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Let \(w=f(x, y, z)\) be a function of three independent variables and write the formal definition of the partial derivative \(\partial f / \partial z\) at \(\left(x_{0}, y_{0}, z_{0}\right)\) . Use this definition to find \(\partial f / \partial z\) at \((1,2,3)\) for \(f(x, y, z)=x^{y z}\).

Short Answer

Expert verified
The partial derivative \(\frac{\partial f}{\partial z}\) at \((1, 2, 3)\) is 0.

Step by step solution

01

Understanding Partial Derivative

The partial derivative of a function with respect to one of the variables represents how the function changes as that specific variable changes while keeping the other variables constant. The notation \(\frac{\partial f}{\partial z}\) indicates the partial derivative of \(f\) with respect to \(z\).
02

Defining Partial Derivative Formally

The formal definition of a partial derivative for a function \(f(x, y, z)\) with respect to \(z\) at a point \((x_0, y_0, z_0)\) is given by: \[ \left. \frac{\partial f}{\partial z} \right|_{(x_0, y_0, z_0)} = \lim_{h \to 0} \frac{f(x_0, y_0, z_0 + h) - f(x_0, y_0, z_0)}{h} \] This definition implies incrementing \(z\) by a small value \(h\) and observing how \(w=f(x, y, z)\) changes, keeping \(x\) and \(y\) constant.
03

Substituting Function Values

For the function \(f(x, y, z) = x^{yz}\), substitute \(x=1\), \(y=2\), and \(z=3\) into the definition of the partial derivative. This gives:\[ f(1, 2, 3) = 1^{2 \times 3} = 1 \] The function simplifies to 1 as \(1\) raised to any power is 1.
04

Calculating Derivative Using Definition

We calculate the partial derivative as follows:\[ \left. \frac{\partial f}{\partial z} \right|_{(1,2,3)} = \lim_{h \to 0} \frac{f(1, 2, 3+h) - f(1, 2, 3)}{h} \] By substitution:\[ f(1, 2, 3+h) = 1^{2(3+h)} = 1 \] Thus, \[ \left. \frac{\partial f}{\partial z} \right|_{(1,2,3)} = \lim_{h \to 0} \frac{1 - 1}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \]
05

Interpreting the Result

The computed result \(0\) indicates that the function \(f(x, y, z) = x^{yz}\) does not change with respect to \(z\) when evaluated at the point \((1,2,3)\), confirming the result from step 4.

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Key Concepts

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

Multivariable Calculus
Multivariable calculus is an extension of single-variable calculus to functions with multiple variables. Instead of dealing with functions of one variable, you can now explore functions that depend on two or more variables. This extension is crucial because many practical problems in the physical and social sciences involve functions of several variables.

In multivariable calculus, you work with concepts like partial derivatives, gradients, and multiple integrals. Partial derivatives, in particular, allow us to understand how a function changes with respect to each individual variable, holding the others constant. This concept is essential as it helps in analyzing surfaces and curves in higher dimensions.

When studying multivariable calculus, one often encounters formulas that include terms with more than one variable, such as 3D shapes. Here, partial derivatives aim to measure the rate of change along one variable while fixing the others, making it a significant tool to dissect complex behaviors in a multivariable context.
Functions of Multiple Variables
Functions of multiple variables are expressions where the function depends on more than one input. For example, a typical function might be something like \(f(x, y, z)\), where it takes in three variables and outputs a value based on these inputs.

These kinds of functions are common in real-world applications. Consider temperature in a room; it may depend on the position in the room (with coordinates \(x\) and \(y\) in a plane) and the height \(z\) above the floor. Therefore, the temperature can be represented as a function of three variables: \(T(x, y, z)\).

When dealing with functions of multiple variables, it helps to visualize them geometrically. In the case of a function \(f(x, y, z)\), you might imagine a three-dimensional surface in a four-dimensional space, which can be difficult to imagine, but diagrams and 3D models can help.
  • The domain is the set of all possible input values \( (x, y, z) \) over which the function is defined.
  • The range is the set of output values the function can produce.
This multi-parameter perspective is why partial derivatives are so useful; they help us see how changing one dimension influences the function while the others stay constant.
Calculus Definitions
To understand calculus, and particularly multivariable calculus, it's important to grasp key definitions and concepts that form the backbone of the subject.

  • Function: A relationship between a set of inputs and a set of permissible outputs, often represented as \(f(x, y, z)\) for three variables.
  • Derivative: In the single-variable context, it represents the rate of change of a function concerning its variable.
  • Partial Derivative: An extension of the derivative to functions of multiple variables. It measures how the function change with respect to one variable while keeping the others constant. For example, \(\frac{\partial f}{\partial z}\) means you examine how \(f\) changes as \(z\) changes, while \(x\) and \(y\) are held fixed.

Formal definition of a partial derivative involves limits:
\[ \left. \frac{\partial f}{\partial z} \right|_{(x_0, y_0, z_0)} = \lim_{h \to 0} \frac{f(x_0, y_0, z_0 + h) - f(x_0, y_0, z_0)}{h} \]

This captures the essence of the derivative by examining the effect of infinitesimally small changes in the variable of interest while others remain unchanged. Understanding these definitions solidifies your foundational knowledge and equips you for more advanced studies in calculus.

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Most popular questions from this chapter

In Exercises \(43-46,\) find the linearization \(L(x, y, z)\) of the function \(f(x, y, z)\) at \(P_{0}\) . Then find an upper bound for the magnitude of the error \(E\) in the approximation \(f(x, y, z) \approx L(x, y, z)\) over the region \(R\) . $$ \begin{array}{l}{f(x, y, z)=x z-3 y z+2 \quad \text { at } \quad P_{0}(1,1,2)} \\\ {R : \quad|x-1| \leq 0.01, \quad|y-1| \leq 0.01, \quad|z-2| \leq 0.02}\end{array} $$

In Exercises \(65-70,\) you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level critical plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. $$ \begin{array}{l}{f(x, y)=2 x^{4}+y^{4}-2 x^{2}-2 y^{2}+3, \quad-3 / 2 \leq x \leq 3 / 2} \\ {-3 / 2 \leq y \leq 3 / 2}\end{array} $$

In Exercises \(65-70,\) you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level critical plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. $$ \begin{array}{ll}{f(x, y)=\left\\{\begin{array}{ll}{x^{5} \ln \left(x^{2}+y^{2}\right),} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)}\end{array}\right.} \\ {-2 \leq x \leq 2,} & {(x, y)=(0,0)}\end{array} $$

Each of Exercises \(59-62\) gives a function \(f(x, y)\) and a positive number \(\epsilon .\) In each exercise, show that there exists a \(\delta>0\) such that for all \((x, y),\) $$ \sqrt{x^{2}+y^{2}}<\delta \Rightarrow|f(x, y)-f(0,0)|<\epsilon $$ $$ f(x, y)=(x+y) /\left(x^{2}+1\right), \quad \epsilon=0.01 $$

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0 .\) b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2} .\) d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z, w)=x^{2}+y^{2}+z^{2}+w^{2}\) subject to the constraints \(2 x-y+z-w-1=0\) and \(x+y-z+w-1=0\)

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