Maximum Rate of Change at a Point on a Function of Several Variables

The Maximum Rate of Change at a Point on a Function of Several Variables

Recall from the Directional Derivatives page that a directional derivative at a point on a function $f$ in the direction of a unit vector $\vec{u}$ measures the rate of change of that function in the direction of $\vec{u}$. Suppose now that we want to figure out which direction has the largest rate of change. We will be able to figure this out with the following theorem.

Theorem 1: Let $z = f(x, y)$ be a differentiable function, and let $\vec{u}$ be a unit vector. The maximum value of the directional derivative $D_{\vec{u}} \: f(x, y)$ at a point $(x, y) \in D(f)$ occurs when $\vec{u}$ is in the same direction as $\nabla f(x, y)$ and the maximum value is the length/magnitude of the gradient vector $\| \nabla f(x, y) \|$. The minimum value of the directional derivative $D_{\vec{u}} \: f(x, y)$ at $(x, y)$ occurs when $\vec{u}$ is in the opposite direction as $\nabla f(x,y)$ and the minimum value is $-\| \nabla f(x, y) \|$.

• Proof: From The Gradient of Functions of Several Variables that the directional derivative of $f$ at $(x, y) \in D(f)$ in the direction of $\vec{u}$ is equal to the dot product of the gradient vector $\nabla f(x, y)$ and $\vec{u}$

(1)

\begin{align} D_{\vec{u}} \: f(x, y) = \nabla f(x,y) \cdot \vec{u} \end{align}

• If we use the alternate formula for The Dot Product of Vectors which says that $\vec{m} \cdot \vec{n} = \| \vec{m} \| \| \vec{n} \| \cos \theta$ where $0 ≤ \theta ≤ \pi$ is the angle between $\vec{m}$ and $\vec{n}$. Noting that $\| \vec{u} \| = 1$ and applying this formula, we have that:

(2)

\begin{align} \quad D_{\vec{u}} \: f(x, y) = \| \nabla f(x, y) \| \| \vec{u} \| \cos \theta \\ \quad D_{\vec{u}} \: f(x, y) = \| \nabla f(x, y) \| \cos \theta \end{align}

• We note that $-1 ≤ \cos \theta ≤ 1$, and so when $\theta = 0$, $\cos \theta = 1$, and so $D_{\vec{u}} \: f(x, y)$ attains its maximal value when $\vec{u}$ is in the same direction as the gradient vector $\nabla f(x, y)$.

• Furthermore, when $\theta = \pi$, that is $\vec{u}$ is in the opposite direction of $\nabla f(x, y)$, then $\cos \theta = -1$ and so $D_{\vec{u}} \: f(x, y)$ attains its minimal value. $\blacksquare$

Theorem 1 can be extended to functions of three variables as well. The proof is analogous to that above.

Theorem 2: Let $w = f(x, y, z)$ be a differentiable function, and let $\vec{u}$ be a unit vector. Then the maximum value of the directional derivative $D_{\vec{u}} \: f(x, y, z)$ occurs when $\vec{u}$ is in the same direction as $\nabla f(x, y, z)$ and the maximum value is the length/magnitude of the gradient vector $\| \nabla f(x, y, z) \|$. The minimum value of the directional derivative $D_{\vec{u}} \: f(x, y, z)$ occurs when $\vec{u}$ is in the opposite direction as $\nabla f(x,y,z)$ and the minimum value is $-\| \nabla f(x, y, z) \|$.

Example 1

Find the maximum rate of change at $(1, 2) \in D(f)$ of the function $z = f(x, y) = 3x^2y^3$.

We first need to compute the gradient of $f$. To do so, let's find the partial derivatives of $f$. We note that $\frac{\partial z}{\partial x} = 6xy^3$ and $\frac{\partial z}{\partial y} = 9x^2y^2$. Therefore $\nabla f(x, y) = (6xy^3, 9x^2y^2)$, and so:

(3)

\begin{align} \nabla f(1, 2) = (48, 36) \\ \| \nabla f(1, 2) \| = \sqrt{48^2 + 36^2} = \sqrt{3600} = 60 \end{align}

By theorem 1, the maximum rate of change is $\| \nabla f(1, 2) \| = 60$.

Example 2

Suppose that a valley is modelled by the function $f(x, y) = \frac{1}{2}xy + \cos x$. A hiker at $(\pi, \pi)$ is coming down to the village centered at $(0, 0)$. Find the direction for which the hiker has the steepest descent and the its rate of change.

We will first find the gradient vector by partial differentiation. We get that:

(4)

\begin{align} \quad \nabla f(x, y) = \left ( \frac{1}{2}y - \sin x, \frac{1}{2}x \right ) \end{align}

To find the direction in which the hiker has the steepest descent, we look in the opposite direction of the gradient vector. Plugging in the point $(\pi, \pi)$ into $-\nabla f(x, y)$ and we have that:

(5)

\begin{align} \quad -\nabla f(\pi, \pi) = \left ( \frac{\pi}{2}, \frac{\pi}{2} \right ) = -\frac{\pi}{2} \vec{i} \frac{\pi}{2} \vec{j} = \left ( -\frac{\pi}{2}, -\frac{\pi}{2} \right ) \end{align}

Thus we see that the steepest descent occurs in the direction of of the vector $\left ( -\frac{\pi}{2}, -\frac{\pi}{2} \right )$. The rate of change is:

(6)

\begin{align} \quad - \biggr \| \left ( -\frac{\pi}{2}, -\frac{\pi}{2} \right ) \biggr \| = - \sqrt{\frac{\pi^2}{4} + \frac{\pi^2}{4}} = -\frac{\pi}{\sqrt{2}} \end{align}