This document is the table of contents for a textbook on classical dynamics. It lists 14 chapters covering topics like matrices and vectors, Newtonian mechanics, oscillations, gravitation, Lagrangian and Hamiltonian dynamics, and special relativity. It also lists the problems solved in the student solutions manual.
1) The document derives both the continuous and discrete forms of hybrid Adams-Moulton methods for step numbers k=1 and k=2. These formulations incorporate off-grid interpolation and off-grid collocation schemes.
2) A matrix inversion technique is used to derive the continuous form. The continuous and discrete coefficients are obtained by solving a matrix equation where the identity matrix equals the product of two other matrices.
3) Error and zero-stability analyses are performed on the derived discrete schemes. The schemes are found to be of good order, with good error constants, implying they are consistent.
Here are the key steps to solve this separable differential equation:
1) Separate the variables: dy/dx = (1-y^2)
2) Integrate both sides: ∫ dy/(1-y^2) = ∫ dx
3) Evaluate the integrals: arctan(y) = x + C
4) Take the inverse tangent of both sides: y = tan(x + C)
This is the general solution.
1.4
Transformations
We can transform a differential equation into a separable one using the following techniques:
So the general solution is:
y = tan(x + C)
1. Change of variables:
This document provides an overview of statistical tests commonly used in neuroimaging such as t-tests, ANOVAs, and regression. It discusses the purposes of these tests and how they are applied. T-tests are used to compare means, for example to determine if the difference between two conditions is statistically significant. ANOVAs examine variances and can be used when comparing more than two groups. Regression allows describing and predicting the relationship between variables and is useful in the general linear model approach used in SPM. Key assumptions and calculations for each method are outlined.
- Quiz 4 will be tomorrow covering sections 3.3, 5.1, and 5.2 of the textbook. It will include 3 problems on Cramer's rule, finding eigenvectors given eigenvalues, and finding characteristic polynomials/eigenvalues of 2x2 and 3x3 matrices. Students must show all work.
- Chapter 6 objectives include extending geometric concepts like length, distance, and perpendicularity to Rn. These concepts are useful for least squares fitting of experimental data to a system of equations.
- The inner product of two vectors u and v in Rn is defined as their dot product, which is the sum of the component-wise products of corresponding elements in u and v.
The document summarizes an investigation into the sum of infinite sequences of the form 1/n! as n approaches infinity. Testing various values for variables a and x, the analysis found that the infinite sum approaches the value of a when x=1, and it approaches the value of x when a is held constant. This was supported by graphs showing the sums leveling off at the respective a or x values as more terms were added. The summary provides the key findings and conclusions from the mathematical analysis in the document.
This document contains notes from a Calculus I class lecture on the derivative. The lecture covered the definition of the derivative and examples of how it can be used to model rates of change in various contexts like velocity, population growth, and marginal costs. It also discussed properties of the derivative like how the derivative of a function relates to whether the function is increasing or decreasing over an interval.
The eigen values of a Hermitian matrix are always real. This is because for a Hermitian matrix A, the quadratic form x*Ax is always real for any vector x. Now, if λ is an eigen value of A corresponding to the eigenvector v, then we have:
λv*v = v*Av
λv*v = v*λv (since Av = λv)
λv*v = λv*v
Therefore, λ must be real. Similarly, for a real symmetric matrix, the quadratic form x'Ax is always real. Hence, the eigen values must be real.
So in summary, the eigen values of both Hermitian and real
1. The document discusses ordinary differential equations and provides definitions and examples of separable, homogeneous, exact, linear, and Bernoulli equations.
2. Methods for solving first order differential equations are presented, including finding acceptable solutions in terms of p, y, or x. Lagrange's and Clairaut's equations are also discussed.
3. Higher order and degree differential equations can be solved using methods like Lagrange's equation, Clairaut's equation, or solving the linear homogeneous and non-homogeneous forms with constant coefficients.
1) The document derives both the continuous and discrete forms of hybrid Adams-Moulton methods for step numbers k=1 and k=2. These formulations incorporate off-grid interpolation and off-grid collocation schemes.
2) A matrix inversion technique is used to derive the continuous form. The continuous and discrete coefficients are obtained by solving a matrix equation where the identity matrix equals the product of two other matrices.
3) Error and zero-stability analyses are performed on the derived discrete schemes. The schemes are found to be of good order, with good error constants, implying they are consistent.
Here are the key steps to solve this separable differential equation:
1) Separate the variables: dy/dx = (1-y^2)
2) Integrate both sides: ∫ dy/(1-y^2) = ∫ dx
3) Evaluate the integrals: arctan(y) = x + C
4) Take the inverse tangent of both sides: y = tan(x + C)
This is the general solution.
1.4
Transformations
We can transform a differential equation into a separable one using the following techniques:
So the general solution is:
y = tan(x + C)
1. Change of variables:
This document provides an overview of statistical tests commonly used in neuroimaging such as t-tests, ANOVAs, and regression. It discusses the purposes of these tests and how they are applied. T-tests are used to compare means, for example to determine if the difference between two conditions is statistically significant. ANOVAs examine variances and can be used when comparing more than two groups. Regression allows describing and predicting the relationship between variables and is useful in the general linear model approach used in SPM. Key assumptions and calculations for each method are outlined.
- Quiz 4 will be tomorrow covering sections 3.3, 5.1, and 5.2 of the textbook. It will include 3 problems on Cramer's rule, finding eigenvectors given eigenvalues, and finding characteristic polynomials/eigenvalues of 2x2 and 3x3 matrices. Students must show all work.
- Chapter 6 objectives include extending geometric concepts like length, distance, and perpendicularity to Rn. These concepts are useful for least squares fitting of experimental data to a system of equations.
- The inner product of two vectors u and v in Rn is defined as their dot product, which is the sum of the component-wise products of corresponding elements in u and v.
The document summarizes an investigation into the sum of infinite sequences of the form 1/n! as n approaches infinity. Testing various values for variables a and x, the analysis found that the infinite sum approaches the value of a when x=1, and it approaches the value of x when a is held constant. This was supported by graphs showing the sums leveling off at the respective a or x values as more terms were added. The summary provides the key findings and conclusions from the mathematical analysis in the document.
This document contains notes from a Calculus I class lecture on the derivative. The lecture covered the definition of the derivative and examples of how it can be used to model rates of change in various contexts like velocity, population growth, and marginal costs. It also discussed properties of the derivative like how the derivative of a function relates to whether the function is increasing or decreasing over an interval.
The eigen values of a Hermitian matrix are always real. This is because for a Hermitian matrix A, the quadratic form x*Ax is always real for any vector x. Now, if λ is an eigen value of A corresponding to the eigenvector v, then we have:
λv*v = v*Av
λv*v = v*λv (since Av = λv)
λv*v = λv*v
Therefore, λ must be real. Similarly, for a real symmetric matrix, the quadratic form x'Ax is always real. Hence, the eigen values must be real.
So in summary, the eigen values of both Hermitian and real
1. The document discusses ordinary differential equations and provides definitions and examples of separable, homogeneous, exact, linear, and Bernoulli equations.
2. Methods for solving first order differential equations are presented, including finding acceptable solutions in terms of p, y, or x. Lagrange's and Clairaut's equations are also discussed.
3. Higher order and degree differential equations can be solved using methods like Lagrange's equation, Clairaut's equation, or solving the linear homogeneous and non-homogeneous forms with constant coefficients.
This document discusses first order differential equations. It defines differential equations and classifies them as ordinary or partial based on whether they involve derivatives with respect to a single or multiple variables. First order differential equations are classified into four types: variable separable, homogeneous, linear, and exact. The document provides examples of each type and explains their general forms and solution methods like separating variables, making substitutions, and integrating.
This document provides an outline and learning objectives for a midterm exam covering vectors and three-dimensional coordinate systems in a Math 21a course. The midterm will cover material up to and including section 11.4 in the textbook. It outlines key topics like three-dimensional coordinate systems, vectors, the dot and cross product, equations of lines and planes, and vector functions. Examples are provided for distance between points in space and rewriting an equation in standard form to identify what surface it represents. Learning objectives are stated for topics like three-dimensional coordinate systems, vectors, and vector addition.
This document provides an overview of algebraic techniques in combinatorics, including linear algebra concepts, partially ordered sets (posets), and examples of problems solved using these techniques. Some key points discussed are:
- Useful linear algebra facts such as rank, determinants, and vector/matrix properties
- Definitions and representations of posets, including Dilworth's theorem relating chains and antichains
- Examples of combinatorial problems solved using linear algebra tools such as vectors/matrices or applying Dilworth's theorem to obtain a divisibility relation poset
Normal density and discreminant analysisVARUN KUMAR
This document provides an overview of Gaussian density and discriminant analysis. It discusses mathematical descriptions of discriminant functions and how they are used in classifiers to select classes. It also covers bi-variate and multi-variate normal density functions and how they relate to dependent and independent random variables. Specifically, it shows how the shape of the loci of bi-variate density functions depends on whether variables are independent or dependent. Finally, it discusses discriminant functions and how to determine decision boundaries between classes.
This document provides examples of constructing the dual problem of a linear programming primal problem and solving it using the two-phase simplex method. It first presents the rules for constructing the dual problem and then works through two examples. The first example derives the dual problem from the primal and solves it using the two-phase method. The second example shows how to find the optimal dual solution given the optimal primal solution using two methods - using the objective coefficients of the primal variables or using the inverse of the primal basic variable matrix.
The document summarizes solutions to homogeneous linear differential equations with constant coefficients. It discusses:
1) Equations with real distinct roots have n distinct exponential solutions.
2) Equations with complex roots have solutions involving sine and cosine of the complex terms.
3) Equations with repeated roots have solutions involving higher powers of the variable.
4) Examples are provided to illustrate the different cases.
This document provides lecture notes for a second semester calculus course. It covers various methods of integration including standard integrals, substitution, integration by parts, and more. It also covers topics like Taylor series, complex numbers, differential equations, vectors, and parametrized curves. The notes were written by Sigurd Angenent and are made freely available under a free documentation license.
This document discusses statistical tools used in quality control laboratories and validation studies, including normal distributions, variance, ranges, coefficients of variation, F-tests, Student's t-tests, and paired t-tests. It provides the formulas and procedures for calculating and applying these statistical concepts to analyze laboratory data and test for significant differences between samples. Examples are given to demonstrate how to perform t-tests to compare averages from independent and paired samples with both known and unknown variances.
This document defines and provides examples of linear differential equations. It discusses:
1) Linear differential equations can be written in the form P(x)y'=Q(x) or P(y)x'=Q(y), where multiplying both sides by an integrating factor μ results in a total derivative.
2) First order linear differential equations of the form P(x)y'=Q(x) have an integrating factor of e∫P(x)dx. The general solution is y(IF)=C.
3) Bernoulli's equation is a differential equation of the form P(x)y'+Q(x)y^n=R(x), where the general solution depends
The document discusses higher order differential equations. It defines nth order differential equations and describes their general forms. For homogeneous equations, the general solution method involves making an operator form, constructing an auxiliary equation, solving for roots, and finding the complementary solution. For non-homogeneous equations, the method of undetermined coefficients is used to find a particular solution and the general solution is the sum of the complementary and particular solutions. Examples are provided to illustrate the solution methods.
Here are the steps to solve this revision exercise:
1. Use the distance formula to find the lengths of each side:
a) AB = √(−1 − 2)2 + (4 − 3)2 = 3.16
b) BC = √(1 − (−1))2 + (−3 − 4)2 = 7.28
c) AC = √(1 − 2)2 + (−3 − 3)2 = 6.08
2. Use the midpoint formula to find the midpoints of each side:
a) MPAB = (0.5, 3.5)
b) MPBC = (0, 0.5)
c) MPAC = (
Lesson 13: Exponential and Logarithmic Functions (Section 021 handout)Matthew Leingang
This document contains lecture notes from a Calculus I class at New York University on October 21, 2010 covering sections 3.1-3.2 on exponential functions. The notes include announcements about an upcoming midterm exam and homework assignment. Statistics on the recent midterm exam are provided, showing the average, median, standard deviation, and what constitutes a "good" or "great" score. The objectives of sections 3.1-3.2 are outlined as understanding exponential functions, their properties, and laws of logarithms. The notes provide definitions and derivations of exponential functions for various exponent values.
Solution of a singular class of boundary value problems by variation iteratio...Alexander Decker
1. The document proposes an effective methodology called the variation iteration method to find solutions to a general class of singular second-order linear and nonlinear boundary value problems.
2. The variation iteration method generates a sequence of correction functionals that converges to the exact solution of the boundary value problem.
3. The author applies the variation iteration method to solve a specific class of boundary value problems and derives the sequence of correction functionals. Convergence of the iterative sequence is also analyzed.
Artificial variable technique big m method (1)ਮਿਲਨਪ੍ਰੀਤ ਔਜਲਾ
The Big-M method is used to handle artificial variables in linear programming problems. It assigns very large coefficients to the artificial variables in the objective function, making them undesirable to include in optimal solutions. This removes the artificial variables from the basis. As an example, the document presents a linear programming problem to minimize an objective function subject to constraints, and shows the steps of converting it to an equivalent problem using artificial variables and applying the Big-M method to arrive at an optimal solution without artificial variables.
11.solution of a singular class of boundary value problems by variation itera...Alexander Decker
1) The document proposes an effective methodology called the variation iteration method to find solutions to singular second order linear and nonlinear boundary value problems.
2) The variation iteration method constructs a sequence of correction functionals to iteratively solve the boundary value problem. It is shown that the limit of the convergent iterative sequence obtained from this method is the exact solution.
3) The convergence of the iterative sequence generated by the variation iteration method is analyzed. It is established that the sequence converges to the exact solution under certain continuity conditions on the problem functions.
A Comparison of Evaluation Methods in Coevolution 20070921Ting-Shuo Yo
This document compares different evaluation methods in coevolution, including averaged score, weighted score, averaged informativeness, weighted informativeness, and a multi-objective approach. It presents results on three test problems: majority function problem, symbolic regression problem, and parity problem. The results show that the multi-objective approach achieves the best performance in terms of objective fitness and correlation between subjective and objective fitness.
This document presents an overview of first order ordinary differential equations and applications. It contains:
1) The standard form of a linear first order differential equation and examples of solving three types of equations.
2) Applications of differential equations to model population growth and finding the equation of a curve given its slope at a point.
3) Solutions to the examples and applications in 3 sentences or less.
This document announces the release of Version 5 educational software containing over 15,000 presentation slides, 1,000 example/student questions, 100 worksheets, 1,200 interactive exercises, and 5,000 mental math questions across two CDs. It provides a 7-minute demo of 20 sample slides and directs users to register for a free account to access additional full presentations.
Modul penggunaan kalkulator sainstifik sebagai ABM dalam MatematikNorsyazana Kamarudin
This document provides information about discriminants of quadratic equations. It defines quadratic equations and explains that the discriminant, which is b^2 - 4ac, provides information about the number and type of roots. A positive discriminant indicates two real roots, a zero discriminant indicates one real root, and a negative discriminant indicates no real roots. Examples of solving quadratic equations with a scientific calculator are provided. Worksheets ask students to determine the type of roots and solutions for different quadratic equations using the discriminant and with or without a calculator.
This presentation explains how the differentiation is applied to identify increasing and decreasing functions,identifying the nature of stationary points and also finding maximum or minimum values.
This document discusses first order differential equations. It defines differential equations and classifies them as ordinary or partial based on whether they involve derivatives with respect to a single or multiple variables. First order differential equations are classified into four types: variable separable, homogeneous, linear, and exact. The document provides examples of each type and explains their general forms and solution methods like separating variables, making substitutions, and integrating.
This document provides an outline and learning objectives for a midterm exam covering vectors and three-dimensional coordinate systems in a Math 21a course. The midterm will cover material up to and including section 11.4 in the textbook. It outlines key topics like three-dimensional coordinate systems, vectors, the dot and cross product, equations of lines and planes, and vector functions. Examples are provided for distance between points in space and rewriting an equation in standard form to identify what surface it represents. Learning objectives are stated for topics like three-dimensional coordinate systems, vectors, and vector addition.
This document provides an overview of algebraic techniques in combinatorics, including linear algebra concepts, partially ordered sets (posets), and examples of problems solved using these techniques. Some key points discussed are:
- Useful linear algebra facts such as rank, determinants, and vector/matrix properties
- Definitions and representations of posets, including Dilworth's theorem relating chains and antichains
- Examples of combinatorial problems solved using linear algebra tools such as vectors/matrices or applying Dilworth's theorem to obtain a divisibility relation poset
Normal density and discreminant analysisVARUN KUMAR
This document provides an overview of Gaussian density and discriminant analysis. It discusses mathematical descriptions of discriminant functions and how they are used in classifiers to select classes. It also covers bi-variate and multi-variate normal density functions and how they relate to dependent and independent random variables. Specifically, it shows how the shape of the loci of bi-variate density functions depends on whether variables are independent or dependent. Finally, it discusses discriminant functions and how to determine decision boundaries between classes.
This document provides examples of constructing the dual problem of a linear programming primal problem and solving it using the two-phase simplex method. It first presents the rules for constructing the dual problem and then works through two examples. The first example derives the dual problem from the primal and solves it using the two-phase method. The second example shows how to find the optimal dual solution given the optimal primal solution using two methods - using the objective coefficients of the primal variables or using the inverse of the primal basic variable matrix.
The document summarizes solutions to homogeneous linear differential equations with constant coefficients. It discusses:
1) Equations with real distinct roots have n distinct exponential solutions.
2) Equations with complex roots have solutions involving sine and cosine of the complex terms.
3) Equations with repeated roots have solutions involving higher powers of the variable.
4) Examples are provided to illustrate the different cases.
This document provides lecture notes for a second semester calculus course. It covers various methods of integration including standard integrals, substitution, integration by parts, and more. It also covers topics like Taylor series, complex numbers, differential equations, vectors, and parametrized curves. The notes were written by Sigurd Angenent and are made freely available under a free documentation license.
This document discusses statistical tools used in quality control laboratories and validation studies, including normal distributions, variance, ranges, coefficients of variation, F-tests, Student's t-tests, and paired t-tests. It provides the formulas and procedures for calculating and applying these statistical concepts to analyze laboratory data and test for significant differences between samples. Examples are given to demonstrate how to perform t-tests to compare averages from independent and paired samples with both known and unknown variances.
This document defines and provides examples of linear differential equations. It discusses:
1) Linear differential equations can be written in the form P(x)y'=Q(x) or P(y)x'=Q(y), where multiplying both sides by an integrating factor μ results in a total derivative.
2) First order linear differential equations of the form P(x)y'=Q(x) have an integrating factor of e∫P(x)dx. The general solution is y(IF)=C.
3) Bernoulli's equation is a differential equation of the form P(x)y'+Q(x)y^n=R(x), where the general solution depends
The document discusses higher order differential equations. It defines nth order differential equations and describes their general forms. For homogeneous equations, the general solution method involves making an operator form, constructing an auxiliary equation, solving for roots, and finding the complementary solution. For non-homogeneous equations, the method of undetermined coefficients is used to find a particular solution and the general solution is the sum of the complementary and particular solutions. Examples are provided to illustrate the solution methods.
Here are the steps to solve this revision exercise:
1. Use the distance formula to find the lengths of each side:
a) AB = √(−1 − 2)2 + (4 − 3)2 = 3.16
b) BC = √(1 − (−1))2 + (−3 − 4)2 = 7.28
c) AC = √(1 − 2)2 + (−3 − 3)2 = 6.08
2. Use the midpoint formula to find the midpoints of each side:
a) MPAB = (0.5, 3.5)
b) MPBC = (0, 0.5)
c) MPAC = (
Lesson 13: Exponential and Logarithmic Functions (Section 021 handout)Matthew Leingang
This document contains lecture notes from a Calculus I class at New York University on October 21, 2010 covering sections 3.1-3.2 on exponential functions. The notes include announcements about an upcoming midterm exam and homework assignment. Statistics on the recent midterm exam are provided, showing the average, median, standard deviation, and what constitutes a "good" or "great" score. The objectives of sections 3.1-3.2 are outlined as understanding exponential functions, their properties, and laws of logarithms. The notes provide definitions and derivations of exponential functions for various exponent values.
Solution of a singular class of boundary value problems by variation iteratio...Alexander Decker
1. The document proposes an effective methodology called the variation iteration method to find solutions to a general class of singular second-order linear and nonlinear boundary value problems.
2. The variation iteration method generates a sequence of correction functionals that converges to the exact solution of the boundary value problem.
3. The author applies the variation iteration method to solve a specific class of boundary value problems and derives the sequence of correction functionals. Convergence of the iterative sequence is also analyzed.
Artificial variable technique big m method (1)ਮਿਲਨਪ੍ਰੀਤ ਔਜਲਾ
The Big-M method is used to handle artificial variables in linear programming problems. It assigns very large coefficients to the artificial variables in the objective function, making them undesirable to include in optimal solutions. This removes the artificial variables from the basis. As an example, the document presents a linear programming problem to minimize an objective function subject to constraints, and shows the steps of converting it to an equivalent problem using artificial variables and applying the Big-M method to arrive at an optimal solution without artificial variables.
11.solution of a singular class of boundary value problems by variation itera...Alexander Decker
1) The document proposes an effective methodology called the variation iteration method to find solutions to singular second order linear and nonlinear boundary value problems.
2) The variation iteration method constructs a sequence of correction functionals to iteratively solve the boundary value problem. It is shown that the limit of the convergent iterative sequence obtained from this method is the exact solution.
3) The convergence of the iterative sequence generated by the variation iteration method is analyzed. It is established that the sequence converges to the exact solution under certain continuity conditions on the problem functions.
A Comparison of Evaluation Methods in Coevolution 20070921Ting-Shuo Yo
This document compares different evaluation methods in coevolution, including averaged score, weighted score, averaged informativeness, weighted informativeness, and a multi-objective approach. It presents results on three test problems: majority function problem, symbolic regression problem, and parity problem. The results show that the multi-objective approach achieves the best performance in terms of objective fitness and correlation between subjective and objective fitness.
This document presents an overview of first order ordinary differential equations and applications. It contains:
1) The standard form of a linear first order differential equation and examples of solving three types of equations.
2) Applications of differential equations to model population growth and finding the equation of a curve given its slope at a point.
3) Solutions to the examples and applications in 3 sentences or less.
This document announces the release of Version 5 educational software containing over 15,000 presentation slides, 1,000 example/student questions, 100 worksheets, 1,200 interactive exercises, and 5,000 mental math questions across two CDs. It provides a 7-minute demo of 20 sample slides and directs users to register for a free account to access additional full presentations.
Modul penggunaan kalkulator sainstifik sebagai ABM dalam MatematikNorsyazana Kamarudin
This document provides information about discriminants of quadratic equations. It defines quadratic equations and explains that the discriminant, which is b^2 - 4ac, provides information about the number and type of roots. A positive discriminant indicates two real roots, a zero discriminant indicates one real root, and a negative discriminant indicates no real roots. Examples of solving quadratic equations with a scientific calculator are provided. Worksheets ask students to determine the type of roots and solutions for different quadratic equations using the discriminant and with or without a calculator.
This presentation explains how the differentiation is applied to identify increasing and decreasing functions,identifying the nature of stationary points and also finding maximum or minimum values.
The document discusses the multiple linear regression model and ordinary least squares (OLS) estimation. It presents the econometric model, where a dependent variable is modeled as a linear function of explanatory variables, plus an error term. It describes the assumptions of the linear regression model, including linearity, independence of observations, exogeneity of regressors, and properties of the error term. It then discusses OLS estimation, goodness of fit, hypothesis testing, confidence intervals, and asymptotic properties of the OLS estimator.
This document provides an overview of the fundamentals of image processing. It begins with an introduction to key mathematical foundations, including vectors, matrices, vector spaces, bases, inner products, projections, and linear transforms. It then covers topics such as discrete-time signals and systems, linear time-invariant systems, sampling continuous signals to discrete and vice versa, digital filter design, image formation via lenses and sensors, point-wise and linear filtering operations, motion estimation, and useful tools like expectation-maximization and principal component analysis. The document serves as a guide to the core concepts and techniques in digital image processing.
The document discusses artificial neural networks and their applications. It covers topics like biological inspiration for ANNs, why they are used, learning strategies and techniques, network architectures like MLP, activation functions, and applications in areas like pattern classification, time series forecasting, control, and optimization. Key applications mentioned include handwritten digit recognition, remote sensing, machine control, and predicting things like hospital stay length and gas prices. References on the topic are also provided.
The document provides instructions on calculating slopes of lines and identifying relationships between parallel and perpendicular lines. It discusses using increments to find the slope of a line using the change in x and y between two points. It also discusses the different forms of linear equations (point-slope, slope-intercept, and standard form) and defines the key rules that parallel lines have the same slope and perpendicular lines have slopes that multiply to -1. The document demonstrates how to perform linear regression on a TI-89 calculator to generate a linear equation from a scatter plot. It provides example problems and homework questions for students to practice these skills.
This document discusses key concepts related to lines and slopes in mathematics including:
- How to calculate the slope of a line using the change in x and y between two points
- The different forms of a line equation (point-slope, slope-intercept, standard)
- That parallel lines have the same slope and perpendicular lines have slopes that are negative reciprocals
- How to perform linear regression using a graphing calculator to find the equation for a line of best fit through data points
The document discusses key concepts in regression analysis including best-fit lines, correlation coefficients, linearization, residuals, weighted means, and using linear approximations and differentials. It provides examples of how to calculate coefficients for best-fit lines, determine correlation direction and strength, find linearized forms of functions, and use Newton's method to find where functions intersect. The goal of regression is to find a formula that minimizes residuals and best represents the relationship between variables.
Scatter diagrams and correlation and simple linear regresssionAnkit Katiyar
The document discusses scatter diagrams, correlation, and linear regression. It defines key terms like predictor and response variables, positively and negatively associated variables, and the correlation coefficient. It also describes how to calculate the linear correlation coefficient and interpret it. The document shows an example of using least squares regression to fit a line to productivity and experience data. It provides formulas to calculate the slope and intercept of the regression line and how to make predictions with the line. However, predictions should stay within the scope of the observed data used to fit the model.
The document describes the Jacobi iterative method for solving systems of linear equations. It begins with an initial estimate for the solution variables, inserts them into the equations to get updated estimates, and repeats this process iteratively until the estimates converge to the desired solution. As an example, it applies the method to a set of 3 equations in 3 unknowns, showing the estimates after each iteration getting progressively closer to the exact solution obtained using Gaussian elimination. A Fortran program implementing the Jacobi method is also presented.
This document provides an introduction to differentiation through a series of lessons and examples:
- Lesson 1 introduces differentiation and its objectives of being able to draw and recognize graphs of quadratic and cubic functions, and use tangents to estimate the gradient of a curve at a point.
- Examples are provided to estimate gradients of curves at given points using tangents, identify graphs based on equations, and determine values and gradients of various functions at different values of x.
- Review questions test the understanding of key concepts like drawing correct tangents, identifying properties of tangents, and determining values and gradients of functions.
This document provides an introduction and overview of Toeplitz and circulant matrices. It discusses how these matrices arise in applications involving time series, signal processing, and discrete time systems. Toeplitz matrices have constant diagonals, while circulant matrices are a special case where each row is a cyclic shift of the row above it. The document outlines the structure and key properties of these matrices and previews the major topics to be covered, including asymptotic behavior, eigenvalues, inverses, and applications to stochastic time series.
The experiment analyzed a sample of plain woven cotton fabric. Various tests were conducted to determine the fabric specifications, including the weave structure, raw materials, thread count, yarn twist, and GSM. The fabric had a plain weave construction with 57 ends per inch and 54 picks per inch using cotton yarns of 28/1 warp count and 30/1 weft count, both with a twist of 23 TPI. The fabric was produced on a tappet loom and is commonly used for apparel and home textiles.
This document discusses 2D and 3D transformations. It begins with an overview of basic 2D transformations like translation, scaling, rotation, and shearing. It then covers representing transformations with matrices and combining transformations through matrix multiplication. Homogeneous coordinates are introduced as a way to represent translations with matrices. The key transformations can all be represented as 3x3 matrices using homogeneous coordinates.
Graphs are used in physics to show relationships between variables. A linear graph indicates a direct proportional relationship between variables. The slope of a linear graph is calculated by taking the rise over the run and represents the ratio of change in the dependent variable to the change in the independent variable. For nonlinear relationships, manipulating the variables, such as squaring or taking reciprocals, can linearize the relationship. The slope of the resulting linear graph then represents a physical quantity defined by the original equation.
Lesson 15: Exponential Growth and Decay (handout)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
This document discusses coordinate system transformations. It introduces three approaches for finding the needed transformations between coordinate systems: direction cosines, Euler angles, and Euler parameters. Direction cosines are presented first as a brute force approach using projections of vector components. The transformation between systems is defined as a matrix multiplication using a direction cosine matrix whose elements are cosines of the angles between axes. Properties of the direction cosine matrix are discussed.
This document discusses coordinate system transformations. It introduces three approaches for finding the needed transformations between coordinate systems: direction cosines, Euler angles, and Euler parameters. Direction cosines use the angles between coordinate system axes and the cosine of those angles to derive transformation matrices. Euler angles express the transformation matrix in terms of three successive rotations about coordinate system axes. Euler parameters provide an elegant alternative to Euler angles by avoiding potential problems with Euler angles.
Here are the key steps:
1) Take out the coefficient of x^2 which is 3:
3(x^2 - 4x + 7/3)
2) Complete the square of the expression in brackets:
3(x^2 - 4x + 4) + 7/3
3) Recognise this is in the form a(x+b)^2 + c where:
a = 3
b = -2
c = 7/3
Therefore, the expression in the required form is:
3(x + 2)2 + 7/3
Sara Saffari: Turning Underweight into Fitness Success at 23get joys
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Cl dys of p and s 5 ed
1. CHAPTER 0
Contents
Preface v
Problems Solved in Student Solutions Manual vii
1 Matrices, Vectors, and Vector Calculus 1
2 Newtonian Mechanics—Single Particle 29
3 Oscillations 79
4 Nonlinear Oscillations and Chaos 127
5 Gravitation 149
6 Some Methods in The Calculus of Variations 165
7 Hamilton’s Principle—Lagrangian and Hamiltonian Dynamics 181
8 Central-Force Motion 233
9 Dynamics of a System of Particles 277
10 Motion in a Noninertial Reference Frame 333
11 Dynamics of Rigid Bodies 353
12 Coupled Oscillations 397
13 Continuous Systems; Waves 435
14 Special Theory of Relativity 461
iii
3. CHAPTER 0
Preface
This Instructor’s Manual contains the solutions to all the end-of-chapter problems (but not the
appendices) from Classical Dynamics of Particles and Systems, Fifth Edition, by Stephen T.
Thornton and Jerry B. Marion. It is intended for use only by instructors using Classical Dynamics
as a textbook, and it is not available to students in any form. A Student Solutions Manual
containing solutions to about 25% of the end-of-chapter problems is available for sale to
students. The problem numbers of those solutions in the Student Solutions Manual are listed on
the next page.
As a result of surveys received from users, I continue to add more worked out examples in
the text and add additional problems. There are now 509 problems, a significant number over
the 4th edition.
The instructor will find a large array of problems ranging in difficulty from the simple
“plug and chug” to the type worthy of the Ph.D. qualifying examinations in classical mechanics.
A few of the problems are quite challenging. Many of them require numerical methods. Having
this solutions manual should provide a greater appreciation of what the authors intended to
accomplish by the statement of the problem in those cases where the problem statement is not
completely clear. Please inform me when either the problem statement or solutions can be
improved. Specific help is encouraged. The instructor will also be able to pick and choose
different levels of difficulty when assigning homework problems. And since students may
occasionally need hints to work some problems, this manual will allow the instructor to take a
quick peek to see how the students can be helped.
It is absolutely forbidden for the students to have access to this manual. Please do not
give students solutions from this manual. Posting these solutions on the Internet will result in
widespread distribution of the solutions and will ultimately result in the decrease of the
usefulness of the text.
The author would like to acknowledge the assistance of Tran ngoc Khanh (5th edition),
Warren Griffith (4th edition), and Brian Giambattista (3rd edition), who checked the solutions of
previous versions, went over user comments, and worked out solutions for new problems.
Without their help, this manual would not be possible. The author would appreciate receiving
reports of suggested improvements and suspected errors. Comments can be sent by email to
stt@virginia.edu, the more detailed the better.
Stephen T. Thornton
Charlottesville, Virginia
v
5. CHAPTER 1
Matrices, Vectors,
and Vector Calculus
1-1.
x2 = x2′
x1′
45˚
x1
45˚
x3
x3′
Axes x′ and x′ lie in the x1 x3 plane.
1 3
The transformation equations are:
x1 = x1 cos 45° − x3 cos 45°
′
x2 = x2
′
x3 = x3 cos 45° + x1 cos 45°
′
1 1
x1 =
′ x1 − x3
2 2
x2 = x2
′
1 1
x3 =
′ x1 − x3
2 2
So the transformation matrix is:
1 1
0 −
2 2
0 1 0
1 1
0
2 2
1
6. 2 CHAPTER 1
1-2.
a)
x3
D
E
γ
β x2
O
α B
A C
x1
From this diagram, we have
OE cos α = OA
OE cos β = OB (1)
OE cos γ = OD
Taking the square of each equation in (1) and adding, we find
2 2 2 2
OE cos 2 α + cos 2 β + cos 2 γ = OA + OB + OD
(2)
But
2 2 2
OA + OB = OC (3)
and
2 2 2
OC + OD = OE (4)
Therefore,
2 2 2 2
OA + OB + OD = OE (5)
Thus,
cos 2 α + cos 2 β + cos 2 γ = 1 (6)
b)
x3
D
E
D′
E′
θ B B′ x2
O
A A′ C C′
x1
First, we have the following trigonometric relation:
OE + OE′ − 2OE OE′ cos θ = EE′
2 2 2
(7)
7. MATRICES, VECTORS, AND VECTOR CALCULUS 3
But,
2 2 2
EE′ = OB′ − OB + OA′ − OA + OD′ − OD
2
2 2
= OE′ cos β ′ − OE cos β + OE′ cos α ′ − OE cos α
2
+ OE′ cos γ ′ − OE cos γ (8)
or,
EE′ = OE′ cos 2 α ′ + cos 2 β ′ + cos 2 γ ′ + OE cos 2 α + cos 2 β + cos 2 γ
2 2 2
− 2OE′ OE cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′
= OE′ 2 + OE2 − 2OE OE′ cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′
(9)
Comparing (9) with (7), we find
cos θ = cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′ (10)
1-3.
x3
A e3 e2′
e3
O x2 e2
e1
e2 e1′
x1 e1 e3′
Denote the original axes by x1 , x2 , x3 , and the corresponding unit vectors by e1 , e2 , e3 . Denote
1 2 3 ′ ′ ′
the new axes by x′ , x′ , x′ and the corresponding unit vectors by e1 , e2 , e3 . The effect of the
rotation is e1 → e3 , e2 → e1 , e3 → e2 . Therefore, the transformation matrix is written as:
′ ′ ′
cos ( e1 , e1 ) cos ( e1 , e2 ) cos ( e1 , e3 ) 0 1 0
′ ′ ′
λ = cos ( e′ , e1 ) cos ( e′ , e2 ) cos ( e′ , e3 ) = 0 0 1
2 2 2
1 0 0
cos ( e′ , e1 ) cos ( e′ , e2 ) cos ( e′ , e3 )
3 3 3
1-4.
a) Let C = AB where A, B, and C are matrices. Then,
Cij = ∑ Aik Bkj (1)
k
(C )
t
ij
= C ji = ∑ Ajk Bki = ∑ Bki Ajk
k k
8. 4 CHAPTER 1
( )
Identifying Bki = Bt ik
( )
and Ajk = At kj
,
(C ) = ∑ (B ) ( A )
t
ij
t
ik
t
kj
(2)
k
or,
C t = ( AB) = Bt At
t
(3)
b) To show that ( AB) = B−1 A−1 ,
−1
( AB) B−1 A−1 = I = ( B−1 A−1 ) AB (4)
That is,
( AB) B−1 A−1 = AIA−1 = AA−1 = I (5)
(B −1
)
A−1 ( AB) = B−1 IB = B−1B = I (6)
1-5. Take λ to be a two-dimensional matrix:
λ11 λ12
λ = = λ11λ 22 − λ12 λ 21 (1)
λ 21 λ 22
Then,
λ = λ11λ 22 − 2λ11λ 22 λ12 λ 21 + λ12 λ 21 + ( λ11λ 21 + λ12 λ 22 ) − ( λ11λ 21 + λ12λ 22 )
2 2 2 2 2 2 2 2 2 2 2 2 2
(2 2
) 2 2 2
( 2 2
) (
= λ 22 λ11 + λ12 + λ 21 λ11 + λ12 − λ11λ 21 + 2λ11λ 22λ12 λ 21 + λ12λ 22
2 2 2
)
(2 2
)(
2 2
)
= λ11 + λ12 λ 22 + λ 21 − ( λ11λ 21 + λ12 λ 22 )
2
(2)
But since λ is an orthogonal transformation matrix, ∑λ λ ij kj = δ ik .
j
Thus,
λ11 + λ12 = λ 21 + λ 22 = 1
2 2 2 2
(3)
λ11λ 21 + λ12 λ 22 = 0
Therefore, (2) becomes
λ =1
2
(4)
1-6. The lengths of line segments in the x j and x ′ systems are
j
L= ∑x 2
j ; L′ = ∑ x′ i
2
(1)
j i
9. MATRICES, VECTORS, AND VECTOR CALCULUS 5
If L = L ′ , then
∑ x = ∑ x′ 2
j i
2
(2)
j i
The transformation is
xi′ = ∑ λ ij x j (3)
j
Then,
∑ x = ∑∑λ
2
j ik xk ∑ λ i x
j i k
(4)
= ∑ xk x ∑ λik λ i
k, i
But this can be true only if
∑λ ik λi = δ k (5)
i
which is the desired result.
1-7.
x3
(0,0,1) (0,1,1)
(1,1,1)
(1,0,1)
x2
(0,0,0) (0,1,0)
(1,0,0) (1,1,0)
x1
There are 4 diagonals:
D1 , from (0,0,0) to (1,1,1), so (1,1,1) – (0,0,0) = (1,1,1) = D1 ;
D 2 , from (1,0,0) to (0,1,1), so (0,1,1) – (1,0,0) = (–1,1,1) = D 2 ;
D 3 , from (0,0,1) to (1,1,0), so (1,1,0) – (0,0,1) = (1,1,–1) = D 3 ; and
D 4 , from (0,1,0) to (1,0,1), so (1,0,1) – (0,1,0) = (1,–1,1) = D 4 .
The magnitudes of the diagonal vectors are
D1 = D 2 = D 3 = D 4 = 3
The angle between any two of these diagonal vectors is, for example,
D1 ⋅ D 2 (1,1,1) ⋅ ( −1,1,1) 1
= cos θ = =
D1 D 2 3 3
10. 6 CHAPTER 1
so that
1
θ = cos−1 = 70.5°
3
Similarly,
D1 ⋅ D 3 D ⋅D D ⋅D D ⋅D D ⋅D 1
= 1 4 = 2 3 = 2 4 = 3 4 =±
D1 D 3 D1 D 4 D2 D3 D2 D 4 D3 D 4 3
1-8. Let θ be the angle between A and r. Then, A ⋅ r = A2 can be written as
Ar cos θ = A2
or,
r cos θ = A (1)
This implies
π
QPO = (2)
2
Therefore, the end point of r must be on a plane perpendicular to A and passing through P.
1-9. A = i + 2j − k B = −2i + 3j + k
a) A − B = 3i − j − 2k
12
A − B = ( 3) + ( −1) + (−2)2
2 2
A − B = 14
b)
B
θ
A
component of B along A
The length of the component of B along A is B cos θ.
A ⋅ B = AB cos θ
A ⋅ B −2 + 6 − 1 3 6
B cos θ = = = or
A 6 6 2
The direction is, of course, along A. A unit vector in the A direction is
1
( i + 2j − k )
6
11. MATRICES, VECTORS, AND VECTOR CALCULUS 7
So the component of B along A is
1
2
( i + 2j − k )
A⋅B 3 3 3
c) cos θ = = = ; θ = cos −1
AB 6 14 2 7 2 7
θ 71°
i j k
2 −1 1 −1 1 2
d) A × B = 1 2 −1 = i −j +k
3 1 −2 1 −2 3
−2 3 1
A × B = 5i + j + 7 k
e) A − B = 3i − j − 2k A + B = −i + 5j
i j k
( A − B ) × ( A + B ) = 3 −1 − 2
−1 5 0
( A − B) × ( A + B) = 10i + 2j + 14k
1-10. r = 2b sin ω t i + b cos ω t j
v = r = 2bω cos ω t i − bω sin ω t j
a)
a = v = −2bω 2 sin ω t i − bω 2 cos ω t j = −ω 2 r
12
speed = v = 4b 2ω 2 cos 2 ω t + b 2ω 2 sin 2 ω t
12
= bω 4 cos 2 ω t + sin 2 ω t
12
speed = bω 3 cos 2 ω t + 1
b) At t = π 2ω , sin ω t = 1 , cos ω t = 0
So, at this time, v = − bω j , a = −2bω 2 i
So, θ 90°
12. 8 CHAPTER 1
1-11.
a) Since ( A × B) i = ∑ ε ijk Aj Bk , we have
jk
(A × B) ⋅ C = ∑∑ ε ijk Aj Bk Ci
i j,k
= C1 ( A2 B3 − A3 B2 ) − C2 ( A1B3 − A3 B1 ) + C3 ( A1B2 − A2 B1 ) (1)
C1 C2 C3 A1 A2 A3 A1 A2 A3
= A1 A2 A3 = − C1 C2 C3 = B1 B2 B3 = A ⋅ ( B × C )
B1 B2 B3 B1 B2 B3 C1 C2 C3
We can also write
C1 C2 C3 B1 B2 B3
(A × B) ⋅ C = − B1 B2 B3 = C1 C2 C3 = B ⋅ ( C × A ) (2)
A1 A2 A3 A1 A2 A3
We notice from this result that an even number of permutations leaves the determinant
unchanged.
b) Consider vectors A and B in the plane defined by e1 , e2 . Since the figure defined by A, B,
C is a parallelepiped, A × B = e3 × area of the base, but e3 ⋅ C = altitude of the parallelepiped.
Then,
C ⋅ ( A × B) = ( C ⋅ e3 ) × area of the base
= altitude × area of the base
= volume of the parallelepiped
1-12.
O
c
C
a h a–c
b c–b
A
b–a
B
The distance h from the origin O to the plane defined by A, B, C is
13. MATRICES, VECTORS, AND VECTOR CALCULUS 9
a ⋅ ( b − a ) × ( c − b)
h=
( b − a ) × ( c − b)
a ⋅ ( b × c − a × c + a × b)
=
b×c−a×c+a×b
a⋅b× c
= (1)
a×b+b×c+c×a
The area of the triangle ABC is:
1 1 1
A= ( b − a ) × ( c − b) = ( a − c ) × ( b − a ) = ( c − b ) × ( a − c ) (2)
2 2 2
1-13. Using the Eq. (1.82) in the text, we have
A × B = A × ( A × X ) = ( X ⋅ A ) A − ( A ⋅ A ) X = φ A − A2 X
from which
( B × A ) + φA
X=
A2
1-14.
1 2 −1 2 1 0 1 −2 1
a) AB = 0 3 1 0 −1 2 = 1 −2 9
2 0 1 1 1 3 5 3 3
Expand by the first row.
−2 9 1 9 1 −2
AB = 1 +2 +1
3 3 5 3 5 3
AB = −104
1 2 −1 2 1 9 7
b) AC = 0 3 1 4 3 = 13 9
2 0 1 1 0 5 2
9 7
AC = 13 9
5 2
14. 10 CHAPTER 1
1 2 −1 8 5
c) ABC = A ( BC ) = 0 3 1 −2 −3
2 0 1 9 4
−5 −5
ABC = 3 −5
25 14
d) AB − Bt At = ?
1 −2 1
AB = 1 −2 9 (from part a )
5 3 3
2 0 1 1 0 2 1 1 5
B A = 1 −1 1 2 3 0 = −2 −2 3
t t
0 2 3 −1 1 1 1 9 3
0 −3 −4
AB − B A = 3 0
t t
6
4 −6 0
1-15. If A is an orthogonal matrix, then
At A = 1
1 0 0 1 0 0 1 0 0
0 a a 0 a − a = 0 1 0
0 − a a 0 a a 0 0 1
1 0 0 1 0 0
0 2a
2
0 = 0 1 0
0 0
2a 2 0 0 1
1
a=
2
15. MATRICES, VECTORS, AND VECTOR CALCULUS 11
1-16.
x3 P
r
θ a r
θ a
x2
x1 r cos θ
r ⋅ a = constant
ra cos θ = constant
It is given that a is constant, so we know that
r cos θ = constant
But r cos θ is the magnitude of the component of r along a.
The set of vectors that satisfy r ⋅ a = constant all have the same component along a; however, the
component perpendicular to a is arbitrary.
Thus the surface represented by r ⋅ a = constant
is a plane perpendicular to a.
1-17.
a
B C
θ
c b
A
Consider the triangle a, b, c which is formed by the vectors A, B, C. Since
C= A−B
C = ( A − B) ⋅ ( A − B)
2
(1)
= A 2 − 2A ⋅ B + B 2
or,
C = A2 + B2 − 2 AB cos θ
2
(2)
which is the cosine law of plane trigonometry.
1-18. Consider the triangle a, b, c which is formed by the vectors A, B, C.
a
α C
B
γ β
c b
A
16. 12 CHAPTER 1
C=A−B (1)
so that
C × B = ( A − B) × B (2)
but the left-hand side and the right-hand side of (2) are written as:
C × B = BC sin α e3 (3)
and
( A − B) × B = A × B − B × B = A × B = AB sin γ e3 (4)
where e3 is the unit vector perpendicular to the triangle abc. Therefore,
BC sin α = AB sin γ (5)
or,
C A
=
sin γ sin α
Similarly,
C A B
= = (6)
sin γ sin α sin β
which is the sine law of plane trigonometry.
1-19.
x2
a2 a
b2 b
α
β
x1
a1 b1
a) We begin by noting that
a − b = a 2 + b 2 − 2ab cos (α − β )
2
(1)
We can also write that
a − b = ( a1 − b1 ) + ( a2 − b2 )
2 2 2
= ( a cos α − b cos β ) + ( a sin α − b sin β )
2 2
( ) ( )
= a 2 sin 2 α + cos 2 α + b 2 sin 2 β + cos 2 β − 2 ab ( cos α cos β + sin α sin β )
= a 2 + b 2 − 2ab ( cos α cos β + sin α sin β ) (2)
17. MATRICES, VECTORS, AND VECTOR CALCULUS 13
Thus, comparing (1) and (2), we conclude that
cos (α − β ) = cos α cos β + sin α sin β (3)
b) Using (3), we can find sin (α − β ) :
sin (α − β ) = 1 − cos 2 (α − β )
= 1 − cos 2 α cos 2 β − sin 2 α sin 2 β − 2cos α sin α cos β sin β
( ) (
= 1 − cos 2 α 1 − sin 2 β − sin 2 α 1 − cos 2 β − 2cos α sin α cos β sin β )
= sin 2 α cos 2 β − 2sin α sin β cos α cos β + cos 2 α sin 2 β
= ( sin α cos β − cos α sin β )2 (4)
so that
sin (α − β ) = sin α cos β − cos α sin β (5)
1-20.
a) Consider the following two cases:
When i ≠ j δ ij = 0 but ε ijk ≠ 0 .
When i = j δ ij ≠ 0 but ε ijk = 0 .
Therefore,
∑ε ijk δ ij = 0 (1)
ij
b) We proceed in the following way:
When j = k, ε ijk = ε ijj = 0 .
Terms such as ε j11 ε 11 = 0 . Then,
∑ε ijk ε jk = ε i12 ε 12 + ε i13 ε 13 + ε i 21 ε 21 + ε i 31 ε 31 + ε i 32 ε 32 + ε i 23 ε 23
jk
Now, suppose i = = 1 , then,
∑= ε 123 ε 123 + ε 132 ε 132 = 1 + 1 = 2
jk
18. 14 CHAPTER 1
for i = = 2 , ∑= ε 213 ε 213 + ε 231 ε 231 = 1 + 1 = 2 . For i = = 3 , ∑= ε 312 ε 312 + ε 321 ε 321 = 2 . But i = 1,
jk jk
= 2 gives ∑ = 0 . Likewise for i = 2, = 1 ; i = 1, = 3 ; i = 3, = 1 ; i = 2, = 3 ; i = 3, = 2.
jk
Therefore,
∑ε ijk ε jk = 2δ i (2)
j,k
c) ∑ε ijk ε ijk = ε 123 ε 123 + ε 312 ε 312 + ε 321 ε 321 + ε 132 ε 132 + ε 213 ε 213 + ε 231 ε 231
ijk
= 1 ⋅ 1 + 1 ⋅ 1 + ( −1) ⋅ ( −1) + ( −1) ⋅ ( −1) + ( −1) ⋅ ( −1) + (1) ⋅ (1)
or,
∑ε ijk ε ijk = 6 (3)
ijk
1-21. ( A × B)i = ∑ ε ijk Aj Bk (1)
jk
( A × B) ⋅ C = ∑ ∑ε ijk Aj Bk Ci (2)
i jk
By an even permutation, we find
ABC = ∑ ε ijk Ai Bj Ck (3)
ijk
1-22. To evaluate ∑ε ijk ε mk we consider the following cases:
k
a) i= j: ∑ε ijk ε mk = ∑ ε iik ε mk = 0 for all i , , m
k k
b) i= : ∑ε ijk ε mk = ∑ ε ijk ε imk = 1 for j = m and k ≠ i , j
k k
= 0 for j ≠ m
c) i = m: ∑ε ijk ε mk = ∑ ε ijk ε ik = 0 for j ≠
k k
= −1 for j = and k ≠ i , j
d) j= : ∑ε ijk ε mk = ∑ ε ijk ε jmk = 0 for m ≠ i
k k
= −1 for m = i and k ≠ i , j
19. MATRICES, VECTORS, AND VECTOR CALCULUS 15
e) j = m: ∑ε ijk ε mk = ∑ ε ijk ε jk = 0 for i ≠
k k
= 1 for i = and k ≠ i , j
f) = m: ∑ε ijk ε mk = ∑ ε ijk ε k = 0 for all i , j , k
k k
g) i ≠ or m : This implies that i = k or i = j or m = k.
Then, ∑ε ijk ε mk = 0 for all i , j , , m
k
h) j ≠ or m : ∑ε ijk ε mk = 0 for all i , j , , m
k
Now, consider δ i δ jm − δ im δ j and examine it under the same conditions. If this quantity
behaves in the same way as the sum above, we have verified the equation
∑ε ijk ε mk = δ i δ jm − δ im δ j
k
a) i = j : δ i δ im − δ im δ i = 0 for all i , , m
b) i = : δ ii δ jm − δ im δ ji = 1 if j = m, i ≠ j , m
= 0 if j ≠ m
c) i = m : δ i δ ji − δ ii δ j = −1 if j = , i ≠ j ,
= 0 if j ≠
d) j = : δi δ m − δ im δ = −1 if i = m, i ≠
= 0 if i ≠ m
e) j = m : δ i δ mm − δ im δ m = 1 if i = , m ≠
= 0 if i ≠
f) = m : δ i δ j − δ il δ j = 0 for all i , j ,
g) i ≠ , m : δ i δ jm − δ im δ j = 0 for all i , j , , m
h) j ≠ , m : δ i δ jm − δ im δ i = 0 for all i , j , , m
Therefore,
∑ε ijk ε mk = δ i δ jm − δ im δ j (1)
k
Using this result we can prove that
A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B) C
20. 16 CHAPTER 1
First ( B × C ) i = ∑ ε ijk Bj Ck . Then,
jk
[ A × (B × C) ] = ∑ ε mn Am ( B × C ) n = ∑ ε mn Am ∑ ε njk BjCk
mn mn jk
= ∑ε mn ε njk Am Bj Ck = ∑ε mn ε jkn Am Bj Ck
jkmn jkmn
= ∑ ∑ ε lmn ε jkn Am Bj Ck
jkm n
( )
= ∑ δ jlδ km − δ k δ jm Am Bj Ck
jkm
= ∑ Am B Cm − ∑ Am BmC = B ∑ AmCm − C ∑ Am Bm
m m m m
= ( A ⋅ C ) B − ( A ⋅ B) C
Therefore,
A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B) C (2)
1-23. Write
( A × B) j = ∑ ε j m A Bm
m
( C × D) k = ∑ ε krs Cr Ds
rs
Then,
21. MATRICES, VECTORS, AND VECTOR CALCULUS 17
[ ( A × B) × (C × D)]i = ∑ ε ijk ∑ ε j m A Bm ∑ ε krs Cr Ds
jk m rs
= ∑ε ijk ε j m ε krs A BmCr Ds
jk mrs
= ∑ε j m ∑ ε ijk ε rsk A BmCr Ds
k
j mrs
= ∑ ε (δ j m ir )
δ js − δ is δ jr A BmCr Ds
j mrs
(
= ∑ ε j m A BmCi Dj − A Bm Di C j
j m
)
= ∑ ε j m Dj A Bm Ci − ∑ ε j m C j A Bm Di
jm jm
= (ABD)Ci − (ABC)Di
Therefore,
[( A × B) × (C × D) ] = (ABD)C − (ABC)D
1-24. Expanding the triple vector product, we have
e × ( A × e) = A ( e ⋅ e) − e ( A ⋅ e) (1)
But,
A ( e ⋅ e) = A (2)
Thus,
A = e ( A ⋅ e) + e × ( A × e) (3)
e(A · e) is the component of A in the e direction, while e × (A × e) is the component of A
perpendicular to e.
22. 18 CHAPTER 1
1-25.
er
eφ
θ
φ
eθ
The unit vectors in spherical coordinates are expressed in terms of rectangular coordinates by
eθ = ( cos θ cos φ , cos θ sin φ , − sin θ )
eφ = ( − sin φ , cos φ , 0 ) (1)
er = ( sin θ cos φ , sin θ sin φ , cos θ )
Thus,
(
eθ = −φ cos θ sin φ − θ sin θ cos φ , φ cos θ cos φ − θ sin θ sin φ , − θ cos θ )
= −θ er + φ cos θ eφ (2)
Similarly,
(
eφ = −φ cos φ , − φ sin φ , 0 )
= −φ cos θ eθ − φ sin θ er (3)
er = φ sin θ eφ + θ eθ (4)
Now, let any position vector be x. Then,
x = rer (5)
( )
x = rer + rer = r φ sin θ eφ + θ eθ + rer
= rφ sin θ eφ + rθ eθ + rer (6)
( ) ( )
x = rφ sin θ + rθφ cos θ + rφ sin θ eφ + rφ sin θ eφ + rθ + rθ eθ + rθ eθ + rer + rer
( ) (
= 2rφ sin θ + 2rθφ cos θ + rφ sin θ eφ + r − rφ 2 sin 2 θ − rθ 2 er )
(
+ 2rθ + rθ − rφ 2 sin θ cos θ eθ ) (7)
or,
23. MATRICES, VECTORS, AND VECTOR CALCULUS 19
1 d 2
x = a = r − rθ 2 − rφ 2 sin 2 θ er +
r dt
( )
r θ − rφ 2 sin θ cos θ eθ
(8)
1
+
d 2
(
r φ sin 2 θ eφ )
r sin θ dt
1-26. When a particle moves along the curve
r = k (1 + cos θ ) (1)
we have
r = − kθ sin θ
(2)
r = − k θ cos θ + θ sin θ
2
Now, the velocity vector in polar coordinates is [see Eq. (1.97)]
v = rer + rθ eθ (3)
so that
v 2 = v = r 2 + r 2θ 2
2
( )
= k 2θ 2 sin 2 θ + k 2 1 + 2 cos θ + cos 2 θ θ 2
= k 2θ 2 2 + 2 cos θ
(4)
and v 2 is, by hypothesis, constant. Therefore,
v2
θ= (5)
2k (1 + cos θ )
2
Using (1), we find
v
θ= (6)
2kr
Differentiating (5) and using the expression for r , we obtain
v 2 sin θ v 2 sin θ
θ= = 2 (7)
4 k (1 + cos θ )
2
4r 2
The acceleration vector is [see Eq. (1.98)]
( )
a = r − rθ 2 er + rθ + 2rθ eθ( ) (8)
so that
24. 20 CHAPTER 1
a ⋅ er = r − rθ 2
( )
= − k θ 2 cos θ + θ sin θ − k (1 + cos θ ) θ 2
θ 2 sin 2 θ
= − k θ 2 cos θ + + (1 + cos θ ) θ 2
2 (1 + cos θ )
1 − cos 2 θ
= − kθ 2 2 cos θ + + 1
2 (1 + cos θ )
3
= − kθ 2 (1 + cos θ ) (9)
2
or,
3 v2
a ⋅ er = − (10)
4 k
In a similar way, we find
3 v 2 sin θ
a ⋅ eθ = − (11)
4 k 1 + cos θ
From (10) and (11), we have
a = ( a ⋅ er ) 2 + ( a ⋅ eθ ) 2 (12)
or,
3 v2 2
a = (13)
4 k 1 + cos θ
1-27. Since
r × (v × r) = (r ⋅ r) v − (r ⋅ v) r
we have
d d
dt
[ r × ( v × r ) ] = dt [ ( r ⋅ r ) v − ( r ⋅ v ) r ]
= (r ⋅ r) a + 2 (r ⋅ v) v − (r ⋅ v) v − ( v ⋅ v) r − (r ⋅ a) r
(
= r 2a + ( r ⋅ v ) v − r v2 + r ⋅ a ) (1)
Thus,
d
dt
(
[ r × ( v × r )] = r 2a + ( r ⋅ v ) v − r r ⋅ a + v 2 ) (2)
25. MATRICES, VECTORS, AND VECTOR CALCULUS 21
∂
1-28. grad ( ln r ) = ∑ ( ln r ) ei (1)
i ∂x i
where
r = ∑x 2
i (2)
i
Therefore,
∂ 1 xi
( ln r ) = r
∂x i ∑x 2
i
i
xi
= 2
(3)
r
so that
1
grad ( ln r ) = 2 ∑ i i
xe (4)
r i
or,
r
grad ( ln r ) = (5)
r2
1-29. Let r 2 = 9 describe the surface S1 and x + y + z 2 = 1 describe the surface S2 . The angle θ
between S1 and S2 at the point (2,–2,1) is the angle between the normals to these surfaces at the
point. The normal to S1 is
( )
grad ( S1 ) = grad r 2 − 9 = grad x 2 + y 2 + z 2 − 9 ( )
= ( 2xe1 + 2 ye2 + 2ze3 ) x = 2, y = 2, z = 1 (1)
= 4e1 − 4e2 + 2e3
In S2 , the normal is:
(
grad ( S2 ) = grad x + y + z 2 − 1 )
= ( e1 + e2 + 2ze3 ) x = 2, y =−2, z = 1 (2)
= e1 + e2 + 2e3
Therefore,
26. 22 CHAPTER 1
grad ( S1 ) ⋅ grad ( S2 )
cos θ =
grad ( S1 ) grad ( S2 )
=
( 4e1 − 4e2 + 2e3 ) ⋅ (e1 + e2 + 2e3 ) (3)
6 6
or,
4
cos θ = (4)
6 6
from which
6
θ = cos −1 = 74.2° (5)
9
3
∂ ( φψ ) ∂ψ ∂ φ
1-30. grad ( φψ ) = ∑ ei = ∑ ei φ + ψ
i =1 ∂x i i ∂x i ∂x i
∂ψ ∂φ
= ∑ ei φ + ∑ ei ψ
i ∂x i i ∂x i
Thus,
grad ( φψ ) = φ grad ψ + ψ grad φ
1-31.
a)
n
∂
12
∂rn
3
grad r = ∑ ei
n
= ∑ ei ∑ xj
2
∂x i ∂x i j
i =1
n
−1
n 2
= ∑ ei 2 xi ∑ x 2
j
i 2 j
n
−1
2
= ∑ ei x i n ∑ x 2
j
i j
= ∑ ei x i n r ( n − 2) (1)
i
Therefore,
grad r n = nr ( n − 2) r (2)
27. MATRICES, VECTORS, AND VECTOR CALCULUS 23
b)
3
∂f ( r ) 3 ∂ f ( r ) ∂ r
grad f ( r ) = ∑ ei = ∑ ei
i =1 ∂x i i =1 ∂ r ∂x i
12
∂ ∂f ( r )
= ∑ ei ∑ xj
2
i ∂x i j ∂r
−1 2
∂f ( r )
= ∑ ei xi ∑ x 2
∂r
j
i j
x i ∂f
= ∑ ei (3)
i r dr
Therefore,
r ∂f ( r )
grad f ( r ) = (4)
r ∂r
c)
∂2
12
∂ 2 ln r
∇ ( ln r ) = ∑
2
= ∑ 2 ln ∑ x j
2
∂xi2 ∂x i j
i
−1 2
1
⋅ 2 xi ∑ x 2
j
∂ 2 j
=∑
i ∂x i
12
∑ xj
2
j
−1
∂
=∑ xi ∑ x 2
i ∂x i
j
j
−2 −1
∂x
= ∑ ( − xi )( 2xi ) ∑ x 2 + ∑ i ∑ x2
i ∂x i j
j j
i j
(
= ∑ −2x 2 r 2
i
j )( ) −2 1
+ 3 2
r
2r 2 3 1
=− 4
+ 2 = 2 (5)
r r r
or,
1
∇ 2 ( ln r ) = (6)
r2
28. 24 CHAPTER 1
1-32. Note that the integrand is a perfect differential:
d d
2ar ⋅ r + 2br ⋅ r = a (r ⋅ r) + b (r ⋅ r) (1)
dt dt
Clearly,
∫ ( 2ar ⋅ r + 2br ⋅ r ) dt = ar + br 2 + const.
2
(2)
1-33. Since
d r rr − rr r rr
= = − 2 (1)
dt r
r2 r r
we have
r rr d r
∫ r − r 2 dt = ∫ dt r dt
(2)
from which
r rr r
∫ r − r 2 dt = r + C
(3)
where C is the integration constant (a vector).
1-34. First, we note that
d
dt
( )
A×A =A×A+A×A (1)
But the first term on the right-hand side vanishes. Thus,
∫ ( A × A) dt = ∫ dt ( A × A) dt
d
(2)
so that
∫ ( A × A) dt = A × A + C (3)
where C is a constant vector.
29. MATRICES, VECTORS, AND VECTOR CALCULUS 25
1-35.
y
x
z
We compute the volume of the intersection of the two cylinders by dividing the intersection
volume into two parts. Part of the common volume is that of one of the cylinders, for example,
the one along the y axis, between y = –a and y = a:
( )
V1 = 2 π a 2 a = 2π a 3 (1)
The rest of the common volume is formed by 8 equal parts from the other cylinder (the one
along the x-axis). One of these parts extends from x = 0 to x = a, y = 0 to y = a 2 − x 2 , z = a to
z = a 2 − x 2 . The complementary volume is then
a a2 − x 2 a2 − x 2
V2 = 8 ∫ dx ∫ dy ∫ dz
0 0 a
= 8 ∫ dx a 2 − x 2 a 2 − x 2 − a
a
0
a
x 3 a3 x
= 8 a2 x − − sin −1
3 2 a 0
16 3
= a − 2π a 3 (2)
3
Then, from (1) and (2):
16 a 3
V = V1 + V2 = (3)
3
30. 26 CHAPTER 1
1-36.
z
d
y
x c2 = x2 + y2
The form of the integral suggests the use of the divergence theorem.
∫ S
A ⋅ da = ∫ ∇ ⋅ A dv
V
(1)
Since ∇ ⋅ A = 1 , we only need to evaluate the total volume. Our cylinder has radius c and height
d, and so the answer is
∫ V
dv = π c 2 d (2)
1-37.
z
R
y
x
To do the integral directly, note that A = R3er , on the surface, and that da = daer .
∫ S
A ⋅ da = R 3 ∫ S
da = R3 × 4π R 2 = 4π R 5 (1)
To use the divergence theorem, we need to calculate ∇ ⋅ A . This is best done in spherical
coordinates, where A = r 3er . Using Appendix F, we see that
1 ∂ 2
∇⋅A =
r 2 ∂r
(
r A r = 5r 2 ) (2)
Therefore,
( )
π 2π R
∫ V
∇ ⋅ A dv = ∫ sin θ dθ ∫
0 0
dφ ∫ r 2 5r 2 dr = 4π R5
0
(3)
Alternatively, one may simply set dv = 4π r 2 dr in this case.
31. MATRICES, VECTORS, AND VECTOR CALCULUS 27
1-38.
z z = 1 – x2 – y2
y
C
x2 + y2 = 1
x
By Stoke’s theorem, we have
∫ (∇ × A) ⋅ da = ∫
S C
A ⋅ ds (1)
The curve C that encloses our surface S is the unit circle that lies in the xy plane. Since the
element of area on the surface da is chosen to be outward from the origin, the curve is directed
counterclockwise, as required by the right-hand rule. Now change to polar coordinates, so that
we have ds = dθ eθ and A = sin θ i + cos θ k on the curve. Since eθ ⋅ i = − sin θ and eθ ⋅ k = 0 , we
have
( − sin θ ) dθ = −π
2π
∫ A ⋅ ds = ∫ 2
(2)
C 0
1-39.
a) Let’s denote A = (1,0,0); B = (0,2,0); C = (0,0,3). Then AB = (−1, 2, 0) ; AC = (−1, 0, 3) ; and
AB × AC = (6, 3, 2) . Any vector perpendicular to plane (ABC) must be parallel to AB × AC , so
the unit vector perpendicular to plane (ABC) is n = (6 7 , 3 7 , 2 7 )
b) Let’s denote D = (1,1,1) and H = (x,y,z) be the point on plane (ABC) closest to H. Then
DH = ( x − 1, y − 1, z − 1) is parallel to n given in a); this means
x−1 6 x −1 6
= =2 and = =3
y−1 3 z−1 2
Further, AH = ( x − 1, y , z) is perpendicular to n so one has 6( x − 1) + 3 y + 2 z = 0 .
Solving these 3 equations one finds
5
H = ( x , y , z) = (19 49 , 34 49 , 39 49) and DH =
7
1-40.
a) At the top of the hill, z is maximum;
∂z ∂z
0= = 2 y − 6 x − 18 and 0= = 2 x − 8 y + 28
∂x ∂y
32. 28 CHAPTER 1
so x = –2 ; y = 3, and the hill’s height is max[z]= 72 m. Actually, this is the max value of z,
because the given equation of z implies that, for each given value of x (or y), z describes an
upside down parabola in term of y ( or x) variable.
b) At point A: x = y = 1, z = 13. At this point, two of the tangent vectors to the surface of the
hill are
∂z ∂z
t1 = (1, 0, ) = (1, 0, −8) and t2 = (0,1, ) = (0,1, 22)
∂x (1,1) ∂y (1,1)
Evidently t1 × t2 = (8, −22,1) is perpendicular to the hill surface, and the angle θ between this
and Oz axis is
(0, 0,1) ⋅ (8, −22,1) 1
cos θ = = so θ = 87.55 degrees.
8 + 22 + 1
2 2 2 23.43
c) Suppose that in the α direction ( with respect to W-E axis), at point A = (1,1,13) the hill is
steepest. Evidently, dy = (tan α )dx and
dz = 2xdy + 2 ydx − 6 xdx − 8 ydy − 18 dx + 28 dy = 22(tan α − 1)dx
then
dx 2 + dy 2 dx cos α −1
tan β = = =
dz 22(tan α − 1)dx 22 2 cos (α + 45)
The hill is steepest when tan β is minimum, and this happens when α = –45 degrees with
respect to W-E axis. (note that α = 135 does not give a physical answer).
1-41.
A ⋅ B = 2a( a − 1)
then A ⋅ B = 0 if only a = 1 or a = 0.
33. CHAPTER 2
Newtonian Mechanics—
Single Particle
2-1. The basic equation is
F = mi xi (1)
a) F ( xi , t ) = f ( xi ) g ( t ) = mi xi : Not integrable (2)
b) F ( xi , t ) = f ( xi ) g ( t ) = mi xi
dxi
mi = f ( xi ) g ( t )
dt
dxi g (t)
= dt : Integrable (3)
f ( xi ) mi
c) F ( xi , xi ) = f ( xi ) g ( xi ) = mi xi : Not integrable (4)
2-2. Using spherical coordinates, we can write the force applied to the particle as
F = Fr er + Fθ eθ + Fφ eφ (1)
But since the particle is constrained to move on the surface of a sphere, there must exist a
reaction force − Fr er that acts on the particle. Therefore, the total force acting on the particle is
Ftotal = Fθ eθ + Fφ eφ = mr (2)
The position vector of the particle is
r = Re r (3)
where R is the radius of the sphere and is constant. The acceleration of the particle is
a = r = Re r (4)
29
34. 30 CHAPTER 2
We must now express er in terms of er , eθ , and eφ . Because the unit vectors in rectangular
coordinates, e1 , e2 , e3 , do not change with time, it is convenient to make the calculation in
terms of these quantities. Using Fig. F-3, Appendix F, we see that
er = e1 sin θ cos φ + e2 sin θ sin φ + e3 cos θ
eθ = e1 cos θ cos φ + e2 cos θ sin φ − e3 sin θ (5)
eφ = −e1 sin φ + e2 cos φ
Then
( ) (
er = e1 −φ sin θ sin φ + θ cos θ cos φ + e2 θ cos θ sin φ + φ sin θ cos φ − e3 θ sin θ )
(6)
= eφ φ sin θ + eθ θ
Similarly,
eθ = −er θ + eφ φ cos θ (7)
eφ = −er φ sin θ − eθ φ cos θ (8)
And, further,
( ) ( ) (
er = −er φ 2 sin 2 θ + θ 2 + eθ θ − φ 2 sin θ cos θ + eφ 2θφ cos θ + φ sin θ ) (9)
which is the only second time derivative needed.
The total force acting on the particle is
Ftotal = mr = mRer (10)
and the components are
(
Fθ = mR θ − φ 2 sin θ cos θ )
(11)
(
Fφ = mR 2θφ cos θ + φ sin θ )
35. NEWTONIAN MECHANICS—SINGLE PARTICLE 31
2-3.
y
v0 P
α
β
x
The equation of motion is
F=ma (1)
The gravitational force is the only applied force; therefore,
Fx = mx = 0
(2)
Fy = my = − mg
Integrating these equations and using the initial conditions,
x ( t = 0 ) = v0 cos α
(3)
y ( t = 0 ) = v0 sin α
We find
x ( t ) = v0 cos α
(4)
y ( t ) = v0 sin α − gt
So the equations for x and y are
x ( t ) = v0 t cos α
(5)
1 2
y ( t ) = v0 t sin α − gt
2
Suppose it takes a time t0 to reach the point P. Then,
cos β = v0 t0 cos α
1 2 (6)
sin β = v0 t0 sin α − gt0
2
Eliminating between these equations,
1 2v sin α 2v0
gt0 t0 − 0 + cos α tan β = 0 (7)
2 g g
from which
36. 32 CHAPTER 2
2v0
t0 =
g
( sin α − cos α tan β ) (8)
2-4. One of the balls’ height can be described by y = y0 + v0 t − gt 2 2 . The amount of time it
takes to rise and fall to its initial height is therefore given by 2v0 g . If the time it takes to cycle
the ball through the juggler’s hands is τ = 0.9 s , then there must be 3 balls in the air during that
time τ. A single ball must stay in the air for at least 3τ, so the condition is 2v0 g ≥ 3τ , or
v0 ≥ 13.2 m ⋅ s −1 .
2-5.
flightpath
N
er
plane mg
point of maximum
acceleration
( )
a) From the force diagram we have N − mg = mv 2 R er . The acceleration that the pilot feels is
( )
N m = g + mv R er , which has a maximum magnitude at the bottom of the maneuver.
2
b) If the acceleration felt by the pilot must be less than 9g, then we have
R≥
v2
=
(
3 ⋅ 330 m ⋅ s −1 ) 12.5 km (1)
8g 8 ⋅ 9.8 m ⋅ s −2
A circle smaller than this will result in pilot blackout.
2-6.
Let the origin of our coordinate system be at the tail end of the cattle (or the closest cow/bull).
a) The bales are moving initially at the speed of the plane when dropped. Describe one of
these bales by the parametric equations
x = x 0 + v0 t (1)
37. NEWTONIAN MECHANICS—SINGLE PARTICLE 33
1 2
y = y0 − gt (2)
2
where y0 = 80 m , and we need to solve for x0 . From (2), the time the bale hits the ground is
τ = 2 y0 g . If we want the bale to land at x (τ ) = −30 m , then x0 = x (τ ) − v0τ . Substituting
v0 = 44.4 m ⋅ s -1 and the other values, this gives x0 −210 m . The rancher should drop the bales
210 m behind the cattle.
b) She could drop the bale earlier by any amount of time and not hit the cattle. If she were late
by the amount of time it takes the bale (or the plane) to travel by 30 m in the x-direction, then
she will strike cattle. This time is given by ( 30 m ) v0 0.68 s .
2-7. Air resistance is always anti-parallel to the velocity. The vector expression is:
1 v 1
W= cw ρ Av 2 − = − cw ρ Avv (1)
2 v 2
Including gravity and setting Fnet = ma , we obtain the parametric equations
x = − bx x 2 + y 2 (2)
y = −by x 2 + y 2 − g (3)
where b = cw ρ A 2m . Solving with a computer using the given values and ρ = 1.3 kg ⋅ m -3 , we
find that if the rancher drops the bale 210 m behind the cattle (the answer from the previous
problem), then it takes 4.44 s to land 62.5 m behind the cattle. This means that the bale
should be dropped at 178 m behind the cattle to land 30 m behind. This solution is what is
plotted in the figure. The time error she is allowed to make is the same as in the previous
problem since it only depends on how fast the plane is moving.
80
60
y (m)
40
20
0
–200 –180 –160 –140 –120 –100 –80 –60 –40
x (m)
With air resistance
No air resistance