chap15

chap15 - Chapter 15 Solution 1 e at e at 2 1 1 1 s L...

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Chapter 15, Solution 1. (a) 2 e e ) at cosh( at - at + = [ ] = + + = a s 1 a s 1 2 1 ) at cosh( L 2 2 a s s (b) 2 e e ) at sinh( at - at = [ ] = + = a s 1 a s 1 2 1 ) at sinh( L 2 2 a s a Chapter 15, Solution 2. (a) ) sin( ) t sin( ) cos( ) t cos( ) t ( f θ ω θ ω = [ ] [ ] ) t sin( ) sin( ) t cos( ) cos( ) s ( F ω θ ω θ = L L = ) s ( F 2 2 s ) sin( ) cos( s ω + θ ω θ (b) ) sin( ) t cos( ) cos( ) t sin( ) t ( f θ ω + θ ω = [ ] [ ] ) t sin( ) cos( ) t cos( ) sin( ) s ( F ω θ + ω θ = L L = ) s ( F 2 2 s ) cos( ) sin( s ω + θ ω θ Chapter 15, Solution 3. (a) [ ] = ) t ( u ) t 3 cos( e -2t L 9 ) 2 s ( 2 s 2 + + + (b) [ ] = ) t ( u ) t 4 sin( e -2t L 16 ) 2 s ( 4 2 + +
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(c) Since [ ] 2 2 a s s ) at cosh( = L [ ] = ) t ( u ) t 2 cosh( e -3t L 4 ) 3 s ( 3 s 2 + + (d) Since [ ] 2 2 a s a ) at sinh( = L [ = ) t ( u ) t sinh( e -4t L ] 1 ) 4 s ( 1 2 + (e) [ ] 4 ) 1 s ( 2 ) t 2 sin( e 2 t - + + = L If ) s ( F ) t ( f → ) s ( F ds d - ) t ( f t → Thus, [ ] ( ) [ ] 1 - 2 t - 4 ) 1 s ( 2 ds d - ) t 2 sin( e t + + = L ) 1 s ( 2 ) 4 ) 1 s (( 2 2 2 + + + = [ ] = ) t 2 sin( e t -t L 2 2 ) 4 ) 1 s (( ) 1 s ( 4 + + + Chapter 15, Solution 4. (a) 16 s se 6 e 4 s s 6 ) s ( G 2 s s 2 2 + = + = (b) 3 s e 5 s 2 ) s ( F s 2 2 + + =
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Chapter 15, Solution 5. (a) [ ] 4 s ) 30 sin( 2 ) 30 cos( s ) 30 t 2 cos( 2 + ° ° = ° + L [ ] + ° = ° + 4 s 1 ) 30 cos( s ds d ) 30 t 2 cos( t 2 2 2 2 L ( ) + = 1 - 2 4 s 1 s 2 3 ds d ds d ( ) ( ) + + = 2 - 2 1 - 2 4 s 1 s 2 3 s 2 4 s 2 3 ds d ( ) ( ) ( ) ( ) ( ) 3 2 2 2 2 2 2 2 2 4 s 1 s 2 3 ) s 8 ( 4 s 2 3 s 2 4 s 1 s 2 3 2 4 s 2s - 2 3 + + + + + = ( ) ( ) 3 2 2 2 2 4 s 1 s 2 3 ) s 8 ( 4 s s 3 2 s 3 s 3 - + + + + = ( ) ( ) 3 2 2 3 3 2 2 4 s s 8 s 3 4 4 s ) 4 2)(s s 3 (-3 + + + + + = [ ] = ° + ) 30 t 2 cos( t 2 L ( ) 3 2 3 2 4 s s 3 s 6 s 3 12 8 + + (b) [ ] = + = 5 t - 4 ) 2 s ( ! 4 30 e t 30 L 5 ) 2 s ( 720 + (c) = = δ ) 0 1 s ( 4 s 2 ) t ( dt d 4 ) t ( u t 2 2 L s 4 s 2 2
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(d) ) t ( u e 2 ) t ( u e 2 -t 1) - -(t = [ ] = ) t ( u e 2 1) - -(t L 1 s e 2 + (e) Using the scaling property, [ ] = = = s 2 1 2 5 ) 2 1 ( s 1 2 1 1 5 ) 2 t ( u 5 L s 5 (f) [ ] = + = 3 1 s 6 ) t ( u e 6 3 t - L 1 s 3 18 + (g) Let f . Then, ) t ( ) t ( δ = 1 ) s ( F = . " = = δ ) 0 ( f s ) 0 ( f s ) s ( F s ) t ( f dt d ) t ( dt d 2 n 1 n n n n n n L L " = = δ 0 s 0 s 1 s ) t ( f dt d ) t ( dt d 2 n 1 n n n n n n L L = δ ) t ( dt d n n L n s Chapter 15, Solution 6. (a) [ ] = δ ) 1 t ( 2 L -s e 2 (b) [ ] = ) 2 t ( u 10 L 2s - e s 10 (c) [ ] = + ) t ( u ) 4 t ( L s 4 s 1 2 + (d) [ ] [ ] = = ) 4 t ( u e e 2 ) 4 t ( u e 2 4) - -(t -4 -t L L ) 1 s ( e e 2 4 -4s +
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Chapter 15, Solution 7. (a) Since [ ] 2 2 4 s s ) t 4 cos( + = L , we use the linearity and shift properties to obtain [ ] = ) 1 t ( u )) 1 t ( 4 cos( 10 L 16 s e s 10 2 s - + (b) Since [ ] 3 2 s 2 t = L , [ ] s 1 ) t ( u = L , [ ] 3 t 2 - 2 ) 2 s ( 2 e t + = L , and [ ] s e ) 3 t ( u s 3 - = L [ ] = + ) 3 t ( u ) t ( u e t t 2 - 2 L s e ) 2 s ( 2 -3s 3 + + Chapter 15, Solution 8.
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