
#1
May 31st, 2017,
00:35
Quote:
The speed of sound $\displaystyle \begin{align*} c \end{align*}$ in an ideal fluid is related to the temperature $\displaystyle \begin{align*} T \end{align*}$ (measured in $\displaystyle \begin{align*} ^{\circ} K \end{align*}$) by
$\displaystyle \begin{align*} c = \sqrt{\gamma \, R \, T} \end{align*}$
where $\displaystyle \begin{align*} \gamma \end{align*}$ and $\displaystyle \begin{align*} R \end{align*}$ are constants.
Suppose that $\displaystyle \begin{align*} T \end{align*}$ increases by 10% from some base value. Use calculus to determine the approximate percentage change in $\displaystyle \begin{align*} c \end{align*}$, according to the ideal fluid model.
If we remember that the derivative is defined by $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{\Delta\,x \to 0} \frac{y\left( x + \Delta\,x \right)  y\left( x \right)}{\Delta \, x} \end{align*}$ then that means that as long as $\displaystyle \begin{align*} \Delta \,x \end{align*}$ is small, then $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} \approx \frac{\Delta \,y}{\Delta\,x} \end{align*}$.
Thus we can say $\displaystyle \begin{align*} \frac{\mathrm{d}c}{\mathrm{d}T} \approx \frac{\Delta\,c}{\Delta\,T} \end{align*}$ and so $\displaystyle \begin{align*} \Delta\,c \approx \frac{\mathrm{d}c}{\mathrm{d}T}\,\Delta\,T \end{align*}$.
Now from $\displaystyle \begin{align*} c = \sqrt{\gamma\,R\,T} \end{align*}$ we have $\displaystyle \begin{align*} \frac{\mathrm{d}c}{\mathrm{d}T} = \frac{\sqrt{\gamma\,R}}{2\,\sqrt{T}} \end{align*}$, and so with a 10\% increase in T, that means $\displaystyle \begin{align*} \Delta\,T = \frac{T}{10} \end{align*}$, thus
$\displaystyle \begin{align*} \Delta\,c &\approx \frac{\sqrt{\gamma\,R}}{2\,\sqrt{T}}\cdot \frac{T}{10} \\ &= \frac{\sqrt{\gamma\,R\,T}}{20} \\ &= \frac{c}{20} \end{align*}$
So that means that the change in c is $\displaystyle \begin{align*} \frac{1}{20} \end{align*}$ of the original c, so a 5% increase.

May 31st, 2017 00:35
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