/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 71 Given that \(\nabla f(4,-5)=2 \m... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 71

Given that \(\nabla f(4,-5)=2 \mathbf{i}-\mathbf{j}\), find the directional derivative of the function \(f\) at the point \((4,-5)\) in the direction of \(\mathbf{a}=5 \mathbf{i}+2 \mathbf{j}\)

Short Answer

Expert verified
The directional derivative is \( \frac{8}{\sqrt{29}} \).

Step by step solution

01

Understand the Directional Derivative Formula

The directional derivative of a function at a given point in a specified direction measures how the function changes as you move in that direction. The formula for the directional derivative of a function \( f \) at a point \( \mathbf{p} \) in the direction of a vector \( \mathbf{a} \) is given by \[ D_{\mathbf{a}} f = abla f(\mathbf{p}) \cdot \mathbf{u} \] where \( \mathbf{u} \) is the unit vector in the direction of \( \mathbf{a} \), and \( abla f(\mathbf{p}) \) is the gradient of \( f \) at point \( \mathbf{p} \).
02

Find the Unit Vector

To find the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{a} = 5 \mathbf{i} + 2 \mathbf{j} \), divide \( \mathbf{a} \) by its magnitude. First, calculate the magnitude of \( \mathbf{a} \): \[ \| \mathbf{a} \| = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29}. \] Thus, the unit vector \( \mathbf{u} \) is \( \mathbf{u} = \frac{1}{\sqrt{29}}(5 \mathbf{i} + 2 \mathbf{j}) \).
03

Calculate the Dot Product

Now compute the dot product of the gradient \( abla f(4, -5) = 2 \mathbf{i} - \mathbf{j} \) with the unit vector \( \mathbf{u} = \frac{1}{\sqrt{29}}(5 \mathbf{i} + 2 \mathbf{j}) \):\[ (2 \mathbf{i} - \mathbf{j}) \cdot \left(\frac{1}{\sqrt{29}}(5 \mathbf{i} + 2 \mathbf{j})\right) = \frac{1}{\sqrt{29}}((2 \times 5) + (-1 \times 2)) = \frac{1}{\sqrt{29}}(10 - 2) = \frac{8}{\sqrt{29}}. \]
04

Simplify the Result

The directional derivative is given by the dot product calculated in Step 3. Therefore, the directional derivative of the function \( f \) at the point \( (4, -5) \) in the direction of \( \mathbf{a} \) is \( \frac{8}{\sqrt{29}} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gradient
When we talk about the gradient, we're referring to a vector that indicates the direction of the steepest ascent of a function. It is denoted by the symbol \( abla f \) and is made up of the partial derivatives of the function with respect to each variable.
  • The gradient is essential in determining how the function changes at any particular point.
  • In the original problem, \( abla f(4,-5)=2 \mathbf{i}-\mathbf{j} \) tells us exactly how the function \( f \) changes at the point \((4,-5)\) based on the gradient vector's components, which point towards the direction of fastest increase.
  • Each component of the gradient vector points in the direction of increase for that specific coordinate.
Think of the gradient as a navigation tool. If you are on a hill and feel the steepest way up, the gradient tells you that direction precisely by pointing towards it. Thus, the gradient greatly aids in calculating directional derivatives, as it represents the change of the function at a specific point.
Unit Vector
A unit vector is a vector with a magnitude of 1. It鈥檚 specifically used to denote direction without considering magnitude. In directional derivatives, converting a direction vector into a unit vector is crucial.
  • The importance of a unit vector is that it allows us to focus solely on direction, removing any scaling factors from the vector's magnitude.
  • In the exercise, to find the unit vector \( \mathbf{u} \) from \( \mathbf{a} = 5 \mathbf{i} + 2 \mathbf{j} \), we first calculate the magnitude of \( \mathbf{a} \), which is \( \sqrt{29} \).
  • Then, we divide each component of \( \mathbf{a} \) by \( \sqrt{29} \) to transform it into a unit vector: \( \mathbf{u} = \frac{1}{\sqrt{29}}(5 \mathbf{i} + 2 \mathbf{j}) \).
Essentially, by ensuring our direction vector is a unit vector, we guarantee that the directional derivative measures the pure rate of change without any extra scaling related to direction length.
Dot Product
The dot product is an operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. It's a crucial concept when calculating the directional derivative.
  • The dot product measures the magnitude of one vector in the direction of another.
  • In our exercise, we computed the dot product between the gradient \( abla f(4,-5) = 2 \mathbf{i} - \mathbf{j} \) and the unit vector \( \mathbf{u} = \frac{1}{\sqrt{29}}(5 \mathbf{i} + 2 \mathbf{j}) \).
  • By using the dot product, we find the directional derivative: \( (2 \mathbf{i} - \mathbf{j}) \cdot \mathbf{u} = \frac{8}{\sqrt{29}} \).
Calculating the dot product here is about discovering how much the function changes as you move in the particular direction defined by the unit vector. The result indicates the rate and direction of change of the function at a given point, thereby providing a powerful insight into the behavior of functions.

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