There is a question that says: Let $f(x,y)=x^2y+y^2$ Fine the direction vector v, such that f increases rapidly in the direction v at the point (1,1). Find the maximum rate of change of f at $(1,1)$
My answer: $$fx = 2xy$$ $$fy = x^2+2y$$
$$\nabla f(x,y) <2xy,x^2+2y> $$ $$\nabla f(1,1) <2,3> $$
Maximum rate will be equal to the magnitude of the gradient which will be $ \sqrt{2^2+3^2} = \sqrt{13}$
Now, this part is confusing.. Why is the directional vector $\vec{v} = <2,3> $
Usually, when the direction of the vector is given to calculate the directional derivate, it always isn't the same with what I get from $\nabla f(x,y)$
Edit: So If I am to calculate the unit vector, it would be $$ \vec{u} = \frac{\vec{v}}{|\vec{|v|}|} = \frac{{<2,3>}}{\sqrt{2^2+3^2}} = < \frac{{2}}{\sqrt{13}} , \frac{{3}}{\sqrt{13}} > $$
And the directional derivative would be $$\nabla f(1,1) * \vec{u} = \frac{{2}}{\sqrt{13}} + 2, \frac{{3}}{\sqrt{13}} * 3$$
Two ways to think about it.
Draw the level curve $x^2y + y^2 = 2.$ Note that $(1,1)$ lies on the curve.
If you want to move away from level as quickly as possible you want to move perpendicular to the level curve. And $\nabla f = (2,3)$ is perpendicular to the level curve.
Alternatively, the directional derivative $=\nabla f\cdot \frac {\bf{u}}{\|\bf {u}\|}$ and $a\cdot b = \|a\|\|b\|\cos \theta$
If we what to maximize $\|\nabla f\|\cos \theta$ we want to chose a $\bf{u}$ such that $\cos \theta = 1$