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Chapter 3: Problem 19

For the following exercises, find the directional derivative of the function in the direction of the unit vector \(\mathbf{u}=\cos \theta \mathbf{i}+\sin \theta \mathbf{j}\).\(w(x, y)=y e^{x}, \theta=\frac{\pi}{3}\)

Short Answer

Expert verified
The directional derivative is \( \frac{y e^{x}}{2} + \frac{\sqrt{3} e^{x}}{2} \).

Step by step solution

01

Find the Gradient of the Function

First, we determine the gradient of the function \( w(x, y) = y e^{x} \). The gradient, denoted as \( abla w \), consists of the partial derivatives of the function with respect to \( x \) and \( y \).\[ abla w = \left( \frac{\partial w}{\partial x}, \frac{\partial w}{\partial y} \right) \] Calculate each partial derivative: - \( \frac{\partial w}{\partial x} = y e^{x} \) - \( \frac{\partial w}{\partial y} = e^{x} \) Therefore, the gradient is \( abla w = (y e^{x}, e^{x}) \).
02

Compute the Unit Vector

The unit vector is given as \( \mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j} \) with \( \theta = \frac{\pi}{3} \). Calculate \( \cos \theta \) and \( \sin \theta \):- \( \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \) - \( \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)Therefore, the unit vector is \( \mathbf{u} = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \).
03

Calculate the Directional Derivative

Using the formula for the directional derivative \( D_\mathbf{u} w = abla w \cdot \mathbf{u} \), substitute the gradient and the unit vector:\[ D_\mathbf{u} w = (y e^{x}, e^{x}) \cdot \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \]Compute the dot product:\[ D_\mathbf{u} w = y e^{x} \cdot \frac{1}{2} + e^{x} \cdot \frac{\sqrt{3}}{2} \]\[ D_\mathbf{u} w = \frac{y e^{x}}{2} + \frac{\sqrt{3} e^{x}}{2} \]This represents the directional derivative of \( w \) at any point \( (x, y) \) in the direction specified by \( \mathbf{u} \).

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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 a fundamental concept in multivariable calculus that describes the direction and rate of the fastest increase of the function. In essence, it's a vector that combines the partial derivatives of a function. For a function like \( w(x, y) = y e^{x} \), the gradient \( abla w \) is calculated by finding the partial derivatives with respect to each variable.
  • The partial derivative with respect to \( x \) is \( \frac{\partial w}{\partial x} = y e^{x} \), indicating how \( w \) changes as \( x \) changes, holding \( y \) constant.
  • The partial derivative with respect to \( y \) is \( \frac{\partial w}{\partial y} = e^{x} \), showing how \( w \) changes as \( y \) changes, keeping \( x \) constant.
Combining these, the gradient vector is \( abla w = (y e^{x}, e^{x}) \), effectively guiding us to where the function grows the most steeply at each point.
Unit Vector
A unit vector is a vector that has a magnitude of 1. It's often used to signify direction without regard to scale. In the context of directional derivatives, the unit vector \( \mathbf{u} \) specifies the direction in which we calculate the derivative of the function.For the unit vector \( \mathbf{u} = \cos \theta \mathbf{i} + \sin \theta \mathbf{j} \), where \( \theta = \frac{\pi}{3} \), we calculate:
  • \( \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \) tells us how much of the direction is along the \( x \)-axis (\( \mathbf{i} \) direction).
  • \( \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \) represents the imbalance towards the \( y \)-axis (\( \mathbf{j} \) direction).
Thus, \( \mathbf{u} = \frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \) confirms that it鈥檚 normalized to 1, making it the perfect candidate to describe directional derivatives truly.
Partial Derivatives
Partial derivatives are crucial when dealing with functions of several variables. They measure how a function's value changes as just one of the input variables changes, keeping the others fixed.Consider the function \( w(x, y) = y e^{x} \). The partial derivatives break down like this:
  • \( \frac{\partial w}{\partial x} = y e^{x} \) shows how \( w \) reacts to small changes in \( x \), viewing \( y \) as constant.
  • \( \frac{\partial w}{\partial y} = e^{x} \) reveals how \( w \) changes as \( y \) varies, keeping \( x \) steady.
These derivatives are the building blocks of the gradient, helping us navigate the multi-dimensional landscape of the function by providing insight into how each variable independently influences the overall picture.

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Most popular questions from this chapter

A rectangular solid is contained within a tetrahedron with vertices at \((1,0,0),(0,1,0),(0,0,1)\), and the origin. The base of the box has dimensions \(x, y\), and the height of the box is \(z .\) If the sum of \(x, y\), and \(z\) is \(1.0\), find the dimensions that maximizes the volume of the rectangular solid.

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