A3 curve meets critical depth line normally

The A3 curve occurs on an adverse slope ($S_0 < 0$) in zone 3, where $y < y_c$ (supercritical). Write the bed slope as $-|S_0|$ in the dynamic equation:

$$\frac{dy}{dx} = \frac{S_0 - S_f}{1 - Fr^2} = \frac{-(|S_0| + S_f)}{1 - Fr^2}$$

In zone 3, $Fr > 1$, so the denominator is negative and $dy/dx > 0$: the depth increases downstream.

(i) Near the critical depth line ($y \to y_c$)

$Fr \to 1$, so the denominator $1 - Fr^2 \to 0$ while the numerator $-(|S_0| + S_f)$ stays finite and non-zero. Hence

$$\frac{dy}{dx} \to \infty$$

The tangent to the curve is vertical, so the curve meets the $y_c$ line normally (at a right angle). (In practice a hydraulic jump forms before this point is reached.)

(ii) Near the channel bottom ($y \to 0$)

For a wide channel, with $S_f = n^2q^2/y^{10/3}$ and $Fr^2 = q^2/(gy^3)$, as $y \to 0$ both $S_f$ and $Fr^2$ become very large and the $|S_0|$ and $1$ terms are negligible:

$$\frac{dy}{dx} \approx \frac{-S_f}{-Fr^2} = \frac{n^2 q^2 / y^{10/3}}{q^2/(g y^3)} = g\,n^2\,y^{-1/3}$$ $$\text{As } y \to 0:\quad \frac{dy}{dx} \to \infty$$

So the tangent is again vertical, and the curve meets the channel bottom normally.

Hence the A3 profile starts vertically at the bed (or at a small depth just after a gate) and ends vertically at the CDL.