For part(a) only

Let S be the surface area and dx a small distance along the pipe, for a fixed point (x,t)

The density of air around this point can be regarded as constant and by the density formula

The acceleration of the mass is

*a = u*_{t}

By the pressure formula

*F = S(p(x, t)- p(x + dx, t))*(7)

By Newtonâ€™s second law, F = ma, plug in (6) and (7), we have

* S(p(x, t)- p(x + dx, t))= ÏSdx u*_{t}

Divide both sides by Sdx

* (p(x, t)- p(x + dx, t)/dx=Ïu*_{t}

which is

Consider a change in mass for a fixed volume in the pipe, it equals to the difference of the mass that flowed in and the mass that flowed out

*dm = dt Ï dx S (u(x, t) â€“ u(x + dx, t))*

Also by the density formula

*dm = (Ï(x, t + dt)- Ï(x, t)) S dx *

From the above two equations

*dt Ï dx S (u(x, t) â€“ u(x + dx, t)) = (Ï(x, t + dt)- Ï(x, t)) S dx*

Divide both sides by dt dx S:

*Ï (u(x, t) â€“ u(x + dx, t)) / dx = (Ï(x, t + dt)- Ï(x, t))/ dt*

Then

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