Problem 10 - Calculus Early Transcendentals

Question

\(7-10\) (a) Find the gradient of \(f\) (b) Evaluate the gradient at the point \(P .\) (c) Find the rate of change of \(f\) at \(P\) in the direction of the vector \mathbf{u} . $$f(x, y, z)=\sqrt{x+y z}, \quad P(1,3,1), \quad \mathbf{u}=\left\langle\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\right\rangle$$

Step-by-Step Solution

Answer

Gradient at \( P \) is \( \left\langle \frac{1}{4}, \frac{1}{4}, \frac{3}{4} \right\rangle \); rate of change in direction of \( \mathbf{u} \) is \( \frac{23}{28} \).

Step 1: Understand the Problem

We need to find the gradient \( abla f \) of the function \( f(x, y, z) = \sqrt{x + yz} \) and evaluate it at the point \( P(1,3,1) \). Additionally, we need to find the directional derivative in the direction of vector \( \mathbf{u} = \left\langle \frac{2}{7}, \frac{3}{7}, \frac{6}{7} \right\rangle \).

Step 2: Calculate the Gradient of f

The gradient is found by taking the partial derivatives of \( f \) with respect to each variable. The gradient is given by \[abla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle\]- \( \frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x + yz}} \)- \( \frac{\partial f}{\partial y} = \frac{z}{2\sqrt{x + yz}} \)- \( \frac{\partial f}{\partial z} = \frac{y}{2\sqrt{x + yz}} \)Thus, \[abla f = \left\langle \frac{1}{2\sqrt{x + yz}}, \frac{z}{2\sqrt{x + yz}}, \frac{y}{2\sqrt{x + yz}} \right\rangle\]

Step 3: Evaluate the Gradient at Point P

Substitute the coordinates of point \( P(1,3,1) \) into the gradient:- \( abla f(1,3,1) = \left\langle \frac{1}{2\sqrt{1 + 3 \cdot 1}}, \frac{1}{2\sqrt{1 + 3 \cdot 1}}, \frac{3}{2\sqrt{1 + 3 \cdot 1}} \right\rangle \)- The expression simplifies as:- Since \( \sqrt{1 + 3 \cdot 1} = \sqrt{4} = 2 \), we have:\[abla f(1,3,1) = \left\langle \frac{1}{4}, \frac{1}{4}, \frac{3}{4} \right\rangle\]

Step 4: Calculate the Directional Derivative

The rate of change in the direction of vector \( \mathbf{u} \) is found using the dot product of \( abla f(1,3,1) \) and \( \mathbf{u} \):\[D_{\mathbf{u}}f(1,3,1) = abla f(1,3,1) \cdot \left\langle \frac{2}{7}, \frac{3}{7}, \frac{6}{7} \right\rangle\]- Calculating the dot product:\[\frac{1}{4} \cdot \frac{2}{7} + \frac{1}{4} \cdot \frac{3}{7} + \frac{3}{4} \cdot \frac{6}{7} = \frac{2}{28} + \frac{3}{28} + \frac{18}{28} = \frac{23}{28}\]

Step 5: Solution Synthesis

- The gradient of the function \( f \) is \( abla f = \left\langle \frac{1}{2\sqrt{x + yz}}, \frac{z}{2\sqrt{x + yz}}, \frac{y}{2\sqrt{x + yz}} \right\rangle \).- At point \( P(1,3,1) \), the gradient evaluates to \( \left\langle \frac{1}{4}, \frac{1}{4}, \frac{3}{4} \right\rangle \).- The rate of change of \( f \) at \( P \) in the direction of vector \( \mathbf{u} \) is \( \frac{23}{28} \).

Key Concepts

Directional Derivative

In multivariable calculus, the directional derivative of a function gives us the rate at which the function changes as we move in a specified direction from a particular point. This is important because while partial derivatives only measure the rate of change along the coordinate axes, the directional derivative measures changes along any direction.
To compute the directional derivative, we need to utilize both the gradient of the function and the direction vector. The process involves finding the dot product between these two components.
Here's how we can find the directional derivative:

• First, calculate the gradient of the function, which involves finding the partial derivatives with respect to each variable.
• Next, ensure that the direction vector is a unit vector by checking its magnitude or normalizing it.
• Finally, take the dot product of the gradient and the direction vector to find the directional derivative.

This result tells us how fast the function changes in the specific direction and is useful in fields like physics and engineering, where direction significantly affects outcomes.

Partial Derivatives

Partial derivatives are fundamental tools in multivariable calculus, helping us understand how a function changes with respect to one variable while keeping the others constant. They are particularly useful when dealing with functions of several variables, like those commonly found in physics and engineering.
For a function of three variables, like our example function \( f(x, y, z) = \sqrt{x + yz} \), we calculate three partial derivatives:

• \( \frac{\partial f}{\partial x} \) measures how \( f \) changes as \( x \) changes while \( y \) and \( z \) are constant.
• \( \frac{\partial f}{\partial y} \) evaluates the change in \( f \) with a variation in \( y \) while keeping \( x \) and \( z \) constant.
• \( \frac{\partial f}{\partial z} \) indicates how \( f \) changes with respect to \( z \) when \( x \) and \( y \) are constant.

These derivatives combine to form the gradient vector, which provides a comprehensive view of the function's changes in all directions.

Rate of Change

The rate of change in multivariable calculus extends our basic understanding of how functions change. Instead of focusing solely on one variable, it considers how a function behaves when multiple variables are at play.
One primary way of determining this is through the gradient and directional derivative. The gradient itself represents the steepest ascent of the function, while the directional derivative gives the rate of change in a specific direction.
For example, at point \( P(1, 3, 1) \), with a direction vector \( \mathbf{u} = \left\langle \frac{2}{7}, \frac{3}{7}, \frac{6}{7} \right\rangle \), our directional derivative measures how fast \( f \) increases when moving along \( \mathbf{u} \).
In practical terms, think of it as finding the slope of a hill (gradient) and checking how steep it is in the direction you want to travel (directional derivative). This is crucial for optimizing functions and ensuring efficient changes in engineering and data analysis.

Vector Calculus

Vector calculus forms the backbone of multivariable calculus, providing the tools to analyze functions that are influenced by multiple variables. It deals with vector fields and various operations on these fields, such as gradients, divergences, and curls.
In the case of a function like \( f(x, y, z) = \sqrt{x + yz} \), vector calculus allows us to uncover a comprehensive picture of the function's behavior. It helps us understand how variables interact and influence changes in different fields.
Key aspects include:

• **Gradient**: Reflects how a function increases most steeply from a point, based on its partial derivatives.
• **Directional Derivative**: Measures the rate of change of the function in a given direction.
• **Vector Fields**: Illustrate how vectors change over space, helpful for depicting fluid flow or magnetic fields.

Vector calculus is essential in numerous applications, from optimizing financial models to designing aerodynamic structures, where multi-variable dependencies abound.