Euler's Equations with auxiliary conditions - "why is $\frac{\delta y}{\delta \alpha}$ and $\frac{\delta z}{\delta \alpha}$ no longer independent?"

Question

let $J(\alpha)$ be a functional of the parameter $\alpha $ such that: \begin{equation}J(\alpha) = \int_{x_1}^{x_2}f\{y,y',z,z';x\}dx \end{equation} and let \begin{equation}f = f\{y,y',z,z';x\} \end{equation} where \begin{equation} y(\alpha,x) = y(0,x) + \alpha\eta_1(x) \end{equation} \begin{equation} z(\alpha,x) = z(0,x) + \alpha\eta_2(x) \end{equation} The constraint is: \begin{equation} g = g\{y_i;x\} = g\{y,z;x\} = 0 \end{equation} (for an example) $g = \sum\limits_{i} x^2_i -\rho^2 =0$ where $\rho = constant$ like a radius of a sphere.

From functions with several dependencies we get \begin{equation} \frac{\delta J}{\delta \alpha} = \int_{x_1}^{x_2}\left[\left(\frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'}\right)\frac{\delta y}{\delta \alpha} + \left(\frac{\delta f}{\delta z} - \frac{d}{dx}\frac{\delta f}{\delta z'}\right)\frac{\delta z}{\delta \alpha}\right] dx \end{equation}

Now originally $\textbf{without}$ a constraint, we will have $\frac{\delta y}{\delta \alpha} = \eta_1(x)$ and $\frac{\delta z}{\delta \alpha} = \eta_2(x)$ each $\eta_i(x) $ is independent thus \begin{equation} \left(\frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'} \right)= 0 \end{equation} and \begin{equation} \left(\frac{\delta f}{\delta z} - \frac{d}{dx}\frac{\delta f}{\delta z'} \right)= 0 \end{equation}

so that $\frac{\delta J}{\delta \alpha} = 0$ when $\alpha =0$, but now because of the constraint, "the variations $\frac{\delta y}{\delta \alpha}$ and $\frac{\delta z}{\delta \alpha}$ are no longer independent, so the expressions in parentheses do not separately vanish at $\alpha = 0$"

$\textbf{Question}$

Why are $\frac{\delta y}{\delta \alpha}$ and $\frac{\delta z}{\delta \alpha}$ no longer independent?

I would really appreciate if you could help me understand it.

Here is the page from text book:(I am not really sure if I need to provide extra information)

Answer 1

The way I see it, if there exists an additional constraint $g{y_i;x}=0$, where $y_i$ are some functions of the parameter $x$ (imagine the $y_i$ to be coordinates that are parametrized by a variable $x$ for argument's sake), then at least one of the $y_i$ functions/coordinates will be expressed as a function of the remaining $y_j,\ j\ne i$ functions/coordinates (Consider for example the equation of the circle, in which $\rho^2=y_1^2+y_2^2$). Therefore, the derivatives with respect to this parametrization will be related (here $0=2y_1\frac{\partial y_1}{\partial a}+2y_2\frac{\partial y_2}{\partial a}$) and thus $\frac{\partial y_1}{\partial a}$ is a function of $\frac{\partial y_2}{\partial a}$. I hope that helps!!