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Verify that \(z_{x y}=z_{y x}\). $$ z=2 x^{3}+5 x^{2} y-6 y^{2}+x y^{4} $$

Short Answer

Expert verified
The mixed partial derivatives \(z_{xy}\) and \(z_{yx}\) are equal, verifying that \(z_{xy} = z_{yx}\).

Step by step solution

01

Understanding the Problem

We have a function \(z\) given by \(z = 2x^3 + 5x^2 y - 6y^2 + xy^4\). We need to show that the mixed partial derivatives \(z_{xy}\) and \(z_{yx}\) are equal.
02

Compute First Partial Derivatives

First, we compute the partial derivatives of \(z\) with respect to \(x\) and \(y\).\[\frac{\partial z}{\partial x} = \frac{d}{dx}(2x^3) + \frac{d}{dx}(5x^2 y) + \frac{d}{dx}(xy^4)\]This results in:\[z_x = 6x^2 + 10xy + y^4\]And then, with respect to \(y\):\[\frac{\partial z}{\partial y} = \frac{d}{dy}(5x^2 y) + \frac{d}{dy}(-6y^2) + \frac{d}{dy}(xy^4)\]This results in:\[z_y = 5x^2 - 12y + 4xy^3\]
03

Compute Mixed Partial Derivative z_{xy}

Next, we compute the mixed partial derivative \(z_{xy}\) by differentiating \(z_x\) with respect to \(y\).\[\frac{\partial}{\partial y}(z_x) = \frac{d}{dy}(6x^2 + 10xy + y^4)\]This results in:\[z_{xy} = 10x + 4y^3\]
04

Compute Mixed Partial Derivative z_{yx}

Similarly, we compute \(z_{yx}\) by differentiating \(z_y\) with respect to \(x\).\[\frac{\partial}{\partial x}(z_y) = \frac{d}{dx}(5x^2 - 12y + 4xy^3)\]This results in:\[z_{yx} = 10x + 4y^3\]
05

Compare the Mixed Partial Derivatives

We find that:\[ z_{xy} = 10x + 4y^3 \]and\[ z_{yx} = 10x + 4y^3 \]Since \(z_{xy} = z_{yx}\), the condition is verified, confirming that the mixed partial derivatives are equal.

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental concept when dealing with functions of several variables, such as the function given in our exercise: \( z = 2x^3 + 5x^2 y - 6y^2 + xy^4 \). In calculus, partial derivatives represent how a function changes with respect to one of its variables while keeping the others constant. This allows us to understand the function's rate of change in a particular direction. For instance, in our exercise, we have partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) indicating how \( z \) changes with \( x \) and \( y \) respectively, while the other variable remains constant.
To compute a partial derivative, differentiate the function with respect to the variable of interest and treat the other variables as constants. This process gives insight into the behavior of multivariable functions, resembling ordinary differentiation but with an additional focus on one variable at a time.
Function of Several Variables
A function of several variables refers to a mathematical expression where the outcome is determined by multiple inputs, just like the given function \( z = 2x^3 + 5x^2 y - 6y^2 + xy^4 \). Such functions are essential in fields like physics, engineering, and economics, where multiple factors influence a result. These functions can depend on variables such as \( x \) and \( y \), and understanding their behavior requires examining how changing any single variable affects the function's value.
Visualizing these functions can be quite complex, often requiring multilayered graphs or contour plots to represent the relationship among the variables. In our specific problem, each term of the function affects \( z \) differently based on the power and relationship of \( x \) and \( y \).
When working with functions of several variables, it is crucial to isolate each variable's effects through partial derivatives, which provide a clearer picture of how each input contributes to the overall output.
Differentiation
Differentiation is a key mathematical tool used to determine the rate at which a function changes. It is an essential part of calculus, particularly when dealing with variables impacting a function in different ways. For functions with multiple variables, differentiation extends into the realm of partial derivatives, where the function's sensitivity to modifications in one variable is explored independently of others.
In the exercise, we calculate the first partial derivatives \( z_x \) and \( z_y \), and then the mixed partial derivatives \( z_{xy} \) and \( z_{yx} \). Differentiation helps us verify that these mixed partial derivatives are equal, indicating that changes in order of differentiation do not affect the result.
This equality holds under the condition of continuous second-order derivatives, reflecting a fundamental property of smooth, well-behaved functions. Differentiation thus offers us powerful insights into how functions behave, providing tools to solve complex problems involving interactions between multiple variables.

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Most popular questions from this chapter

Find the directional derivative of \(f\) at \(P\) in the direction of \(\mathbf{v}\); that is, find \(D_{\mathbf{u}} f(P) . \quad\) where \(\quad \mathbf{u}=\frac{\mathbf{v}}{|\mathbf{v}|}\). $$ f(x, y, z)=\sqrt{x y z}: \quad P(2,-1,-2) \cdot \mathbf{v}=\mathbf{i}+2 \mathbf{j}-2 \mathbf{k} $$

Find the maximum directional derivative of \(f\) at \(P\) and the direction in which it occurs. $$ f(x, y, z)=\sqrt{2 x+4 y+6 z}: \quad P(7,5,5) $$

You are standing at the point \((30,20,5)\) on a hill with the shape of the surface $$ z=100 \exp \left(-\frac{x^{2}+3 y^{2}}{701}\right) $$ (a) In what direction (with what compass heading) should you proceed in order to climb the most steeply? At what angle from the horizontal will you initially be climbing? (b) If, instead of climbing as in part (a), you head directly west (the negative \(x\) -direction). then at what angle will you be climbing initially?

A rectangular open-topped box is to have volume \(700 \mathrm{in} .{ }^{3}\) The material for its bottom costs \(7 \mathrm{c} / \mathrm{in} .^{2}\), and the material for its four vertical sides costs \(5 \mathrm{c} / \mathrm{in} .^{2}\) Use the method of Lagrange multipliers to find what dimensions will minimize the cost of the material used in constructing this box.

The function \(z=f(x, y)\) describes the shape of a hill: \(f(P)\) is the altitude of the hill above the point \(P(x, y)\) in the \(x y\) -plane. If you start at the point \((P, f(P))\) of this hill, then \(D_{u} f(P)\) is your rate of climb (rise per unit of horizontal distance) as you proceed in the horizontal direction \(\mathbf{u}=a \mathbf{i}+b \mathbf{j}\). And the angle at which you climb while you walk in this direction is \(\gamma=\tan ^{-1}\left(D_{\mathrm{u}} f(P)\right)\), as shown in Fig. 13.8.11. You are standing at the point \((-100,-100,430)\) on a hill that has the shape of the graph of $$ z=500-(0.003) x^{2}-(0.004) y^{2} $$ with \(x, y\), and \(z\) given in feet. (a) What will be your rate of climb (rise over rum) if you head northwest? At what angle from the horizontal will you be climbing? (b) Repeat part (a). except now you head northeast.

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