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Chain Rule with several independent variables. Find the following derivatives. \(z_{s}\) and \(z_{r},\) where \(z=x y-x^{2} y, x=s+t,\) and \(y=s-t\)

Short Answer

Expert verified
Question: Find the partial derivatives of the function \(z = xy - x^2y\) with respect to the independent variables s and r, where the relationships between the variables are: \(x = s + t\) and \(y = s - t\). Answer: The partial derivatives of the function z with respect to the independent variables s and r are: \(z_s = (s-t) - 2(s+t)(s-t) + (s+t) - (s+t)^2\) \(z_r = (s+t) - (s+t)^2\)

Step by step solution

01

Rewrite z as a function of s and t

Given the relationships between x, y, and z, \(z = xy - x^2y\) and \(x = s + t\) \(y = s - t\) Substitute the expressions for x and y in terms of s and t into the equation for z: \(z(s,t) = (s+t)(s-t) - (s+t)^2(s-t)\)
02

Apply the Chain Rule to find the partial derivatives

Since z is a function of s and t, we need to find the partial derivatives of z with respect to s and r, denoted as \(z_s\) and \(z_r\). To find \(z_s\), apply the chain rule by taking the derivative of z with respect to s, keeping t constant: \(z_s = \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}\) To find \(z_r\), apply the chain rule by taking the derivative of z with respect to r, keeping s constant: \(z_r = \frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}\)
03

Partially differentiate z with respect to x and y

Differentiate z with respect to x and y: \(\frac{\partial z}{\partial x} = y - 2xy\) \(\frac{\partial z}{\partial y} = x - x^2\)
04

Partially differentiate x and y with respect to s and r

Differentiate x and y with respect to s and r: \(\frac{\partial x}{\partial s} = 1\) \(\frac{\partial x}{\partial r} = 0\) \(\frac{\partial y}{\partial s} = 1\) \(\frac{\partial y}{\partial r} = 1\)
05

Calculate \(z_s\) and \(z_r\)

Substitute the results of Step 3 and Step 4 into the expressions for \(z_s\) and \(z_r\) derived in Step 2: \(z_s = (y - 2xy)(1) + (x - x^2)(1) = (s-t) - 2(s+t)(s-t) + (s+t) - (s+t)^2\) \(z_r = (y - 2xy)(0) + (x - x^2)(1) = 0 + (s+t) - (s+t)^2\) Thus, the derivatives of z with respect to the independent variables s and r are: \(z_s = (s-t) - 2(s+t)(s-t) + (s+t) - (s+t)^2\) \(z_r = (s+t) - (s+t)^2\)

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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 multivariable calculus, used to find the rate at which a function changes with respect to one of its variables, while keeping other variables constant. Imagine you have a function, like in the given exercise, that depends on multiple variables. In this scenario, you would use partial derivatives to examine how changing one specific variable affects the outcome of the function.

In the problem, we have a function \(z\) that depends on \(x\) and \(y\). However, \(x\) and \(y\) themselves are expressed in terms of \(s\) and \(t\). By using the chain rule for partial derivatives, we simplify the function into expressions solely in terms of \(s\) and \(t\), making it easier to differentiate.

Partial derivatives are often denoted by the symbol \(\partial\). For example, \(\frac{\partial z}{\partial x}\) represents the partial derivative of \(z\) with respect to \(x\). In other words, it describes how \(z\) changes as \(x\) changes, with all other variables held constant. Understanding and calculating partial derivatives give insight into the behavior of multivariable functions, making them an indispensable tool in calculus.
Independent Variables
Independent variables are crucial in defining the behavior of a function. In the context of the exercise, we are dealing with functions of several variables. Here, \(s\) and \(t\) are considered independent variables as they define the values of \(x\) and \(y\), and subsequently \(z\).

When dealing with independent variables in calculus, it's essential to understand that the change in a dependent variable, such as \(z\), is affected by how each independent variable contributes to its overall behavior. The independence implies that changing one variable doesn't directly change the others, which is why they are examined separately in derivative calculations.

In multivariable calculus, analyzing how functions respond to changes in their independent variables is key to understanding their nature and potential applications. This approach helps identify trends or patterns in data, optimize functions, and solve real-world problems, such as those found in physics, engineering, and economics.
Multivariable Calculus
Multivariable calculus extends basic calculus concepts to functions of several variables. It involves tools and methods for evaluating and understanding functions with two or more input variables. This area of study is vital, especially in fields that require modeling complex systems, such as fluid dynamics, thermodynamics, and various engineering disciplines.

In our exercise, multivariable calculus allows us to work with more intricate relationships between variables. The function \(z = xy - x^2y\) involves two variables, \(x\) and \(y\), which in turn depend on \(s\) and \(t\). Through multivariable calculus, we can apply the chain rule to simplify expressions and find derivatives concerning each independent variable.

The tools used in multivariable calculus, such as partial derivatives and the chain rule, play a crucial role in identifying maximum or minimum values of functions and optimizing processes. It enhances the basic calculus toolkit, enabling us to solve more complex and applicable problems, vital in advancing scientific and technological solutions.

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Most popular questions from this chapter

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