/*! 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 43 First note that $$d(\sqrt{x^{2... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 43

First note that $$d(\sqrt{x^{2}+y^{2}})=\frac{x}{\sqrt{x^{2}+y^{2}}} d x+\frac{y}{\sqrt{x^{2}+y^{2}}} d y$$ Then \(x d x+y d y=\sqrt{x^{2}+y^{2}} d x\) becomes $$\frac{x}{\sqrt{x^{2}+y^{2}}} d x+\frac{y}{\sqrt{x^{2}+y^{2}}} d y=d(\sqrt{x^{2}+y^{2}})=d x$$ The left side is the total differential of \(\sqrt{x^{2}+y^{2}}\) and the right side is the total differential of \(x+c\). Thus \(\sqrt{x^{2}+y^{2}}=x+c\) is a solution of the differential equation.

Short Answer

Expert verified
The solution is \(\sqrt{x^2 + y^2} = x + C\).

Step by step solution

01

Understand Given Information

We start with the given identity for the differential of \[d(\sqrt{x^{2}+y^{2}}) = \frac{x}{\sqrt{x^{2}+y^{2}}} \, dx + \frac{y}{\sqrt{x^{2}+y^{2}}} \, dy.\]This represents the total differential of the function \( \sqrt{x^2 + y^2} \).
02

Reshape Equation

The equation \( x \, dx + y \, dy = \sqrt{x^2 + y^2} \, dx \) is given. We want to express it in a form similar to a differential. Divide through by \( \sqrt{x^2 + y^2} \):\[ \frac{x}{\sqrt{x^2 + y^2}} \, dx + \frac{y}{\sqrt{x^2 + y^2}} \, dy = dx. \]
03

Compare with Known Differential

The left side of our reshaped equation is now identical to the given total differential \( d(\sqrt{x^2 + y^2}) \). This implies\[d(\sqrt{x^2 + y^2}) = dx.\]
04

Integrate Both Sides

Integrate both sides to express it in terms of original functions:\[ \sqrt{x^2 + y^2} = x + C, \]where \( C \) is the constant of integration.
05

State the Solution

The problem implies that \( \sqrt{x^2 + y^2} = x + C \) is a solution to the differential equation. Hence, this is the relationship between the variables that satisfies our initial equation.

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.

Total Differential
The concept of a total differential is crucial in multivariable calculus. It provides a way to approximate the change in a function with respect to all its variables. For a function like \ \(f(x, y) = \sqrt{x^2 + y^2}\ \), the total differential \ \(df\ \) represents the change in \ \(f\ \) due to small changes in \ \(x\ \) and \ \(y\ \).
In our exercise, \ \(d(\sqrt{x^2+y^2}) = \frac{x}{\sqrt{x^2+y^2}}dx + \frac{y}{\sqrt{x^2+y^2}}dy\ \) is the total differential of \ \(\sqrt{x^2+y^2}\ \).
This expression essentially tells how the function \ \(\sqrt{x^2+y^2}\ \) changes when you make small changes to \ \(x\ \) and \ \(y\ \).
The differential terms \ \(dx\ \) and \ \(dy\ \) represent infinitesimally small increments in \ \(x\ \) and \ \(y\ \), and their coefficients give the rate of change of \ \(\sqrt{x^2+y^2}\ \) with respect to these variables.
This type of differential analysis facilitates understanding how multivariable functions can change and helps in solving problems that involve small changes across different dimensions.
Integration
Integration is a key process in calculus used to find solutions to differential equations or to determine areas under curves. In the context of our exercise, integration appears when determining the solution to a differential equation.
Once we have written the differential equation \ \(d(\sqrt{x^2+y^2}) = dx\ \), the next step is to integrate both sides to find the function from its differential.
This process helps us to revert the effects of taking the differential, effectively calculating the original function, possibly up to a constant.
When we perform the integration on both sides:
  • On the left side, integrating \ \(d(\sqrt{x^2+y^2})\ \) gives us back \ \(\sqrt{x^2+y^2}\ \).
  • On the right side, integrating \ \(dx\ \) yields \ \(x + C\ \), where \ \(C\ \) represents the constant of integration.
    This constant is crucial because it accounts for the indefinite nature of integration. Every family of curves that differ by a constant is a possible solution to the differential equation.
Understanding integration not only helps solve specific differential equations like this one, but it also exposes how broader patterns in integration can describe a range of potential scenarios.
Equation Solving
Solving a differential equation involves finding a function that satisfies the equation. In this exercise, the challenge is to find a function that fits the given relationships between differentials.
The process begins by recognizing equivalent forms using differentials. We observed that \ \( \frac{x}{\sqrt{x^2+y^2}}dx + \frac{y}{\sqrt{x^2+y^2}}dy = dx\ \) matches the known total differential form.
This insight allows us to equate \ \(d(\sqrt{x^2+y^2})\ =\ dx\ \), leading directly to integration.
After integrating, the solution \ \(\sqrt{x^2+y^2}\ =\ x\ +\ C\ \) offers a relationship that governs \ \(x\ \) and \ \(y\ \) based on their geometric interpretation as a distance. Each \ \(C\ \) determines a specific curve, representing distinct solutions that depend on initial conditions or specific scenarios.
Equation solving in this context showcases how identifiable patterns allow us to derive solutions that satisfy complex differential relations. It's all about transforming, recognizing known forms, and employing calculus methods to unravel solutions that elaborate the behavior of involved variables.

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

From \(d A / d t=10-A / 100\) we obtain \(A=1000+c e^{-t / 100} .\) If \(A(0)=0\) then \(c=-1000\) and \(A(t)=\) \(1000-1000 e^{-t / 100}\).

We are looking for a function \(y(x)\) such that \\[y^{2}+\left(\frac{d y}{d x}\right)^{2}=1\\]. Using the positive square root gives \\[\frac{d y}{d x}=\sqrt{1-y^{2}} \Longrightarrow \frac{d y}{\sqrt{1-y^{2}}}=d x \Longrightarrow \sin ^{-1} y=x+c\\]. Note that when \(c=c_{1}=0\) and when \(c=c_{1}=\pi / 2\) we obtain the well known particular solutions \(y=\sin x\) \(y=-\sin x, y=\cos x,\) and \(y=-\cos x .\) Note also that \(y=1\) and \(y=-1\) are singular solutions.

(a) From \(m d v / d t=m g-k v\) we obtain \(v=m g / k+c e^{-k t / m} .\) If \(v(0)=v_{0}\) then \(c=v_{0}-m g / k\) and the solution of the initial-value problem is $$v(t)=\frac{m g}{k}+\left(v_{0}-\frac{m g}{k}\right) e^{-k t / m}$$. (b) As \(t \rightarrow \infty\) the limiting velocity is \(m g / k\) (c) From \(d s / d t=v\) and \(s(0)=0\) we obtain $$s(t)=\frac{m g}{k} t-\frac{m}{k}\left(v_{0}-\frac{m g}{k}\right) e^{-k t / m}+\frac{m}{k}\left(v_{0}-\frac{m g}{k}\right)$$.

(a) \((i)\) Using Newton's second law of motion, \(F=m a=m d v / d t,\) the differential equation for the velocity \(v\) is $$m \frac{d v}{d t}=m g \sin \theta \quad \text { or } \quad \frac{d v}{d t}=g \sin \theta$$ where \(m g \sin \theta, 0< \theta< \pi / 2,\) is the component of the weight along the plane in the direction of motion. \((i i)\) The model now becomes $$m \frac{d v}{d t}=m g \sin \theta-\mu m g \cos \theta$$ where \(\mu m g \cos \theta\) is the component of the force of sliding friction (which acts perpendicular to the plane) along the plane. The negative sign indicates that this component of force is a retarding force which acts in the direction opposite to that of motion. \((i i i)\) If air resistance is taken to be proportional to the instantaneous velocity of the body, the model becomes $$m \frac{d v}{d t}=m g \sin \theta-\mu m g \cos \theta-k v$$ where \(k\) is a constant of proportionality. (b) \((i)\) With \(m=3\) slugs, the differential equation is $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2} \quad \text { or } \quad \frac{d v}{d t}=16$$ Integrating the last equation gives \(v(t)=16 t+c_{1} .\) since \(v(0)=0,\) we have \(c_{1}=0\) and so \(v(t)=16 t\). \((i i)\) With \(m=3\) slugs, the differential equation is $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2}-\frac{\sqrt{3}}{4} \cdot(96) \cdot \frac{\sqrt{3}}{2} \quad \text { or } \quad \frac{d v}{d t}=4$$ In this case \(v(t)=4 t\). (iii) When the retarding force due to air resistance is taken into account, the differential equation for velocity \(v\) becomes $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2}-\frac{\sqrt{3}}{4} \cdot(96) \cdot \frac{\sqrt{3}}{2}-\frac{1}{4} v \quad \text { or } \quad 3 \frac{d v}{d t}=12-\frac{1}{4} v$$ The last differential equation is linear and has solution \(v(t)=48+c_{1} e^{-t / 12} .\) since \(v(0)=0,\) we find \(c_{1}=-48,\) so \(v(t)=48-48 e^{-t / 12}\).

From \(\frac{1}{1-2 y} d y=d t\) we obtain \(-\frac{1}{2} \ln |1-2 y|=t+c\) or \(1-2 y=c_{1} e^{-2 t} .\) Using \(y(0)=5 / 2\) we find \(c_{1}=-4\). The solution of the initial-value problem is \(1-2 y=-4 e^{-2 t}\) or \(y=2 e^{-2 t}+\frac{1}{2}\).
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Kinematics Physics

Read Explanation

Electricity

Read Explanation

Radiation

Read Explanation

Mechanics and Materials

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.