Understanding Integrals and the Importance of Constants

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Homework Help Overview

The discussion revolves around an RC series circuit with a variable resistor, where the resistance is defined as R = a + bt. The charge on the capacitor is described by a differential equation involving capacitance, which is questioned for its constancy. The original poster seeks clarification on whether capacitance can be treated as a constant in this context.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of treating capacitance as a constant and discuss the validity of integrating the equation. Questions arise regarding the simplification of terms and the proper handling of variables in integrals.

Discussion Status

Some participants express agreement on the constancy of capacitance, while others engage in a deeper examination of integration techniques and variable handling. There is an acknowledgment of misunderstandings regarding integration methods, and guidance is offered to clarify these points.

Contextual Notes

Participants note the lack of explicit instructions regarding the treatment of capacitance and the assumptions that may affect the evaluation of the integral. The original poster reflects on gaps in their understanding of integration, indicating a need for further study.

keyermoond
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Homework Statement


Suppose an RC series circuit has a variable resistor. If the resistance at time t is given by by R = a + bt, where a and b are known positive constants then the charge q(t) on the capacitor satisfies

(a+bt) q' + (1/C)q = V

where V is some constant. Also q(0) = q_0
Find q(t) as an explicit function of t.

Homework Equations



Now I have obtained the answer, however my main question is: am I allowed to treat C (capacitance) as a constant in this equation. It doesn't specify in the question, but to my knowledge (unless I am wrong of course) capacitance is a constant value and only depends on material and physical parameters of the capacitor itself (how it is build).

If I can't treat C as a constant then I believe there is no way to evaluate the integral in integrating factor and I'd have to leave it as it is.

Process is simple from there, I rewrite the equation in standart form, find the integrating factor and obtain a formula for q(t), evaluate an integration constant with q(0) = q_0 and obtain the overall solution q(t).

The answer looks quite frightening btw

3. My solution:
attached pdf file
 

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The capacitance must be constant during the all process, which just depends on its material and geometric shape, as what you say above,
 
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Is there some reason you're not simplifying ##\frac{(a+bt)^k}{a+bt}## to ##(a+bt)^{k-1}## and integrating the righthand side? Also, the ##t## in the integral is a dummy variable, so it's not correct to set it to 0.
 
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That's what my prof does. By setting it from 0 to t, we ensure evaluated integral is equal to the true value with t as a variable, how would setting it to something arbitary as t_o (I'm assuming that's what you mean)be any better? My understanding of it is that we need to choose some "convenient" interval, what's wrong with 0 to t?

And thank you for pointing out I can actually simplify it further, I completely missed that.
 
keyermoond said:
That's what my prof does. By setting it from 0 to t, we ensure evaluated integral is equal to the true value with t as a variable, how would setting it to something arbitary as t_o (I'm assuming that's what you mean)be any better? My understanding of it is that we need to choose some "convenient" interval, what's wrong with 0 to t?
What I'm saying is you can't do something like this:
$$\int (a+bt)^2\,dt = \int a\,dt$$ by claiming you're setting ##t=0##. The ##t## inside the integral isn't the same ##t## that appears elsewhere.
 
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vela said:
What I'm saying is you can't do something like this:
$$\int (a+bt)^2\,dt = \int a\,dt$$ by claiming you're setting ##t=0##. The ##t## inside the integral isn't the same ##t## that appears elsewhere.
I see what you mean, my mistake, thank you for pointing it out. I see holes in my knowledge about understanding of basic integration now, will have to fill them in :)
 

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