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Let $\rho = K* \frac{P}{T} $ where K is s real constant number. So $\rho$ is related to $\frac{P}{T}$. Someone said that we can conclude from this:

$\frac{\Delta\rho}{\rho} = \frac{\Delta P}{P} - \frac{\Delta T}{T}$ Equation (I)

Delta is the common delta we use in physics or math. $f(x2) - f(x1) = \Delta f(x)$

My problem is how can we conclude Equation (I)? Is it true?

Irregardless
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titansarus
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  • In physics, first it takes logarithm and then derivate, and finally changing $d\rho$ with $\Delta\rho$. – Nosrati Aug 11 '17 at 19:08

4 Answers4

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Hint:   differentiate $\,\log \rho = \log K + \log P - \log T\,$.

dxiv
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You can't because it's generally not true.

However, there is a similar relation between the differentials: the trick here is "logarithmic differentiation": for positive $z$, you have

$$ \mathrm{d} \log(z) = \frac{\mathrm{d}z}{z} $$

and logarithms turn products into sums. Here, it compues

$$ \mathrm{d} \log(\rho) = \mathrm{d} \log(KP/T) = \mathrm{d}\log(K) + \mathrm{d}\log(P) - \mathrm{d}\log(T) $$

Since $\mathrm{d}K = 0$, this equation leads to

$$ \frac{\mathrm{d}\rho}{\rho} = \frac{\mathrm{d}P}{P} - \frac{\mathrm{d}T}{T} $$

If you're not integrating the differentials to obtain the difference, then the things you can say about the differences $\Delta \rho$ and such are mainly through differential approximation: given a small variation $v$, you have $\Delta z \approx \nabla_v z$.

If you apply such a $v$ to the above equation and then substitute in the approximation, you get

$$ \frac{\Delta \rho}{\rho} \approx \frac{\Delta P}{P} - \frac{\Delta T}{T} $$

Note the approximation sign here; unlike the equation between differentials, this equation is generally not exactly true.

1

We have $\rho T=kP$ and the rate of changes is $$(\rho+\Delta\rho)(T+\Delta T)=k(P+\Delta P)$$ $k$ is a constant. When varieties of $\Delta\rho$ and $\Delta T$ are small, then $\Delta\rho.\Delta T$ is very small and we can ignore it, so \begin{eqnarray} \rho T+\Delta\rho T+\Delta T\rho+\Delta\rho\Delta T &=& kP+k\Delta P\\ \Delta\rho T+\Delta T\rho+\Delta\rho\Delta T &=& k\Delta P\\ \dfrac{\Delta\rho T+\Delta T\rho}{\rho T} &=& \dfrac{k\Delta P}{kP}\\ \dfrac{\Delta\rho}{\rho}+\dfrac{\Delta T}{T} &=& \dfrac{\Delta P}{P} \end{eqnarray}

Nosrati
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We have (note that each of these is some function of $x$ or whatever the input is): $$\rho = K_0\times \frac{P}{T}$$

subtract $\rho (x+x_2)$ from both sides ($x_2$ is the lag in question).

$$\Delta \rho = K_0 \times \frac{P}{T} -\rho(x+x_2) = K_0 \times \frac{P}{T} - K_0 \frac{P (x+x_2)}{T(x+x_2)} $$

Now divide by $\rho$

$$\frac{\Delta \rho}{\rho} = \frac{K_0 \times\frac{P}{T} - K_0 \frac{P (x+x_2)}{T(x+x_2)}}{\rho} =1-\frac{P(x+x_2)T}{T(x+x_2)P}$$

So NOT true generally.

Tony
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