Toronto Math Forum
APM3462016F => APM346Lectures => Chapter 2 => Topic started by: Shaghayegh A on September 18, 2016, 07:45:39 PM

Section 2.1 of the textbook states $$u_t a+u_x b$$ is the directional derivative of u in the direction l=(a,b). But there's an extra factor of $$\frac{1}{\sqrt{a^2+b^2}}$$ right? (which disappears if we set $$u_t a+u_x b$$ to 0). As in:
$$\nabla_{l}u \;. \frac{\bar{l}}{\bar{l}}=(\partial u/ \partial t \;\hat{t}\;+\;\partial u/ \partial x \; \hat{x}).\frac{ (a\hat{t}+b\hat{x})}{\sqrt{a^2+b^2}}=(u_t a+u_x b)\frac{1}{\sqrt{a^2+b^2}}$$?

Often in the courses by directional derivative mean that the direction is normalized, which means made of the length $1$, as you wrote $\frac{\vec{l}}{\vec{l}}$. We do not do this (normalize). Then we have a derivative along vector field. I will change in the Textbook.