Graphical intuition behind directional derivative interpreted as a(∂f/∂x)+b(∂f/∂y), where <a, b> is directional vector of unit length?

Question

Given a function f(x, y), the directional derivative is defined to be a(∂f/∂x)+b(∂f/∂y), with being a vector of unit length designating the direction in which we wish to take the derivative. I understand how we go from the limit definition of a derivative in the direction of unit vector to a(∂f/∂x)+b(∂f/∂y), but I do not understand how to interpret this theorem (Df(x, y) = a(∂f/∂x)+b(∂f/∂y)) GRAPHICALLY. It is hard for me to visualize the derivative at a given point (x, y) in the direction of being a(∂f/∂x)+b(∂f/∂y). For example, evaluating f(x, y) at directional vector <1, 0> yields the derivative ∂f/∂x. I find it difficult visualizing what is happening when we evaluate the directional derivative at, say, f(1, 1) in the direction <1/√2, 1/√2>. I can't quite visualize why the partial derivative with respect to x at point f(1, 1) multiplied by the x component plus the partial derivative with respect to y at point f(1, 1) multiplied by the y component ends up being derivative at f(1, 1) in the direction of the vector. Can you explain what is happening?

Answer 1

To visualize what is happening, think about the graph of the function in 3D space as a wiggly sheet above (and possibly below) the x-y plane, which is like the floor. Picking a point $(x_0,y_0)$ on the floor, the height of the sheet above that point is given by $f(x_0,y_0)$. If you want to know the directional derivative at that point in the direction of a vector $(a,b)$, what you are doing is essentially slicing the graph above the point $(x_0,y_0)$ in the direction of the vector $(a,b)$ (i.e. cutting through x-y plane along the line given by $y-y_0 = \frac{b}{a}(x-x_0)$). When we restrict our attention to just this 2D slice, we have what looks like the graph of a single-variable function, and the directional derivative gives the slope of the tangent line of this function above the point in question.

The reason the directional derivative ends up being a linear combination of the directional derivatives in the two standard directions is because the differentiability of the function $f$ at $(x_0, y_0)$ means that the function can be locally approximated by a plane at that point (this is called the tangent plane). So if we look at the tangent line of our "slice" in the direction of the x-axis and the tangent line of the "slice" in the direction of the y-axis, these form two lines in 3D space that intersect at the point $(x_0, y_0, f(x_0,y_0))$ on our wiggly sheet. There is only one plane that contains two intersecting lines, and this plane is the tangent plane. So these two tangent lines completely determine the tangent plane, and hence any other tangent line of any other slice. If we slice the function in a direction almost parallel to the x-axis, then the slope of the tangent line will be almost equal to $\partial f/\partial x$, whereas if we slice it in a direction almost parallel to the y-axis, it will be almost $\partial f/\partial y.$ The unit vector $(a,b)$ basically tells you what proportion of the slope is determined by the slope in the x-direction (a) and what proportion is determined by the y-direction (b).

As an interesting aside: if the function $f$ is not differentiable at $(x_0, y_0)$, it may still be possible to find the directional derivatives $\partial f/\partial x$ and $\partial f/\partial y$, however the directional derivative in a different direction might not exist, or if it does, might not be given by the formula you state. This is because there is no well-defined tangent plane to the function at that point.