Problem 21
Question
Determine \(d f\). $$ f(x, y, z)=x e^{y^{2}-z^{2}} $$
Step-by-Step Solution
Verified Answer
The differential is \(df = e^{y^2 - z^2} dx + 2xy e^{y^2 - z^2} dy - 2xz e^{y^2 - z^2} dz\).
1Step 1: Understand the Problem
We are given a function \(f(x, y, z) = x e^{y^2 - z^2}\) and need to find its differential \(df\). This involves finding the partial derivatives of the function with respect to each variable and then applying the formula for the total differential.
2Step 2: Find Partial Derivatives
First, we find the partial derivative of \(f\) with respect to \(x\): \[\frac{\partial f}{\partial x} = e^{y^2 - z^2}\] Then, the partial derivative with respect to \(y\): \[\frac{\partial f}{\partial y} = x \cdot e^{y^2 - z^2} \cdot 2y = 2xy e^{y^2 - z^2}\] Finally, the partial derivative with respect to \(z\): \[\frac{\partial f}{\partial z} = x \cdot e^{y^2 - z^2} \cdot (-2z) = -2xz e^{y^2 - z^2}\]
3Step 3: Write the Differential
Using the formula for the total differential: \[ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz\] Substituting the partial derivatives: \[ df = e^{y^2 - z^2} dx + 2xy e^{y^2 - z^2} dy - 2xz e^{y^2 - z^2} dz\] Thus, \(df\) is fully expressed in terms of \(dx\), \(dy\), and \(dz\).
Key Concepts
Partial DerivativesMultivariable CalculusDifferential Calculus
Partial Derivatives
In multivariable calculus, a partial derivative is a derivative where you fix all but one of the variables. This means you differentiate with respect to one variable while treating the others as constants.
This concept is especially important when dealing with functions of multiple variables, such as in the given function, \(f(x, y, z) = x e^{y^2 - z^2}\).
We have three variables: \(x\), \(y\), and \(z\). The partial derivative of \(f\) with respect to each variable gives the rate at which \(f\) changes as that particular variable changes, while the others remain constant.
This concept is especially important when dealing with functions of multiple variables, such as in the given function, \(f(x, y, z) = x e^{y^2 - z^2}\).
We have three variables: \(x\), \(y\), and \(z\). The partial derivative of \(f\) with respect to each variable gives the rate at which \(f\) changes as that particular variable changes, while the others remain constant.
- Partial Derivative with respect to \(x\): Here, \(y\) and \(z\) are constants. Differentiation yields \(\frac{\partial f}{\partial x} = e^{y^2 - z^2}\).
- Partial Derivative with respect to \(y\): Now, \(x\) and \(z\) are constants, and we find \(\frac{\partial f}{\partial y} = 2xy e^{y^2 - z^2}\).
- Partial Derivative with respect to \(z\): Here, \(x\) and \(y\) are constants, resulting in \(\frac{\partial f}{\partial z} = -2xz e^{y^2 - z^2}\).
Multivariable Calculus
Multivariable calculus is an extension of single-variable calculus and involves more than one variable. It is particularly useful for dealing with functions that depend on multiple inputs.
In the function \(f(x, y, z) = x e^{y^2 - z^2}\), we are considering how each of the three input variables affects the output. Each variable can influence the outcome differently, and multivariable calculus provides the tools to analyze these effects precisely.
In contrast to single-variable calculus, where we find derivatives with respect to one variable, in multivariable calculus, we use gradients and partial derivatives to understand changes in multiple directions. This helps in applications like physics, engineering, and more, where systems often depend on various simultaneous factors.
In the function \(f(x, y, z) = x e^{y^2 - z^2}\), we are considering how each of the three input variables affects the output. Each variable can influence the outcome differently, and multivariable calculus provides the tools to analyze these effects precisely.
In contrast to single-variable calculus, where we find derivatives with respect to one variable, in multivariable calculus, we use gradients and partial derivatives to understand changes in multiple directions. This helps in applications like physics, engineering, and more, where systems often depend on various simultaneous factors.
- Total Differential: This is a significant tool in multivariable calculus. It approximates the value of the function based on small changes in the input variables, using partial derivatives.
- Geometrical Interpretation: Functions of multiple variables can often be visualized as surfaces in three-dimensional space. The total differential helps to approximate slopes or gradients of these surfaces, contributing to understanding their geometry.
Differential Calculus
Differential calculus is the branch of calculus revolving around the concept of a derivative, which represents how a function changes as its inputs change.
For multivariable functions, like \(f(x, y, z) = x e^{y^2 - z^2}\), this means understanding how changes in each variable affect the function's output. The total differential \(df\) encapsulates these changes in consolidated form.
For multivariable functions, like \(f(x, y, z) = x e^{y^2 - z^2}\), this means understanding how changes in each variable affect the function's output. The total differential \(df\) encapsulates these changes in consolidated form.
- Calculating the Total Differential: Once you have partial derivatives, they are combined to form the total differential. For instance, \(df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz\).
- Practical Application: This is useful for estimating how outputs change in response to small changes in any or all of the variable inputs. It gives insight into the function's sensitivity and response to varying conditions.
Other exercises in this chapter
Problem 21
Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ f(x, y)=\sin x+
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Find \(\partial w / \partial u\) and \(\partial w / \partial v\). $$ w=\frac{y z}{x^{2}+x y} ; x=u^{2}, y=v^{2}, z=u^{2}-v^{2} $$
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From (1) it follows that the directional derivative of a function \(f\) at a point is smallest in the direction opposite to the gradient of \(f\) at that point.
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Find the first partial derivatives of \(f\) at the given point. $$ f(x, y)=\sqrt{4 x^{2}+y^{2}} ;(2,-3) $$
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