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.

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.

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.

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.

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.

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).

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.

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.

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.

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.