Solving a Question on L'Hopital's Rule

[Kurdt]

Having trouble with a question on L'Hopital's rule. I have never come across it must have misseda lecture. From what I understand the rule approximates values at a limit. Here's what I have anyway.

I've derived a velocity gradient for a spherically symmetric, isothermal stellar wind as follows.

$$\frac{dv}{dr}=\frac{2a^2}{r}(1-\frac{r_c}{r})\frac{v}{(v^2-a^2)}$$

where a is the sound speed and $$v=a$$ at the critical point $$r=r_c$$

So apparently when applying L'Hopital's rule I can approximate the velocity gradient as

$$(\frac{dv}{dr})_r_c=+or-\frac{a}{r_c}$$

When I apply the limits all I can come up with is the following and I do not understand what Ihave to do to get the quoted answer.

$$(\frac{dv}{dr})_r_c=\frac{2a^2}{r_c}$$

Thanks in advance.

 

[matt grime]

Firsrtly limit of what as what tends to where?

L'hopital states that if as x->k f(x) and g(x) both tend to zero that lim as x tends to k of f/g is the lim as x tends to k of f'/g'

how does that relate to your question?

 

[Kurdt]

Basically I'm just trying to estimate the velocity gradient at the critical point (i.e. when v->a and r->rc). The use of l'hopital's rule was suggested in the question and the answer that is produced from application of the rule was the second equation I wrote down there.

I just have trouble applying the rule as I have never heard of it until the question was set.

 

[Falcon]

Yeah, l'hopitals rule is used for evaluating limits which turn out to be 0/0 or inf/inf when it can't be done by factoring and cancelling or by using a taylor polynomial.

I'm not sure I see the application.. suppose I should leave it to someone with a bit more knowledge.

 

[HallsofIvy]

At first I was a bit puzzled by the statement that L'Hopital's rule "approximates" anything- then I figured out what was going on. As Matt Grime said, L'Hopital's rule says that if g-> 0 and h-> 0 then h/g has the same limit as h'/g' (' meaning derivative). That's not an "approximation", that's exactly true. HOWEVER, what is going on here is that you are using the limit to approximate the VALUE of the function close to the critical point.
Your equation is:
$$\frac{dv}{dr}=\frac{2a^2}{r}(1-\frac{r_c}{r})\frac{v}{(v^2-a^2)}$$
We can rewrite the right hand side as
$$\frac{2a^2v}{r^2}\frac{r-r_c}{v^2-a^2}$$
and now as r-> r_c, v->a so that last fraction is "0/0".

Using L'Hopital on just that last fraction, its limit as r-> r_c would b
$$\frac{1}{2vv'}$$
(Since we are taking the limit as r-> r_c, we take the derivative with respect to r- hence that "v'" in the denominator.)

Using that limit as an approximation for the actual value of the fraction near r_c, the differential equation becomes
$$\frac{dv}{dr}= \frac{2a^2v}{r^2}\frac{1}{2v\frac{dv}{dr}}$$
or
$$(\frac{dv}{dr})^2= \frac{a^2}{r^2}$$
resulting in
$$\frac{dv}{dv}= +/- \frac{a}{r}$$

Cute!

 

[matt grime]

Thanks for finishing that off - I was really concerned because I didn't understand where the o had come from in $$+or-\frac{a}{r}$$ and my idea of how to work it out was not going to produce something that looked correct with an r term and a 1/r term.

It just didn't click that it was the word or.

 

[Kurdt]

Thanks very much HallsofIvy and matt. I thought I'd gotten past the difficult bit when I had to derive the velocity gradient but my understanding of the rule was somewhat lacking. It is all clear now though and I apologise for any ambiguity in my explanation.

 

[matt grime]

It's just a side effect of typesetting in tex that it treats all letters as variables and guesses how to space things accordingly. It was just compounded with there being an r in the original equation.