# Toronto Math Forum

## APM346-2016F => APM346--Lectures => Chapter 2 => Topic started by: Shaghayegh A on September 18, 2016, 07:45:39 PM

Title: question from 2.1 of textbook
Post 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:

$$\nabla_{l}u \;. \frac{\bar{l}}{|\bar{l}|}=(\partial u/ \partial t \;\hat{t}\;+\;\partial u/ \partial x \; \hat{x}).\frac{ (a\hat{t}+b\hat{x})}{\sqrt{a^2+b^2}}=(u_t a+u_x b)\frac{1}{\sqrt{a^2+b^2}}$$?
Title: Re: question from 2.1 of textbook
Post by: Victor Ivrii on September 20, 2016, 10:57:18 AM
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.