/*! 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 3 Use an appropriate form of the c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 3

Use an appropriate form of the chain rule to find $d z / d t $$ z=3 \cos x-\sin x y ; x=1 / t, y=3 t $$

Short Answer

Expert verified
\( \frac{dz}{dt} = \frac{3\sin(1/t)}{t^2} + \frac{3t \cos(1/t)}{t^2} - \frac{3 \cos(1/t)}{t} \).

Step by step solution

01

Determine Variables and Their Dependencies

We know that the function \( z \) is dependent on two variables \( x \) and \( y \). Specifically, we have the equations \( z = 3\cos x - \sin xy\), where \( x = \frac{1}{t} \) and \( y = 3t \). This means both \( x \) and \( y \) are functions of \( t \).
02

Find Partial Derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \)

Compute the partial derivative of \( z \) with respect to \( x \):\[\frac{\partial z}{\partial x} = \frac{d}{d x}(3\cos x - \sin xy) = -3\sin x - y \cos x\]Now compute the partial derivative of \( z \) with respect to \( y \):\[\frac{\partial z}{\partial y} = \frac{d}{d y}(3\cos x - \sin xy) = -x \cos x\]
03

Calculate \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \)

We need \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \) to use the chain rule:\( x = \frac{1}{t} \Rightarrow \frac{d x}{d t} = -\frac{1}{t^2} \)\( y = 3t \Rightarrow \frac{d y}{d t} = 3 \)
04

Apply the Chain Rule to Find \( \frac{dz}{dt} \)

Using the chain rule, we can express the derivative \( \frac{dz}{dt} \) as:\[\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \]Substituting our earlier results:\[\frac{dz}{dt} = \left(-3\sin x - y \cos x\right) \left(-\frac{1}{t^2}\right) + \left(-x \cos x\right) (3)\]Simplify and substitute \( x = \frac{1}{t} \) and \( y = 3t \) into the equation.
05

Simplify the Expression

Inserting \( x = \frac{1}{t} \) and \( y = 3t \) into our differentiated equation:\[\frac{dz}{dt} = (3\sin(\frac{1}{t}) + 3t \cos(\frac{1}{t}))\frac{1}{t^2} + \frac{1}{t} \cos(\frac{1}{t}) \cdot (-3) \]Finally simplifying:\[\frac{dz}{dt} = \frac{3\sin(\frac{1}{t})}{t^2} + \frac{3t \cos(\frac{1}{t})}{t^2} - \frac{3 \cos(\frac{1}{t})}{t} \]Combine terms to obtain the final expression for \( \frac{dz}{dt} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 which is especially useful when a function depends on multiple variables. Instead of taking the derivative of a function with respect to one independent variable, partial derivatives allow us to see how a function changes as we alter just one of several variables, keeping the others constant. For our given function, \( z = 3 \cos x - \sin xy \), it relies on two variables: \( x \) and \( y \). When we find the partial derivative \( \frac{\partial z}{\partial x} \), we're looking at how \( z \) changes as \( x \) changes, considering \( y \) constant, and vice versa for \( \frac{\partial z}{\partial y} \).
  • To compute \( \frac{\partial z}{\partial x} \), differentiate with respect to \( x \), keeping \( y \) unchanged: \( \frac{\partial z}{\partial x} = -3\sin x - y \cos x \).
  • For \( \frac{\partial z}{\partial y} \), differentiate with respect to \( y \), keeping \( x \) constant: \( \frac{\partial z}{\partial y} = -x \cos x \).
This process is crucial when applying the chain rule for variables that depend on other parameters, like \( x \) and \( y \) depending on \( t \) in this problem. By mastering partial derivatives, we set the groundwork for more complex multivariable calculus tasks.
Functional Dependency
Functional dependency refers to how variables in a function are reliant on one another. In the context of calculus, understanding this dependency is vital for applying techniques like the chain rule correctly. For example, in our exercise, the function \( z \) depends on \( x \) and \( y \), which in turn depend on \( t \). Therefore, changes in \( t \) directly affect \( z \) via changes in \( x \) and \( y \). Knowing this dependency chain allows us to use the chain rule effectively.
  • If \( x \) changes because \( t \) changes, \( z \) gets affected too.
  • The same applies to \( y \), as it follows any modification of \( t \).
In practical applications, recognizing how each variable is dependent on others simplifies the process of determining how one variable's change impacts another, which is particularly important in disciplines involving complex systems or simulations.
Implicit Differentiation
Implicit differentiation is a powerful tool for finding the derivative of expressions where one variable is not isolated on one side of the equation. Instead of manipulating equations to solve for a single variable, we embrace the relationships and differentiate as they are. This approach is especially useful if dealing with functions that are not easily rearranged.
In our specific problem, neither \( x \) nor \( y \) is isolated; instead, they form a relationship through dependence on \( t \). Implicit differentiation comes into play when we take the chain rule to compute \( \frac{dz}{dt} \):
  • By treating \( x \) and \( y \) as functions of \( t \), we can find their respective derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
  • By substituting these derivatives into the chain rule, we implicitly differentiate with respect to \( t \) to find how \( z \) changes.
This technique simplifies handling situations where directly solving for derivatives in explicit function form can be highly complex, showcasing implicit differentiation's strength in calculus.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(f\) denote a differentiable function of two variables. Although we have defined what it means to say that \(f\) is differentiable, we have not defined the "derivative" of \(f\) Write a short paragraph that discusses the merits of defining the derivative of \(f\) to be the gradient \(\nabla f\).

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 \(m \text { and } b .]\)

Let \(u(w, x, y, z)=x e^{y w} \sin ^{2} z .\) Find $$ \begin{array}{ll}{\text { (a) } \frac{\partial u}{\partial x}(0,0,1, \pi)} & {\text { (b) } \frac{\partial u}{\partial y}(0,0,1, \pi)} \\ {\text { (c) } \frac{\partial u}{\partial w}(0,0,1, \pi)} & {\text { (d) } \frac{\partial u}{\partial z}(0,0,1, \pi)} \\ {\text { (e) } \frac{\partial^{4} u}{\partial x \partial y \partial w \partial z}} & {\text { (f) } \frac{\partial^{4} u}{\partial w \partial z \partial y^{2}}}\end{array} $$

Confirm that the mixed second-order partial derivatives of \(f\) are the same. $$ f(x, y)=e^{x} \cos y $$

Prove: If \(f, f_{x},\) and \(f_{y}\) are continuous on a circular region containing \(A\left(x_{0}, y_{0}\right)\) and \(B\left(x_{1}, y_{1}\right),\) then there is a point \(\left(x^{*}, y^{*}\right)\) on the line segment joining \(A\) and \(B\) such that \(f\left(x_{1}, y_{1}\right)-f\left(x_{0}, y_{0}\right)\) $$ =f_{x}\left(x^{*}, y^{*}\right)\left(x_{1}-x_{0}\right)+f_{y}\left(x^{*}, y^{*}\right)\left(y_{1}-y_{0}\right) $$ This result is the two-dimensional version of the Mean Value Theorem. [Hint: Express the line segment joining \(A\) and \(B\) in parametric form and use the Mean-Value Theorem for functions of one variable. \(]\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.