# Math Help - Theory of proportions

1. ## Theory of proportions

My textbook says:

We write the equation in the form:

$k_{(n)}=\frac{e+2f \lambda + g \lambda^2}{E+2F \lambda + G \lambda^2}$ where $\lambda = \frac{dv}{du}$.

From the theory of proportions we can write:

$k_{(n)}=\frac{(e+f \lambda) + \lambda(f+g \lambda)}{(E+F \lambda) + \lambda (F+G \lambda)}=\frac{f+g\lambda}{F+G\lambda}=\frac{e+f\ lambda}{E+F\lambda}$

Can anyone explain this use of the "theory of proportions" to me? Intuitively it makes some sense as large values of lambda give one solution and small values of lambda give the other solution. But then why not go the the extreme of saying:

$k_{(n)}=\frac{g}{G}=\frac{e}{E}$

You may recognise K as "Normal curvature" or the ratio of the first and second fundamental forms of a surface.

2. ## Theory of Proportions

Hello Dave,

$\frac{\alpha a + \beta b}{\alpha A + \beta B} = \frac{a + \beta b} { A + \beta B} = \frac{ \alpha a + b}{\alpha A + B} =$ provided $\frac{a}{A} = \frac{b} {B}$

Proof:

$\frac{a}{A} = \frac{b}{B}$ means that $\exists m$ such that $a= m b$ and $A = m B.$

Substituting for $a$ and $A$,
$\frac{\alpha mb + \beta b}{\alpha mB + \beta B} = \frac{mb + \beta b} { mB + \beta B} = \frac{ \alpha mb + b}{\alpha mB + B} = \frac{ ( \alpha m + \beta ) b}{( \alpha m + \beta ) B} = \frac{(m + \beta ) b} { (m + \beta )B} = \frac{ ( \alpha m + 1)b}{( \alpha m + 1) B}$
$=\frac b B$.

Similarly substituting for $a= \frac b m$ and $B= \frac A m$ we can see that

$\frac{\alpha a + \beta b}{\alpha A + \beta B} = \frac{a + \beta b} { A + \beta B} = \frac{ \alpha a + b}{\alpha A + B} = \frac a A$.

Setting $\alpha = 1$, $\beta = \lambda$, $a= e + f \lambda$, $b=f + g \lambda$,
$A= E + F \lambda$ and $B = F + G \lambda$ shows the desired result.
It appears you are on page 135 of the book I am reading! I was myself puzzled by 1.5.36 when I first saw it!

3. Thanks for your help, and welcome to the forum!

That is a great book, and if you take a look at some of my other posts you will find that most of my questions have come from there.