Why is $\lambda$ called the *instantaneous* rate of change in the exponential distribution? In the following paramterisation of the exponential distribution
$${\displaystyle f(t;\lambda )={\begin{cases}\lambda e^{-\lambda t}&t\geq 0,\\0&t<0.\end{cases}}},$$
$\lambda$ is called the "rate" parameter. If $T \sim \text{Exp}(\lambda)$, I think I understand the intuition behind why it's called a (average) rate - because it's the average number of arrivals per unit time $\left( \lambda = \frac{1}{\mathbb E(T)}\right)$; On average, there is 1 arrival in $\mathbb E (T)$ amount of time.
However, in some places (for example, in continuous time markov chains), this $\lambda$ is called the instantaneous rate of change.
How is $\lambda$ an instantaneous rate of change (what makes it instantaneous?)?
 A: Some motivation for the term "instantaneous" in this context is that the transition rate matrix for a continuous-time Markov chain is actually an infinitesimal generator. Let $\{X(t):t\geqslant 0\}$ be a CTMC. Define the jump times of the process by $J_0=0$ and
$$
J_{n+1} = \inf\{t>J_n: X_t\ne X_{J_n}\},\ n\geqslant 1,
$$
the holding times by
$$
S_n = J_n-J_{n-1},\ n\geqslant 0,
$$
and the jump process by
$$
Y_n = X_{J_n},\ n\geqslant 0.
$$
We are mostly interested in CTMCs that have right-continuous sample paths, that is for any $t\geqslant 0$, there exists $\varepsilon>0$ such that $X(t+s)=X(t)$ for all $0\leqslant s\leqslant\varepsilon$. This ensures that the holding times are strictly positive. There is also the matter of "explosion," where there can exist a random time $\xi$ such that
$$
\xi:= \sup_n J_n =\sum_{n=1}^\infty S_n <\infty.
$$
Note that this can only happen in CTMCs on countably infinite state spaces, as boundedness of the transition rates implies that $\xi=+\infty$. This is a rather pathological case, however, since it means there are infinite transitions in a finite amount of time - and it is not clear how to define the process after that time!
Now, for times $s,t>0$ and states $i,j$ we can write $P_t:= \mathbb P(X(t+s)=j\mid X_s=i)$ due to homogeneity. The collection of matrices $\{P_t:t\geqslant 0\}$ determine the transient behavior of the process and in fact form a semigroup, as $P_{t+s}=P_tP_s$ (a semigroup is a set with a binary operation that is associative). Morever, since $P_\varepsilon\to P_0=I$ (the identity matrix) as $\varepsilon\downarrow0$, this semigroup is right-continuous for all $t$.
Some important results are the following:

For any states $i$ and $j$, the the following limits exist and are nonnegative:
\begin{align} q_i:&=\lim_{\varepsilon\downarrow0}\frac{(1-P_\varepsilon(i,i))}\varepsilon\\
q_{ij} :&= \lim_{\varepsilon\downarrow0}\frac{P_\varepsilon(i,j))}\varepsilon.
 \end{align}

Set $q_{ii}=-q_i$ and $q_{ij}$ as defined above, then the matrix $A=(q_{ij})$ is the infinitesimal generator of the semigroup. An interesting example of this is a discrete time Markov chain subordinated to a Poisson process. Let $\{\hat X_n:n=0,1,\ldots\}$ be a Markov chain with transition matrix $Q$ and $\{N(t):t\geqslant0\}$ an independent Poisson process with intensity $\lambda>0$. Define
$$
X_t := \hat X_{N_t},\ t\geqslant 0.
$$
Then $\{X_t:t\geqslant 0\}$ is a continuous time Markov chain with generator $A=\lambda(Q-I)$.
The infinitesimal generator also happens to be the unique solution to the backward Kolmogorov differential equations $$P'(t)=AP(t),$$ where we can explicitly write $P$ as the matrix exponential of $A$: $$P(t) = e^{Qt} := \sum_{n=0}^\infty \frac{Q^n}{n!}. $$ It also turns out that explosiveness becomes an issue here - the backward equations are well-defined for any CTMC, but the analogous forward equation $P'(t)=P(t)A$ cannot be rigorously justified for explosive processes.
I hope this answer sheds some light as to why the transition rates in a continuous time Markov chain are called "instantaneous."
A: Consider a inhomogenous poisson process with rate function $\lambda(t)$. For any given interval $[0,t]$ the count distribution is given by $P(N(t)=n) = \frac{\Lambda(t)^n}{n!}e^{-\Lambda(t)}$, where $\Lambda(t) = \int_0^t \lambda(t) \;dt$
From this perspective, it is hopefully clearer why $\lambda$ is a rate. For continuous-time Markov chains, the probability of a transition from state $i$ to state $j$ after time interval $\delta$ is also a poisson process, with rate interpretation as above.
When dealing with a standard poisson process, the rate does not change and so $\lambda$ can be interpreted as an average accumulation rate: $\Lambda(t) = \lambda t$
