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Magazine Chapter 11: Problem 3
Use the Chain Rule to find \(d z / d t\) or \(d w / d t\) $$w=x e^{y / z}, \quad x=t^{2}, \quad y=1-t, \quad z=1+2 t$$
Short Answer
Expert verified
\( \frac{d w}{d t} = e^{(1-t)/(1+2t)} \left(2t - \frac{t^2}{1+2t} - \frac{2t^2(1-t)}{(1+2t)^2}\right) \).
Step by step solution
01
Differentiate w with respect to t
First, note that we are trying to find \( \frac{d w}{d t} \). Using functions \( w = x e^{y/z} \), \( x = t^2 \), \( y = 1 - t \), and \( z = 1 + 2t \), apply the chain rule to obtain the expression for \( \frac{d w}{d t} \): \[\frac{d w}{d t} = \frac{\partial w}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial w}{\partial y} \cdot \frac{d y}{d t} + \frac{\partial w}{\partial z} \cdot \frac{d z}{d t}\].
02
Partial derivatives
Calculate partial derivatives of \( w \): - \(\frac{\partial w}{\partial x} = e^{y/z}\) since \( w = x e^{y/z} \). - \(\frac{\partial w}{\partial y} = \frac{x}{z}e^{y/z}\) using the chain rule for exponential functions. - \(\frac{\partial w}{\partial z} = -\frac{xy}{z^2}e^{y/z}\).
03
Calculate derivatives with respect to t
Calculate the derivatives of each function with respect to \( t \): - \( \frac{d x}{d t} = 2t \). - \( \frac{d y}{d t} = -1 \). - \( \frac{d z}{d t} = 2 \).
04
Substitute and simplify
Now substitute the partial derivatives and the derivatives into the chain rule formula: \[\frac{d w}{d t} = e^{y/z} \cdot 2t + \frac{x}{z} e^{y/z} \cdot (-1) + \left(-\frac{xy}{z^2} e^{y/z}\right) \cdot 2\].Substitute \( x = t^2 \), \( y = 1-t \), \( z = 1+2t \) into the equation: \[\frac{d w}{d t} = e^{(1-t)/(1+2t)} \cdot 2t - \frac{t^2}{1+2t}e^{(1-t)/(1+2t)} - \frac{2t^2(1-t)}{(1+2t)^2}e^{(1-t)/(1+2t)}\].
05
Final expression for dw/dt
Combine the terms in the expression:\[\frac{d w}{d t} = e^{(1-t)/(1+2t)} \left(2t - \frac{t^2}{1+2t} - \frac{2t^2(1-t)}{(1+2t)^2}\right)\].Simplify by distributing and combining terms if necessary. This gives the derivative \( \frac{d w}{d t} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
When we embark on finding partial derivatives, it's essential to understand that we are considering the rate of change of a function with respect to one variable while keeping the other variables constant. This concept is crucial, especially in multivariable calculus where a function can have more than one input.
In the provided exercise, we have a function, \(w = x e^{y/z}\). Here, partial derivatives denoted as \(\frac{\partial w}{\partial x}\), \(\frac{\partial w}{\partial y}\), and \(\frac{\partial w}{\partial z}\) allow us to analyze how \(w\) changes with each individual variable, keeping the others constant.
In the provided exercise, we have a function, \(w = x e^{y/z}\). Here, partial derivatives denoted as \(\frac{\partial w}{\partial x}\), \(\frac{\partial w}{\partial y}\), and \(\frac{\partial w}{\partial z}\) allow us to analyze how \(w\) changes with each individual variable, keeping the others constant.
- \(\frac{\partial w}{\partial x} = e^{y/z}\): Changes in \(x\) directly affect \(w\), illustrating a linear relationship with the exponential term remaining constant.
- \(\frac{\partial w}{\partial y} = \frac{x}{z}e^{y/z}\): Here, we see how \(w\) responds to changes in \(y\), factoring in the division by \(z\).
- \(\frac{\partial w}{\partial z} = -\frac{xy}{z^2}e^{y/z}\): This expresses the impact of \(z\) on \(w\), highlighting an inverse squared relationship which can result in more significant changes for small increments in \(z\).
Function Differentiation
Function differentiation, particularly using the chain rule, aids in finding derivatives of composite functions. The chain rule is a formula used to differentiate compositions of two or more functions. In simpler terms, if you have a function inside another, the chain rule helps determine how a change in one variable affects the final outcome.
In our exercise, the function \(w = x e^{y/z}\) is dependent on multiple other functions. We differentiate with respect to \(t\) and need to track how these nested functions interact. This scenario requires us to utilize the chain rule efficiently.
In our exercise, the function \(w = x e^{y/z}\) is dependent on multiple other functions. We differentiate with respect to \(t\) and need to track how these nested functions interact. This scenario requires us to utilize the chain rule efficiently.
- Identify the roles of each function component: Know which functions depend on others and hierarchically determine the order of differentiation.
- Apply the chain rule: For \(\frac{d w}{d t}\), we calculate \(\frac{\partial w}{\partial x} \cdot \frac{d x}{d t}\), \(\frac{\partial w}{\partial y} \cdot \frac{d y}{d t}\), and \(\frac{\partial w}{\partial z} \cdot \frac{d z}{d t}\) to unmask dependencies.
Exponential Functions
In our exploration of functions, exponential functions like \(e^{y/z}\) present unique characteristics. An exponential function involves a constant base raised to a variable exponent, leading to powerful growth patterns.
These functions are prevalent across various mathematical models, notably in calculating compound interest and population growth, but they play a significant role in calculus as well due to their unique derivative properties.
These functions are prevalent across various mathematical models, notably in calculating compound interest and population growth, but they play a significant role in calculus as well due to their unique derivative properties.
- Derivative of an exponential function: The derivative of \(e^{u}\) with respect to \(u\) remains \(e^{u}\), showing exponential growth or decay remains consistent in magnitude.
- Application in partial differentiation: In our case, \(e^{y/z}\) isn't simplified as the exponential exponent term stays complex. Yet, exponential functions retain their structure through the differentiation process as seen in \(\frac{\partial w}{\partial x} = e^{y/z}\).
- Exponential impact: Despite complex variables, simplifying where possible or factoring ensures clarity in the computations of nested functions.
