Table of Contents

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 realvalued 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)
The norm of the gradient is thus:
(2)
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)
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)
The norm of the gradient of $f$ is:
(5)
The norm of the gradient of $f$ at the point $(1, 1)$ is therefore:
(6)
Therefore the maximum rate of change is $\sqrt{20}$ and the minimum rate of change is $\sqrt{20}$.