Derive lowest order (linear) approximation

In summary: I mean is log(K) and log(PA) are both multiplied by...1/a. Yes, sorry so what I mean is log(K) and log(PA) are both multiplied by...1/a.
  • #1
Jen2114
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Homework Statement


For a single mechanical unit lung, assume that the relationship among pressure, volume, and number of moles of ideal gas in the ling is given by PA((VL)/(NL)a = K, where a = 1 and K is a constant. Derive the lowest-order (linear approximation to the relationship among changes in pressure, changes in volume, and changes in moles of gas within the lung.

Homework Equations

The Attempt at a Solution


I know that to solve this you need to take the total derivative. However, I am having trouble arriving at the correct answer because I'm not really sure what to do with the derivative. So my calculations are the ones below:
PA ((VL)/(NL)a = K --> dPA/PA(PA(VL/NL)a ) = dPA/PA(K) --> dPA/PA(VL/NL)a-1 = 1
 
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  • #2
Jen2114 said:

Homework Statement


For a single mechanical unit lung, assume that the relationship among pressure, volume, and number of moles of ideal gas in the ling is given by PA((VL)/(NL)a = K, where a = 1 and K is a constant. Derive the lowest-order (linear approximation to the relationship among changes in pressure, changes in volume, and changes in moles of gas within the lung.

Homework Equations

The Attempt at a Solution


I know that to solve this you need to take the total derivative. However, I am having trouble arriving at the correct answer because I'm not really sure what to do with the derivative. So my calculations are the ones below:
PA ((VL)/(NL)a = K --> dPA/PA(PA(VL/NL)a ) = dPA/PA(K) --> dPA/PA(VL/NL)a-1 = 1
I think you might find this problem easier if you took the log of both sides of the equation first before taking the derivative.
 
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  • #3
tnich said:
I think you might find this problem easier if you took the log of both sides of the equation first before taking the derivative.
Thanks, so doing that I got the following:
log(PA(VL/NL))a = log(K) --> a*log(PAVL/NL) = log (K) --> a[log(PA) + log(VL) - log(NL)] = log(K) using log properties for multiplication and division. Is this good so far?
 
  • #4
Jen2114 said:
Thanks, so doing that I got the following:
log(PA(VL/NL))a = log(K) --> a*log(PAVL/NL) = log (K) --> a[log(PA) + log(VL) - log(NL)] = log(K) using log properties for multiplication and division. Is this good so far?
That depends on what you mean by PA((VL)/(NL)a = K in the problem statement. Is that ##(\frac{P_AV_L}{N_L})^a = K##, ##P_A(\frac{V_L}{N_L})^a = K##, or something else?
 
  • #5
tnich said:
That depends on what you mean by PA((VL)/(NL)a = K in the problem statement. Is that ##(\frac{P_AV_L}{N_L})^a = K##, ##P_A(\frac{V_L}{N_L})^a = K##, or something else?
It should be the second one where only VL and NL are raised to exponent variable a. So, in that case is
the result a[log(PA) + log(VL) - log(NL) = log (K) correct?
 
  • #6
Jen2114 said:
It should be the second one where only VL and NL are raised to exponent variable a. So, in that case is
the result a[log(PA) + log(VL) - log(NL) = log (K) correct?
No. Try doing it step-by-step.
 
  • #7
tnich said:
No. Try doing it step-by-step.
So these are the steps I took
1. Divide both sides by PA --> PA(VL/NL)a = K -->
(VL/NL)a = K/PA
2. Take log of both sides so a in exponent can be brought down --> log ( (VL/NL)a )= log(K/PA) --> a*log(VL/NL) = log((K/PA))
3. Using log properties rewrite VL/NL and K/PA as differences to get -->
log(VL) - log(NL) = 1/a *(log(K) - log(PA) ... not sure what I am doing wrong so far
 
  • #8
Are you familiar with the product rule and the quotient rule for differentiation?
 
  • #9
Jen2114 said:
So these are the steps I took
1. Divide both sides by PA --> PA(VL/NL)a = K -->
(VL/NL)a = K/PA
2. Take log of both sides so a in exponent can be brought down --> log ( (VL/NL)a )= log(K/PA) --> a*log(VL/NL) = log((K/PA))
3. Using log properties rewrite VL/NL and K/PA as differences to get -->
log(VL) - log(NL) = 1/a *(log(K) - log(PA) ... not sure what I am doing wrong so far
As you did in the problem statement, you have left out the closing parenthesis in step 3. That notation error has apparently prompted you to misinterpret which terms are multiplied by 1/a.
 
  • #10
tnich said:
As you did in the problem statement, you have left out the closing parenthesis in step 3. That notation error has apparently prompted you to misinterpret which terms are multiplied by 1/a.
Yes, sorry so what I mean is log(K) and log(PA) are both multiplied by 1/a
 
  • #11
Chestermiller said:
Are you familiar with the product rule and the quotient rule for differentiation?
Yes, but I am not sure how it works for a total derivative of an expression.
 
  • #12
Jen2114 said:
Yes, sorry so what I mean is log(K) and log(PA) are both multiplied by 1/a
Right. Now differentiate your equation.
 
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  • #13
tnich said:
Right. Now differentiate your equation.
Ok so then -->1. d/VL(log(VL)) - d/NL(log(NL)) = 1/a*(d/K(log(K))) - 1/a*(d/PA(log(PA)))
2. dVL/VL - dNL/NL = -1/a*dPL/PL and since K is a constant the derivative is 0 so it this correct? Thanks.
 
  • #14
Jen2114 said:
Yes, but I am not sure how it works for a total derivative of an expression.
$$d\left[P\left(\frac{V_L}{N_L}\right)^a\right]=\left(\frac{V_L}{N_L}\right)^adP+aP\left(\frac{V_L}{N_L}\right)^{a-1}d\left(\frac{V_L}{N_L}\right)$$
$$d\left(\frac{V_L}{N_L}\right)=\frac{N_LdV_L-V_LdN_L}{N_L^2}$$
 
  • #15
Chestermiller said:
$$d\left[P\left(\frac{V_L}{N_L}\right)^a\right]=\left(\frac{V_L}{N_L}\right)^adP+aP\left(\frac{V_L}{N_L}\right)^{a-1}d\left(\frac{V_L}{N_L}\right)$$
$$d\left(\frac{V_L}{N_L}\right)=\frac{N_LdV_L-V_LdN_L}{N_L^2}$$
Oh ok this makes sense to me according to the product and quotient rule but the final solution given to us in class should be
dPA/aPA = -dVL/VL + dNL/NL so maybe I just can't see it but I'm not sure if the result you showed me simplifies to this so I know what the solution should be but I'm not sure how to arrive to that final solution.
 
  • #16
Jen2114 said:
Oh ok this makes sense to me according to the product and quotient rule but the final solution given to us in class should be
dPA/aPA = -dVL/VL + dNL/NL so maybe I just can't see it but I'm not sure if the result you showed me simplifies to this so I know what the solution should be but I'm not sure how to arrive to that final solution.
It does reduct to that final solution. Just work on it a little. This is just algebra.
 
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  • #17
Jen2114 said:
Ok so then -->1. d/VL(log(VL)) - d/NL(log(NL)) = 1/a*(d/K(log(K))) - 1/a*(d/PA(log(PA)))
2. dVL/VL - dNL/NL = -1/a*dPL/PL and since K is a constant the derivative is 0 so it this correct? Thanks.
Yes. That looks right.
 
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  • #18
Chestermiller said:
It does reduct to that final solution. Just work on it a little. This is just algebra.
Ok, thanks! will do.
 
  • #19
tnich said:
Yes. That looks right.
Thank you! , this really helped.
 
  • #20
Jen2114 said:
Thank you! , this really helped.
You are welcome. You are not quite done, yet. You still need to state your equation in terms of ΔPA, ΔNL, and ΔVL. I am assuming your teacher has demonstrated something similar and you not required to give a formal proof. Otherwise, you would need to write out a Taylor series and do some more work.
 

FAQ: Derive lowest order (linear) approximation

What is the lowest order (linear) approximation?

The lowest order (linear) approximation is a mathematical technique used to approximate a complex function by simplifying it into a linear equation. This is done by taking the first order Taylor series of the function and keeping only the terms with the highest degree.

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The lowest order (linear) approximation is important in scientific research because it allows for the simplification of complex functions, making them easier to analyze and understand. It can also provide a good starting point for further analysis and more accurate approximations.

What are the limitations of the lowest order (linear) approximation?

The limitations of the lowest order (linear) approximation include its inability to accurately represent functions with high variability or non-linear behavior. It also becomes less accurate as the distance from the point of approximation increases.

How is the lowest order (linear) approximation different from higher order approximations?

The lowest order (linear) approximation only takes into account the first order Taylor series of a function, while higher order approximations include more terms and therefore provide a more accurate representation of the function. However, higher order approximations can be more complex and difficult to work with.

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The lowest order (linear) approximation has many practical applications, including in physics, engineering, and economics. It is commonly used to solve differential equations, approximate non-linear systems, and model real-world phenomena with high accuracy. It is also used in machine learning and data analysis to make predictions and draw insights from large datasets.

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