The Maximum Rate of Change at a Point on a Function Examples 1

Recall from The Maximum Rate of Change at a Point on a Function of Several Variables page that if $z = f(x, y)$ is a two variable real-valued function and $\vec{u}$ is a unit vector then the maximum rate of change at any point $(x, y) \in D(f)$ is the magnitude of the gradient at $(x, y)$, $\| \nabla f(x, y) \|$, and the minimum rate of change at any point $(x, y) \in D(f)$ is in the opposite direction of the gradient and is $- \| \nabla f(x, y) \|$.

We will now look at some examples of applying this theorem.

Example 1

Let $f(x, y) = xy$, and consider the point $(2, 0, 0)$. What is the maximum and minimum rates of change at this point on $f$? In what direction is the rate of change equal to $-1$?

The gradient of $f$ is:

(1)

\begin{align} \quad \nabla f(x, y) = \left ( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right ) = (y, x) \end{align}

The norm of the gradient is thus:

(2)

\begin{align} \quad \| \nabla f(x, y) \| = \sqrt{y^2 + x^2} \end{align}

The norm of the gradient at the point $(2, 0)$ is therefore $\| \nabla (2, 0) \| = \sqrt{0 + 4} = \sqrt{4} = \pm 2$. The maximum rate of change is therefore $2$ and occurs in the direction of the gradient, $\nabla f(2, 0) = (0, 2)$, and the minimum rate of change is $-2$ and occurs in the direction opposite of the gradient, that is $-\nabla f(2, 0) = (0, -2)$.

Let $\vec{u} = (a, b)$ be a unit vector - that is $a^2 + b^2 = 1$. The rate of change is equal to $-1$ in the direction of the vector $\vec{u}$ that satisfies:

(3)

\begin{align} \quad -1 = D_{\vec{u}} \: f(2, 0) = (a, b) \cdot \nabla f (2, 0) = (a, b) \cdot (0, 2) = 2b \end{align}

Therefore $b = -\frac{1}{2}$. Since $\vec{u} = (a, b)$ is a unit vector, we must have that $a^2 + b^2 = 1$, so then $a = \pm \frac{\sqrt{3}}{2}$.

Therefore, $D_{\vec{u}}$ has a rate of change of $-1$ in the directions of the vectors $\left ( \pm \frac{\sqrt{3}}{2}, -\frac{1}{2} \right )$.

Example 2

Find the maximum and minimum rate of change of the function $f(x, y) = x^2 - 2y^2$ at the point $(1, 1) \in D(f)$.

The gradient of $f$ is:

(4)

\begin{align} \quad \nabla f(x, y) = (2x, -4y) \end{align}

The norm of the gradient of $f$ is:

(5)

\begin{align} \quad \| \nabla f(x, y) \| = \sqrt{4x^2 + 16y^2} \end{align}

The norm of the gradient of $f$ at the point $(1, 1)$ is therefore:

(6)

\begin{align} \quad \| \nabla f(1, 1) \| = \sqrt{4 + 16} = \sqrt{20} \end{align}

Therefore the maximum rate of change is $\sqrt{20}$ and the minimum rate of change is $-\sqrt{20}$.