Polchinski's toy model of renormalization group flow: significance of main steps In "Renormalization and Effective Lagrangians" (Nucl. Phys. B, 231 p269, 1984; preprint), Polchinski begins section 2 with a toy model to demonstrate the renormalization group with a relevant and irrelevant coupling.
He starts with a scalar theory with a 4-point and 6-point interaction. The dimensionless couplings for these interactions are $\lambda_4$ and $\lambda_6 = \Lambda^2 g_6$ where $\Lambda$ is the scale of the theory.
The $\beta$-function equations are:

If $\bar\lambda_4, \bar\lambda_6)$ are a solution to these equations, then we can examine small deviations from this trajectory: $\varepsilon_i \equiv \lambda_i - \bar\lambda_i$.
The equations to $\mathcal O(\varepsilon)$ are:

Now here's where I'm a bit confused: Polchinski explains the $\lambda_6$ will term affects the relative flow of the $\lambda_4$ between nearby trajectories. For example, the following figure: two nearby points A1 and A2 have the same $\lambda_4$ but flow to points B1 and B2 which now have quite different $\lambda_4$s. 

Polchinski then says that even though $\varepsilon_4$ is big, there is a point B2' on trajectory 2 that is close to A2. In order to illuminate this, he defines this particular combination of couplings:

Question 1: I don't quite see the motivation for writing this. However, Polchinski says that $(\xi_4,\xi_6)$ is a vector that points from B1 to B2', which I think I can see if I think of
$$ 
\frac{d\bar\lambda_6/d\Lambda}{d\bar\lambda_4/d\Lambda} 
=
\frac{d\bar\lambda_6}{d\bar\lambda_4} .
$$
Is that reasonable? Then I'm thinking of $\bar\lambda_6$ as a function of $bar\lambda_4$: $\bar\lambda_6 = \bar\lambda_6(\bar\lambda_4)$. 
In this way, to quote Polchinski above eq (8), we are "subtracting off a multiple of the tangent vector to the trajectory."
Question 2: I'm having a hard time showing that the flow equation for $\xi_6$ is:

Is it obvious how this follows from the earlier flow equations?
Thanks!
 A: Ah, I think I've sorted it out. I share the key steps here. 
For simplicity, let me write $t = \ln \Lambda$ so that $d/dt = \Lambda (d/d\Lambda)$. I further use the shorthand where $\dot\lambda = d\lambda/dt$. 
For question 1, I believe an explanation is to annotate Polchinski's figure as follows:

Here the green line is the tangent line of the lower trajectory. The expression for the separation between the trajectories, $\xi_6$ is approximated by looking at the vertical separation between B2 and A2, $\varepsilon_6 =\lambda_6(t+\delta t)-\bar\lambda_6(t+\delta t)$, minus $\Delta \varepsilon_6$, which is the "rise" from the tangent line approximation. 
For question 2, one can arrive at this result from simply calculating $\dot\xi_6$. It is useful to note that $\beta_i$ depends on $t$ through $\lambda_4(t)$ and $\lambda_6(t)$. This means, for example, 
$$
\dot \beta_4 = 
+ \frac{\partial \beta_4}{\partial_4}\dot\lambda_4
+ \frac{\partial \beta_4}{\partial_6}\dot\lambda_6
$$
Thus the following corollaries may be useful:
(1) $$\frac{\ddot \lambda_4}{\dot\lambda_4} = \frac{d\ln \beta_4}{dt}$$
(2) $$\ddot \lambda_6
=
2\dot\lambda_6  + \frac{\partial \beta_6}{\partial_4}\dot\lambda_4 + \frac{\partial \beta_6}{\partial_6}\dot\lambda_6
$$
(3) $$ \frac{\partial \beta_4}{\partial_6}
=
\frac{1}{\dot\lambda_6}
\left(
\dot\beta_4 
- \frac{\partial \beta_4}{\partial_\lambda 4} \dot\lambda_4
\right)
$$
Using these relations, I think it is simply a matter of applying the chain rule and grouping terms together.
