Suppose we have a multivariable function $f$ and I want to find it's directional derivative along the direction $\vec u$. Then the formula tells us that the directional derivative will be $$\nabla f\cdot \vec u = |\nabla f||u|cosθ$$ where $θ$ is the angle between these two vectors. That means that, for the directional derivative to be $0$, then $cosθ = 0 \to θ= π/2$.
Is there an intuitive way to understand this? Why the gradient needs to be perpendicular to my chosen direction in order for the directional derivative, i.e.the slope of the function in the particual point along my chosen direction, to be $0$ ?
Thanks in advance.
EDIT : I can maybe understand something related to the contours. https://en.wikipedia.org/wiki/Directional_derivative The first picture here shows a contour plot, where we see that the gradient is perpendicular to the contours and we also see a dir. derivative. If these 2 vectors were perpendicular,then the dir. derivative would have to be tangent to the contour and therefore, our unit vector $u$ would be tangent to it. That means that our direction is tangent to the contour. So for small steps, the function wouldn't change value. So our rate of change would be zero, i.e. the dir. derivative would be zero. Is that correct?
Best Answer
In order for the gradient $\nabla f$ to exist at a given point, $f$ has to be differentiable at that point.
For a two-variable function $f,$ this implies that you can graph $f$ in three dimensions ($x$ and $y$ for the input variables, $z$ for the output value) and then there will be a plane tangent to $f$ at the point where you found the gradient. If the plane is horizontal the gradient is zero; otherwise the plane is tilted by rotating around some horizontal line. The fastest way to climb on the plane is to go upward perpendicular to that horizontal line; the gradient is simply the projection of that direction back down onto the $x,y$ plane.
Since the gradient is perpendicular to the horizontal line, the horizontal line is perpendicular to the gradient. So your angle $\theta = \pi/2$ is simply selecting a direction along the horizontal line.
The situation for $f$ with more than two variables is analogous, though harder to visualize graphically.
Some related questions:
Why tangent surface is a plane
Trouble with gradient intuition
Why is the gradient the vector of strongest slope?