/*! 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 1 (a) Find all the first partial d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 1

(a) Find all the first partial derivatives of the following functions \(f(x, y)\) : (i) \(x^{2} y\), (ii) \(x^{2}+y^{2}+4\), (iii) \(\sin (x / y)\), (iv) \(\tan ^{-1}(y / x)\), (v) \(r(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}\). (b) For (i), (ii) and (v), find \(\partial^{2} f / \partial x^{2}, \partial^{2} f / \partial y^{2}\) and \(\partial^{2} f / \partial x \partial y\). (c) For (iv) verify that \(\partial^{2} f / \partial x \partial y=\partial^{2} f / \partial y \partial x\).

Short Answer

Expert verified
First partial derivatives and second partial derivatives are calculated as shown. Mixed partial derivatives for given functions are equal.

Step by step solution

01

Compute First Partial Derivatives

To compute the first partial derivatives, take the derivative of each function with respect to each variable. (i) For the function \( f(x,y) = x^2 y \):\[ \frac{\partial f}{\partial x} = 2xy \; \text{and} \; \frac{\partial f}{\partial y} = x^2 \](ii) For the function \( f(x,y) = x^2 + y^2 + 4 \):\[ \frac{\partial f}{\partial x} = 2x \; \text{and} \; \frac{\partial f}{\partial y} = 2y \](iii) For the function \( f(x,y) = \sin(x/y) \):\[ \frac{\partial f}{\partial x} = \frac{1}{y} \cos{\frac{x}{y}} \; \text{and} \; \frac{\partial f}{\partial y} = -\frac{x}{y^2} \cos{\frac{x}{y}} \](iv) For the function \( f(x,y) = \tan^{-1}(y/x) \):\[ \frac{\partial f}{\partial x} = \frac{-y}{x^2 + y^2} \; \text{and} \; \frac{\partial f}{\partial y} = \frac{x}{x^2 + y^2} \](v) For the function \( r(x,y,z) = \sqrt{x^2 + y^2 + z^2} \):\[ \frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}} \; \frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}} \; \frac{\partial r}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}} \]
02

Compute Second Partial Derivatives for (i)

For \( f(x,y) = x^2 y \):\[ \frac{\partial^2 f}{\partial x^2} = \frac{\partial(2xy)}{\partial x} = 2y \]\[ \frac{\partial^2 f}{\partial y^2} = \frac{\partial(x^2)}{\partial y} = 0 \]\[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial(x^2)}{\partial x} = 2x \]
03

Compute Second Partial Derivatives for (ii)

For \( f(x,y) = x^2 + y^2 + 4 \):\[ \frac{\partial^2 f}{\partial x^2} = \frac{\partial(2x)}{\partial x} = 2 \]\[ \frac{\partial^2 f}{\partial y^2} = \frac{\partial(2y)}{\partial y} = 2 \]\[ \frac{\partial^2 f}{\partial x \partial y} = 0 \]
04

Compute Second Partial Derivatives for (v)

For \( r(x,y,z) = (x^2 + y^2 + z^2)^{1/2} \):\[ \frac{\partial^2 r}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{x}{\sqrt{x^2 + y^2 + z^2}} \right) = \frac{y^2 + z^2}{(x^2 + y^2 + z^2)^{3/2}} \]\[ \frac{\partial^2 r}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{y}{\sqrt{x^2 + y^2 + z^2}} \right) = \frac{x^2 + z^2}{(x^2 + y^2 + z^2)^{3/2}} \]\[ \frac{\partial^2 r}{\partial x \partial y} = \frac{\partial}{\partial y} \left( \frac{x}{\sqrt{x^2 + y^2 + z^2}} \right) = -\frac{xy}{(x^2 + y^2 + z^2)^{3/2}} \]
05

Verify Mixed Partial Derivatives for (iv)

For \( f(x,y) = \tan^{-1}(y/x) \):First compute \( \frac{\partial^2 f}{\partial x \partial y} \)\[ \frac{\partial}{\partial y} \left( \frac{-y}{x^2 + y^2} \right) = \frac{-x^2 + y^2}{(x^2 + y^2)^2} \]Then compute \( \frac{\partial^2 f}{\partial y \partial x} \)\[ \frac{\partial}{\partial x} \left( \frac{x}{x^2 + y^2} \right) = \frac{x^2 - y^2}{(x^2 + y^2)^2} \]Since \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \), they are equal.

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.

first partial derivatives
First partial derivatives are the derivatives of a function taken with respect to one of its variables, while keeping the other variables constant. They provide the rate of change of the function with respect to a particular variable. For example, if we have a function of two variables, like \( f(x, y) \), the first partial derivatives are \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).

To compute these, we treat other variables as constants and differentiate accordingly:

1. For \( f(x, y) = x^2 y \):
\[ \frac{\partial f}{\partial x} = 2xy \; \text{and} \; \frac{\partial f}{\partial y} = x^2 \]
2. For \( f(x, y) = x^2 + y^2 + 4 \):
\[ \frac{\partial f}{\partial x} = 2x \; \text{and} \; \frac{\partial f}{\partial y} = 2y \]

First partial derivatives help us understand how the function changes along different axes in a multi-dimensional space.
second partial derivatives
Second partial derivatives are the derivatives of the first partial derivatives. They measure how the rate of change, given by the first partial derivatives, changes with respect to the variables. For instance, for a function \( f(x, y) \), there are several possible second partial derivatives, such as \( \frac{\partial^2 f}{\partial x^2} \), \( \frac{\partial^2 f}{\partial y^2} \), and mixed partial derivatives like \( \frac{\partial^2 f}{\partial x \partial y} \).

Here's how you compute them for some functions:

1. For \( f(x, y) = x^2 y \):
\[ \frac{\partial^2 f}{\partial x^2} = 2y \]
\[ \frac{\partial^2 f}{\partial y^2} = 0 \]

2. For \( f(x, y) = x^2 + y^2 + 4 \):
\[ \frac{\partial^2 f}{\partial x^2} = 2 \]
\[ \frac{\partial^2 f}{\partial y^2} = 2 \]

Second partial derivatives provide deeper insights into the curvature or concavity of the function's surface.
mixed partial derivatives
Mixed partial derivatives are the second partial derivatives where the differentiation is with respect to different variables. For a function of two variables \( f(x, y) \), this means, for example, taking the partial derivative of \( \frac{\partial f}{\partial x} \) with respect to \( y \), and vice versa.

Consider the following examples:

1. For \( f(x, y) = x^2 y \):
\[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial (2xy)}{\partial y} = 2x \]
2. For \( f(x, y) = x^2 + y^2 + 4 \):
\[ \frac{\partial^2 f}{\partial x \partial y} = 0 \]

They highlight how the function changes as you simultaneously adjust multiple variables. Mixed partial derivatives are essential in multivariable calculus and have important applications, such as in optimization problems.
verification of mixed partial derivatives
Verification of mixed partial derivatives involves checking if different orders of mixed partial derivatives of a function are equal. According to Schwarz's theorem (also known as Clairaut's theorem), if the mixed partial derivatives are continuous, they are equal. That is:
\[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \]

Let鈥檚 verify this for \( f(x, y) = \tan^{-1}(y/x) \):

1. Compute \( \frac{\partial^2 f}{\partial x \partial y} \):
\[ \frac{\partial}{\partial y} \left( \frac{-y}{x^2 + y^2} \right) = \frac{-x^2 + y^2}{(x^2 + y^2)^2} \]

2. Compute \( \frac{\partial^2 f}{\partial y \partial x} \):
\[ \frac{\partial}{\partial x} \left( \frac{x}{x^2 + y^2} \right) = \frac{x^2 - y^2}{(x^2 + y^2)^2} \]

Both mixed partial derivatives are equal, thus verifying Schwarz's theorem. This process ensures the consistency and correctness of calculating mixed partial derivatives in multivariable functions.

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

As in the previous two exercises on the thermodynamics of a simple gas, the quantity \(d S=T^{-1}(d U+P d V)\) is an exact differential. Use this to prove that $$ \left(\frac{\partial U}{\partial V}\right)_{T}=T\left(\frac{\partial P}{\partial T}\right)_{V}-P $$ In the van der Waals model of a gas, \(P\) obeys the equation $$ P=\frac{R T}{V-b}-\frac{a}{V^{2}} $$ where \(R, a\) and \(b\) are constants. Further, in the limit \(V \rightarrow \infty\), the form of \(U\) becomes \(U=c T\), where \(c\) is another constant. Find the complete expression for \(U(V, T)\)

Prove that the envelope of the circles whose diameters are those chords of a given circle that pass through a fixed point on its circumference, is the cardioid $$ r=a(1+\cos \theta) $$ Here \(a\) is the radius of the given circle and \((r, \theta)\) are the polar coordinates of the envelope. Take as the system parameter the angle \(\phi\) between a chord and the polar axis from which \(\theta\) is measured.

A barn is to be constructed with a uniform cross-sectional area \(A\) throughout its length. The cross-section is to be a rectangle of wall height \(h\) (fixed) and width \(w\), surmounted by an isosceles triangular roof that makes an angle \(\theta\) with the horizontal. The cost of construction is \(\alpha\) per unit height of wall and \(\beta\) per unit (slope) length of roof. Show that, irrespective of the values of \(\alpha\) and \(\beta\), to minimise costs \(w\) should be chosen to satisfy the equation $$ w^{4}=16 A(A-w h) $$ and \(\theta\) made such that \(2 \tan 2 \theta=w / h\).

Find and evaluate the maxima, minima and saddle points of the function $$ f(x, y)=x y\left(x^{2}+y^{2}-1\right) $$

The function \(f(x)\) is differentiable and \(f(0)=0\). A second function \(g(y)\) is defined by $$ g(y)=\int_{0}^{y} \frac{f(x) d x}{\sqrt{y-x}} $$ Prove that $$ \frac{d g}{d y}=\int_{0}^{y} \frac{d f}{d x} \frac{d x}{\sqrt{y-x}} $$ For the case \(f(x)=x^{n}\), prove that $$ \frac{d^{n} g}{d y^{n}}=2(n !) \sqrt{y} $$
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Medical Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Translational Dynamics

Read Explanation

Waves Physics

Read Explanation

Kinematics Physics

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.