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Calculus: Paula adjustment question?

No cardioid and find the intersection Its front line must be created to represent how Stage in the form of the opposite? It to suppress the noise of the audience When recording a live show because the sound engineers Cardioid pickup pattern, and often use a microphone. Let the 3-position the microphone Best pick-up area and front of the stage at the boundary of M, this study Is measured in meters, you are in a microphone pole cardioid is given by. They want to know the musicians area Mike's pick of the best stage, is within the range. Their Answer questions. Start a hint from the front line of the stage to form the Arctic A written representation and find its intersection with cardioid. Stage Install the image http://farm3.static.flickr.com/2140/2329344862_36145d769d.jpg?v=0, thank you

Go On the vertical axis and the y-axis horizontal X-Sketch Number: coordinates. In these Cartesian system in the front stage, given the line y = 3 [Unit m] you can change it in terms of coordinates r, θ, and the definition of y Using polar coordinate system: sin (θ), so the line of the Arctic And = Y) two-coordinate θ (sin is a system in a given study · = 3 <=> R a = 3/sin (θ) is a cardioid Toucheng'll give you an equation to calculate A study by = 4 ° (1 + sin (θ)) [Unit] m where the cross angles: 4 ° (1 + Sin (θ)) = 3/sin (θ) <=> 4 ° (sin (θ) + sin ² (θ)) = 3 <=> sin ² (θ) + Sin (θ) - (3 / 4) = 0) / 4) θ (sin + of the equation (3 fix the car) => Sin (θ) = (-1 / 2) ± √ a ((1 / 4) => sin (θ) = 1 / 2 (sin, another solution (Θ) =- 2 분의 3 is infeasible), the angle that satisfies the above conditions are: θ ₁ = 30 ° = (sixth) · π θ ₂ = 150 ° = (5 / 6) and the integration limits You can find the area between the lines of π: 3/sin (θ) ≤ ≤ r in the 4 ° (1 + Sin (θ)) θ ₁ ≤ θ ≤ θ ₂ dθ So remember that Dr. Polar coordinates, the R DA =: = ∫ θ ₁ → θ ₂ ∫ 3/sin (θ) → (1 + sin (θ)) [R] in the drdθ (because ∫ rdr = (1 / 2), research ² = (1 / 2) · ∫ θ ₁ → θ ₂ [(4 ° (1 + sin (θ))) ² - (3/sin (θ)) ²] dθ = (1 / 2) · ∫ θ ₁ → θ ₂ [16 + 32 ° sin (θ) + 16 ° sin ² (θ) - 9 · csc ² (θ)] dθ on the reasons: ∫ sin (θ) dθ = - Cause (θ) ∫ sin ² (θ) dθ = (1 / 2) ° (θ - sin (θ) · Cause (Θ)) ∫ csc ² (θ) dθ = - b (θ) Cause (θ) / sin = (θ) => ∫ [16 + 32 ° sin (θ) + 16 ° sin ² (θ) - 9 · csc ² a (θ)] dθ = 16 and θ - 32 ° Cause (Θ) + 8 ° (θ - sin (θ) · Cause (θ)) + 9 · Cause (θ) / Sin (θ) = 24 ° θ - Because (θ) ° (32 + 8 · sin (θ) - 9/sin (θ)), therefore : = (1 / 2) · ∫ θ ₁ → θ ₂ [16 + 32 ° sin (θ) + 16 ° sin ² (θ) - 9 · csc ² (θ)] dθ = 12 ° (θ ₂ - θ ₁) - Cause (θ ₂) ° (16 + 4 ° sin (θ ₂) - (9 / 2) / sin (θ ₂)) + Cause (θ ₁) ° (16 + 4 ° sin (θ ₁) - (9 / 2) / Sin (θ ₁)) sin (θ ₁) = sin (θ ₂) = 1 / 2; Cause (θ ₂) = (- √ 3) / 2 Cause (θ ₁) = (√ 3) / 2 = 12 ° ((6 분의 5) · π - (a sextant A) · π) + (√ 3) / 2 ° (16 + 4 ° (1 / 2) - (9 / 2) / (1 / 2)) + (√ 3) / 2 ° (16 + 4 ° ( 1 / 2) - (9 / 2) / (1 / 2)) = 8 · π + 9 · √ 3 ≈ 40.72 (per ㎡)

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Eden Prairie School Board to grapple with contentious boundary plan tonight | Learning Curve
Courtesy of Eden Prairie Schools The Eden Prairie School Board listens to input from community members. In 1998, almost two decades after he immigrated to the United States to attend college, Ahmed Jama moved to Eden Prairie for the same reason many people do. Jama and his wife were starting a family and wanted to live in a community with good schools. So far, the experiences of their son and ...