Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 40: Problem 2
If \(z=f(u, v, w)\) and \(z=3 u^{2}-2 v+4 w^{3} v^{2}\) find the total differential, \(\mathrm{d} z\)
Short Answer
Expert verified
The total differential is \( dz = 6u \, du + (-2 + 8w^3v) \, dv + 12w^2v^2 \, dw \).
Step by step solution
01
Identify the Function
The given function is a multivariable function: \[ z = 3u^2 - 2v + 4w^3v^2 \] where \(z\) depends on \(u\), \(v\), and \(w\). To find the total differential, we need to differentiate with respect to each variable.
02
Differentiate with Respect to u
Partial differentiate \(z\) with respect to \(u\):\[ \frac{\partial z}{\partial u} = 6u \] This is because the derivative of \(3u^2\) with respect to \(u\) is \(6u\), and the terms \(-2v\) and \(4w^3v^2\) are constants with respect to \(u\).
03
Differentiate with Respect to v
Partial differentiate \(z\) with respect to \(v\):\[ \frac{\partial z}{\partial v} = -2 + 8w^3v \] This is derived by differentiating \(-2v\) to get \(-2\) and \(4w^3v^2\) to get \(8w^3v\).
04
Differentiate with Respect to w
Partial differentiate \(z\) with respect to \(w\):\[ \frac{\partial z}{\partial w} = 12w^2v^2 \] This results from differentiating the term \(4w^3v^2\) with respect to \(w\), while \(3u^2\) and \(-2v\) are constant relative to \(w\).
05
Substitute into Total Differential Formula
The total differential \(dz\) is given by:\[ dz = \frac{\partial z}{\partial u} du + \frac{\partial z}{\partial v} dv + \frac{\partial z}{\partial w} dw \] Substitute the partial derivatives into this formula:\[ dz = 6u \, du + (-2 + 8w^3v) \, dv + 12w^2v^2 \, dw \] This expression represents the total differential of \(z\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept in calculus that deals with functions of several variables. When you have a function like \( z = f(u, v, w) \), you can take the derivative with respect to each of those variables, treating all others as constants during differentiation.
- To find the partial derivative with respect to \( u \), you differentiate assuming \( v \) and \( w \) are constants. In our case, the partial derivative of \( 3u^2 \) with respect to \( u \) is \( 6u \), while the terms \(-2v\) and \(4w^3v^2\) disappear, as they do not involve \( u \).
- Similarly, when differentiating with respect to \( v \), only terms containing \( v \) contribute to the derivative. This involves separating their contributions, leading us to solve \( -2v \) and \( 4w^3v^2 \), providing \( -2 + 8w^3v \).
- For partial differentiation with respect to \( w \), only \( 4w^3v^2 \) is relevant, contributing \( 12w^2v^2 \) to the derivative.
Multivariable Calculus
Multivariable calculus extends single-variable calculus concepts to functions with more than one variable. It opens up exploration into how changes in various dimensions affect function outcomes. Think about a surface shaped by a function \( f(u, v, w) \) where each variable can be independently varied.
- In multivariable calculus, gradients and directional derivatives become crucial. They help determine how a function changes across different paths within the variable space.
- The total differential is a specific application of these principles, providing a linear approximation for how small changes in inputs (\( du, dv, dw \)) affect the output \( dz \).
- Understanding multivariable systems is essential for fields like physics, engineering, economics, and more, where systems naturally involve multiple variables and dimensions.
Differential Equations
Differential equations involve equations that relate a function to its derivatives, encapsulating how functions change and evolve. In the context of total differentials, differential equations often play a role when we model how changes in variables influence other variables.
- In our exercise, constructing the total differential gives insight into small changes in variables \( u, v, \) and \( w \) and their impact on \( z \). This forms a basis to formulate differential equations for dynamic systems.
- By equating total differential expressions to a zero or another function, one can generate differential equations capturing real-world phenomena, allowing predictions and deeper analysis of system behaviors.
- Solving differential equations often requires understanding both initial conditions and the specific relations among variables, reflected through their derivatives.
