Problem 266 - Calculus Volume 3

Question

For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(\mathbf{v}\). $$f(x, y)=x y, P(1,1), \mathbf{u}=\left\langle\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle$$

Step-by-Step Solution

Answer

The directional derivative is \( \sqrt{2} \).

Step 1: Compute the Gradient

Find the gradient \( abla f(x, y) \) of the function \( f(x, y) = xy \). The gradient is \( abla f(x, y) = \left< \frac{\partial}{\partial x}xy, \frac{\partial}{\partial y}xy \right> \). Compute the partial derivatives: \( \frac{\partial}{\partial x}xy = y \) and \( \frac{\partial}{\partial y}xy = x \). Therefore, \( abla f(x, y) = \langle y, x \rangle \).

Step 2: Evaluate the Gradient at a Specific Point

Substitute the point \( P(1, 1) \) into the gradient \( abla f(x, y) = \langle y, x \rangle \). This gives \( abla f(1, 1) = \langle 1, 1 \rangle \).

Step 3: Verify the Direction Vector

Ensure that the given direction vector \( \mathbf{u} = \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle \) is a unit vector. A vector is a unit vector if its magnitude is 1. Check this by calculating the magnitude: \( \sqrt{ \left( \frac{\sqrt{2}}{2} \right)^2 + \left( \frac{\sqrt{2}}{2} \right)^2 } = \sqrt{ \frac{1}{2} + \frac{1}{2} } = \sqrt{1} = 1 \). The vector is a unit vector.

Step 4: Compute the Dot Product

Calculate the dot product of the gradient at \( P \) and the direction vector \( \mathbf{u} \). That is, find \( abla f(1, 1) \cdot \mathbf{u} = \langle 1, 1 \rangle \cdot \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle = 1 \cdot \frac{\sqrt{2}}{2} + 1 \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \).

Step 5: Conclude the Directional Derivative

The directional derivative of \( f \) at point \( P(1,1) \) in the direction of \( \mathbf{u} \) is \( \sqrt{2} \).

Key Concepts

Gradient

The gradient of a function is like a compass pointing in the direction of the greatest rate of increase of the function. For a multivariable function, the gradient is a vector composed of all the partial derivatives of the function. It tells us how to move in the direction where the function grows the fastest.

In this context, the gradient of the function \( f(x, y) = xy \) is determined by computing its partial derivatives. The gradient \( abla f(x, y) \) is given by \( \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle \). For the function \( xy \), the partial derivative with respect to \( x \) is \( y \) and with respect to \( y \) is \( x \).

Thus, \( abla f(x, y) = \langle y, x \rangle \). At the point \( P(1, 1) \), substitute \( x = 1 \) and \( y = 1 \) into the gradient, resulting in \( \langle 1, 1 \rangle \).

Understanding gradients is crucial in fields such as physics and machine learning, where they are used to optimize functions by moving in directions indicated by the gradient.

Partial Derivatives

Partial derivatives provide the rate at which a function changes as one of the variables is altered, while keeping others constant. They are an extension of the concept of a derivative from single-variable calculus to functions of multiple variables.

For the function \( f(x, y) = xy \), calculating the partial derivatives is straightforward. When taken with respect to \( x \), treat \( y \) as a constant, leading to the partial derivative \( \frac{\partial f}{\partial x} = y \). Conversely, when differentiating with respect to \( y \), treat \( x \) as constant, resulting in \( \frac{\partial f}{\partial y} = x \).

These derivatives form the components of the gradient vector, providing key information on how the function behaves in each direction. Partial derivatives are essential for understanding and analyzing surfaces in three dimensions, greatly impacting fields such as economics and engineering.

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar. It's significant in the calculation of directional derivatives because it combines the effects of a vector's direction with another vector's magnitude.

When we have a gradient vector \( abla f(1,1) = \langle 1, 1 \rangle \) and a direction vector \( \mathbf{u} = \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle \), the dot product is found by multiplying corresponding components and adding them up: \( 1 \cdot \frac{\sqrt{2}}{2} + 1 \cdot \frac{\sqrt{2}}{2} = \sqrt{2} \).

This scalar value \( \sqrt{2} \) represents the directional derivative of the function in the direction specified by \( \mathbf{u} \). The dot product effectively measures how aligned two vectors are and is crucial in various applications, including physics and computer graphics.