Deriving Hamiltonian with Lagrange multiplier method

Question

This is probably a basic concept, but I am having a hard time understanding why in deriving the Hamiltonian with the Lagrange multiplier method 1) we introduce a time-varying multiplier for optimization and 2) we take derivatives with respect to functions as if they were constants.

I've looked through a few chapters in some optimal control textbooks and convex optimization, and am still a bit confused.

During class, my professor showed us this proof:

$$ \begin{gathered} \max _{\{c(t)\}_{t \geq 0}} \int_0^{\infty} u(c(t)) e^{-\rho t} dt \text {, s.t., } \\ \dot{k}(t)=f(k(t))-c(t), \text { for all } t \geq 0 . \end{gathered} $$

Where $k(t)$ is our state variable, $u, f$ are continuous, strictly concave functions, and $\rho$ is a positive constant.

My professor first starts with a Lagrangian and uses some algebra: $$ \mathcal{L}= \int_0^{\infty} u(c(t)) e^{-\rho t} d t+\int_0^{\infty} \mu(t)[f(k(t))-c(t)-\dot{k}(t)] d t $$ $$ \mathcal{L}= \int_0^{\infty} u(c(t)) e^{-\rho t}+\mu(t)[f(k(t))-c(t)] d t-\int_0^{\infty} \mu(t) \dot{k}(t) d t $$ Integration by parts: $$ \int_0^{\infty} \mu(t) \dot{k}(t) d t=\lim _{t \rightarrow \infty} \mu(t) k(t)-\mu(0) k_0-\int_0^{\infty} \dot{\mu}(t) k(t) d t . $$ Define the function $$ \mathcal{H} \equiv u(c(t)) e^{-\rho t}+\mu(t)[f(k(t))-c(t)], $$ so that we have $$ \mathcal{L}=\mu(0) k_0-\lim _{t \rightarrow \infty} \mu(t) k(t)+\int_0^{\infty}[\mathcal{H}+\dot{\mu}(t) k(t)] d t $$

Then, he says that the first-order condition for consumption is $$ u^{\prime}(c(t)) e^{-\rho t}=\mu(t) \text {, for all } t \geq 0 . $$

But, I'm still quite confused about

1. Why is it necessary for the multiplier $\mu(t)$ to be time-varying? In the discrete-time case, we still were able to solve the problem with only 1 multiplier. Since it's possible to derive the integral by using an infinite sum, wouldn't a single multiplier suffice in this case too?

2. When, in general, can we treat a function as a constant and take derivatives with respect to that function? In this above first order condition, we obtained it by taking the derivative of the integrand with respect to $c(t)$. But, in doing so, it seems we took $c(t)$ as a constant. Does this require any assumptions about the function $\mathcal{H} - \dot{\mu}(t)k(t)$?

Thanks in advance!