$$f(x,y) = 4-x^2-y^2$$ $$u=<\frac{3}{5},-\frac{4}{5}>$$
How do I know that they're increasing in the same direction at point (1,1)? How do I determine that $\nabla f(1,1)$ isn't parallel to the level curve of $f$ at (1,1)?
$$\nabla f(x,y) = <-2x,-2y>$$ $$\nabla f(1,1) = <-2,-2>$$ $$\hat u = <\frac{3}{5},-\frac{4}{5}>$$
I could dot product them for $D_u$ but does that tell me if they're increasing in the same direction?
The directional derivative of $f$ along $u$ at $(1,1)$ is \begin{align} \nabla_uf(1,1) &= \nabla f(1,1)\cdot u\\ &= (-2,-2)\cdot(3/5,-4/5) \\ &= 2/5 \end{align} Since $2/5>0$, $f$ increases in the direction of $u$ at $(1,1)$.
The point $(1,1)$ is on the level curve $f=2$ (since $4-1^2-1^2=2)$. That level curve is $$ x^2+y^2=2 $$ That's the circle of radius $\sqrt{2}$ centered at the origin. A parametrization for it is $$ (x(t),y(t))=\sqrt{2}\,(\cos(t),\sin(t))\quad(0\leq t\leq 2\pi) $$ A tangent vector to that curve at $(1,1)$ is \begin{align} (x'(\pi/4),y'(\pi/4)) &= \sqrt{2}\,(-\sin(\pi/4),\cos(\pi/4))\\ &= (-1,1) \end{align} Since there is no $\alpha\in\mathbb{R}$ such that $$ \alpha\nabla f(1,1) = \alpha(-2,-2) = (-1,1) $$ we conclude that $\nabla f(1,1)$ isn't parallel to the level curve of $f$ at $(1,1)$.
Note: However, as we know, $\nabla f(1,1)$ is perpendicular to the level curve of $f$ at $(1,1)$. In effect, $$ \nabla f(1,1) \cdot (-1,1) = (-2,2) \cdot (-1,1) = 0 $$
Clearly the only way that a vector $v$ can be both perpendicular and parallel to another vector $w$ is if one of them is the $0$ vector.