Proving P1>Po: A Mathematical Analysis

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SUMMARY

The discussion centers on proving the inequality P1 > Po using the equations dp = -rho(g)dy and the integral relationships between pressures and heights. The user derives the equation P1 - Po = (rho)(g)(del(y)), concluding that since rho(g) is a positive constant, it follows that P1 is indeed greater than Po. The mathematical steps taken confirm the validity of the inequality, establishing a clear relationship between pressure differences and height changes in a fluid context.

PREREQUISITES
  • Understanding of fluid mechanics principles, particularly hydrostatic pressure.
  • Familiarity with calculus, specifically integration and differential equations.
  • Knowledge of the variables involved: density (rho), gravitational acceleration (g), and height (y).
  • Ability to interpret mathematical proofs and inequalities.
NEXT STEPS
  • Study hydrostatic pressure equations in fluid mechanics.
  • Learn about the implications of pressure differences in fluid dynamics.
  • Explore advanced integration techniques relevant to physics problems.
  • Review mathematical proofs involving inequalities in calculus.
USEFUL FOR

Students of physics and engineering, particularly those focusing on fluid mechanics, as well as educators seeking to understand mathematical proofs related to pressure differentials.

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Homework Statement


Prove that P1>Po

Homework Equations


dp=-rho(g)dy
integral(Po,P1)dp=-integral(Y2,Y1) rho(g)dy

The Attempt at a Solution


dp=-rho(g)dy
intgrl(Po,P1) dp=-integral(Y2,Y1) rho(g)dy
Po-P1=-(rho(g))(Y2-Y1)
Po-P1=(rho(g))(Y1-Y2)
Y2-Y1-->Y1-Y2<0 defined as -del(y)
Po-P1=-(rho)(g)(del(y))
P1-Po=+(rho)(g)(del(y))

not sure where to proceed or if this proves that P1>Po. Please help.

Oh and the attached file has a diagram on it.
 

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Last edited:
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Since rho(g) is a positive constant, the equation P1-Po=+(rho)(g)(del(y)) implies that P1 is bigger than Po. Thus, P1>Po.
 

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