How can $\frac{\partial x}{\partial t}$ be zero even though $x$ is a function of coordinates dependent on $t$ i.e., $x(q_1(t),\ldots, q_n(t))$? I was reading about the generalised coordinates in Mechanics by Keith Symon; at the concerned page, he was explaining about how to express the the velocity in Cartesian coordinates in terms of the generalised coordinates $q_1, q_2,\ldots, q_n$ viz.
$$\dot x_i~=~ \sum_{k~=~1}^{3N} \frac{\partial x_i}{\partial q_k}~\dot q_k + \frac{\partial x_i}{\partial t};$$ using this he could then write the kinetic energy as:
$$T ~=~ \underbrace{\sum_{k~=~ 1}^{3N}\sum_{l~=~1}^{3N}\sum_{i~=~ 1}^N ~m_i\left(\frac{\partial x_i}{\partial q_k}\frac{\partial x_i}{\partial q_l}+ \frac{\partial y_i}{\partial q_k}\frac{\partial y_i}{\partial q_l}+\frac{\partial z_i}{\partial q_k}\frac{\partial z_i}{\partial q_l}\right)\dot q_k\dot q_l}_{T_2} + \underbrace{\sum_{k~=~1}^{3N}\sum_{i~=~1}^N~\frac12 m_i\left(\frac{\partial x_i}{\partial q_k}\frac{\partial x_i}{\partial t}+\frac{\partial y_i}{\partial q_k}\frac{\partial y_i}{\partial t}+ \frac{\partial z_i}{\partial q_k}\frac{\partial z_i}{\partial t}\right)\dot q_k}_{T_1} + \underbrace{\sum_{i~=~1}^N \frac12m_i \left(\left(\frac{\partial x_i}{\partial t}\right)^2+ \left(\frac{\partial y_i}{\partial t}\right)^2+ \left(\frac{\partial z_i}{\partial t}\right)^2\right)}_{T_0}$$
He then remarked that when $x_i$ doesn't depend on time $t$ explicitly, $T_1$ and $T_0$ are zero. These terms appear only in "moving coordinate system".
That means $\frac{\partial x_i}{\partial t}~=~ 0$  when the terms $T_1$ and $T_0$ are zero.
What I'm not getting is why $\frac{\partial x_i}{\partial t}$ is zero even though it depends on $q_1, q_2, \ldots, q_n$ which are themselves functions of time $t\,.$
$\frac{\partial x_i}{\partial t}$ means keeping the other variables $q_i$ constant, we check whether $x(q_1(t), \ldots, q_n(t), t+ \mathrm dt) -x(q_1(t), \ldots, q_n(t), t) $ is zero or not.
But $q_i$s are also the functions of $t,$ so won't by taking $t+\mathrm dt$ change their values and ultimately $x_i$s?
In a word, I'm not getting why $x_i$ has to depend on $t$ explicitly so as to have a non-zero $\frac{\partial x_i}{\partial t}$ even though the other variables $q_i$s it depends on are themselves functions of $t\,.$
Also, I was surprised a bit when he said the terms $T_1$ and $T_0$ are zero when 

$t$ does not appear explicitly in \begin{align}q_1 & = q_1(x_1, y_1, z_1, \ldots,x_n, y_n, z_n; t )\\ q_2 & = q_2(x_1, y_1, z_1, \ldots,x_n, y_n, z_n; t )\\ \vdots & ~\\ q_{3N} & = q_{3N}(x_1, y_1, z_1, \ldots,x_n, y_n, z_n; t )\end{align}

I thought for $T_1$ and $T_0$ to be zero, $\frac{\partial x_i}{\partial t}$ must be zero which means $t$ must not appear explicitly when $x_i$ is expressed in terms of the generalised coordinates $q_i$s but not the other way as said by the author above.
So, does that mean the explicit dependence of $x_i$ on time $t$ depend on the fact whether the generalised coordinates themselves depend explicitly on time $t$? If so, how? I'm simply failing to visualise that.
 A: This is the usual confusion caused by using the same symbol for a physical quantity as for the function which says how it's related to other quantities.
Consider some function $f(z_1,\dots,z_{3N},s)$. It has partial derivatives $\partial f/\partial z_k$ and $\partial f/\partial s$. Nothing strange so far.
Now say that the quantity $x$ depends on the quantities $q_1, \dots, q_{3N}$ and on time $t$ in a way which is described by this function:
$$
x = f(q_1,\dots,q_{3N},t)
.
$$
And suppose also that the quantities $q_i$ depend on time $t$ in a way which is described by some functions $g_i$:
$$
q_i = g_i(t)
.
$$
Then $x$ can be seen as a function of $t$:
$$
x = f\bigl( g_1(t),\dots,g_{3N}(t),t \bigr)
= h(t)
.
$$
To compute $h'(t)$, you use the chain rule:
$$
h'(t)
= \sum_{k=1}^{3N} \frac{\partial f}{\partial z_k}\bigl( g_1(t),\dots,g_{3N}(t),t \bigr) \cdot g_k'(t)
+ \frac{\partial f}{\partial s}\bigl( g_1(t),\dots,g_{3N}(t),t \bigr) \cdot 1
.
$$
Of course, it may happen that the function $f$ that we started with is independent of $s$, so that $\partial f/\partial s=0$, and then the last term is absent.
