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Calculus Project

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Josseline Herrera

on 24 May 2013

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Transcript of Calculus Project

Sound Engineering Uses calculus to study the way sound changes in order to produce effective designs of structures that can transmit sound efficiently without distorting it . Shape Conclusion Sound Intensity How to Calculate Sound Intensity found an efficient shape and dimensions for optimizing the acoustics in the music hall
determined if the shape and dimensions allowed the sound intensity to remain at a high level of quality at the back of the theater
determined that as area increases, sound intensity decreases so changing the dimensions to make it wider, taller or longer, would increase the area and therefore lessen the intensity. Coned shaped
Reflects the Ancient Greek design
Seats elevated and expanded outward
Steps and part of the seats made of limestone How is Calculus Applied in the Real World? By: Josseline Herrera, Tanya Piña, and Diana Del Castillo Other Factors grooved walls made of sound absorbing panels to eliminate any echo that would have been produced
hard plastic panels on the ceiling of the stage to reflect back any sound waves that were to go upwards Sound waves travel in a circular motion I=P/A
I= intensity
P= power in watts
A= area in meters sq The decibel (dB) is used to measure sound intensity
0 dB= 1 x 10^-12 watts/m^2 To find the area, we used
calculus in this manner:


A= 188.56 m^2 Our desired sound intensity was
98 dB (6.3 x 10^-3 watts/m^2) because that is the average amount of decibels produced by a large orchestra. The theater produces a parabola that is expressed by f(x)= -x^2/20 +10 After determining the area and our desired sound intensity, we calculated for the power:

6.3 x 10^-3= P/ 188.56
P= 1.19 watts The cross sections are expressed by these three parabolas. We determined the sound intensity at three different cross sections in the theater to analyze the behavior of sound waves and how area affects them. Using the power that we found from the first cross section, we can determine what the sound intensity will be when the theater is 20 meters high. A=

A= 754. 25 m^2 I=(1.19)/(754.25)
= 1.58 x 10^-3 watts/m^2 Converting Watts/m^2 Into Decibels (10 dB) log(I/Io) when Io= 1 x 10^-12 watts/m^2 (10 dB) log((1.58 x 10^-3)/(1 x 10^-12)) =91.99dB. The third cross section would be 30 meters high and is expressed by the equation f(x)= -x^2 /60 +30 A=

A= 1697.06 m^2 I=(1.19)/(1697.06)
= 7.0 x 10^-4 watts/m^2 (10 dB) log(7.0 x 10^-4)/(1 x 10^-12) = 88.45 dB 98 dB 91.99 dB 88.45dB Decibels? We have: What We Did: Optimized the intensity of sound throughout the theater
Found the ideal dimensions for the theater to maintain the intensity of sound.
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