Gregg said:
In the notes it says that
\text{v}\cdot \nabla \text{u} = |\text{v}|\frac{du}{dl}
\text{v} = (a(x,y), b(x,y))
l is the arclength in the v-direction.
Why is this?
The LHS is the projection of v onto the gradient of u, the other thing is the magnitude of v, multiplied by the du/dl.
l is the arclength of
what in the v direction? And you haven't told us what u is. Nor what v represents. What we have here is an example of "guess the question". Since I have a bit of time on my hands, I will expound a bit about a question I think might be relevant.
Let ##u(x,y)## be a scalar field, perhaps the temperature at ##(x,y)##. Let a curve ##C## be given by ##\vec R(t) = \langle x(t), y(t)\rangle## represent the location of a moving particle. What if we want the rate of change of ##u## as the particle moves along ##C##? Well we have$$
\frac{du}{ds}=\frac{du}{dt}\frac{dt}{ds}=
(u_x\frac {dx}{dt}+ u_y\frac{dy}{dt})\frac{dt}{ds}$$Multiply both sides by$$
\frac{ds}{dt}=\frac 1 {\frac{dt}{ds}}$$to get$$
\frac{du}{ds}\frac{ds}{dt}=u_x\frac {dx}{dt}+ u_y\frac{dy}{dt}=
\nabla u\cdot \frac {d\vec R}{dt}= \nabla u\cdot \vec V$$
where ##\vec V## is the velocity of the particle. Since ##v = \frac{ds}{dt}=|\vec V|## is its speed, this can be written$$
|\vec V|\frac{du}{ds}=\nabla u\cdot \vec V$$This looks a lot like your result, using ##s## instead of ##l## for arc length. There may be a more direct way of getting the result but, hey, I'm not even sure I worked the problem you were thinking about.