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Chapter 11: Problem 22

Find the differential of the function. $$ T=\frac{v}{1+u v w} $$

Short Answer

Expert verified
The differential of \( T \) is \( dT = \frac{1}{(1 + u v w)^2} dv - \frac{v^2 w}{(1 + u v w)^2} du - \frac{u v^2}{(1 + u v w)^2} dw \).

Step by step solution

01

Identify the Function

The given function is defined as the variable \( T \) in terms of \( v \), \( u \), and \( w \), specifically: \[ T = \frac{v}{1 + u v w} \].
02

Find the Partial Derivatives

To find the differential of a multivariable function, we need the partial derivatives with respect to each variable. Start by differentiating \( T \) with respect to \( v \), \( u \), and \( w \).
03

Differentiate with Respect to v

Using the quotient rule, differentiate \( T \) with respect to \( v \):\[ \frac{\partial T}{\partial v} = \frac{(1 + u v w) - v (u w)}{(1 + u v w)^2} \]which simplifies to:\[ \frac{\partial T}{\partial v} = \frac{1 + u v w - u v w}{(1 + u v w)^2} = \frac{1}{(1 + u v w)^2} \].
04

Differentiate with Respect to u

Differentiate \( T \) with respect to \( u \):\[ \frac{\partial T}{\partial u} = -\frac{v^2 w}{(1 + u v w)^2} \].
05

Differentiate with Respect to w

Differentiate \( T \) with respect to \( w \):\[ \frac{\partial T}{\partial w} = -\frac{u v^2}{(1 + u v w)^2} \].
06

Construct the Differential dT

Combine the partial derivatives to write the differential \( dT \):\[ dT = \frac{1}{(1 + u v w)^2} dv - \frac{v^2 w}{(1 + u v w)^2} du - \frac{u v^2}{(1 + u v w)^2} dw \].

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental tool in differential calculus, especially when dealing with functions of multiple variables. When you have a function like \( T = \frac{v}{1 + uvw} \), which depends on more than one variable鈥擻( v \), \( u \), and \( w \)鈥攗nderstanding how changes in each individual variable affect the function is crucial. Here's what makes partial derivatives unique:
  • They measure the rate of change of the function with respect to one variable, keeping the others constant.
  • The notation most commonly used is \( \frac{\partial T}{\partial v} \) for partial derivative with respect to \( v \).
For the given function, calculating partial derivatives with respect to each variable helps us understand the sensitivity of \( T \) to changes in \( v \), \( u \), and \( w \). Ultimately, these individual expressions combine to form the overall differential of the function.
Multivariable Functions
Multivariable functions are those that contain more than one independent variable. In this exercise, the function \( T = \frac{v}{1 + uvw} \) is dependent on three variables: \( v \), \( u \), and \( w \).Key features of multivariable functions include:
  • Evaluation of how each variable individually influences the function's outcome.
  • Greater complexity in derivative computation compared to single-variable functions, due to interactions between variables.
When dealing with multivariable functions, it's essential to analyze how changes in any single variable affect the outcome while others remain constant. This is where partial derivatives play a vital role. By carefully exploring the changes in function value caused by changes in each variable separately, we can sketch a more comprehensive understanding of the function's behavior.
Quotient Rule
The quotient rule is a technique used in calculus to differentiate functions expressed as the quotient of two other functions. For the function \( T = \frac{v}{1 + uvw} \), applying the quotient rule allows us to differentiate with respect to each individual variable such as \( v \).The quotient rule formula is expressed as:\[ \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \]where \( f \) is the numerator and \( g \) is the denominator.Let's break it down:
  • For \( \frac{\partial T}{\partial v} \), you consider \( f = v \) and \( g = 1 + uvw \).
  • Using the quotient rule, you differentiate \( f \) and \( g \) separately, and substitute into the formula.
This strategy simplifies the differentiation process when dealing with quotients of functions, offering clear steps to follow for precise results.
Differentials
Differentials provide a way to understand the infinitesimal changes in multivariable functions. In this example, our task is to find the differential \( dT \) of \( T = \frac{v}{1 + uvw} \). Here's why differentials are important:
  • Differentials are essentially linear approximations of how a function changes when its variables change very slightly.
  • They aid in predicting the behavior of complex functions at particular points by considering small increments.
The differential \( dT \) for this function is represented as a linear combination of changes in all variables:\[ dT = \frac{1}{(1 + uvw)^2} dv - \frac{v^2 w}{(1 + uvw)^2} du - \frac{u v^2}{(1 + uvw)^2} dw \]Each component explains how a minor change in \( v \), \( u \), or \( w \) impacts \( T \), providing a detailed picture of the function's local behavior.

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Most popular questions from this chapter

Find the maximum and minimum values of \(f\) subject to the given constraints. Use a computer algebra system to solve the system of equations that arises in using Lagrange multipliers. (If your CAS finds only one solution, you may need to use additional commands.) $$ f(x, y, z)=x+y+z ; \quad x^{2}-y^{2}=z, x^{2}+z^{2}=4 $$

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=\left(x^{2}+y^{2}\right) e^{y^{2}-x^{2}}$$

At what points does the normal line through the point \((1,2,1)\) on the ellipsoid \(4 x^{2}+y^{2}+4 z^{2}=12\) intersect the sphere \(x^{2}+y^{2}+z^{2}=102 ?\)

The plane \(x+y+2 z=2\) intersects the paraboloid \(z=x^{2}+y^{2}\) in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

(a) Maximize \(\sum_{i=1}^{n} x_{i} y_{i}\) subject to the constraints \(\sum_{i=1}^{n} x_{i}^{2}=1\) and \(\sum_{i=1}^{n} y_{i}^{2}=1\) (b) Put $$ x_{i}=\frac{a_{i}}{\sqrt{\sum a_{j}^{2}}} \quad \text { and } \quad y_{i}=\frac{b_{i}}{\sqrt{\sum b_{j}^{2}}} $$ to show that $$ \sum a_{i} b_{i} \leqslant \sqrt{\sum a_{j}^{2}} \sqrt{\Sigma b_{j}^{2}} $$ for any numbers \(a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n} .\) This inequality is known as the Cauchy-Schwarz Inequality.
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