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Chapter 11: Problem 1

Find the directional derivative of \(f\) at the given point in the direction indicated by the angle \(\theta .\) $$f(x, y)=y e^{-x}, \quad(0,4), \quad \theta=2 \pi / 3$$

Short Answer

Expert verified
The directional derivative of \( f \) at point \( (0, 4) \) in the direction of \( \theta = \frac{2\pi}{3} \) is \( 2 + \frac{\sqrt{3}}{2} \).

Step by step solution

01

Compute Gradient of f

First, find the gradient of the function \( f(x, y) = y e^{-x} \). The gradient \( abla f(x, y) \) is calculated by finding the partial derivatives with respect to \( x \) and \( y \). For \( \frac{\partial f}{\partial x} = -y e^{-x} \) and for \( \frac{\partial f}{\partial y} = e^{-x} \), thus the gradient is \( abla f(x, y) = (-y e^{-x}, e^{-x}) \).
02

Evaluate Gradient at the Given Point

Substitute the point \( (0, 4) \) into the gradient function. Compute \( abla f(0, 4) = (-4 e^{0}, e^{0}) = (-4, 1) \).
03

Determine the Unit Vector in Direction Theta

Convert the given angle \( \theta = \frac{2\pi}{3} \) to a unit vector. The unit vector is \( \mathbf{u} = (\cos(\frac{2\pi}{3}), \sin(\frac{2\pi}{3})) = (-\frac{1}{2}, \frac{\sqrt{3}}{2}) \).
04

Calculate the Directional Derivative

Using the gradient at the point and the unit direction vector, calculate the directional derivative using the dot product formula: \( D_{\mathbf{u}}f = abla f \cdot \mathbf{u} \). Compute: \[ D_{\mathbf{u}}f = (-4, 1) \cdot \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) = -4 \left(-\frac{1}{2}\right) + 1 \cdot \frac{\sqrt{3}}{2} = 2 + \frac{\sqrt{3}}{2} \].
05

Final Result Simplification

Combine the terms to express the directional derivative in a simplified form: \[ D_{\mathbf{u}}f = 2 + \frac{\sqrt{3}}{2} \].

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Key Concepts

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

Gradient
The gradient of a function is like a map that shows which direction to go in to experience the fastest increase in the function's value. Consider it a vector that points you precisely in that direction. For a function of two variables, such as our given example, the gradient is a vector composed of the partial derivatives with respect to each of those variables.
  • The gradient of a function \( f(x, y) \) is represented as \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
  • This vector tells you in which direction the function, \( f \), increases most rapidly from any point \( (x, y) \).
For our function, \( f(x, y) = y e^{-x} \), the gradient is \( abla f(x, y) = (-y e^{-x}, e^{-x}) \). By substituting the point \( (0, 4) \), we specifically know how the function behaves at that exact location on its surface.
Partial Derivatives
Partial derivatives are the building blocks for understanding changes in a function of several variables. They are analogous to regular derivatives but specifically focused on how a function changes as one variable changes, while keeping others constant.
  • For a function \( f(x, y) \), the partial derivative \( \frac{\partial f}{\partial x} \) captures how \( f \) changes as \( x \) changes, with \( y \) kept constant.
  • Similarly, \( \frac{\partial f}{\partial y} \) shows how \( f \) changes with a change in \( y \), with \( x \) remaining the same.
By finding these derivatives, you get a sense of the function's slope in different directions. In our example, \( \frac{\partial f}{\partial x} = -y e^{-x} \) and \( \frac{\partial f}{\partial y} = e^{-x} \), which together form the gradient vector. This step is crucial in determining the directional derivative.
Unit Vector
A unit vector is a vector with a magnitude of one. It's primarily used to indicate direction without concerning magnitude. When we discuss directions in space, unit vectors play a critical role. They ensure that when multiplying with another vector, such as a gradient, they only influence the direction and not the magnitude.
  • A unit vector \( \mathbf{u} \) can often be represented as \( \mathbf{u} = (\cos(\theta), \sin(\theta)) \), where \( \theta \) is the angle indicating direction.
  • In our exercise, the angle \( \theta = \frac{2\pi}{3} \) results in the unit vector \( \mathbf{u} = (-\frac{1}{2}, \frac{\sqrt{3}}{2}) \).
Utilizing unit vectors ensures that when we compute the directional derivative, the focus remains on the direction specified by \( \theta \) while keeping the impact on magnitude neutral.
Dot Product
The dot product serves as a crucial tool in calculating the directional derivative. It's a method of multiplying vectors that results in a scalar (a single number). It effectively combines the magnitudes of two vectors and the cosine of the angle between them to determine some form of combined effect along a direction.
  • The dot product of two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \) is given by \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \).
  • In the context of directional derivatives, the dot product \( abla f \cdot \mathbf{u} \) allows us to combine the gradient and the unit vector to find out how much of the gradient's direction is aligned with the specified direction \( \mathbf{u} \).
In our problem, the calculation \( (-4, 1) \cdot \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \) yielded \( 2 + \frac{\sqrt{3}}{2} \), which represents how strongly the function \( f \) escalates in the direction defined by \( \mathbf{u} \) from the point \( (0, 4) \).

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Most popular questions from this chapter

Find the extreme values of \(f\) subject to both constraints. $$ f(x, y, z)=x^{2}+y^{2}+z^{2} ; \quad x-y=1, \quad y^{2}-z^{2}=1 $$

Suppose that the directional derivatives of \(f(x, y)\) are known at a given point in two nonparallel directions given by unit vectors \(\mathbf{u}\) and \(\mathbf{v} .\) Is it possible to find \(\nabla f\) at this point? If so, how would you do it?

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=x^{2}+x y+y^{2}+y$$

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=y^{3}+3 x^{2} y-6 x^{2}-6 y^{2}+2$$

Find the extreme values of \(f\) subject to both constraints. $$ f(x, y, z)=y z+x y ; \quad x y=1, \quad y^{2}+z^{2}=1 $$
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