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.

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.

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 $$\theta = \frac{3\pi}{2}$$.

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

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, $$\frac{d^2y}{dx^2} = -1$$.

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 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 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.