/*! 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 5 Find \begin{equation}\quad \te... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \begin{equation}\quad \text { a. }\left(\frac{\partial w}{\partial y}\right)_{x} \quad \text { b. }\left(\frac{\partial w}{\partial y}\right)_{z}\end{equation} at the point \((w, x, y, z)=(4,2,1,-1)\) if \begin{equation}w=x^{2} y^{2}+y z-z^{3} \quad \text { and } \quad x^{2}+y^{2}+z^{2}=6.\end{equation}

Short Answer

Expert verified
Both partial derivatives are 7 at the given point.

Step by step solution

01

Understand the Problem

We need to find the partial derivatives \( \left(\frac{\partial w}{\partial y}\right)_{x} \) and \( \left(\frac{\partial w}{\partial y}\right)_{z} \) at a specific point. These expressions require different conditions to be held constant: the first derivative holds \( x \) constant while evaluating the change in \( w \) with respect to \( y \), and the second derivative holds \( z \) constant.
02

Differentiate with Respect to y Keeping x Constant

Given the equation \( w = x^2 y^2 + yz - z^3 \), we differentiate \( w \) with respect to \( y \), holding \( x \) constant. This gives:\[\left(\frac{\partial w}{\partial y}\right)_{x} = 2x^2 y + z\]Substitute \( x = 2 \), \( y = 1 \), and \( z = -1 \) into this equation, yielding:\[\left(\frac{\partial w}{\partial y}\right)_{x} = 2(2)^2(1) + (-1) = 8 - 1 = 7\]
03

Establish Relationship for z being Constant

Use the constraint \( x^2 + y^2 + z^2 = 6 \) to find another expression for \( y \) as a function of \( x \) and \( z \) while keeping \( z \) constant. Since \( z = -1 \), substitute back into the equation to find \( y^2 = 6 - x^2 - 1^2 = 5 - x^2 \).
04

Differentiate with Respect to y Keeping z Constant

Using implicit differentiation from the constraint equation, apply the chain rule:- Interaction between these variables implies \( \left(\frac{\partial w}{\partial y}\right)_{z} = \frac{\partial w}{\partial y} \) because the relationship given by the constraint allows us to solve directly.The expression we found in Step 2 is still valid, so we have:\[\left(\frac{\partial w}{\partial y}\right)_{z} = 2x^2 y + z\]Thus, substituting back the values \( x = 2 \), \( y = 1 \), and \( z = -1 \) again results in:\[2 \times 2^2 \times 1 + (-1) = 8 - 1 = 7\]
05

Combine Results

Both derivatives under different conditions yield the same numerical value at the given point. Thus, both derivatives \( \left(\frac{\partial w}{\partial y}\right)_{x} \) and \( \left(\frac{\partial w}{\partial y}\right)_{z} \) are equal to 7 at \( (w, x, y, z) = (4, 2, 1, -1) \).

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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 calculus that deals with functions of multiple variables. Unlike single-variable calculus, where we focus on a function with one independent variable, multivariable calculus handles functions depending on two or more variables. These functions take the form \( f(x, y, z, ...) \), allowing us to explore higher-dimensional spaces.

In multivariable calculus, partial derivatives are a fundamental concept. A partial derivative represents how a function changes as one of its variables changes, while all other variables remain constant. This is crucial in understanding how multi-variable functions behave in different directions.

Let's take a function \( w = f(x, y, z) \). The partial derivative \( \left( \frac{\partial w}{\partial y} \right)_x \) refers to the rate of change of \( w \) with respect to \( y \), while keeping \( x \) constant. If our function includes relationships like \( w = x^2 y^2 + yz - z^3 \), it provides a window into how these variables interconnect and influence each other in a multi-variable context.
Implicit differentiation
Implicit differentiation is a technique used when we differentiate equations that define one or more variables implicitly rather than solving one variable in terms of others explicitly. In multivariable contexts, it is often not feasible to express one variable solely in terms of others due to complex relationships between variables.

For example, in the provided constraint \( x^2 + y^2 + z^2 = 6 \), we cannot easily solve for one variable. Instead, implicit differentiation allows us to differentiate both sides of an equation with respect to one variable, while treating the other variables as functions.
  • First, differentiate the entire equation with respect to \( y \).
  • Treat other variables as constant or as functions that depend on \( y \).
  • Apply the chain rule where necessary to account for these dependencies.
This technique is particularly useful to find partial derivatives when constraints link variables as it allows us to account for changes without needing explicit functional forms.
Constraint equations
Constraint equations are relationships that the variables in a problem must satisfy. They restrict the values the variables can take. In a multivariable calculus context, these equations enable or restrict the paths through which we explore a function's behavior.

For the problem at hand, the constraint is \( x^2 + y^2 + z^2 = 6 \). This spherical constraint limits the combination of \( x, y, \) and \( z \) values that the function can consider.
  • These constraints are essential to solving real-world problems since they reflect real limitations or conditions.
  • Constraint equations assist in understanding permissible variable configurations.
  • They offer critical insight while performing differentiation since they may redefine our view of variable independence.
Effectively working with constraint equations involves recognizing these limitations and interpreting them into our calculations, as seen in this problem where we first use the constraint to redefine \( y \) in terms of \( x \) and \( z \).

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