a) Directly implied by definition of $\rho(x,t)$ is $N(t,a,b)=\int_{a}^{b}\rho(t,x) dx$.

b) by definition of $q$ and conservation of cars we have:

$$ \frac{\partial N}{\partial t}(t,a,b)=\lim_{h \rightarrow 0} \frac{N(t+h,a,b)-N(t,a,b)}{h} \\

=\lim_{h \rightarrow 0} \frac{h(q(t,a)-q(t,b))}{h} \\

=q(t,a)-q(t,b) $$

c) Differentiating integral form of $N(t,a,b)$ with respect to $t$

$$

\frac{\partial N}{\partial t}=\int_{a}^{b}\rho_t(t,x) dx$$

making it equal to result of part (b) we get the integral form of "conservation of cars":

$$

\int_{a}^{b}\rho_t(t,x) dx=q(t,a)-q(t,b) $$

d) RHS of above equation can be expressed as $\int_{a}^{b}-q_x(t,x) dx$. Therefore

\begin{equation} \int_{a}^{b}\rho_t(t,x) dx=\int_{a}^{b}-q_x(t,x) dx

\end{equation}

Since $a$ and $b$ are arbitrary, $(1)$ implies $\rho_t=-q_x$. The PDE $\rho_t+q_x=0$ is conservation of cars equation.

e) Letting $q=c\rho$, we will get a first order linear PDE with constant coeffiecient:

$$\rho_t+c\rho_x=0 \\

\rightarrow \rho(t,x)=\phi(x-ct)

$$

The last equation means density of cars is moving at some constant pace $c$ in time, i.e. all cars are driving at speed $c$.

Intuitively we know that driver's speed has negative correlation with traffic density. A more realistic choice for $c$ is to let it be a monotone decreasing function of $\rho$. In this case however, conservation of cars equation is not linear anymore. This is discussed in detail by Prof. Ivrii in

last year's forum.