Cardioid Boundary
Calculus: Polar Coordinate Question?
Not sure how to write the line representing the front of the stage in polar form and finding its intersection with the cardioid?
When recording live performances, sound engineers often use a microphone with a cardioid pickup pattern because it suppresses noise from the audience. Suppose the microphone is placed 3 m from the front of the stage and the boundary of the optimal pickup region is given by the cardioid , where r is measured in meters and the microphone is at the pole. The musicians want to know the area they will have on stage within the optimal pickup range of the microphone. Answer their question.
Hint: Start by writing the line representing the front of the stage in polar form and finding its intersection with the cardioid.
Image of stage setup:
http://farm3.static.flickr.com/2140/2329344862_36145d769d.jpg?v=0
Thanks A Lot
Let x be the horizontal axis and y the vertical axis in your sketch.
In these Cartesian coordinate system the front stage line is given by:
y = 3 [unit m]
You can transform this to polar coordinate system using the definition of y in terms of coordinates r and θ:
y = r·sin(θ)
So the line in polar coordinates system is given by
r·sin(θ) = 3
<=>
r = 3/sin(θ)
Your cardioid is given by
r = 4·(1 + sin(θ)) [unit m]
Equate the two line equations to calculate the angle where they intersect:
4·(1 + sin(θ)) = 3/sin(θ)
<=>
4·(sin(θ) + sin²(θ)) = 3
<=>
sin²(θ) + sin(θ) – (3/4) = 0
solve this quadratic equation for sin(θ)
=>
sin(θ) = (-1/2) ± √((1/4) + (3/4))
=>
sin(θ) = 1/2
(the other solution sin(θ)=-3/2 is infeasible)
The two angles satisfy condition above:
θ₁ = 30° = (1/6)·π
θ₂ = 150° = (5/6)·π
The area between the lines can be found by integrating with the limits:
3/sin(θ) ≤ r ≤ 4·(1 + sin(θ))
θ₁ ≤ θ ≤ θ₂
Remember that in polar coordinates
dA = r dr dθ
Hence:
A = ∫ θ₁→θ₂ ∫ 3/sin(θ)→(1 + sin(θ)) [ r ]drdθ
(because ∫ rdr = (1/2)·r²
= (1/2) · ∫ θ₁→θ₂ [ (4·(1 + sin(θ)))² - (3/sin(θ))² ] dθ
= (1/2) · ∫ θ₁→θ₂ [ 16 + 32·sin(θ) + 16·sin²(θ) - 9·csc²(θ) ] dθ
because:
∫ sin(θ) dθ = -cos(θ)
∫ sin²(θ) dθ = (1/2)·(θ – sin(θ)·cos(θ))
∫ csc²(θ) dθ = -cot(θ) = -cos(θ)/sin(θ)
=>
∫ [16 + 32·sin(θ) + 16·sin²(θ) - 9·csc²(θ) ] dθ
= 16·θ – 32·cos(θ) + 8·(θ – sin(θ)·cos(θ)) + 9·cos(θ)/sin(θ)
= 24·θ – cos(θ)·( 32 + 8·sin(θ) – 9/sin(θ) )
Hence:
A
= (1/2) · ∫ θ₁→θ₂ [ 16 + 32·sin(θ) + 16·sin²(θ) - 9·csc²(θ) ] dθ
= 12·(θ₂-θ₁) – cos(θ₂)·( 16 + 4·sin(θ₂) – (9/2)/sin(θ₂) )
+ cos(θ₁)·(16 + 4·sin(θ₁) – (9/2)/sin(θ₁) )
with
sin(θ₁)=sin(θ₂)=1/2 ; cos(θ₂)=(-√3)/2 cos(θ₁)=(√3)/2
A=
12·((5/6)·π – (1/6)·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 ( unit m² )