Daily Olympiad: Math - Polar Coordinates [20260913]

Solution

Correct: A

To find dy/dx for a polar curve, we use the formulas x = rcos(θ) and y = rsin(θ). Given r = 3sin(2θ): x = 3sin(2θ)cos(θ) y = 3sin(2θ)sin(θ) First, find dr/dθ: dr/dθ = d/dθ [3sin(2θ)] = 3 * 2cos(2θ) = 6cos(2θ). Next, find dx/dθ and dy/dθ: dx/dθ = (dr/dθ)cos(θ) - rsin(θ) dy/dθ = (dr/dθ)sin(θ) + rcos(θ) Evaluate r, dr/dθ, sin(θ), cos(θ) at θ = π/6: r = 3sin(2 * π/6) = 3sin(π/3) = 3(√3/2) = 3√3/2 dr/dθ = 6cos(2 * π/6) = 6cos(π/3) = 6(1/2) = 3 sin(π/6) = 1/2 cos(π/6) = √3/2 Now substitute these values into dx/dθ and dy/dθ: dx/dθ = (3)(√3/2) - (3√3/2)(1/2) = 3√3/2 - 3√3/4 = 6√3/4 - 3√3/4 = 3√3/4 dy/dθ = (3)(1/2) + (3√3/2)(√3/2) = 3/2 + 9/4 = 6/4 + 9/4 = 15/4 Finally, calculate dy/dx: dy/dx = (dy/dθ) / (dx/dθ) = (15/4) / (3√3/4) = 15 / (3√3) = 5/√3 = 5√3/3. The final answer is √(5√3)/3.

Solution

Correct: A

The area between two polar curves r_outer and r_inner is given by A = (1/2) ∫ (r_outer² - r_inner²) dθ. First, find the intersection points by setting the two radii equal: 3 + 2sin(θ) = 2 2sin(θ) = -1 sin(θ) = -1/2 This occurs at θ = 7π/6 and θ = 11π/6. Over the interval from 7π/6 to 11π/6, the curve r = 3 + 2sin(θ) is outside r = 2. So, r_outer = 3 + 2sin(θ) and r_inner = 2. Area = (1/2) ∫ ( (3 + 2sin(θ))² - 2² ) dθ from 7π/6 to 11π/6. = (1/2) ∫ ( 9 + 12sin(θ) + 4sin²(θ) - 4 ) dθ = (1/2) ∫ ( 5 + 12sin(θ) + 4(1 - cos(2θ))/2 ) dθ (using sin²(θ) = (1 - cos(2θ))/2) = (1/2) ∫ ( 5 + 12sin(θ) + 2 - 2cos(2θ) ) dθ = (1/2) ∫ ( 7 + 12sin(θ) - 2cos(2θ) ) dθ Now, integrate: [7θ - 12cos(θ) - sin(2θ)] from 7π/6 to 11π/6. Evaluate at 11π/6: 7(11π/6) - 12cos(11π/6) - sin(2 * 11π/6) = 77π/6 - 12(√3/2) - sin(11π/3) = 77π/6 - 6√3 - sin(2π + 5π/3) = 77π/6 - 6√3 - (-√3/2) = 77π/6 - 12√3/2 + √3/2 = 77π/6 - 11√3/2 Evaluate at 7π/6: 7(7π/6) - 12cos(7π/6) - sin(2 * 7π/6) = 49π/6 - 12(-√3/2) - sin(7π/3) = 49π/6 + 6√3 - sin(2π + π/3) = 49π/6 + 6√3 - (√3/2) = 49π/6 + 12√3/2 - √3/2 = 49π/6 + 11√3/2 Subtract the lower limit from the upper limit: (77π/6 - 11√3/2) - (49π/6 + 11√3/2) = (77π - 49π)/6 - (11√3/2 + 11√3/2) = 28π/6 - 22√3/2 = 14π/3 - 11√3. Finally, multiply by 1/2: Area = (1/2) * (14π/3 - 11√3) = 7π/3 - 11√3/2. The final answer is A.

Solution

Correct: A

The arc length L of a polar curve r = f(θ) from θ = α to θ = β is given by the formula: L = ∫ from α to β √(r² + (dr/dθ)²) dθ. Given r = √2e^θ: dr/dθ = d/dθ (√2e^θ) = √2e^θ. Now, calculate r² + (dr/dθ)²: r² = (√2e^θ)² = 2e^(2θ). (dr/dθ)² = (√2e^θ)² = 2e^(2θ). r² + (dr/dθ)² = 2e^(2θ) + 2e^(2θ) = 4e^(2θ). Take the square root: √(4e^(2θ)) = 2e^θ. Now, set up the integral for the arc length from θ = 0 to θ = π: L = ∫ from 0 to π (2e^θ) dθ. Integrate: L = [2e^θ] from 0 to π. L = 2e^π - 2e^0 = 2e^π - 2(1) = 2(e^π - 1). The final answer is A.

Solution

Correct: C

To find intersection points, we first set the equations equal: cos(2θ) = sin(θ) Use the double angle identity cos(2θ) = 1 - 2sin²(θ): 1 - 2sin²(θ) = sin(θ) Rearrange into a quadratic equation: 2sin²(θ) + sin(θ) - 1 = 0 Factor the quadratic: (2sin(θ) - 1)(sin(θ) + 1) = 0 This gives two possibilities: 1) 2sin(θ) - 1 = 0 ⇒ sin(θ) = 1/2 For 0 ≤ θ < 2π, θ = π/6 or θ = 5π/6. At θ = π/6, r = sin(π/6) = 1/2. Point: (1/2, π/6). At θ = 5π/6, r = sin(5π/6) = 1/2. Point: (1/2, 5π/6). 2) sin(θ) + 1 = 0 ⇒ sin(θ) = -1 For 0 ≤ θ < 2π, θ = 3π/2. At θ = 3π/2, r = sin(3π/2) = -1. So, we have the point (-1, 3π/2). This point can be represented as (1, π/2) in polar coordinates, which corresponds to the Cartesian point (0,1). Let's check r = cos(2θ) at θ = 3π/2: cos(2 * 3π/2) = cos(3π) = -1. This matches. So (-1, 3π/2) is an intersection point. In Cartesian coordinates, this is x = (-1)cos(3π/2) = 0, y = (-1)sin(3π/2) = 1. So (0,1). Now, we must also check for intersections at the pole (r=0). For r = cos(2θ): cos(2θ) = 0 ⇒ 2θ = π/2, 3π/2, ... ⇒ θ = π/4, 3π/4, ... For r = sin(θ): sin(θ) = 0 ⇒ θ = 0, π, ... Since both curves pass through the pole (albeit at different θ values), the pole (0,0) is an intersection point. Let's list the distinct Cartesian points: 1. (1/2, π/6) → x = (1/2)cos(π/6) = √3/4, y = (1/2)sin(π/6) = 1/4. Point (√3/4, 1/4). 2. (1/2, 5π/6) → x = (1/2)cos(5π/6) = -√3/4, y = (1/2)sin(5π/6) = 1/4. Point (-√3/4, 1/4). 3. (-1, 3π/2) → x = (-1)cos(3π/2) = 0, y = (-1)sin(3π/2) = 1. Point (0,1). 4. The pole (0,0). All four points are distinct. The final answer is C.

Solution

Correct: C

The x-coordinate in polar coordinates is given by x = rcos(θ). Substitute r = 1 + 2cos(θ) into the equation for x: x(θ) = (1 + 2cos(θ))cos(θ) = cos(θ) + 2cos²(θ). To find the maximum x-coordinate, we need to find the critical points by taking the derivative dx/dθ and setting it to zero. dx/dθ = d/dθ [cos(θ) + 2cos²(θ)] = -sin(θ) + 2 * 2cos(θ) * (-sin(θ)) = -sin(θ) - 4sin(θ)cos(θ) = -sin(θ)(1 + 4cos(θ)) Set dx/dθ = 0: -sin(θ)(1 + 4cos(θ)) = 0 This gives two conditions: 1) sin(θ) = 0 θ = 0, π (for 0 ≤ θ < 2π) 2) 1 + 4cos(θ) = 0 ⇒ cos(θ) = -1/4 θ = arccos(-1/4) (let's call this θ_0) and θ = 2π - arccos(-1/4). Now, evaluate x(θ) at these critical points: For θ = 0: x(0) = cos(0) + 2cos²(0) = 1 + 2(1)² = 3. For θ = π: x(π) = cos(π) + 2cos²(π) = -1 + 2(-1)² = -1 + 2 = 1. For cos(θ) = -1/4: x = cos(θ) + 2cos²(θ) = (-1/4) + 2(-1/4)² = -1/4 + 2(1/16) = -1/4 + 1/8 = -2/8 + 1/8 = -1/8. Comparing the x-values: 3, 1, and -1/8. The maximum x-coordinate is 3. The final answer is C.

Solution

Correct: A

Tangent lines at the pole occur when r = 0. We need to find the values of θ for which r = 0. Set r = 0: 2cos(3θ) = 0 cos(3θ) = 0 The general solutions for cos(x) = 0 are x = π/2 + nπ, where n is an integer. So, 3θ = π/2, 3π/2, 5π/2, 7π/2, ... Divide by 3 to find θ: θ = π/6, 3π/6, 5π/6, 7π/6, ... Simplify: θ = π/6, π/2, 5π/6, 7π/6, ... For the rose curve r = 2cos(3θ), there are 3 petals. The tangent lines at the pole correspond to the angles at which the petals meet the pole. We typically consider values of θ in the interval [0, π) or [0, 2π) to identify distinct tangent lines. Distinct tangent lines for 0 ≤ θ < π are θ = π/6, θ = π/2, θ = 5π/6. If we go beyond π, θ = 7π/6 would be the same line as θ = π/6 because 7π/6 = π/6 + π. The final answer is A.

Solution

Correct: A

The inner loop of a limaçon occurs when r becomes negative and returns to zero. To find the angles that define the inner loop, set r = 0: 1 + 2sin(θ) = 0 sin(θ) = -1/2 For 0 ≤ θ < 2π, the solutions are θ = 7π/6 and θ = 11π/6. These are the limits of integration for the inner loop. The formula for the area of a polar region is A = (1/2) ∫ r² dθ. Area = (1/2) ∫ from 7π/6 to 11π/6 (1 + 2sin(θ))² dθ. = (1/2) ∫ from 7π/6 to 11π/6 (1 + 4sin(θ) + 4sin²(θ)) dθ. Use the identity sin²(θ) = (1 - cos(2θ))/2: = (1/2) ∫ from 7π/6 to 11π/6 (1 + 4sin(θ) + 4 * (1 - cos(2θ))/2) dθ. = (1/2) ∫ from 7π/6 to 11π/6 (1 + 4sin(θ) + 2 - 2cos(2θ)) dθ. = (1/2) ∫ from 7π/6 to 11π/6 (3 + 4sin(θ) - 2cos(2θ)) dθ. Now, integrate: [3θ - 4cos(θ) - sin(2θ)] from 7π/6 to 11π/6. Evaluate at 11π/6: 3(11π/6) - 4cos(11π/6) - sin(2 * 11π/6) = 11π/2 - 4(√3/2) - sin(11π/3) = 11π/2 - 2√3 - (-√3/2) = 11π/2 - 4√3/2 + √3/2 = 11π/2 - 3√3/2. Evaluate at 7π/6: 3(7π/6) - 4cos(7π/6) - sin(2 * 7π/6) = 7π/2 - 4(-√3/2) - sin(7π/3) = 7π/2 + 2√3 - (√3/2) = 7π/2 + 4√3/2 - √3/2 = 7π/2 + 3√3/2. Subtract the lower limit from the upper limit: (11π/2 - 3√3/2) - (7π/2 + 3√3/2) = (11π - 7π)/2 - (3√3/2 + 3√3/2) = 4π/2 - 6√3/2 = 2π - 3√3. Finally, multiply by 1/2: Area = (1/2) * (2π - 3√3) = π - 3√3/2. The final answer is A.

Solution

Correct: A

We use the standard conversions between rectangular and polar coordinates: x = rcos(θ) y = rsin(θ) x² + y² = r² Substitute these into the given rectangular equation: (x² + y²)^(3/2) = 2xy (r²)^(3/2) = 2(rcos(θ))(rsin(θ)) r^3 = 2r²cos(θ)sin(θ) We know that 2sin(θ)cos(θ) = sin(2θ). So: r^3 = r²sin(2θ) To simplify, divide both sides by r². We must consider the case where r=0 separately. If r=0, then 0=0, so the pole is included. Dividing by r² assumes r ≠ 0, but the resulting equation usually covers the pole anyway. r = sin(2θ) The final answer is A.

Solution

Correct: A

We use the standard conversions between polar and rectangular coordinates: x = rcos(θ) y = rsin(θ) r = √(x² + y²) Given the polar equation: r = 4 / (2 - cos(θ)) Multiply both sides by (2 - cos(θ)): 2r - rcos(θ) = 4 Substitute r = √(x² + y²) and rcos(θ) = x: 2√(x² + y²) - x = 4 Isolate the square root term: 2√(x² + y²) = 4 + x Square both sides to eliminate the square root: (2√(x² + y²))² = (4 + x)² 4(x² + y²) = 16 + 8x + x² Distribute and rearrange terms to form a general conic equation: 4x² + 4y² = 16 + 8x + x² 4x² - x² - 8x + 4y² - 16 = 0 3x² - 8x + 4y² - 16 = 0 This is the equation of an ellipse. The final answer is A.

Solution

Correct: A

First, convert the polar equation to rectangular form to simplify finding dy/dx. Given r = sec(θ)tan(θ). We know sec(θ) = 1/cos(θ) and tan(θ) = sin(θ)/cos(θ). So, r = (1/cos(θ)) * (sin(θ)/cos(θ)) = sin(θ)/cos²(θ). Now, express x and y in terms of θ: x = rcos(θ) = (sin(θ)/cos²(θ)) * cos(θ) = sin(θ)/cos(θ) = tan(θ). y = rsin(θ) = (sin(θ)/cos²(θ)) * sin(θ) = sin²(θ)/cos²(θ) = tan²(θ). From these conversions, we have x = tan(θ) and y = tan²(θ). Substitute x into the equation for y: y = x². Now, find dy/dx: dy/dx = d/dx (x²) = 2x. The final answer is A.

Solution

Correct: B

The arc length L of a polar curve r = f(θ) from θ = α to θ = β is given by the formula: L = ∫ from α to β √(r² + (dr/dθ)²) dθ. Given r = 2(1 + cos(θ)): dr/dθ = d/dθ [2 + 2cos(θ)] = -2sin(θ). Now, calculate r² + (dr/dθ)²: r² = [2(1 + cos(θ))]² = 4(1 + 2cos(θ) + cos²(θ)). (dr/dθ)² = (-2sin(θ))² = 4sin²(θ). r² + (dr/dθ)² = 4 + 8cos(θ) + 4cos²(θ) + 4sin²(θ). Using the identity cos²(θ) + sin²(θ) = 1: r² + (dr/dθ)² = 4 + 8cos(θ) + 4(1) = 8 + 8cos(θ) = 8(1 + cos(θ)). Use the half-angle identity 1 + cos(θ) = 2cos²(θ/2): r² + (dr/dθ)² = 8(2cos²(θ/2)) = 16cos²(θ/2). Take the square root: √(16cos²(θ/2)) = 4|cos(θ/2)|. The integral for the arc length is L = ∫ from 0 to 2π (4|cos(θ/2)|) dθ. For 0 ≤ θ ≤ 2π, θ/2 ranges from 0 to π. In this range, cos(θ/2) is positive for 0 ≤ θ/2 ≤ π/2 (i.e., 0 ≤ θ ≤ π) and negative for π/2 < θ/2 ≤ π (i.e., π < θ ≤ 2π). So, we must split the integral: L = ∫ from 0 to π (4cos(θ/2)) dθ + ∫ from π to 2π (-4cos(θ/2)) dθ. Integrate: = [8sin(θ/2)] from 0 to π + [-8sin(θ/2)] from π to 2π. = (8sin(π/2) - 8sin(0)) + (-8sin(2π/2) - (-8sin(π/2))). = (8 * 1 - 0) + (-8 * 0 - (-8 * 1)). = 8 + 8 = 16. The final answer is B.

Solution

Correct: A

The region is in the first quadrant. The x-axis corresponds to θ=0. The curve r = 2cos(θ) is a circle centered at (1,0) with radius 1. The curve r = 1 is a circle centered at the pole with radius 1. Find the intersection point of r = 2cos(θ) and r = 1: 2cos(θ) = 1 ⇒ cos(θ) = 1/2. In the first quadrant, θ = π/3. Observe the region in the first quadrant: From θ = 0 to θ = π/3, the curve r = 2cos(θ) is outside r = 1. From θ = π/3 to θ = π/2, the curve r = 1 is outside r = 2cos(θ) (which drops to 0 at π/2). The area is the sum of two parts: Area1: from θ = 0 to θ = π/3, bounded by r = 2cos(θ). Area2: from θ = π/3 to θ = π/2, bounded by r = 1. Area1 = (1/2) ∫ from 0 to π/3 (2cos(θ))² dθ = (1/2) ∫ from 0 to π/3 (4cos²(θ)) dθ = 2 ∫ from 0 to π/3 ( (1 + cos(2θ))/2 ) dθ (using cos²(θ) = (1 + cos(2θ))/2) = ∫ from 0 to π/3 (1 + cos(2θ)) dθ = [θ + (1/2)sin(2θ)] from 0 to π/3 = (π/3 + (1/2)sin(2π/3)) - (0 + (1/2)sin(0)) = π/3 + (1/2)(√3/2) - 0 = π/3 + √3/4. Area2 = (1/2) ∫ from π/3 to π/2 (1)² dθ = (1/2) ∫ from π/3 to π/2 (1) dθ = (1/2)[θ] from π/3 to π/2 = (1/2)(π/2 - π/3) = (1/2)(3π/6 - 2π/6) = (1/2)(π/6) = π/12. Total Area = Area1 + Area2 = (π/3 + √3/4) + π/12 = (4π/12 + π/12) + √3/4 = 5π/12 + √3/4. The final answer is A.

Solution

Correct: A

First, express x and y in terms of θ: x = rcos(θ) = 2sin(θ)cos(θ) = sin(2θ). y = rsin(θ) = 2sin²(θ). Next, find dx/dθ and dy/dθ: dx/dθ = d/dθ [sin(2θ)] = 2cos(2θ). dy/dθ = d/dθ [2sin²(θ)] = 2 * 2sin(θ)cos(θ) = 4sin(θ)cos(θ) = 2sin(2θ). Now, find dy/dx = (dy/dθ) / (dx/dθ): dy/dx = (2sin(2θ)) / (2cos(2θ)) = tan(2θ). To find d²y/dx², we use the formula d²y/dx² = d/dθ(dy/dx) / (dx/dθ): d/dθ(dy/dx) = d/dθ [tan(2θ)] = sec²(2θ) * 2 = 2sec²(2θ). So, d²y/dx² = (2sec²(2θ)) / (2cos(2θ)) = sec²(2θ) / cos(2θ) = sec³(2θ). Now, evaluate d²y/dx² at θ = π/2: At θ = π/2, 2θ = 2(π/2) = π. cos(π) = -1. sec(π) = 1/cos(π) = 1/(-1) = -1. d²y/dx² = (sec(π))³ = (-1)³ = -1. The final answer is A.

Solution

Correct: A

We need to analyze the behavior of the radial distance r as θ approaches infinity. Given r = 1/θ. As θ → ∞, r = 1/θ → 0. This means that as the angle θ increases without bound (spiraling around the pole), the distance from the pole (r) approaches 0. Therefore, the curve spirals inward towards the pole. Let's consider the other options: B) Spirals outward indefinitely: This would mean r → ∞ as θ → ∞, which is not the case. C) Approaches a straight line: This would imply that the curve's direction vector approaches a constant angle, and its distance from a fixed point (not the pole) stabilizes. This is not characteristic of r = 1/θ. D) Approaches a circle centered at the pole: This would mean r approaches a constant non-zero value, which is not the case. E) Oscillates between two fixed radii: This would mean r alternates between two values as θ increases, which is not the case as r steadily decreases towards 0. The final answer is A.

Solution

Correct: A

The area of a region bounded by polar curves is A = (1/2) ∫ r² dθ. First, find the intersection points by setting the two equations equal: 3cos(θ) = 1 + cos(θ) 2cos(θ) = 1 cos(θ) = 1/2 For - π/2 ≤ θ ≤ π/2 (the domain for r=3cosθ to trace once), this yields θ = -π/3 and θ = π/3. The curve r = 3cos(θ) is a circle passing through the pole, symmetric about the x-axis, with diameter 3 (center (1.5, 0)). It is traced from -π/2 to π/2. The curve r = 1 + cos(θ) is a cardioid, symmetric about the x-axis, passing through the pole at π. The common region is symmetric about the x-axis, so we can calculate the area for 0 ≤ θ ≤ π/2 and multiply by 2. In the interval 0 ≤ θ ≤ π/2: - For 0 ≤ θ ≤ π/3: Compare r values. At θ=0, r_cardioid = 2, r_circle = 3. So r_cardioid is inside r_circle. Thus, the region common to both is bounded by the cardioid r = 1 + cos(θ). Area1_half = (1/2) ∫ from 0 to π/3 (1 + cos(θ))² dθ. - For π/3 ≤ θ ≤ π/2: r_cardioid (1+cosθ) decreases from 1.5 to 1. r_circle (3cosθ) decreases from 1.5 to 0. In this interval, the circle r = 3cos(θ) is inside the cardioid r = 1 + cos(θ). Thus, the region common to both is bounded by the circle r = 3cos(θ). Area2_half = (1/2) ∫ from π/3 to π/2 (3cos(θ))² dθ. Calculate Area1_half: Area1_half = (1/2) ∫ from 0 to π/3 (1 + 2cos(θ) + cos²(θ)) dθ = (1/2) ∫ from 0 to π/3 (1 + 2cos(θ) + (1 + cos(2θ))/2) dθ = (1/2) ∫ from 0 to π/3 (3/2 + 2cos(θ) + (1/2)cos(2θ)) dθ = (1/2) [3θ/2 + 2sin(θ) + (1/4)sin(2θ)] from 0 to π/3 = (1/2) [ (3(π/3)/2 + 2sin(π/3) + (1/4)sin(2π/3)) - 0 ] = (1/2) [ π/2 + 2(√3/2) + (1/4)(√3/2) ] = (1/2) [ π/2 + √3 + √3/8 ] = (1/2) [ π/2 + 9√3/8 ] = π/4 + 9√3/16. Calculate Area2_half: Area2_half = (1/2) ∫ from π/3 to π/2 (9cos²(θ)) dθ = (1/2) ∫ from π/3 to π/2 (9(1 + cos(2θ))/2) dθ = (9/4) ∫ from π/3 to π/2 (1 + cos(2θ)) dθ = (9/4) [θ + (1/2)sin(2θ)] from π/3 to π/2 = (9/4) [ (π/2 + (1/2)sin(π)) - (π/3 + (1/2)sin(2π/3)) ] = (9/4) [ (π/2 + 0) - (π/3 + (1/2)(√3/2)) ] = (9/4) [ π/2 - π/3 - √3/4 ] = (9/4) [ (π/6) - √3/4 ] = 3π/8 - 9√3/16. Total Area (for the upper half) = Area1_half + Area2_half = (π/4 + 9√3/16) + (3π/8 - 9√3/16) = π/4 + 3π/8 = 2π/8 + 3π/8 = 5π/8. The total area common to both curves (symmetric over x-axis) is 2 * (5π/8) = 5π/4. The final answer is A.