My post is to ask why in some applications it keeps only 1st order of taylor equation. The question is from computer vision (LK optical flow). The assumption is $$I(x,y,t) = I(x+dy, y+dy, t+dt)$$ which means the intensity of the same pixel in different time is the same. (x,y) is the location of a pixel in an image at time $t$ and $dx$, $dy$ and $dt$ indicate the increment of each of them at time $t+dt$.
To establish the relation between two adjacent images, it applys Taylor equation onto the right part of the equation above and keeps only the first order $$I(x+dx, y+dy, t+dt) \approx I(x,y,t) + \frac{\partial I}{\partial x}dx + \frac{\partial I}{\partial y}dy+\frac{\partial I}{\partial t}dt $$. I don't understand why it uses only the first order of the taylor equation of this function. A second equation is whether it is a wrong typing that it intended to apply the Taylor equation onto the left part of the equation. Otherwise, I don't understand why the partial derivative of $I(x+dx, y+dy, t+dt)$ gives $\frac{\partial I}{\partial x}dx + \frac{\partial I}{\partial y}dy+\frac{\partial I}{\partial t}dt$.
Regarding why first order instead of higher order, the answer in this case is most likely the same as the answer in every other application of Taylor polynomials: In order to minimize computation time, use the smallest order Taylor polynomial for which the approximation error is acceptably small.
When you are reading some work that uses the first order Taylor approximation, you are free to question the author's (often) unstated but implicit assumption that the error of the first order Taylor approximation is acceptably small.