 3 - cos θ using polar differentiation. The formula is dy/dx = (dr/dθ·sin θ + r·cos θ)/(dr/dθ·cos θ - r·sin θ). With dr/dθ = sin θ, at θ = 0: r = 3 - 1 = 2, dr/dθ = 0, sin(0) = 0, cos(0) = 1. Substituting: dy/dx = (0·0 + 2·1)/(0·1 - 2·0) = 2/0, which is undefined. A common error is assuming dy/dx = 0 when the numerator is finite but the denominator is zero. When the denominator equals zero, dy/dx is undefined, indicating a vertical tangent line at that point.

Question 9

1. $-2$
2. $\dfrac{1}{2}$
3. $2$ (correct answer)
4. $0$
5. $-\dfrac{1}{2}$

Explanation: This problem requires the skill of differentiating in polar coordinates to find the slope dy/dx. To compute dy/dx for a polar curve, use the formula dy/dx = [ (dr/dθ) sin θ + r cos θ ] / [ (dr/dθ) cos θ - r sin θ ], derived from parametric equations x = r cos θ and y = r sin θ. For r = 1 + 2 cos θ, first compute dr/dθ = -2 sin θ. At θ = π/2, r = 1 and dr/dθ = -2, so the numerator is -2 · 1 + 1 · 0 = -2 and the denominator is -2 · 0 - 1 · 1 = -1, yielding dy/dx = 2. A tempting distractor like 0 might come from mistakenly setting the numerator to zero when cos θ = 0, without computing dr/dθ properly. Always verify the signs in the numerator and denominator of the polar dy/dx formula to capture the correct direction of the tangent line.

Question 10

1. $0$
2. $\dfrac{1}{2}$ (correct answer)
3. $2$
4. undefined
5. $-\dfrac{1}{2}$

Explanation: This problem requires finding dy/dx for r = 2θ + 1 using polar differentiation. The formula is dy/dx = (dr/dθ·sin θ + r·cos θ)/(dr/dθ·cos θ - r·sin θ). With dr/dθ = 2, at θ = 0: r = 0 + 1 = 1, dr/dθ = 2, sin(0) = 0, cos(0) = 1. Substituting: dy/dx = (2·0 + 1·1)/(2·1 - 1·0) = 1/2. A tempting error is to differentiate r = 2θ + 1 as if it were a Cartesian function, yielding 2. Remember that polar differentiation requires the complete conversion formula, accounting for how both r and θ change along the curve.

Question 11

1. $0$ (correct answer)
2. $1$
3. $-1$
4. $2$
5. $-2$

Explanation: This problem requires finding dy/dx for r = csc θ = 1/sin θ using polar differentiation. The polar derivative formula is dy/dx = (dr/dθ·sin θ + r·cos θ)/(dr/dθ·cos θ - r·sin θ). For r = csc θ, dr/dθ = -csc θ cot θ, and at θ = π/4: r = √2, dr/dθ = -√2·1 = -√2, sin(π/4) = cos(π/4) = 1/√2. Substituting: dy/dx = (-√2·(1/√2) + √2·(1/√2))/(-√2·(1/√2) - √2·(1/√2)) = (-1 + 1)/(-1 - 1) = 0/(-2) = 0. Students might incorrectly treat csc θ as a constant rather than differentiating it properly. Always differentiate r(θ) completely and evaluate all trigonometric functions at the specified angle.

Question 12

1. $0$
2. $-\frac{1}{2}$
3. $\frac{1}{2}$ (correct answer)
4. $-1$
5. $2$

Explanation: This problem requires finding dy/dx for a polar curve using the skill of polar differentiation. For polar curves, we use the formula dy/dx = (dy/dθ)/(dx/dθ), where x = r cos θ and y = r sin θ. With r = 2sin θ + cos θ, we find dr/dθ = 2cos θ - sin θ, so at θ = π/2, we have r = 2 and dr/dθ = -1. Computing the derivatives: dy/dθ = (dr/dθ)sin θ + r cos θ = (-1)(1) + (2)(0) = -1, and dx/dθ = (dr/dθ)cos θ - r sin θ = (-1)(0) - (2)(1) = -2. The common error is forgetting the product rule when differentiating x and y with respect to θ. Therefore, dy/dx = -1/(-2) = 1/2, confirming that the key strategy is to carefully apply the product rule to both x = r cos θ and y = r sin θ.

Question 13

1. $0$
2. $1$ (correct answer)
3. $-1$
4. $\frac{1}{2}$
5. $-\frac{1}{2}$

Explanation: This problem asks for dy/dx of the cardioid r = 1 + sin θ using polar differentiation. We apply the formula dy/dx = (dr/dθ·sin θ + r·cos θ)/(dr/dθ·cos θ - r·sin θ). Given r = 1 + sin θ, we find dr/dθ = cos θ. At θ = 0: r = 1 + sin(0) = 1, dr/dθ = cos(0) = 1, and dy/dx = (1·0 + 1·1)/(1·1 - 1·0) = 1/1 = 1. A common mistake is using just dr/dθ as the slope without the proper conversion. For polar curves, always use the complete dy/dx formula that accounts for both radial and angular changes.

Question 14

1. $-\frac{1}{2}$
2. $0$ (correct answer)
3. $\frac{1}{2}$
4. $-1$
5. $1$

Explanation: This problem asks for dy/dx of r = 2 - sin θ using polar differentiation. We use dy/dx = (dr/dθ·sin θ + r·cos θ)/(dr/dθ·cos θ - r·sin θ). Given r = 2 - sin θ, we find dr/dθ = -cos θ. At θ = π/2: r = 2 - sin(π/2) = 2 - 1 = 1, dr/dθ = -cos(π/2) = 0, and dy/dx = (0·1 + 1·0)/(0·0 - 1·1) = 0/(-1) = 0. A common mistake is computing dr/dθ incorrectly or mixing up signs in the formula. For polar differentiation, carefully evaluate each component and maintain proper signs throughout the calculation.

Question 15

1. $-1$ (correct answer)
2. $\dfrac{dr}{d\theta}\Big|_{0}$
3. $0$
4. $1$
5. undefined

Explanation: This problem requires the skill of differentiation in polar coordinates to find the slope of the tangent line in Cartesian coordinates. To find $dy/dx$ for a polar curve $r = f(\theta)$, use the formula $\dfrac{dy}{dx} = \dfrac{f'(\theta) \sin \theta + f(\theta) \cos \theta}{f'(\theta) \cos \theta - f(\theta) \sin \theta}$. For $r = 1 - \sin \theta$, $f'(\theta) = -\cos \theta$, so at $\theta = 0$, $f'(0) = -1$ and $f(0) = 1$. Plugging in, the numerator is $-1 \cdot 0 + 1 \cdot 1 = 1$ and the denominator is $-1 \cdot 1 - 1 \cdot 0 = -1$, yielding $dy/dx = 1/(-1) = -1$. A tempting distractor is choice B, $dr/d\theta$ at 0 which is $-1$, but this fails because it matches only coincidentally and does not apply the full formula for the slope. When computing polar derivatives, always remember to use the complete formula to account for the coordinate transformation, a strategy applicable to any polar curve.

Question 16

1. $-1$ (correct answer)
2. $0$
3. $1$
4. $-\dfrac{1}{2}$
5. $\dfrac{1}{2}$

Explanation: This problem requires the skill of differentiating in polar coordinates to find the slope dy/dx. To compute dy/dx for a polar curve, use the formula dy/dx = [ (dr/dθ) sin θ + r cos θ ] / [ (dr/dθ) cos θ - r sin θ ], derived from parametric equations x = r cos θ and y = r sin θ. For r = 1/(1 + sin θ), first compute dr/dθ = -cos θ / (1 + sin θ)². At θ = 0, r = 1 and dr/dθ = -1, so the numerator is -1 · 0 + 1 · 1 = 1 and the denominator is -1 · 1 - 1 · 0 = -1, yielding dy/dx = -1. A tempting distractor like 0 might arise from incorrectly setting terms to zero without full computation. Always verify the signs in the numerator and denominator of the polar dy/dx formula to capture the correct direction of the tangent line.

Question 17

1. $0$ (correct answer)
2. $\dfrac{1}{3}$
3. $\dfrac{3}{2}$
4. $1$
5. $\dfrac{dr}{d\theta}\big|_{\theta=\pi/2}=0$

Explanation: This problem requires finding dy/dx for a polar curve using polar differentiation formulas. For polar curves, we use dy/dx = (dy/dθ)/(dx/dθ) where x = r cos θ and y = r sin θ. With r = 2 + sin θ, we get dx/dθ = -r sin θ + r' cos θ = -(2 + sin θ) sin θ + cos θ cos θ and dy/dθ = r cos θ + r' sin θ = (2 + sin θ) cos θ + cos θ sin θ. At θ = π/2, we have r = 3, r' = 0, so dx/dθ = -3(1) + 0 = -3 and dy/dθ = 3(0) + 0 = 0. Many students mistakenly compute dr/dθ instead of dy/dx, but these are different quantities. The key strategy is to always use the parametric chain rule formula dy/dx = (dy/dθ)/(dx/dθ) for polar curves.

Question 18

1. $-\dfrac{2}{\pi}$ (correct answer)
2. $\dfrac{\pi}{2}$
3. $0$
4. $-\dfrac{\pi}{2}$
5. $\dfrac{2}{\pi}$

Explanation: This problem involves finding dy/dx for the polar spiral r = θ using polar differentiation. For polar curves, we apply dy/dx = (dy/dθ)/(dx/dθ) where x = r cos θ and y = r sin θ. With r = θ and r' = 1, we get dx/dθ = -θ sin θ + cos θ and dy/dθ = θ cos θ + sin θ. At θ = π/2, we have dx/dθ = -(π/2)(1) + 0 = -π/2 and dy/dθ = (π/2)(0) + 1 = 1. Therefore dy/dx = 1/(-π/2) = -2/π. Students often confuse the reciprocal relationship and write π/2 instead of -2/π. Remember that polar differentiation requires careful application of the product rule to both x(θ) and y(θ).

Question 19

1. $-\frac{\pi}{2}$
2. $-\frac{2}{\pi}$ (correct answer)
3. $0$
4. $3$

Explanation: To find the slope of the tangent line, we compute $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$. The Cartesian coordinates are $x = r\cos\theta = 3\theta\cos\theta$ and $y = r\sin\theta = 3\theta\sin\theta$. Using the product rule, we find the derivatives with respect to $\theta$: $\frac{dx}{d\theta} = 3\cos\theta - 3\theta\sin\theta$ and $\frac{dy}{d\theta} = 3\sin\theta + 3\theta\cos\theta$. Evaluating at $\theta = \frac{\pi}{2}$ gives $\frac{dx}{d\theta} = 3(0) - 3(\frac{\pi}{2})(1) = -\frac{3\pi}{2}$ and $\frac{dy}{d\theta} = 3(1) + 3(\frac{\pi}{2})(0) = 3$. The slope is $\frac{dy}{dx} = \frac{3}{-3\pi/2} = -\frac{2}{\pi}$.

Question 20

1. $\sqrt{3}$
2. $\frac{1}{\sqrt{3}}$ (correct answer)
3. $-\frac{1}{\sqrt{3}}$
4. $-\sqrt{3}$

Explanation: We use the formula for the slope in polar coordinates: $\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}$. For $r = 4\cos\theta$, we have $\frac{dr}{d\theta} = -4\sin\theta$. Substituting these into the formula gives $\frac{dy}{dx} = \frac{(-4\sin\theta)\sin\theta + (4\cos\theta)\cos\theta}{(-4\sin\theta)\cos\theta - (4\cos\theta)\sin\theta} = \frac{4(\cos^2\theta - \sin^2\theta)}{-8\sin\theta\cos\theta} = \frac{4\cos(2\theta)}{-4\sin(2\theta)} = -\cot(2\theta)$. At $\theta = \frac{\pi}{3}$, the slope is $- \cot(\frac{2\pi}{3}) = -(-\frac{1}{\sqrt{3}}) = \frac{1}{\sqrt{3}}$.