Gradient Vector

Gradient Vector

The gradient vector of $f$ is defined to be:

$$\begin{array}{rl}\nabla f& =⟨f_x,f_y,\dots ⟩\\ & \\ & =\left[\begin{array}{c}\frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\\ \dots \end{array}\right]\end{array}$$

Or if we want to use the standard basis vectors the gradient is:

$$\nabla f=\frac{\partial f}{\partial x}\vec{i}+\frac{\partial f}{\partial y}\vec{j}+\frac{\partial f}{\partial y}\vec{k}$$

The above definitions are for functions of 3 variables, but of course this extends to any number of variables

Some nice facts about the gradient vector:

Theorem: Maximum Rate of Change

The maximum value of $D_{\vec{u}}f(\vec{x})$ (and hence the maximum rate of change of the function $f(\vec{x})$) is given by $\mid \mid \nabla f(\vec{x})\mid \mid$ and will occur in the direction given by $\nabla f(\vec{x})$.

Proof: Maximum Rate of Change

Simple proof based on the definition of the cross product.

$$\begin{array}{rl}{D}_{\vec{u}}f& =\nabla f\cdot \vec{u}\\ & =\mid \mid \nabla f\mid \mid \Vert |\vec{u}||\cos \theta \\ & =\mid \mid \nabla f\mid \mid \cos \theta \end{array}$$

where $\theta$ is the angle between the gradient and $\vec{u}$.

The largest possible value of $\cos \theta$ is $1$ which occurs at $\theta =0$. Therefore, the maximum value of $D_{\vec{u}}f(x)$ is $\mid \mid \nabla f(\vec{x})\mid \mid$.

The maximum value occurs when the angle between gradient and $\vec{u}$ is $0$; in other words, when $\vec{u}$ is pointing in the same direction as the gradient, $\nabla f(\vec{x})$.

Another one:

Fact

The gradient vector $\nabla f(x_0,y_0)$ is orthogonal to the level curve $f(x,y)=k$ at the point $(x_0,y_0)$. Likewise, the gradient vector $\nabla f(x_0,y_0,z_0)$ is orthogonal to the level surface $f(x,y,z)=k$ at the point $(x_0,y_0,z_0)$.

Proof: Gradient Vector Orthogonality

Let’s prove the ${\mathbb{R}}^3$ case. Let $S$ be the level surface given by $f(x,y,z)=k$ and let $P=(x_0,y_0,z_0)$. Note that $P$ will be on $S$.

Now, let $C$ be any curve on $S$ that contains $P$. Let $\vec{r}(t)=⟨x(t),y(t),z(t)⟩$ be the vector equation for $C$ and suppose that $t_0$ be the value of $t$ such that $\vec{r}(t_0)=⟨x_0,y_0,z_0⟩$. In other words, $t_0$ is the value of $t$ that gives $P$.

Because $C$ lies on $S$ we know that points on $C$ must satisfy the equation for $S$:

$$f(x(t),y(t),z(t))=k$$

With the chain rule, we get:

$$\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}=0$$

Note that $\nabla f=⟨f_x,f_y,f_z⟩$ and ${\vec{r}}^{\prime }(t)=⟨{\vec{x}}^{\prime }(t),{\vec{y}}^{\prime }(t),{\vec{z}}^{\prime }(t)⟩$, so this becomes:

$$\nabla f\cdot {\vec{r}}^{\prime }(t)=0$$

At $t=t_0$, this is

$$\nabla f(x_0,y_0,z_0)\cdot {\vec{r}}^{\prime }(t_0)=0$$

This tells us that the gradient vector at $P$, $\nabla f(x_0,y_0,z_0)$, is orthogonal to the tangent vector, ${\vec{r}}^{\prime }(t_0)$, to any curve $C$ that passes through $P$ and on the surface $S$ and so must also be orthogonal to the surface $S$.