Where does the 4 come from in the derivative of an integral?

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Homework Help Overview

The discussion revolves around taking the derivative of an integral involving a function V and constants u and P. The original poster is trying to understand the derivation of a specific term in the solution, particularly the factor of 4 in the expression for V.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive V from the given equation but questions the appearance of the factor 4 in the correct solution. Some participants inquire about the nature of the variables involved, specifically whether u is a function of r and the status of u and P as constants. Others suggest rewriting the equation for clarity and propose integration steps.

Discussion Status

The discussion is ongoing, with participants providing clarifications and exploring different interpretations of the equation. There is a focus on understanding the mathematical expressions and the implications of constants versus variables, but no consensus has been reached regarding the derivation of the factor 4.

Contextual Notes

Participants are navigating the complexities of the problem, including the notation used for derivatives and the assumptions about the variables involved. The original poster has acknowledged a correction in their initial equation, which may influence the discussion.

juice34

Homework Statement


Can someone help me take the derivative of the integral
u(1/r)((d/dr)[(r)(dV/dr)])=P


Homework Equations





The Attempt at a Solution


my attempt yields V=(Pr^2)/(2u)+C(1), which is not right. The actual answer is V=(Pr^2)/4u+C(1)ln(r)+C(2). I am having trouble finding out where the 4 comes from could someone please explain to me what is going on. Thank YOU
 
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Is u a function of r? If not why write u(1/r) instead of just (u/r)? Are u and P constants? I think I know the answers to those questions but you should pose your question more carefully. Rewriting your equation using ' for d/dr:

(rV')' = Pr/u

Integrate:

rV' = Pr2/(2u) + C

Divide both sides by r and integrate again.
 
Last edited:
LCKurtz u and P are constant. And I am not sure i follow this part, (rV')' = r/u. Is it equal to d/dx(r(dV/dr))
 
Last edited by a moderator:
I accidently left off the P, which I just edited to correct.

(rV')' is d/dr (r dV/dr)

No x in there. Its much neater to write with primes.
 
LCKurtz i cannot thank you enough, you are a excellent help!
 

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