Problem 19 In Exercises $19-22,$ function... [FREE SOLUTION]

Chapter 12: Problem 19

In Exercises $19-22,$ functions $z=f(x, y), x=g(s, t)$ and $y=h(s, t)$ are given. (a) Use the Multivariable Chain Rule to compute $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ (b) Evaluate $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ at the indicated $s$ and $t$ values.$ 12. $$ z=x^{2} y, \quad x=s-t, \quad y=2 s+4 t ; \quad s=1, t=0 $$

Short Answer

Expert verified

$ \frac{\partial z}{\partial s} = 6 $, $ \frac{\partial z}{\partial t} = 0 $.

Step by step solution

Understanding the Multivariable Chain Rule

The Multivariable Chain Rule states that if a function $ z = f(x, y) $ and both $ x $ and $ y $ are functions of two other variables $ s $ and $ t $, then the partial derivative of $ z $ with respect to $ s $ is given by $ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s} $. Similarly, $ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} $.

Calculate Partial Derivatives of z with respect to x and y

For the function $ z = x^2 y $, calculate the partial derivatives:$ \frac{\partial z}{\partial x} = 2xy $ and $ \frac{\partial z}{\partial y} = x^2 $.

Calculate Partial Derivatives of x and y with respect to s and t

Using the given expressions $ x = s-t $ and $ y = 2s + 4t $:$ \frac{\partial x}{\partial s} = 1 $ and $ \frac{\partial x}{\partial t} = -1 $.For $ y $, $ \frac{\partial y}{\partial s} = 2 $ and $ \frac{\partial y}{\partial t} = 4 $.

Apply the Chain Rule for $ \frac{\partial z}{\partial s} $

Using the chain rule, \[ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s} = 2xy(1) + x^2(2) \]

Substitute Functions for x and y into the Partial Derivatives

Substitute $ x = s - t $ and $ y = 2s + 4t $ into the expressions for $ \frac{\partial z}{\partial x} $ and $ \frac{\partial z}{\partial y} $:$ \frac{\partial z}{\partial x} = 2(s-t)(2s + 4t) $$ \frac{\partial z}{\partial y} = (s-t)^2 $

Evaluate Partial Derivatives at given s and t

Substitute $ s = 1 $ and $ t = 0 $:For $ \frac{\partial z}{\partial s} $:- $ x = 1 - 0 = 1 $, $ y = 2(1) + 4(0) = 2 $- $ \frac{\partial z}{\partial s} = 2 \cdot 1 \cdot 2 + 1^2 \cdot 2 = 4 + 2 = 6 $

Apply the Chain Rule for $ \frac{\partial z}{\partial t} $

Using the chain rule, \[ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} = 2xy(-1) + x^2(4) \]

Evaluate Second Partial Derivative at given s and t

Substitute $ s = 1 $ and $ t = 0 $:For $ \frac{\partial z}{\partial t} $:- $ \frac{\partial z}{\partial t} = 2 \cdot 1 \cdot 2 \cdot (-1) + 1^2 \cdot 4 = -4 + 4 = 0 $

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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 concept in calculus, especially in dealing with functions of more than one variable. When we say we are taking a partial derivative, we're focusing on how a function changes with respect to one variable while keeping the others constant. Let's break this down further:

If we have a function, say, $z = f(x, y)$, it means $z$ depends on two variables: $x$ and $y$.
The partial derivative of $z$ with respect to $x$, denoted $\frac{\partial z}{\partial x}$, measures how $z$ changes as $x$ varies, while $y$ remains fixed.
Similarly, $\frac{\partial z}{\partial y}$ shows how $z$ changes with a change in $y$, keeping $x$ constant.

For a practical example, consider the function $z = x^2y$. Here:

$\frac{\partial z}{\partial x} = 2xy$ meaning as $x$ changes, $z$ changes proportionally to $2xy$.
$\frac{\partial z}{\partial y} = x^2$ highlights that $z$ changes linearly with $y$ and is weighted by $x^2$.

Partial derivatives allow us to understand and predict the behavior of complex systems described by multivariable functions.

Functions of Multiple Variables

Functions of multiple variables are essential in modeling situations where a quantity depends on two or more variables. For instance, weather modeling, economics, and engineering often involve such functions.To comprehend functions of multiple variables:

A multivariable function like $z = f(x, y)$ suggests that $z$ results from a combination of both $x$ and $y$.
Each input variable can influence the output $z$ in different ways, which is where partial derivatives come into play.
Visualizing these functions involves imagining a surface in a three-dimensional space, where each point $(x, y, z)$ corresponds to different values of $x$ and $y$.

In this context, understanding how $x$ and $y$ contribute to $z$'s behavior becomes crucial. For example:

If $x = s - t$ and $y = 2s + 4t$, as given, then changes in $s$ and $t$ will define changes in $x$ and $y$, respectively.
This transforms our earlier 2-variable problem into a 4-variable one but boils down to the specific parameter path $s$ and $t$ determine.

This transformation highlights how interconnected systems and parameters result in complex but insightful relationships.

Calculus with Parameters

Calculus with parameters extends the realm of basic calculus to scenarios where variables themselves depend on other parameters. In real-world applications, many elements change with respect to time or other conditions, and parameters help account for this.Here's how this works:

Consider a function like $x = g(s, t)$ and $y = h(s, t)$. Here, $x$ and $y$ change with $s$ and $t$, which are parameters.
The multivariable chain rule helps us determine how these changes affect the function $z = f(x, y)$.
The chain rule for partial derivatives gives a way to account for how $z$ evolves due to changes in $s$ and $t$.

When evaluating at specific points, such as $s = 1$ and $t = 0$, parameters provide exact values, revealing concrete insights about the system. For instance, by plugging in specific $s$ and $t$ values, we see direct impacts on $x$ and $y$, and therefore on $z$. This forms an essential part of predictive modeling in various fields, ranging from physics to finance.

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Short Answer

Step by step solution

Understanding the Multivariable Chain Rule

Calculate Partial Derivatives of z with respect to x and y

Calculate Partial Derivatives of x and y with respect to s and t

Apply the Chain Rule for \( \frac{\partial z}{\partial s} \)

Substitute Functions for x and y into the Partial Derivatives

Evaluate Partial Derivatives at given s and t

Apply the Chain Rule for \( \frac{\partial z}{\partial t} \)

Evaluate Second Partial Derivative at given s and t

Key Concepts

Partial Derivatives

Functions of Multiple Variables

Calculus with Parameters

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