Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 32

Use the transformations relating polar and Cartesian coordinates to prove that $$ \frac{d \theta}{d t}=\frac{1}{r^{2}}\left[x \frac{d y}{d t}-y \frac{d x}{d t}\right] $$

Short Answer

Expert verified
The provided identity is indeed correct. It's proven by leveraging the relationships between Cartesian and polar coordinates and applying the chain rule of differentiation.

Step by step solution

01

Recall relationship between cartesian and polar coordinates

Polar and Cartesian coordinates can be related through the following equations: \( x = r cos(\theta) \), \( y = r sin(\theta) \)
02

Calculate dx/dt and dy/dt

By taking the derivative of \( x = r cos(\theta) \) with respect to \( t \), we get: \(\frac{dx}{dt} = \frac{dr}{dt} cos(\theta) - r sin(\theta) \frac{d\theta}{dt}\). Similarly, taking the derivative of \( y = r sin(\theta) \) with respect to t, we get: \(\frac{dy}{dt} = \frac{dr}{dt} sin(\theta) + r cos(\theta) \frac{d\theta}{dt}.\
03

Substitute dx/dt and dy/dt in given expression

Substitute the expressions for dx/dt and dy/dt from step 2 into the given expression \( x \frac{dy}{dt} - y \frac{dx}{dt} \). This becomes \( r cos(\theta) * [ \frac{dr}{dt} sin(\theta) + r cos(\theta) \frac{d\theta}{dt} ] - r sin(\theta) * [ \frac{dr}{dt} cos(\theta) - r sin(\theta) \frac{d\theta}{dt} ] \).
04

Simplify expression

On simplifying the above expression, \(r^2 \frac{d\theta}{dt}\) terms cancel out and we get \(r^2 \frac{d\theta}{dt}\)
05

Divide by r^2

Finally, divide the whole expression by \(r^2\), bears out the required result, \(\frac{d\theta}{dt} = \frac{1}{r^{2}} [ x \frac{dy}{dt} - y \frac{dx}{dt} ]\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A spring fixed at its upper end is stretched 6 inches by a 10-pound weight attached at its lower end. The spring-mass system is suspended in a viscous medium so that the system is subjected to a damping force of \(5 \frac{d x}{d t}\) lbs. Describe the motion of the system if the weight is drawn down an additional 4 inches and released. What would happen if you changed the coefficient " 5 " to "4"? [You may need to consult your introductory physics text.]

Consider the flight of a tennis ball with mass \(57 \mathrm{~g}\) and a diameter of \(66.0 \mathrm{~mm}\). Assume the ball is served \(6.40\) meters from the net at a speed of \(50.0 \mathrm{~m} / \mathrm{s}\) down the center line from a height of \(2.8 \mathrm{~m}\). It needs to just clear the net \((0.914 \mathrm{~m})\). a. Ignoring air resistance and spin, analytically find the path of the ball assuming it just clears the net. Determine the angle to clear the net and the time of flight. b. Find the angle to clear the net assuming the tennis ball is given a topspin with \(w=50 \mathrm{rad} / \mathrm{s}\) c. Repeat part b assuming the tennis ball is given a bottom spin with \(\omega=50 \mathrm{rad} / \mathrm{s}\) d. Repeat parts \(\mathrm{a}, \mathrm{b}\), and \(\mathrm{c}\) with a drag force, taking \(C_{D}=0.55\).

Find all the solutions of the second-order differential equations. When an initial condition is given, find the particular solution satisfying that condition. a. \(y^{\prime \prime}-9 y^{\prime}+20 y=0\). b. \(y^{\prime \prime}-3 y^{\prime}+4 y=0, \quad y(0)=0, \quad y^{\prime}(0)=1\). c. \(x^{2} y^{\prime \prime}+5 x y^{\prime}+4 y=0, \quad x>0\). d. \(x^{2} y^{\prime \prime}-2 x y^{\prime}+3 y=0, \quad x>0\)

Use the Method of Variation of Parameters to determine the general solution for the following problems. a. \(y^{\prime \prime}+y=\tan x\). b. \(y^{\prime \prime}-4 y^{\prime}+4 y=6 x e^{2 x}\).

17\. A piece of a satellite falls to the ground from a height of \(10,000 \mathrm{~m}\). Ignoring air resistance, find the height as a function of time. [Hint: For free fall from large distances, $$ \ddot{h}=-\frac{G M}{(R+h)^{2}} $$ Multiplying both sides by \(\dot{h}\), show that $$ \frac{d}{d t}\left(\frac{1}{2} \dot{h}^{2}\right)=\frac{d}{d t}\left(\frac{G M}{R+h}\right) $$ Integrate and solve for \(\dot{h}\). Further integrating gives \(h(t) .]\)
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

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.

Sign-up for free