Answer: $\frac{1}{2}\int_{\pi/6}^{5\pi/6} \left[(3\sin\theta)^2 - (1+\sin\theta)^2\right] d\theta$

Find the intersection points: Setting $3\sin\theta = 1 + \sin\theta$ gives $2\sin\theta = 1$, so $\sin\theta = \frac{1}{2}$, yielding $\theta = \pi/6$ and $\theta = 5\pi/6$. Determine which curve is outer: Between $\theta = \pi/6$ and $\theta = 5\pi/6$, $3\sin\theta > 1 + \sin\theta$, so the circle is the outer curve. Write the correct area formula: Area between polar curves: $A = \frac{1}{2}\int_{\pi/6}^{5\pi/6}\left[(3\sin\theta)^2 - (1+\sin\theta)^2\right]d\theta$. Why distractors fail: Option A uses $[0, \pi]$, which includes regions where the cardioid is outside the circle. Option C reverses outer and inner curves and uses wrong limits. Option D subtracts $r$-values instead of $r^2$-values and omits the $\frac{1}{2}$ factor.

Answer: $\frac{1}{2}\int_{\alpha}^{\beta} r^2 \, d\theta$

Identify the polar area formula: The area enclosed by a polar curve is $A = \frac{1}{2}\int_{\alpha}^{\beta} [f(\theta)]^2 \, d\theta$. Why the correct answer works: Option C matches the standard formula with both the $\frac{1}{2}$ factor and the $r^2$ integrand. Why distractors fail: Option A is missing the $\frac{1}{2}$ factor. Option B uses $r$ instead of $r^2$. Option D is the formula for polar arc length, not area.

Answer: $-\frac{1}{4}$

Find dy/dx: $\frac{dx}{dt} = 1 - \cos t$, $\frac{dy}{dt} = \sin t$. So $\frac{dy}{dx} = \frac{\sin t}{1 - \cos t}$. Differentiate dy/dx with respect to t: Let $u = \frac{\sin t}{1-\cos t}$. Using the quotient rule: $\frac{du}{dt} = \frac{\cos t(1-\cos t) - \sin t(\sin t)}{(1-\cos t)^2} = \frac{\cos t - \cos^2 t - \sin^2 t}{(1-\cos t)^2} = \frac{\cos t - 1}{(1-\cos t)^2} = \frac{-1}{1-\cos t}$. Compute d²y/dx²: $\frac{d^2y}{dx^2} = \frac{du/dt}{dx/dt} = \frac{-1/(1-\cos t)}{1-\cos t} = \frac{-1}{(1-\cos t)^2}$. At $t = \pi$: $(1-\cos\pi)^2 = (1-(-1))^2 = 4$. So $\frac{d^2y}{dx^2} = -\frac{1}{4}$. Why distractors fail: Option B has the wrong sign. Option C ($-1/2$) forgets to square $(1-\cos t)$ in the denominator. Option D ($0$) may come from incorrectly evaluating $\sin\pi = 0$ and stopping prematurely.

Answer: A thin circular sector of radius $r$ and angle $d\theta$

Understand the geometric derivation: The area of a circular sector with radius $r$ and central angle $d\theta$ is $\frac{1}{2}r^2 \, d\theta$. Integrating these infinitesimal sectors yields the polar area formula. Why the correct answer works: Option B correctly identifies the sector model, whose area formula directly explains the $\frac{1}{2}$ factor. Why distractors fail: Option A describes a rectangle, which would give area $r^2 \, d\theta$ (no $\frac{1}{2}$). Option C describes a triangle, but the infinitesimal element is curved, not triangular with two straight legs. Option D invokes a semicircle, which is not the correct model.

Answer: $r = 4\cos\theta$

Apply the polar substitutions: Substitute $x = r\cos\theta$, $y = r\sin\theta$, and $x^2 + y^2 = r^2$: $r^2 = 4r\cos\theta$. Simplify: Dividing both sides by $r$ (for $r \neq 0$): $r = 4\cos\theta$. Why distractors fail: Option A uses $\sin\theta$, which would correspond to $x^2 + y^2 = 4y$. Option B ($r = 2$) would represent $x^2+y^2 = 4$, a centered circle. Option D keeps $r^2$ on the left, forgetting to divide by $r$.

Answer: $\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$

Recall the parametric arc length formula: The arc length is $L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$, derived from the Pythagorean theorem applied to infinitesimal displacements. Why the correct answer works: Option A presents the correct formula with squared derivatives summed under a square root. Why distractors fail: Option B simply adds the derivatives without squaring or taking a square root. Option C places unsquared derivatives under a square root. Option D sums the squares but omits the square root.

Answer: $6$

Compute the derivatives: $\frac{dx}{dt} = -3\cos^2 t \sin t$ and $\frac{dy}{dt} = 3\sin^2 t \cos t$. So $\sqrt{(dx/dt)^2 + (dy/dt)^2} = 3|\sin t \cos t|$. Use symmetry to simplify: The astroid has four-fold symmetry. In the first quadrant ($0 \le t \le \pi/2$), both $\sin t$ and $\cos t$ are nonnegative, so $|\sin t \cos t| = \sin t \cos t$. The arc length in one quadrant is $\int_0^{\pi/2} 3\sin t \cos t \, dt$. Multiplying by 4 gives $L = 4 \cdot 3\int_0^{\pi/2} \sin t \cos t \, dt = 12\int_0^{\pi/2} \sin t \cos t \, dt$. Evaluate the integral: $12\int_0^{\pi/2}\sin t\cos t\,dt = 12 \cdot \frac{1}{2}\left[\sin^2 t\right]_0^{\pi/2} = 12 \cdot \frac{1}{2}(1 - 0) = 6$. Why distractors fail: Option A ($3$) is the arc length of two quadrants, not all four. Option C ($3\pi$) incorrectly introduces $\pi$. Option D ($\frac{3}{2}$) is the arc length of a single quadrant.

Answer: The student integrated over the full period $[0, 2\pi]$, which gives the total area of all four petals, not one loop.

Identify the curve structure: $r = \sin(2\theta)$ is a four-petaled rose. Each petal occupies an interval of length $\pi/2$. Determine correct limits for one loop: One petal (e.g., in the first quadrant) has $r \ge 0$ for $\theta \in [0, \pi/2]$. So one petal's area is $\frac{1}{2}\int_0^{\pi/2}\sin^2(2\theta)\,d\theta$. Why the correct answer works: Integrating from $0$ to $2\pi$ covers all four petals. The student's integral gives four times the area of one petal. Why distractors fail: Option A: the $\frac{1}{2}$ factor is present. Option B: the curve uses $\sin$, not $\cos$. Option D: the formula correctly uses $r^2$.

Answer: $6\pi$

Compute the derivatives: $\frac{dx}{dt} = -3\sin t$ and $\frac{dy}{dt} = 3\cos t$. Apply the arc length formula: $L = \int_0^{2\pi} \sqrt{(-3\sin t)^2 + (3\cos t)^2} \, dt = \int_0^{2\pi} \sqrt{9\sin^2 t + 9\cos^2 t} \, dt = \int_0^{2\pi} 3 \, dt = 6\pi$. Verify with geometry: This is a circle of radius 3, so its circumference is $2\pi(3) = 6\pi$, confirming the answer. Why distractors fail: Option A ($3\pi$) gives half the circumference. Option C ($9\pi$) likely results from incorrectly using $r^2$ instead of $r$. Option D ($2\pi$) is the circumference of a unit circle.

Answer: $\int_0^{2\pi} \sqrt{(1-\cos t)^2 + \sin^2 t} \, dt$

Compute the derivatives: $\frac{dx}{dt} = 1 - \cos t$ and $\frac{dy}{dt} = \sin t$. Write the arc length integrand: $L = \int_0^{2\pi} \sqrt{(1 - \cos t)^2 + \sin^2 t} \, dt$, which matches Option A. Why distractors fail: Option B swaps sine and cosine in the derivatives. Option C incorrectly simplifies the integrand to $\sqrt{\cos^2 t + \sin^2 t} = 1$, ignoring the derivative of $-\sin t$ in $dx/dt$. Option D adds the derivatives instead of using the Pythagorean combination.

Answer: $3$

Find the parametric derivatives: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2$. Compute dy/dx: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3t}{2}$. Evaluate at t = 2: $\frac{dy}{dx}\bigg|_{t=2} = \frac{3(2)}{2} = 3$. Why distractors fail: Option A ($2$) may come from computing $dx/dt$ at $t=2$ divided by something incorrectly. Option C ($6$) results from evaluating $3t$ at $t=2$ without dividing by 2. Option D ($\frac{2}{3}$) inverts the derivative ratio.

Answer: Both are correct; the cardioid $r = 1 + \cos\theta$ is symmetric about the polar axis, so both methods yield $\frac{3\pi}{2}$.

Identify the symmetry: The cardioid $r = 1 + \cos\theta$ is symmetric about the polar axis ($\theta = 0$). The curve is traced once as $\theta$ goes from $0$ to $2\pi$, and $r \ge 0$ throughout. Verify both approaches: Student A: $A = \frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2 d\theta = \frac{3\pi}{2}$. Student B: $A = 2 \cdot \frac{1}{2}\int_0^{\pi}(1+\cos\theta)^2 d\theta = \frac{3\pi}{2}$. Both give the same result. Why distractors fail: Option A is wrong because the cardioid is symmetric. Option B is wrong because integrating over the full period $[0, 2\pi]$ does not double-count since the curve is traced exactly once. Option D is wrong because $r = 1 + \cos\theta$ has no inner loop.

Answer: $\frac{\pi}{8}$

Identify the integration limits: For $r = \cos(2\theta)$, one petal in the first quadrant runs from $\theta = -\pi/4$ to $\theta = \pi/4$ (where $r \ge 0$). Set up the polar area integral: $A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \cos^2(2\theta) \, d\theta$. Evaluate using the double-angle identity: Using $\cos^2(2\theta) = \frac{1 + \cos(4\theta)}{2}$: $A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \frac{1+\cos(4\theta)}{2} \, d\theta = \frac{1}{4}\left[\theta + \frac{\sin(4\theta)}{4}\right]_{-\pi/4}^{\pi/4} = \frac{1}{4}\left(\frac{\pi}{2}\right) = \frac{\pi}{8}$. Why distractors fail: Option A ($\pi/2$) and Option B ($\pi/4$) overestimate by omitting the $\frac{1}{2}$ factor or making algebraic errors. Option D ($\pi/16$) results from an extra spurious factor of $\frac{1}{2}$.

Answer: The correct formula is $\frac{d^2y}{dx^2} = \frac{d}{dt}\!\left(\frac{dy/dt}{dx/dt}\right) \div \frac{dx}{dt}$, which involves the derivative of the first derivative ratio with respect to $t$.

Recall the parametric second derivative formula: The correct formula is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$. One must first compute $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, differentiate this with respect to $t$, and then divide by $\frac{dx}{dt}$ again. Why the correct answer works: Option B accurately describes the two-step process: differentiate the first derivative quotient with respect to $t$, then divide by $dx/dt$. Why distractors fail: Option A is partially right about not using $d^2x/dt^2$ in the denominator, but it incorrectly suggests simply dividing $d^2y/dt^2$ by $dx/dt$, which is also wrong. Option C is false—parametric second derivatives do exist. Option D is false—the formula works for non-constant derivatives.

Answer: $r = 2 + 3\cos\theta$

Recall limaçon classification: For $r = a + b\cos\theta$: if $|b| > |a|$, the limaçon has an inner loop; if $|b| = |a|$, it is a cardioid; if $|b| < |a|$, there is no inner loop. Classify each option: Option A: $a=2, b=3$, so $|b|>|a|$ → inner loop. Option B: $a=3, b=3$, so $|b|=|a|$ → cardioid. Option C: $a=4, b=3$ → dimpled, no loop. Option D: $a=3, b=2$ → no loop. Why the correct answer works: Only in Option A is $|b| > |a|$, producing negative values of $r$ and thus an inner loop.

Answer: $\frac{dy/dt}{dx/dt}$

Recall the chain rule for parametric derivatives: By the chain rule, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$, provided $\frac{dx}{dt} \neq 0$. Why the correct answer works: Option B directly states this chain rule relationship: we divide $dy/dt$ by $dx/dt$. Why distractors fail: Option A inverts the fraction. Option C multiplies the two derivatives rather than dividing. Option D uses the second derivative of $y$ with respect to $t$, which is part of the formula for $\frac{d^2y}{dx^2}$, not $\frac{dy}{dx}$.

Answer: $x = r\cos\theta, \quad y = r\sin\theta$

Recall the conversion formulas: From the unit circle definitions, $x = r\cos\theta$ and $y = r\sin\theta$. Why the correct answer works: Option B matches the standard conversion identities exactly. Why distractors fail: Option A swaps sine and cosine. Option C uses tangent and secant, which are not the standard conversion formulas. Option D incorrectly squares $r$.

Answer: $\frac{\sin t + \cos t}{\cos t - \sin t}$

Compute the parametric derivatives: $\frac{dx}{dt} = e^t\cos t - e^t\sin t = e^t(\cos t - \sin t)$ and $\frac{dy}{dt} = e^t\sin t + e^t\cos t = e^t(\sin t + \cos t)$. Form the ratio: $\frac{dy}{dx} = \frac{e^t(\sin t + \cos t)}{e^t(\cos t - \sin t)} = \frac{\sin t + \cos t}{\cos t - \sin t}$. The $e^t$ factors cancel. Why distractors fail: Option B inverts the numerator and denominator. Option C forgets to differentiate and simply divides $y$ by $x$. Option D ($\tan t$) would only be correct if $x = \cos t$ and $y = \sin t$ without the $e^t$ factor, and even then the sign would differ.

Answer: $\sqrt{2}\,(e^{2\pi} - 1)$

Compute the integrand: For $r = e^{\theta}$, $\frac{dr}{d\theta} = e^{\theta}$. So $\sqrt{r^2 + (dr/d\theta)^2} = \sqrt{e^{2\theta} + e^{2\theta}} = e^{\theta}\sqrt{2}$. Evaluate the integral: $L = \int_0^{2\pi} \sqrt{2}\,e^{\theta}\,d\theta = \sqrt{2}\left[e^{\theta}\right]_0^{2\pi} = \sqrt{2}(e^{2\pi} - 1)$. Why distractors fail: Option A omits the $\sqrt{2}$ factor. Option C uses $2$ instead of $\sqrt{2}$. Option D squares $e^{\theta}$ in the antiderivative, obtaining $e^{2\theta}$ instead of $e^{\theta}$.

Answer: The first student is correct; $\frac{dy}{dx} = -\cot t$ follows directly from $(\cos t)/(-\sin t)$.

Compute the derivatives: $\frac{dx}{dt} = -\sin t$ and $\frac{dy}{dt} = \cos t$. Form the ratio: $\frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t$. Verify geometrically: At $t = \pi/4$ on the unit circle, the slope should be $-1$. Indeed $-\cot(\pi/4) = -1$, while $-\tan(\pi/4) = -1$ also. But at $t = \pi/3$: $-\cot(\pi/3) = -1/\sqrt{3}$ while $-\tan(\pi/3) = -\sqrt{3}$. The point is $(\cos(\pi/3), \sin(\pi/3)) = (1/2, \sqrt{3}/2)$, and the slope of the unit circle at that point is $-x/y = -1/\sqrt{3}$, confirming $-\cot t$. Why distractors fail: Option B inverts the ratio. Option C is false—the slope on a circle is generally nonzero. Option D is false because $\cot t \neq \tan t$ in general.

Answer: $t = \pm\frac{1}{\sqrt{3}}$

Find where dy/dt = 0: $\frac{dy}{dt} = 3t^2 - 1 = 0 \implies t^2 = \frac{1}{3} \implies t = \pm\frac{1}{\sqrt{3}}$. Check that dx/dt ≠ 0: $\frac{dx}{dt} = 2t$. At $t = \pm\frac{1}{\sqrt{3}}$, $\frac{dx}{dt} = \pm\frac{2}{\sqrt{3}} \neq 0$. So both are valid horizontal tangent points. Why distractors fail: Option A: at $t=0$, $\frac{dy}{dt} = -1 \neq 0$. Option C: at $t = \pm1$, $\frac{dy}{dt} = 2 \neq 0$. Option D includes $t = 0$, which does not satisfy $dy/dt = 0$.

Answer: Area: $\frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2 d\theta$; Perimeter: $\int_0^{2\pi}\sqrt{(1+\cos\theta)^2 + \sin^2\theta}\,d\theta$

Recall the polar area formula: Area $= \frac{1}{2}\int_0^{2\pi} r^2 \, d\theta = \frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2 d\theta$. Recall the polar arc length formula: Perimeter $= \int_0^{2\pi}\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta$. Here $\frac{dr}{d\theta} = -\sin\theta$, so the integrand is $\sqrt{(1+\cos\theta)^2 + \sin^2\theta}$. Why Option C is correct: It has the correct $\frac{1}{2}$ in the area formula with $r^2$, and the correct polar arc length integrand. Why distractors fail: Option A uses $r$ (not $\sqrt{r^2+(dr/d\theta)^2}$) for perimeter. Option B omits the $\frac{1}{2}$ in the area formula. Option D uses $r$ instead of $r^2$ in the area integral.