I'm not sure I understand your questions either.
50.09 is the end of the video; there is nothing after 50.09.
r is a vector-valued function. Specifically, the values of the function r are position vectors. It takes a real number, s, as its input. Its output is a vector r(s) = <r
1(s),r
2(s),r
3(s)>, one vector for each number s. We can think of position vectors as points. If you're picturing them as arrows, think of the points at the tips of the arrows. The other end of the arrows is at one fixed point, namely <0,0,0>, the zero vector.
The points which are values of the function r form a curve in space.
dr/ds is the derivative of r. It too is a vector-valued function, but of a slightly different kind. For each number s, it outputs a vector dr/ds (s). You can imagine this vector, dr/ds (s), as an arrow with its blunt end not at <0,0,0> but at the point r(s). The vector dr/ds (s) is tangent to the curve at this point, r(s). We call such vectors tangent vectors to the curve, or, synonymously, velocity vectors.
By convention, we use the letter s for the independent variable here when the curve is parameterized by arc length. This means that, if a and b are specific values that s can take, specific real numbers, then the length of the curve between r(a) and r(a+b) is |b - a| (the absolute value of "a minus b"). It happens that when we describe a curve by a function for which this rule about arc length holds, each vector dr/ds (s) has unit length; that is ||dr/ds (s)|| = 1.
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Why is this? This derivative is, by definition, the limit of the change in r divided by the change in s, as the change in s approaches zero. Intuitively, the curve becomes more and more like a straight line when seen at higher magnitude.
We could have used some other function to describe the same curve, a function for which this rule about the length of the curve doesn't hold. In that case, we'd probably use a different letter for the independent variable, such as t. Then the length of the tangent vector could be any real number, not necessarily 1.
We can make some other, arbitrary parameter, t, a function of s, and vice versa. Then, by the chain rule,
dr/ds = dr/dt * dt/ds.
But dt/ds = (ds/dt)
-1, the reciprocal of speed (speed being the magnitude of the velocity vector). So dr/ds is a velocity vector divided by its own magnitude. That is, a
unit vector.
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w is a scalar field. It takes a position vector as its input, and gives as its output a real number. That is, for each point in space (each position vector), <x,y,z>, the function w associates a real number, w(x,y,z).
The gradient of w is a vector field, whose values are gradient vectors [itex]\nabla w (x,y,z)[/itex] for each position vector <x,y,z>. This is a function which takes a point in space, the position vector <x,y,z>, as its input, and gives as its output another kind of vector, the gradient vector, associated with that point. The gradient vectors point in the direction of the greatest rate of increase (the steepest rise) of the values of the scalar field w. The rate of increase in that direction, at the point <x,y,z>, is the magnitude of the gradient vector at <x,y,z>, that is, the magnitude of [itex]\nabla w (x,y,z)[/itex].
What if we want to know the rate of increase of w, at some point <x,y,z>, in some other direction, an arbitrary direction, not necessarily the direction with the greatest rate of increase? Then we can take the dot product of [itex]\nabla w (x,y,z)[/itex] with a unit vector, say u, pointing in the desired direction:
[tex]\nabla w (x,y,z) \; \cdot \; \textbf{u} = \left \| \nabla w (x,y,z) \right \| \; \left \| \textbf{u} \right \| \; \cos(\theta)[/tex]
where [itex]\theta[/itex] (theta) is the angle between these two vectors. This gives us the projection of [itex]\nabla w (x,y,z)[/itex] onto the line through u, in other words, the component of [itex]\nabla w (x,y,z)[/itex] in the direction of u. The right side of this equation denotes multiplication of three real numbers, the magnitude of the gradient vector at <x,y,z>, a unit vector, and the cosine of the angle between them.
Is that any help?
