/*! 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 Find the total differential. \... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the total differential. \(z=\frac{x^{2}}{y}\)

Short Answer

Expert verified
The total differential of the function \(z=\frac{x^{2}}{y}\) is \(dz = \frac{2x}{y} dx - \frac{x^{2}}{y^{2}} dy\).

Step by step solution

01

Find the Partial Derivative with respect to x

Firstly, let's find the partial derivative of \(z\) with respect to \(x\). This means treating \(y\) as a constant and differentiating \(z\) in terms of \(x\). So, \(\frac{\partial z}{\partial x} = \frac{2x}{y}\). This partial derivative gives the rate of change of \(z\) with respect to \(x\) holding \(y\) constant.
02

Find the Partial Derivative with respect to y

Now, let's find the partial derivative of \(z\) with respect to \(y\). This means treating \(x\) as a constant and differentiating \(z\) in terms of \(y\). So, \(\frac{\partial z}{\partial y} = -\frac{x^{2}}{y^{2}}\). This partial derivative gives the rate of change of \(z\) with respect to \(y\) holding \(x\) constant.
03

Find the Total Differential

Replace the found partial derivatives into the total differential formula. We get: \(dz = \frac{2x}{y} dx - \frac{x^{2}}{y^{2}} dy\). Hence, the total differential of the function \(z=\frac{x^{2}}{y}\) is \(dz = \frac{2x}{y} dx - \frac{x^{2}}{y^{2}} dy\).

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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 essential when dealing with multivariable functions. They help us understand how a function changes when one variable changes, while keeping others constant. In the problem provided, we're dealing with a function\(z = \frac{x^2}{y}\).

To find the partial derivative with respect to \(x\), we treat \(y\) as constant. This simplifies the differentiation process. The result, \(\frac{\partial z}{\partial x} = \frac{2x}{y}\), gives the rate of change of \(z\) as \(x\) varies and \(y\) is fixed. Similarly, the partial derivative with respect to \(y\) is \(\frac{\partial z}{\partial y} = -\frac{x^2}{y^2}\), which shows how \(z\) changes when \(y\) varies and \(x\) is held constant.

These derivatives are the building blocks for constructing the total differential, showing the overall change in a function as all variables vary.
Rate of Change
The rate of change is a concept that explains how one quantity changes in relation to another. In single-variable calculus, we'd describe how a function \(f(x)\) changes as \(x\) changes. However, in multivariable calculus, we deal with functions of more than one variable, like our given function \(z = \frac{x^2}{y}\).

Each partial derivative gives the rate of change of the function with respect to a single variable.
- The rate of change of \(z\) with respect to \(x\) is \(\frac{2x}{y}\), indicating how \(z\) responds as \(x\) changes (with \(y\) constant).- Similarly, the rate of change with respect to \(y\) is \(-\frac{x^2}{y^2}\), showing the impact on \(z\) as \(y\) varies, keeping \(x\) constant.

The total differential combines these partial changes to estimate the total rate of change in \(z\) due to small changes in both \(x\) and \(y\).
Multivariable Calculus
Multivariable calculus extends calculus concepts to functions with two or more variables. It opens the door to exploring and understanding complex systems in fields like physics, engineering, and economics. Our function, \(z = \frac{x^2}{y}\), is an example of how multivariable calculus operates.

This area of math involves calculating derivatives for functions with multiple inputs, called partial derivatives, and using them to analyze changes in multi-variable systems.
- Understanding behavior changes: How one variable can affect the output while others remain constant.- Finding optimal solutions: Identifying maximum or minimum values in complex surfaces.

The total differential \(dz = \frac{2x}{y} dx - \frac{x^2}{y^2} dy\) summarizes the combination of changes across multiple dimensions. Multivariable calculus enables us to predict how a small change in \(x\) or \(y\) will influence \(z\). By mastering these concepts, we unlock tools for solving real-world problems that single-variable calculus can't address.

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Most popular questions from this chapter

In Exercises \(39-42,\) find \(\partial w / \partial r\) and \(\partial w / \partial \theta\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(r\) and \(\boldsymbol{\theta}\) before differentiating. \(w=x^{2}-2 x y+y^{2}, x=r+\theta, \quad y=r-\theta\)

Heat-Seeking Path In Exercises 57 and \(58,\) find the path of a heat-seeking particle placed at point \(P\) on a metal plate with a temperature field \(T(x, y)\). $$ T(x, y)=400-2 x^{2}-y^{2}, \quad P(10,10) $$

The temperature at the point \((x, y)\) on a metal plate is modeled by \(T(x, y)=400 e^{-\left(x^{2}+y\right) / 2}, x \geq 0, y \geq 0\) (a) Use a computer algebra system to graph the temperature distribution function. (b) Find the directions of no change in heat on the plate from the point (3,5) . (c) Find the direction of greatest increase in heat from the point (3,5)

Find the gradient of the function and the maximum value of the directional derivative at the given point. $$ \frac{\text { Function }}{f(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}}} \frac{\text { Point }}{(1,4,2)} $$

Investigation \(\quad\) In Exercises \(\mathbf{3 3}\) and \(\mathbf{3 4}\), (a) use the graph to estimate the components of the vector in the direction of the maximum rate of increase of the function at the given point. (b) Find the gradient at the point and compare it with your estimate in part (a). (c) In what direction would the function be decreasing at the greatest rate? Explain. $$ \begin{array}{l} f(x, y)=\frac{1}{10}\left(x^{2}-3 x y+y^{2}\right), \\ (1,2) \end{array} $$

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