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Chapter 14: Problem 34

\(\begin{array}{l}{\text { Find } \partial w / \partial v \quad \text { when } \quad u=-1, v=2 \quad \text { if } \quad w=x y+\ln z} \\ {x=v^{2} / u, y=u+v, z=\cos u}\end{array}\)

Short Answer

Expert verified
The value of \(\partial w / \partial v\) when \(u = -1, v = 2\) is \(-8\).

Step by step solution

01

Express w in terms of u and v

Given the expressions for \(x\), \(y\), and \(z\) in terms of \(u\) and \(v\), we need to write \(w\) as a function of \(u\) and \(v\):\[ w = xy + \ln z = \left(\frac{v^2}{u}\right)(u+v) + \ln(\cos u).\]
02

Simplify the expression for w

Substitute the expressions for \(x\), \(y\), and simplify: \[w = \frac{v^2}{u}(u+v) + \ln(\cos u) = (v^2 + \frac{v^3}{u}) + \ln(\cos u). \] This gives:\[w = v^2 + \frac{v^3}{u} + \ln(\cos u).\]
03

Differentiate w with respect to v

Use partial differentiation to find \(\frac{\partial w}{\partial v}\):\[\frac{\partial w}{\partial v} = \frac{\partial}{\partial v}\left(v^2 + \frac{v^3}{u} + \ln(\cos u)\right).\] The terms simplify to:\[\frac{\partial w}{\partial v} = 2v + \frac{3v^2}{u}.\]
04

Evaluate the derivative at given points

Substitute \(u = -1\) and \(v = 2\) into \(\frac{\partial w}{\partial v} = 2v + \frac{3v^2}{u}\): \[\frac{\partial w}{\partial v}\Big|_{u=-1, v=2} = 2(2) + \frac{3(2)^2}{-1} = 4 - \frac{12}{1}.\]This simplifies to \(4 - 12 = -8\).

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Key Concepts

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

Functions of Several Variables
Understanding functions of several variables is fundamental in calculus when dealing with equations involving more than one input. In this context, you have expressions like \(w = xy + \ln z\), where \(x, y,\) and \(z\) are themselves dependent on other variables such as \(u\) and \(v\).
  • A function of several variables takes multiple inputs and outputs a single value.
  • The variables can depend on other parameters, making them functions themselves.
  • You can think of \(w\) as "changing" if you vary \(u\) or \(v\); hence, \(w\)'s behavior is influenced by these underlying variables.
Expressing \(w\) fully in terms of \(u\) and \(v\) involves substituting these variables into their respective expressions, such as \(x = \frac{v^2}{u}, y = u+v, z = \cos u\). This clarity is essential for further differentiation operations.
Derivatives
Derivatives are a measure of how a function changes as its input changes. In the context of partial derivatives, we focus on how a function changes with respect to one particular variable, keeping others constant.
  • Partial derivatives allow us to explore the rate of change of multivariable functions with respect to some but not all variables.
  • In our exercise, differentiating \(w\) with respect to \(v\) requires holding \(u\) constant.
  • Applying the derivative formula to \(w = v^2 + \frac{v^3}{u} + \ln(\cos u)\) involves calculating \(\frac{\partial w}{\partial v} = 2v + \frac{3v^2}{u}\).
The result \(\frac{\partial w}{\partial v} = 2v + \frac{3v^2}{u}\) illustrates how \(w\) changes as \(v\) varies, while \(u\) remains unchanged. Thus, partial differentiation can provide nuanced insights into how different variables influence a function.
Chain Rule
The chain rule is a crucial tool when taking derivatives of composite functions, or when variables are functions of other variables. Essentially, it provides a method to differentiate complex expressions by proceeding through each nested function.
  • Whenever functions are composed of other functions, and you need to differentiate them, the chain rule comes into play.
  • For partial differentiation, the chain rule helps manage situations where each variable involved is a function of several other variables.
  • In our task, while \(w\) is directly expressed as a function of \(u\) and \(v\), understanding \(x, y, z\) connections also exemplifies the use of nested functions.
Utilizing the chain rule effectively enables us to systematically handle the different layers of dependency among variables. This helps simplify and correctly compute the overall derivative when complex interdependencies exist, as seen with \(w = xy + \ln z\).

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Most popular questions from this chapter

In Exercises \(53-60,\) sketch a typical level surface for the function. $$ f(x, y, z)=x^{2}+y^{2}+z^{2} $$

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 \(f(x, y, z)=x y+y z\) subject to the constraints \(x^{2}+y^{2}-\) \(2=0\) and \(x^{2}+z^{2}-2=0.\)

Find all the second-order partial derivatives of the functions in Exercises \(41-50 .\) $$r(x, y)=\ln (x+y)$$

Which order of differentiation will calculate \(f_{x y}\) faster: \(x\) first or \(y\) first? Try to answer without writing anything down. $$\begin{array}{l}{\text { a. } f(x, y)=x \sin y+e^{y}} \\ {\text { b. } f(x, y)=1 / x} \\ {\text { c. } f(x, y)=y+(x / y)} \\ {\text { d. } f(x, y)=y+x^{2} y+4 y^{3}-\ln \left(y^{2}+1\right)} \\ {\text { e. } f(x, y)=x^{2}+5 x y+\sin x+7 e^{x}} \\ {\text { f. } f(x, y)=x \ln x y}\end{array}$$

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 curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$f(x, y)=\left\\{\begin{array}{ll}{x^{5} \ln \left(x^{2}+y^{2}\right),} & {(x, y) \neq(0,0)} \\ {0,} & {(x, y)=(0,0)}\end{array}\right.$$ $$-2 \leq x \leq 2, \quad-2 \leq y \leq 2$$
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