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ds 53. dt = 8 sin" (+ + 12). s (0) = 8

ds 53. dt = 8 sin" (+ + 12). s (0) = 8

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Answer question 53 picture attached below
Exercises 4.3 297 Initial Value Problems Solve the initial value problems in Exercises 51-56. ds 51. = 12r (3r2 – 1)', s (1) = 3 dt cos t dy = 4x (x + 8)-/3, y(0)=0 sin dr z tan z 52. dx ds Edz sec z ² (t +5), s (0) = 8 53. = 8 sin dt cos(Vi +3) dt dr 54. = 3 cos de d's 55. dr? = -4 sin (21 -). s'(0) = 100, s(0) = 0 50. dx? d²y = 4 sec 2x tan 2x, y'(0) = 4, y(0) = -1 57. The velocity of a particle moving back and forth on a line is v = ds/dt = 6 sin 21 m/sec for all t. If s = 0 when 1 = 0, find the value of s when t =n/2 sec. 58. The acceleration of a particle moving back and forth on a line is a = d's/d? = x' cos nt m/sec? for all t. If s = 0 and v = 8 m/sec when t =0, find s when t = 1 sec. Theory and Examples 59. It looks as if we can integrate 2 sin x cos x with respect to x in three different ways: - /udu np u= sin r. a) sin x cos x dx y reducing the integral y the integral a bit and ill see what we mean = u? + C, = sin? x+C ercises 47 and 48. b) I sin x cos x dx = - 2u du u= cos x. = -u? + C2 = - cos? x + C by w = 2+ v c) I sin x cos x dx = sin 2x dx 2 sin x cosx = sin 2x cos 2x +C3. )dx Can all three integrations be correct? Give reasons for your an- en by w = 1+ v? swer. 60. The substitution u tan x gives tan² x + C. sec x tan x dx = udu= +C = --/ - The substitution u = secx gives sec x tan x dx = u du = +C= +C. Can both integrations be correct? Give reasons for your answer.
Transcribed Image Text:Exercises 4.3 297 Initial Value Problems Solve the initial value problems in Exercises 51-56. ds 51. = 12r (3r2 – 1)', s (1) = 3 dt cos t dy = 4x (x + 8)-/3, y(0)=0 sin dr z tan z 52. dx ds Edz sec z ² (t +5), s (0) = 8 53. = 8 sin dt cos(Vi +3) dt dr 54. = 3 cos de d's 55. dr? = -4 sin (21 -). s'(0) = 100, s(0) = 0 50. dx? d²y = 4 sec 2x tan 2x, y'(0) = 4, y(0) = -1 57. The velocity of a particle moving back and forth on a line is v = ds/dt = 6 sin 21 m/sec for all t. If s = 0 when 1 = 0, find the value of s when t =n/2 sec. 58. The acceleration of a particle moving back and forth on a line is a = d's/d? = x' cos nt m/sec? for all t. If s = 0 and v = 8 m/sec when t =0, find s when t = 1 sec. Theory and Examples 59. It looks as if we can integrate 2 sin x cos x with respect to x in three different ways: - /udu np u= sin r. a) sin x cos x dx y reducing the integral y the integral a bit and ill see what we mean = u? + C, = sin? x+C ercises 47 and 48. b) I sin x cos x dx = - 2u du u= cos x. = -u? + C2 = - cos? x + C by w = 2+ v c) I sin x cos x dx = sin 2x dx 2 sin x cosx = sin 2x cos 2x +C3. )dx Can all three integrations be correct? Give reasons for your an- en by w = 1+ v? swer. 60. The substitution u tan x gives tan² x + C. sec x tan x dx = udu= +C = --/ - The substitution u = secx gives sec x tan x dx = u du = +C= +C. Can both integrations be correct? Give reasons for your answer.
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