Arc-length integral with curve giving extremal value

[Leb]

Homework Statement

View attachment B3.bmp
For whoever does not want to read the attached problem:

Firstly, I need to express the arc-length from given x=r\cos\theta, y=r\sin\theta z = f(r), \text{ where } f(r) \text{ is an infinitely differentiable function and } r=r(\theta) \text{ i.e. parameter is } \theta I begin with expressing it as ds = \sqrt{ r^{2} + r'^{2} + f'^{2} } is that correct ?

I then need to show that the solutions of those curves are given by:
r ' ^{2} = r^{2}\frac{\left( \frac{r^{2}}{C} - 1\right)}{1 + f ' ^{2)}}

Homework Equations

Lagrangian is independent of theta in this case, so I think I can use
\frac{d}{d\theta}\left( L - r' \frac{\partial{L}}{\partial{r'}} \right) = 0

The Attempt at a Solution

Using the above, the term in parenthesis is equal to a constant. Differentiating L w.r.t. r' and rearranging I get r'^{2} = \frac{r^{4}+2f'^{2}r^{2} + f'^{4}-Br^{2} - Bf'{2}}{B} where B is a constant. I thought that this uses the good old multiply by "1" trick, but when I tried it it got no cancellations.

I can see the r^{2}\left( \frac{r^{2}}{C} - 1\right)<br /> term in there, but do not know how to get the rest...

 

[voko]

For the line element ds the following is true $ds^2 = dx^2 + dy^2 + dz^2$. Since your curve is on a surface parametrized by $r$ and $\theta$, and since $r$ itself is a function of $\theta$, you should end up with $ds = g(\theta)d\theta$.

 

[Leb]

Thanks for the reply !
Do you mean I should express \frac{df}{d\theta} \text{ as } \frac{df}{dr}\frac{dr}{d\theta} ?

 

[voko]

What I mean is that the expression for ds you got is incorrect. You can derive the correct equation from the expression for ds in the Cartesian coordinates by using the chain rule.

 

[Leb]

Hmm, that's strange, I followed this http://tutorial.math.lamar.edu/Classes/CalcII/PolarArcLength.aspx

 

[voko]

<br /> ds^2 = dx^2 + dy^2 + dz^2<br /> = (d(r \cos \theta))^2 + (d(r \sin \theta))^2 + (d(f(r)))^2<br /> = (dr \cos \theta - r \sin \theta d\theta)^2<br /> + (dr \sin \theta + r \cos \theta d\theta)^2<br /> + (f&#039;dr)^2<br /> \\<br /> = dr^2 + r^2 d\theta^2 + f&#039;^2 dr^2<br /> = (1 + f&#039;^2)dr^2 + r^2d\theta^2<br /> \\<br /> = (1 + f&#039;^2)(dr(\theta))^2 + r^2d\theta^2<br /> = (1 + f&#039;^2)(r&#039;d\theta)^2 + r^2d\theta^2<br /> = ((1 + f&#039;^2)r&#039;^2 + r^2)d\theta^2<br /> \\<br /> ds = \sqrt {(1 + f&#039;^2)r&#039;^2 + r^2}d\theta<br />

 

[Leb]

Thanks for taking the time to type that, but if we parameterize w.r.t. theta, why do we have d \theta ?

 

[voko]

Parametrized by $\theta$ means that the length is a function of $\theta$: $s = s(\theta)$, thus $ds = s'd\theta$.

 

[Leb]

So I take it that the previous link is rubbish ? Thanks again !

 

[voko]

No, it is not rubbish. If you look carefully, the formula I derived equals their formula if f = const. Your problem has three dimensions, not two as they have.

 

[Leb]

My problem is that we are parametrizing w.r.t. theta and usualy everyone is parametrizing w.r.t. some other parameter e.g. t http://phdmath.wordpress.com/2010/02/24/arc-length-2/
I mean, if we are differentiating w.r.t. \theta, would it not just disappear ?

 

[voko]

Why should it disappear? Take a very simple example: $y = x^2$. You can say that "y is parametrized by x". Then $dy = 2x dx$; x does not disappear. Even if $y = x$, you still have $dy = dx$.

 

[Leb]

Well, I am a numpty and always had particular trouble with notation of derivatives and using the "implicit" chain rule.

Anyway, thanks very much for bothering to reply !

 

[SteamKing]

Or you could try:

http://tutorial.math.lamar.edu/Classes/CalcIII/VectorArcLength.aspx

 

[Leb]

Hehe, reparametrization... I always viewed the parameters as an "arbitrary" choice when we are talking about simple curvy curves (not circles or ellipses).

Thanks for the link !