To determine To derive: The equations x = υ 0 k ( 1 − e − k t ) cos α and y = υ 0 k ( 1 − e − k t ) ( sin α ) + g k 2 ( 1 − k t − e − k t ) by solving the given initial value problem. Explanation The given differential equation is d 2 r d t 2 = − g j − k d r d t , with the initial conditions r ( 0 ) = 0 and d r d t | t = 0 = ( υ 0 cos α ) i + ( υ 0 sin α ) j . Compare the equation d 2 r d t 2 + k d r d t = − g j with that of d 2 r d t 2 + P ( t ) d r d t = Q ( t ) and obtain P ( t ) = k and Q ( t ) = − g j . Integrate both sides of the differential equation d 2 r d t = − g j − k d r d t as follows. ∫ d 2 r d t = 1 v ( t ) ∫ v ( t ) Q ( t ) d t = 1 e ∫ P ( t ) d t ∫ e ∫ P ( t ) d t Q ( t ) d t = 1 e ∫ k d t ∫ e ∫ k d t ( − g j ) d t = − g e k t ( ∫ e k t d t ) j On further simplification, d r d t = − g e − k t ( e k t k j + C 1 ) = − g k j − g C 1 e − k t = − g k j + C e − k t where C = − g C 1 Now evaluate the constant C using the initial condition as follows. d r d t | t = 0 = − g k j + C e ( 0 ) ( υ 0 cos α ) i + ( υ 0 sin α ) j = − g k j + C C = ( υ 0 cos α ) i + ( υ 0 sin α + g k ) j Substitute the constant C and obtain d r d t as shown below. d r d t = − g k j + ( ( υ 0 cos α ) i + ( υ 0 sin α + g k ) j ) e − k t = ( υ 0 e − k t cos α ) i + ( e − k t ( υ 0 sin α + g k ) − g k ) j Again integrate d r d t and obtain r ( t ) as follows. ∫ d r d t = ∫ ( ( υ 0 e − k t cos α ) i + ( e − k t ( υ 0 sin α + g k ) − g k ) j ) d t r ( t ) = ( ∫ ( υ 0 e − k t cos α ) d t ) i + ( ∫ ( e − k t ( υ 0 sin α + g k ) − g k ) d t ) j = ( − υ 0 e − k t cos α k ) i + ( − e − k t k ( υ 0 sin α + g k ) − g t k ) j + C 2 Now evaluate the constant C 2 using the initial condition r ( 0 ) = 0 as follows

14th Edition

ISBN: 9780134438986

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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Chapter 13.2, Problem 41E

To determine

To derive: The equations x=υ0k(1−e−kt)cosα and y=υ0k(1−e−kt)(sinα)+gk2(1−kt−e−kt) by solving the given initial value problem.

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Chapter 13.2, Problem 40E

Chapter 13.2, Problem 42E

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