For a differentiable scalar field \(f(x,y)\) or \(f(x,y,z)\), the gradient collects the partial derivatives into a vector.
It points in the direction of steepest increase of \(f\).
Gradient (2 variables)
\[
\nabla f(x,y)=\left\langle \frac{\partial f}{\partial x},\,\frac{\partial f}{\partial y}\right\rangle
\]
Gradient (3 variables)
\[
\nabla f(x,y,z)=\left\langle \frac{\partial f}{\partial x},\,\frac{\partial f}{\partial y},\,\frac{\partial f}{\partial z}\right\rangle
\]
Directional derivative
The directional derivative of \(f\) at a point \(P\) in the direction of a unit vector \(\hat u\) is the dot product:
\[
D_{\hat u}f(P)=\nabla f(P)\cdot \hat u
\]
Why \(\hat u\) must be unit
\[
\hat u=\frac{u}{\lVert u\rVert}
\]
Using a unit vector makes \(D_{\hat u}f(P)\) represent the rate of change of \(f\) per 1 unit of distance in that direction.
Steepest ascent
Among all unit directions, the maximum directional derivative occurs in the gradient direction.
\[
\begin{aligned}
\max_{\lVert u\rVert=1} D_{u}f(P) &= \lVert \nabla f(P)\rVert \\
\text{Direction of steepest ascent} &= \frac{\nabla f(P)}{\lVert \nabla f(P)\rVert}
\end{aligned}
\]
University extension: normals to level surfaces
If \(f(x,y,z)=c\) defines a level surface, then \(\nabla f(P)\) is normal (perpendicular) to the surface at \(P\).
\[
\nabla f(P)\perp \text{tangent plane at }P
\]
Worked example
Let \(f(x,y)=x^2+y^2\), \(P=(1,1)\), and \(u=\left\langle \frac{1}{\sqrt2},\frac{1}{\sqrt2}\right\rangle\).
\[
\begin{aligned}
\nabla f(x,y) &= \left\langle 2x,\,2y\right\rangle \\
\nabla f(1,1) &= \left\langle 2,\,2\right\rangle \\
D_{u}f(1,1) &= \nabla f(1,1)\cdot u \\
&= \left\langle 2,2\right\rangle \cdot \left\langle \frac{1}{\sqrt2},\frac{1}{\sqrt2}\right\rangle \\
&= \frac{4}{\sqrt2}=2\sqrt2
\end{aligned}
\]
Tip: If \(\nabla f(P)=\vec 0\), then all directional derivatives at \(P\) are \(0\) (flat to first order).
