Explanation for speed of an electrical impulse Our calculus book, Stewart,  has a problem where they claim that for a metal cable (inner radius $r$) encased in insulation (outer radius $R$), the speed of an electrical impulse is given by
$$v = - k \left(\frac{r}{R}\right)^2 \ln \left(\frac{r}{R}\right)$$
where $k$ is a positive constant.
My question
What I would like to know is the physical justification for their claim.
My thoughts
There claim is somewhat surprising, since for sufficiently high insulation R, with r fixed, the speed of the impulse decreases (by L'Hospital) with more insulation.

EDIT: I received this email after contacting Brooks/Cole, the publisher of the textbook. The response didn't really help unfortunately.

Hi Professor ...,
I just heard back from the author regarding your query: “I can understand why Professor
... thinks this equation is counterintuitive, but it is in fact correct. I have been >trying to track down the source that I used in devising this problem, but unfortunately I >can’t seem to find it right now.” I will certainly let you know if he is able to track >down the source information. I’m sorry I can’t give you a more concrete answer at this >time. Best, ...
[JIRA] (KYTS-1199) Content Feedback from Instructor for ISBN: 0495014281 Essential >Calculus: Early Transcendentals 1st edition.

 A: I think i came to the origins of this equation.  In all likelihood, this equation describes not a speed of an electrical impulse but a direct current power transmitted via a superconducting coaxial cable.  
A proof:
Consider a simple transmission  DC coaxial cable.  To eliminate the energy losses due to Joule heating in the cable, the inner(of radius $r$) and outer(of radius$R$) conductors are made from a superconductor. The inner conductor is insulated by a dielectric material. How much power can be transferred through the cable?
Let the maximum allowable magnetic field induction on the surface of the superconductor be $B_\text{max}$ and the maximum electric field in the insulating interlayer be $E_\text{max}$.  Let a current through the cable be $I$. 
Then the following holds:  
$$B_\text{max}=\frac{\mu_0I}{2\pi r}\Rightarrow I=\frac{2\pi}{\mu_0}rB_\text{max}$$    
Since we are dealing with superconductors they keep the potential on the surface(as well as the linear charge density λ) constant.  
That means the following holds:  
$$E_\text{max}=\frac{\lambda}{2\pi\epsilon_0r}$$   
The potential difference between inner and outer conductors:  
$$U=\frac{\lambda}{2\pi\epsilon_0}\int_{r}^{R}\frac{dl}{l}=\frac{\lambda}{2\pi\epsilon_0}\ln\frac{R}{r}=rE_\text{max}\ln\frac{R}{r} $$   
So, the power transmission:  
$$P=UI=\frac{2\pi}{\mu_0}E_\text{max}B_\text{max}r^2\ln\frac{R}{r}$$  
This is the same function as in the question, only with different constant $k$ 
To analyze the result let's introduce the ratio $x=\frac{R}{r}$:  
$$P=UI=\frac{2\pi}{\mu_0}E_\text{max}B_\text{max}R^2\left(\frac{\ln x}{x^2}\right)$$    
One can see that at $R=\text{const}$, $P$ as a function of $x$ has a maximum. This happens at $x=\sqrt{e}$   
So the maximum power transfer in the DC superconducting transmission cable:  
$$P_\text{max}=\frac{\pi}{\mu_0e}E_\text{max}B_\text{max}R^2$$
