Problem 301 - Calculus 3

Question

For the following exercises, find the maximum rate of change of \(f\) at the given point and the direction in which it occurs. $$ f(x, y)=\cos (3 x+2 y),\left(\frac{\pi}{6},-\frac{\pi}{8}\right) $$

Step-by-Step Solution

Answer

Maximum rate of change is \(\sqrt{\frac{13}{2}}\) in direction \(\left(\frac{-3}{\sqrt{13}}, \frac{-2}{\sqrt{13}}\right)\).

Step 1: Compute the Gradient of the Function

The first step is to find the gradient of the function \(f(x, y) = \cos(3x + 2y)\). The gradient is the vector of partial derivatives, \(abla f(x, y)\). Compute the partial derivative with respect to \(x\): \(\frac{\partial f}{\partial x} = -3\sin(3x + 2y)\). Compute the partial derivative with respect to \(y\): \(\frac{\partial f}{\partial y} = -2\sin(3x + 2y)\). Thus, \(abla f(x, y) = (-3\sin(3x + 2y), -2\sin(3x + 2y))\).

Step 2: Evaluate the Gradient at the Given Point

Next, evaluate \(abla f(x, y)\) at the given point \(\left(\frac{\pi}{6}, -\frac{\pi}{8}\right)\). Calculate \(3x + 2y\) at this point: \(3 \cdot \frac{\pi}{6} + 2 \cdot \left(-\frac{\pi}{8}\right) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}\). Substitute into \(abla f\): \(abla f\left(\frac{\pi}{6}, -\frac{\pi}{8}\right) = (-3\sin(\frac{\pi}{4}), -2\sin(\frac{\pi}{4})) = \left(-3 \cdot \frac{\sqrt{2}}{2}, -2 \cdot \frac{\sqrt{2}}{2}\right)\). Thus, \(abla f\left(\frac{\pi}{6}, -\frac{\pi}{8}\right) = \left(-\frac{3\sqrt{2}}{2}, -\sqrt{2}\right)\).

Step 3: Calculate the Magnitude of the Gradient

The magnitude of the gradient gives the maximum rate of change of the function. Use the formula \(|abla f| = \sqrt{\left(-\frac{3\sqrt{2}}{2}\right)^2 + (-\sqrt{2})^2}\). Simplify: \(|abla f| = \sqrt{\frac{9}{2} + 2}\). So, \(|abla f| = \sqrt{\frac{13}{2}}\).

Step 4: Identify the Direction of Maximum Rate of Change

The direction of maximum rate of change is in the direction of the gradient vector. Compute the unit vector in the direction of \(abla f\) by dividing each component by \(|abla f|\). The unit vector is \(\left(\frac{-\frac{3\sqrt{2}}{2}}{\sqrt{\frac{13}{2}}}, \frac{-\sqrt{2}}{\sqrt{\frac{13}{2}}}\right)\). Simplify each component: \(\left(\frac{-3}{\sqrt{13}}, \frac{-2}{\sqrt{13}}\right)\).

Key Concepts

Maximum Rate of Change

When working with functions of multiple variables, we often seek the maximum rate of change at a given point. This represents the steepest slope or the fastest increase or decrease of the function value. The maximum rate of change is determined by the magnitude of the gradient vector.

Here’s how it works:

• The gradient vector, discovered through partial derivatives, tells us the direction and rate of the steepest ascent of the function.
• The larger this magnitude, the faster the function changes in that particular direction.

When evaluating a function at a point like \(rac{\pi}{6}, -\frac{\pi}{8}\), the calculated magnitude of the gradient vector acts as our maximum rate of change. In this problem, it simplifies to \(\sqrt{\frac{13}{2}}\). That tells us how quickly \(f(x, y) = \cos(3x + 2y)\) changes in its steepest direction at that specific point.

Partial Derivatives

Partial derivatives are the building blocks of the gradient. They provide insights into how a function changes as each variable is altered independently. By holding other variables constant, partial derivatives focus on one-variable change.

For the function \(f(x, y) = \cos(3x + 2y)\):

• The partial derivative with respect to \(x\) is \(-3\sin(3x + 2y)\).
• The partial derivative with respect to \(y\) is \(-2\sin(3x + 2y)\).

These expressions tell us how sensitive the function is to changes in \(x\) and \(y\). If you thought of the surface of this function as a mountain, then these derivatives indicate how steeply the surface rises or falls as you move in the \(x\) or \(y\) direction. Evaluating these at a specific point gives us part components of the gradient.

Gradient Vector

The gradient vector, written as \(abla f(x, y)\), is a crucial concept in multivariable calculus. It combines the partial derivatives into a vector form. This vector essentially points in the direction of the steepest ascent of the function.

For \(f(x, y) = \cos(3x + 2y)\), the gradient is calculated by:

• Using the partial derivatives to form a vector \( (-3\sin(3x + 2y), -2\sin(3x + 2y)) \).

At the specified point \(\left(\frac{\pi}{6}, -\frac{\pi}{8}\right)\), this becomes:

• \((-\frac{3\sqrt{2}}{2}, -\sqrt{2})\).

Why is this important? Because this vector direction is the path where the change is most rapid, synonymous with the maximum rate of change. Make it a unit vector to understand the pure direction, thus giving a directional insight independent of magnitude.