/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Let \(z=e^{2 x} \sin y .\) Find ... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(z=e^{2 x} \sin y .\) Find $$\begin{array}{llll}{\text { (a) } \partial z / \partial x} & {\text { (b) } \partial z / \partial y} & {\text { (c) } \partial z / \partial x(0, y)} & {}\end{array}$$ $$\text { (d) } \partial z /\left.\partial x\right|_{(x, 0)} \quad \text { (e) } \partial z /\left.\partial y\right|_{(0, y)} \quad \text { (f) } \partial z /\left.\partial y\right|_{(x, 0)}$$ $$\text { (g) } \partial z /\left.\partial x\right|_{(\ln 2,0)} \quad \text { (h) } \partial z /\left.\partial y\right|_{(\ln 2,0)}$$

Short Answer

Expert verified
(a) \(2e^{2x} \sin y\), (b) \(e^{2x} \cos y\); (c) \(2 \sin y\), (d) 0, (e) \(\cos y\), (f) \(e^{2x}\); (g) 0, (h) 4.

Step by step solution

01

Partial Derivative with respect to x

To find \( \frac{\partial z}{\partial x} \), we treat \( y \) as a constant. The derivative of \( e^{2x} \) is \( 2e^{2x} \), so:\[\frac{\partial z}{\partial x} = 2e^{2x} \sin y.\]
02

Partial Derivative with respect to y

To find \( \frac{\partial z}{\partial y} \), we treat \( x \) as a constant. The derivative of \( \sin y \) is \( \cos y \), so:\[\frac{\partial z}{\partial y} = e^{2x} \cos y.\]
03

Evaluate \( \frac{\partial z}{\partial x} \) at \((0,y)\)

Using the expression found in Step 1, substitute \( x = 0 \):\[\frac{\partial z}{\partial x}(0, y) = 2e^0 \sin y = 2 \sin y.\]
04

Evaluate \( \frac{\partial z}{\partial x} \) at \((x,0)\)

Using the expression found in Step 1, substitute \( y = 0 \):\[\frac{\partial z}{\partial x}(x, 0) = 2e^{2x} \sin 0 = 0.\]
05

Evaluate \( \frac{\partial z}{\partial y} \) at \((0,y)\)

Using the expression found in Step 2, substitute \( x = 0 \):\[\frac{\partial z}{\partial y}(0, y) = e^0 \cos y = \cos y.\]
06

Evaluate \( \frac{\partial z}{\partial y} \) at \((x,0)\)

Using the expression found in Step 2, substitute \( y = 0 \):\[\frac{\partial z}{\partial y}(x, 0) = e^{2x} \cos 0 = e^{2x}.\]
07

Evaluate \( \frac{\partial z}{\partial x} \) at \((\ln 2,0)\)

Using the expression found in Step 4, substitute \( x = \ln 2 \):\[\frac{\partial z}{\partial x}(\ln 2, 0) = 0.\]
08

Evaluate \( \frac{\partial z}{\partial y} \) at \((\ln 2,0)\)

Using the expression found in Step 6, substitute \( x = \ln 2 \):\[\frac{\partial z}{\partial y}(\ln 2, 0) = e^{2\ln 2} = (e^{\ln 2})^2 = 4.\]

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Key Concepts

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

Calculus
Calculus is a branch of mathematics that studies how things change. It's a tool used to analyze changes between values that are related by a function. Two main operations in calculus are differentiation and integration. In basic terms, differentiation deals with finding the rate at which a quantity changes. In our exercise, we focus on the differentiation aspect, specifically partial differentiation. Calculus helps in understanding how different variables affect a function and how sensitive the function is to change in those variables.When dealing with functions of multiple variables, such as our function \(z = e^{2x} \sin(y)\), calculus allows us to explore the concept of partial derivatives. A partial derivative shows how a function changes in one direction around a specific point, holding other variables constant. This analysis is foundational for many fields, such as physics, engineering, and economics, where systems depend on multiple factors. Understanding calculus helps us not only to solve problems but also to model real-world scenarios accurately.
Function of Multiple Variables
A function of multiple variables is like a machine that takes inputs and gives one output, but with more complexity. In our exercise, the function \( z = e^{2x} \sin(y) \) takes two inputs: \(x\) and \(y\). The beauty of such functions is that they can describe surfaces or curves in higher dimensions. This becomes very useful in real-life applications like 3D graphics, where we try to map surfaces accurately.For functions of multiple variables, we often explore how changes in one variable affect the function while keeping others constant. This leads us to the concept of partial derivatives, which are crucial in multivariable calculus. These derivatives help us calculate rates of change along specific axes. By analyzing these changes, we get insights into the geometry and behavior of complex systems. When tackling problems involving such functions, you often use matrix algebra and vector calculus to compute solutions.
Differentiation
Differentiation is a process used in calculus to determine how a function changes as its input changes. It's essentially about finding the slope or gradient of a function. When applied to functions of a single variable, this is straightforward. However, for functions of multiple variables, the process involves finding partial derivatives.Understanding the concept of differentiation is key when dealing with multivariable functions like \( z = e^{2x} \sin(y) \). Here, differentiation allows us to find how changing \(x\) or \(y\) individually influences \(z\). Partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) capture this influence. Studying these partial derivatives involves:
  • Considering each variable independently as the function changes.
  • Keeping other variables constant to see the isolated effects.
Differentiation gives us the tools to surface these relationships clearly, which can then be applied in fields that require precise optimization and calibration.

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Most popular questions from this chapter

Find \(\partial w / \partial x, \partial w / \partial y,\) and \(\partial w / \partial z\) using implicit differentiation. Leave your answers in terms of \(x, y, z,\) and \(w .\) $$ \ln \left(2 x^{2}+y-z^{3}+3 w\right)=z $$

A common problem in experimental work is to obtain a mathematical relationship \(y=f(x)\) between two variables \(x\) and \(y\) by "fitting" a curve to points in the plane that correspond to experimentally determined values of \(x\) and \(y,\) say $$ \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right) $$ The curve \(y=f(x)\) is called a mathematical model of the data. The general form of the function \(f\) is commonly determined by some underlying physical principle, but sometimes it is just determined by the pattern of the data. We are concerned with fitting a straight line \(y=m x+b\) to data. Usually, the data will not lie on a line (possibly due to experimental error or variations in experimental conditions), so the problem is to find a line that fits the data "best" according to some criterion. One criterion for selecting the line of best fit is to choose \(m\) and \(b\) to minimize the function $$ g(m, b)=\sum_{i=1}^{n}\left(m x_{i}+b-y_{i}\right)^{2} $$ This is called the method of least squares, and the resulting line is called the regression line or the least squares line of best fit. Geometrically, \(\left|m x_{i}+b-y_{i}\right|\) is the vertical distance between the data point \(\left(x_{i}, y_{i}\right)\) and the line \(y=m x+b\) These vertical distances are called the residuals of the data points, so the effect of minimizing \(g(m, b)\) is to minimize the sum of the squares of the residuals. In these exercises, we will derive a formula for the regression line. The purpose of this exercise is to find the values of \(m\) and \(b\) that produce the regression line. (a) To minimize \(g(m, b),\) we start by finding values of \(m\) and \(b\) such that \(\partial g / \partial m=0\) and \(\partial g / \partial b=0 .\) Show that these equations are satisfied if \(m\) and \(b\) satisfy the conditions $$ \left(\sum_{i=1}^{n} x_{i}^{2}\right) m+\left(\sum_{i=1}^{n} x_{i}\right) b=\sum_{i=1}^{n} x_{i} y_{i} $$ \(\left(\sum_{i=1}^{n} x_{i}\right) m+n b=\sum_{i=1}^{n} y_{i}\) (b) Let \(\bar{x}=\left(x_{1}+x_{2}+\cdots+x_{n}\right) / n\) denote the arithmetic average of \(x_{1}, x_{2}, \ldots, x_{n} .\) Use the fact that $$ \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} \geq 0 $$ to show that $$ n\left(\sum_{i=1}^{n} x_{i}^{2}\right)-\left(\sum_{i=1}^{n} x_{i}\right)^{2} \geq 0 $$ with equality if and only if all the \(x_{i}\) 's are the same. (c) Assuming that not all the \(x_{i}\) 's are the same, prove that the equations in part (a) have the unique solution $$ \begin{aligned} m=& \frac{n \sum_{i=1}^{n} x_{i} y_{i}-\sum_{i=1}^{n} x_{i} \sum_{i=1}^{n} y_{i}}{n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^{n} x_{i}\right)^{2}} \\ b=& \frac{1}{n}\left(\sum_{i=1}^{n} y_{i}-m \sum_{i=1}^{n} x_{i}\right) \end{aligned} $$ [Note: We have shown that \(g\) has a critical point at these values of \(m\) and \(b\). In the next exercise we will show that \(g\) has an absolute minimum at this critical point. Accepting this to be so, we have shown that the line \(y=m x+b\) is the regression line for these values of \(m \text { and } b .]\)

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