/*! 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 37 Find the gradient of the functio... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the gradient of the function and the maximum value of the directional derivative at the given point. $$ \begin{array}{ll} \underline{\text { Function}} & \underline{\text {Point}} \\ f(x, y, z)=x e^{y z} &\quad (2,0,-4) \end{array} $$

Short Answer

Expert verified
The gradient of the function at the point (2,0,-4) is [1, 0, 0] and the maximum value of the directional derivative at this point is 1.

Step by step solution

01

Calculate the Partial Derivatives

Begin by obtaining the partial derivatives of the given function with respect to each variable:The partial derivative of \( f \) with respect to \( x \) is : \( f_x(x, y, z) = e^{y z} \)The partial derivative of \( f \) with respect to \( y \) is : \( f_y(x, y, z) = z x e^{y z} \)The partial derivative of \( f \) with respect to \( z \) is : \( f_z(x, y, z) = y x e^{y z} \)
02

Evaluate the Partial Derivatives at the Given Point

Substitute the given point \((2,0,-4)\) into the partial derivatives:\( f_x(2,0,-4) = e^{0 * -4} = 1 \)\( f_y(2,0,-4) = -4 * 2 * e^{0 * -4} = 0 \)\( f_z(2,0,-4) = 0 * 2 * e^{0 * -4} = 0 \)
03

Determine the Gradient Vector

The gradient of \( f \) is a vector of its partial derivatives, thus:\( \nabla f(2,0,-4) = [ f_x(2,0,-4), f_y(2,0,-4), f_z(2,0,-4) ] = [1,0,0] \)
04

Calculate the Maximum Value of the Directional Derivative

The directional derivative reaches its maximum value in the direction of the gradient, which, in this case, is [1,0,0]. The magnitude of the gradient vector determines the maximum value of the directional derivative. In this case, the gradient vector is simply [1,0,0], so the maximum value of the directional derivative is \( ||\nabla f(2,0,-4)|| = \sqrt{1^2 + 0^2 + 0^2} = 1 \).

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Key Concepts

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

Directional Derivative
To understand what a directional derivative is, think of it as measuring the rate at which a function changes as you move in a specified direction. It extends the idea of a derivative from one-dimensional functions to multi-dimensional functions. When we calculate a directional derivative, we are essentially asking: "If I were to move a small step in a certain direction from a given point, how does the function value change?" This is essential for understanding the behavior of the function in the surrounding space.
If we have a function, for instance, the directional derivative in the direction of a vector \( \vec{u} \) at a point \( \vec{p} \) can be expressed mathematically as:
\[ D_{\vec{u}} f(\vec{p}) = abla f(\vec{p}) \cdot \vec{u} \]
Where:
  • \( abla f \) represents the gradient of the function \( f \).
  • \( \cdot \) denotes the dot product.
  • \( \vec{u} \) is a unit vector indicating the direction.
This value lets us know how steeply our function is climbing or falling as we go in the given direction. A key point is that the directional derivative reaches its maximum when \( \vec{u} \) aligns with the gradient \( abla f \), which tells us the function increases the most quickly in the gradient's direction.
Partial Derivatives
Partial derivatives help you to understand how a function changes when you vary just one of its variables, keeping the others constant. It is similar to slicing through a multi-dimensional object to examine one dimension at a time. Consider the function \( f(x, y, z) = x e^{yz} \), where changes in \( x \), \( y \), and \( z \) affect the outcome.
To find partial derivatives, follow these steps:
  • To compute \( \frac{\partial f}{\partial x} \), differentiate \( f \) as if \( y \) and \( z \) were constants, giving us \( e^{yz} \).
  • To compute \( \frac{\partial f}{\partial y} \), differentiate \( f \) as if \( x \) and \( z \) were constants, resulting in \( xz e^{yz} \).
  • To compute \( \frac{\partial f}{\partial z} \), do the same with \( x \) and \( y \) taken as constants, yielding \( xy e^{yz} \).
Partial derivatives are foundational for calculating the gradient, which tells us how a multi-variable function varies in all dimensions. They provide the building blocks for understanding more complex changes in multivariable functions.
Calculus
Calculus is the mathematical study of continuous change and is often considered the magic tool for comprehending the world of change. It is divided into two main branches: differential calculus and integral calculus. Differential calculus focuses on rates of change, which is where derivatives come in, solving problems like understanding the slope of a curve at a point. Integral calculus, meanwhile, deals with accumulation of quantities, such as areas under a curve.
Derivatives and gradients come into play frequently in differential calculus, especially when dealing with functions of several variables. One important aspect is the gradient, a vector field that points in the direction of the steepest ascent of the function.
In our case, calculus shows us how to compute derivatives to understand a function's behavior at any given point. This includes:
  • Calculating partial derivatives to understand changes in multi-variable functions.
  • Using the gradient to find directions in which the function increases or decreases most rapidly.
  • Determining the directional derivative to figure out the rate of change in specific directions.
By mastering calculus, you can tackle not only academic problems but also real-world scenarios needing precision and understanding of how variables interact and change.

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Most popular questions from this chapter

Modeling Data The table shows the net sales \(x\) (in billions of dollars), the total assets \(y\) (in billions of dollars), and the shareholder's equity \(z\) (in billions of dollars) for Wal-Mart for the years 1998 through 2003. (Source: 2003 Annual Report for Wal-Mart ) \begin{tabular}{|l|c|c|c|c|c|c|} \hline Year & 1998 & 1999 & 2000 & 2001 & 2002 & 2003 \\ \hline\(x\) & \(118.0\) & \(137.6\) & \(165.0\) & \(191.3\) & \(217.8\) & \(244.5\) \\ \hline\(y\) & \(45.4\) & \(50.0\) & \(70.3\) & \(78.1\) & \(83.5\) & \(94.7\) \\ \hline\(z\) & \(18.5\) & \(21.1\) & \(25.8\) & \(31.3\) & \(35.1\) & \(39.3\) \\ \hline \end{tabular} A model for these data is \(z=f(x, y)=0.156 x+0.031 y-1.66\) (a) Use a graphing utility and the model to approximate \(z\) for the given values of \(x\) and \(y\). (b) Which of the two variables in this model has the greater influence on shareholder's equity? (c) Simplify the expression for \(f(x, 55)\) and interpret its meaning in the context of the problem.

Use Lagrange multipliers to find the minimum distance from the curve or surface to the indicated point. [Hint: In Exercise 23, minimize \(f(x, y)=x^{2}+y^{2}\) subject to the constraint \(2 x+3 y=-1 .]\) $$ \begin{array}{ll} \underline{\text { Curve }} & \underline{\text {Point}} \\ \text { Line: } 2 x+3 y=-1 \quad (0,0) \end{array} $$

A cargo container (in the shape of a rectangular solid) must have a volume of 480 cubic feet. The bottom will cost \(\$ 5\) per square foot to construct and the sides and the top will cost \(\$ 3\) per square foot to construct. Use Lagrange multipliers to find the dimensions of the container of this size that has minimum cost.

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The utility function \(U=f(x, y)\) is a measure of the utility (or satisfaction) derived by a person from the consumption of two products \(x\) and \(y .\) Suppose the utility function is \(U=-5 x^{2}+x y-3 y^{2}\) (a) Determine the marginal utility of product \(x\). (b) Determine the marginal utility of product \(y\). (c) When \(x=2\) and \(y=3\), should a person consume one more unit of product \(x\) or one more unit of product \(y\) ? Explain your reasoning. (d) Use a computer algebra system to graph the function. Interpret the marginal utilities of products \(x\) and \(y\) graphically.

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