Solve Proportionality Problems with Ease: Get Proportionality Help Now!

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Homework Help Overview

The discussion revolves around a proportionality problem related to fluid pressure and pipe diameter. The original poster seeks assistance in calculating the multiplicative and percentage changes in pressure when the diameter of a pipe decreases by 20%, based on the inverse relationship between pressure and the sixth power of the diameter.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster presents their calculations and expresses confusion regarding the percentage change result. Some participants suggest rechecking specific steps in the calculations, particularly the interpretation of the diameter decrease.

Discussion Status

Participants are actively engaging with the calculations, with some providing guidance on re-evaluating steps. There is a recognition of a potential mistake in the original calculations, and a revised multiplicative change and percentage change have been proposed, though no consensus has been reached on the correctness of these values.

Contextual Notes

The problem involves a specific relationship between pressure and diameter, and assumptions about the decrease in diameter are being discussed. The original poster's calculations are based on the assumption that a 20% decrease in diameter corresponds to a specific mathematical representation.

DB
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proportionality help needed please

stuck on a proportionality problem...
1) the pressure exerted by a fluid in a pipe varies inversely as the sixth power of the diameter. what is the multiplicative change (delta X_m) and percentage change (delta X%) in the pressure is the diameter of the pipe decreases by 20%?
[tex]p\propto \frac{1}{d^6}[/tex]

[tex]\frac{p_2}{p_1}=\frac{1}{0.2^6}[/tex]

[tex]\frac{p_2}{p_1}=15625[/tex]

therefore

[tex]\frac{15625}{1}\equiv \frac{1}{0.2^6}[/tex]

so

[tex]\Delta X_m=15625[/tex]

and

[tex]\Delta X_m -1 = \Delta X_{relative}[/tex]

and

[tex]\Delta X_{rel.} * 100 = Delta X_%[/tex]

so

[tex]15625-1=15624*100=\Delta X_%[/tex]

[tex]\Delta X_% = 1.5624 * 10^6[/tex]?
i must of done something wrong...any help appreciated :smile:
 
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if u can't see my last three step i got the percentage change as 1.5624*10^6%
 
DB said:
[tex]p\propto \frac{1}{d^6}[/tex]

[tex]\frac{p_2}{p_1}=\frac{1}{0.2^6}[/tex]
Recheck this step. When the diameter decreases by 20%: [itex]D_2 = 0.8 D_1[/itex]
 
thanks doc
shame on me, stupid mistake, but this seems right, the multiplicative change is 3.18 and the percentage change is 218%. that right?
 
Sounds good to me.
 
thanks doc
 

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