Program Explanation Given information: The hailstone starts from rest with density δ = 1 and a negligible small radius r = 0 . After the radius grows by accretion, as r = k t and the net force F equals the time rate of change d p d t of momentum p = m v . The differential equation is, d d t ( m v ) = m g with v ( 0 ) = 0 Here, m is the variable mass of hailstone. Concept used: The linear first order equation is, d y d x + P ( x ) y = Q ( x ) The solution of above equation is, y ( x ) e ∫ P ( x ) d x = ∫ ( Q ( x ) e ∫ P ( x ) d x ) d x + C Here, e ∫ P ( x ) d x is the integrating factor and C is the constant of integration. Explanation: From the Newton’s second law the differential equation is, d d t ( m v ) = m g ....... (1) Here, g is with the influence of gravity. The radius of hailstone is, r ( t ) = k t The volume of spherical hailstone is, V = 4 3 π { r ( t ) } 3 Substitute k t for r ( t ) in above equation as, V = 4 3 π ( k t ) 3 = 4 3 π k 3 t 3 The density of the hailstone is, δ = m V Since, the density of hailstone is 1, substitute 1 for δ and 4 3 π k 3 t 3 for V in above equation as, 1 = m 4 3 π k 3 t 3 m = 4 3 π k 3 t 3 The time rate of change of momentum p = m v can be written as, d d t ( m v ) = v ⋅ d m d t + m d v d t Substitute 4 3 π k 3 t 3 for m in above equation as, d d t ( m v ) = v ⋅ d d t ( 4 3 π k 3 t 3 ) + m d v d t = v ⋅ 4 3 π k 3 ( 3 t 2 ) + m d v d t = 4 v π k 3 t 2 + m d v d t Substitute 4 v π k 3 t 2 + m d v d t for d d t

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

5th Edition

ISBN: 9780321816252

Author: C. Henry Edwards, David E. Penney, David Calvis

Publisher: PEARSON

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Chapter 1.5, Problem 42P

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To explain:The solution of differential equation ddt(mv)=mg and to show that the time rate of change of velocity of hailstone is dvdt=g4 .

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Chapter 1.5, Problem 41P

Chapter 1.5, Problem 43P

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Exercise 1 Function and Structure [30 pts] Please debug the following program and answer the following questions. There is a cycle in a linked list if some node in the list can be reached again by continuously following the next pointer. #include typedef struct node { int value; struct node *next; } node; int 11_has_cycle (node *first) if (first == node *head = { NULL) return 0; first; while (head->next != NULL) { } if (head first) { return 1; } head = head->next; return 0; void test ll_has_cycle () { int i; node nodes [6]; for (i = 0; i < 6; i++) { nodes [i] .next = NULL; nodes [i].value = i; } nodes [0] .next = &nodes [1]; nodes [1] .next = &nodes [2]; nodes [2] .next = &nodes [3]; nodes [3] .next nodes [4] .next &nodes [4]; NULL; nodes [5] .next = &nodes [0]; printf("1. Checking first list for cycles. \n Function 11_has_cycle says it has s cycle\n\n", 11_has_cycle (&nodes [0])?"a":"no"); printf("2. Checking length-zero list for cycles. \n Function 11_has_cycle says it has %s…

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Chapter 1 Solutions

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

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The solution of differential equation d d t ( m v ) = m g and to show that the time rate of change of velocity of hailstone is d v d t = g 4 .

Solution Summary: The author explains the solution of the differential equation dt(mv)=mg and the time rate of change of velocity of hailstone.

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To explain:The solution of differential equation ddt(mv)=mg and to show that the time rate of change of velocity of hailstone is dvdt=g4 .

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Chapter 1 Solutions