1. Directional Derivatives

Let f(x,y) - x^2y + xy^2. Find all vectors u such that Duf(1, -2) is exactly one-half the maximal possible directional derivative.

2. Originally Posted by kiddopop
Let f(x,y) - x^2y + xy^2. Find all vectors u such that Duf(1, -2) is exactly one-half the maximal possible directional derivative.
I assume you mean f(x,y)=$\displaystyle x^2y+ xy^2$

First find $\displaystyle \nabla f(x,y)$ and $\displaystyle ||\nabla f(x,y)||$. That norm [b]is[/tex] the "maximal possible directional derivative[/tex].

The directional derivative in the direction of vector $\displaystyle \vec{u}$ is $\displaystyle \nabla f(x,y)\cdot \frac{\vec{u}}{||\vec{u}||}$

Of course, $\displaystyle \frac{\vec{u}}{||\vec{u}||}$ is just the unit vector in the direction of $\displaystyle \vec{u}$ and that can always be written as $\displaystyle cos(\theta)\vec{i}+ sin(\theta)\vec{j}$ where $\displaystyle \theta$ is the angle $\displaystyle \vec{u}$ makes with the x-axis.

That is, $\displaystyle \vec{u}$ is any vector making angle $\displaystyle \theta$ with the x-axis where $\displaystyle \nabla f \cdot cos(\theta)\vec{i}+ sin(\theta)\vec{j}= \frac{1}{2}||\nabla f||$ which is the same as $\displaystyle f_x cos(\theta)+ f_y sin(\theta)= \frac{1}{2}\sqrt{(f_x)^2+ (f_y)^2}$.