traffic flow

**galactus** · Dec 3rd 2008, 12:02 PM

Hello all:

I have been studying a little traffic flow because of having been in road construction I never dreamed it was so involved, mathematically.

Here's a problem:

Suppose we are interested in the change in the number of cars N(t) between two observers, one fixed at x=a and the other moving in some prescribed manner, x=b(t).

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

The derivative of an integral with a variable limit is

$\frac{dN}{dt}=\frac{db}{dt}{\rho}(b,t)+\int_{a}^{b (t)}\frac{{\partial}{\rho}}{{\partial}t}dx$

Show this result using either considering:

$\lim_{h\to 0}\frac{N(t+h)-N(t)}{h}$ or by using the chain rule for derivatives.

Also, Using $\frac{{\partial}{\rho}}{{\partial}t}=\frac{{-\partial}}{{\partial}x}({\rho}u)$ , show that

$\frac{dN}{dt}={-\rho}(b,t)\left[u(b,t)-\frac{db}{dt}\right]+{\rho}(a,t)u(a,t)$

I must admit, I am rather stymied. I ahve been going thorugh a rough time and I am having a difficult time concentrating.

**shawsend** · Dec 3rd 2008, 02:38 PM

The first part is just Leibniz's Rule right? For the second, if:

$\frac{dN}{dt}=\frac{db}{dt}{\rho}(b,t)+\int_{a}^{b (t)}\frac{{\partial}{\rho}}{{\partial}t}dx$

and:

$\frac{{\partial}{\rho}}{{\partial}t}=\frac{{-\partial}}{{\partial}x}({\rho}u)$

then:

$\frac{dN}{dt}=\frac{db}{dt}{\rho}(b,t)-\int_{a}^{b(t)}\frac{\partial}{\partial x}\left(\rho u\right)dx$

$\frac{dN}{dt}=\frac{db}{dt}\rho(b,t)-\rho u\Biggr|_a^{b(t)}$

$=\frac{db}{dt}\rho(b,t)-\rho(b,t)u(b,t)+\rho(a,t)u(a,t)$

**galactus** · Dec 3rd 2008, 05:25 PM

Thank you. It is rather obvious now.I haven't been myself lately

.

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