hw 3 sol 2019.pdf

# hw 3 sol 2019.pdf - HW3 Solution P1 = 0−(0− = 2 = 2...

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Unformatted text preview: HW3 Solution P1 = 0− , (0− ) = 2 = 2 1ℎ (0− ) = 0 ℎ ℎ = 0+ + +) (0 = (0 −) 1 0 + ∫ () = (2 + 0) = 2 0− , () ℎ 0− 0+ . + +) (0 −) = (0 1 0 + ∫ () = (0 + 0) = 0 0− () = () (2) 0− 0+ . ℎ, (0+ ) = (0+ ) = 0 (0+ ) 2 |0+ = = = 4/ 1 2 P2 = 0− , (0− ) = 0 (0− ) = 0 . = 0+ , + (0 +) −) = (0 1 0 + ∫ () = (0 + 0) 0− , () ℎ 0− 0+ ( 1 ). + (0 +) −) = (0 1 0 + ∫ () = (0 + 0) 0− , () ℎ 0− 0+ ( 1 &2 ). ℎ, = 0+ , 2 (0+ ) = (0+ ) × 2 = 0 1 (0+ ) = (0+ ) − 2 (0+ ) = 0 . = ∞, ℎ ℎ, 1 (∞) = (∞) = 0 2 (∞) = 2 () 1 + 2 . = 0+ , , { = 1 + = + 2 = 1 + 2 , = + { 1 2 = ℎ , 1 − = ( − ) 1 2 1 ⟹ = 2 1 2 = − { ()&(), ⇒ 1 | + { 0 2 | + = 0 / 0 |0+ = = (/) 1 . = 0+ , (), , ℎ ℎ 2 . , = + 2 = 1 + 2 , 2 = ℎ , = + 2 2 ⇒ 2 2 2 = 2 { 2 2 2 1 2 = ( − ) 2 ⇒ 2 2 2 = 2 { 2 2 (), ()& (), ⇒ 2 2 2 2 1 | 0+ = 2 ( − ) |0+ = (/ 2 ) 2 1 P3 . = 0− , , (0− ) = 1 (0− ) = (0− ) = (0− ) = 0 1 (0− ) = (0− ) = 0 (0− ) = (0− ) = 2 (0− ) = () 2 = 0+ , , { = 1 + 2 + + = 1 + , = 1 = 2 = = , + +) (0 +) = (0 = (0 −) 1 0 () + ∫ () = (0− ) = 0− 2 () 0− 0+ ℎ . + (0+ ) = 1 (0+ ) + (0+ ) = 1 (0+ ) × 1 + (0− ) + 1 0 ∫ () 0− = (0+ ) × 1 + (0− ) = (0+ ) × 1 = 1 () 2 () = () 0− 0+ ℎ . . , = 1 + ℎ , ⇒ = × 1 + ⇒ ⇒ 1 ∫ () −∞ 1 = 1 + 1 1 |0+ = (1 + ) |0+ = − 1 P4. = 0− , = 1 = 1 = 0+ , , = + + , = = = , + +) (0 = (0 −) 1 0 + ∫ () = (1 + 0) 0− , () ℎ 0− 0+ ( ). + +) (0 −) = (0 1 0 + ∫ () = (1 + 0) 0− , () ℎ 0− 0+ ( ). (0+ ) = × (0+ ) = 1 ℎ & , 1 = ∫ () + × + −∞ ⇒ 2 = + + 2 { 1 = ( − − ) ⇒ 2 1 = ( − − ) 2 { & = 2 1 = = − − = − − 2 2 2 2 { = 2 − 2 − 2 ℎ ⇒ 2 | + { 2 0 1 |0+ = ( − − )|0+ = 3 / 1 = ( − − ) |0+ = −50 − 1 − 1 × 3 = −3050.001 / 2 & + (0+ ) = (0− ) + 2 |0+ { 1 0 ∫ () = 1 0− |0+ = |0+ = 1 / 2 1 | += | + = 3 / 2 0 0 |0+ = ( − − )|0+ = 3 | +=( − − ) | + = −50 − 1 − 3 = −3050.001 / 0 0 2 2 2 =( 2 − − ) | + = 500 − 3 + 1 × 3050.001 = 3050001 + 497 = 3050498 / 2 2 2 0 P5 Only (c) and (h) are required; TA did (c) thru (h) . 2 + 7 + 12 = 0 2 ℎ 2 + 7 + 12 = 0 1 = −7 + 1 = −3 2 2 = −7 − 1 = −4 2 () = 1 −3 + 2 −4 . 2 + 5 + 4 = 0 2 ℎ 2 + 5 + 4 = 0 1 = −5 + 3 = −1 2 2 = −5 − 3 = −4 2 () = 1 − + 2 −4 . 2 + + 6 = 0 2 ℎ 2 + + 6 = 0 1 = −1 + √23 2 2 = −1 − √23 2 1 √23 2 () = −2 (1 − . + 2 √23 2 ) 2 + + 2 = 0 2 ℎ 2 + + 2 = 0 1 = −1 + √7 2 2 = −1 − √7 2 1 √7 2 () = −2 (1 − . + 2 2 +2 + = 0 2 ℎ 2 + 2 + 1 = 0 1 = −1 2 = −1 () = 1 − + 2 − ℎ. 2 + 4 + 4 = 0 2 ℎ 2 + 4 + 4 = 0 1 = −2 2 = −2 √7 2 ) () = 1 −2 + 2 −2 P6 Only (c) and (h) are required; TA did (c) thru (h) : (0+ ) = 2, . | +=1 0 2 + 7 + 12 = 0 2 , () = 1 −3 + 2 −4 , { 1 + 2 = 2 −31 − 42 = 1 ⇒{ 1 = 9 2 = −7 , () = 9 −3 − 7 −4 . 2 + 5 + 4 = 0 2 , () = 1 − + 2 −4 , + 2 = 2 { 1 −1 − 42 = 1 ⇒{ 1 = 3 2 = −1 , () = 3 − − −4 . 2 + + 6 = 0 2 , 1 √23 2 () = −2 (1 − + 2 √23 2 ) , 1 + 2 = 2 = ∗ 1 √23 ( ) ( ) − { 2 1 + 2 + 2 2 − 1 = 1 =1 ⇒{ 1 2 = 1 , 1 √23 () = 2 −2 cos( ) 2 . 2 + + 2 = 0 2 , 1 √7 2 () = −2 (1 − + 2 √7 2 ) , 1 + 2 = 2 = ∗ 1 √7 ( ) ( ) {− 2 1 + 2 + 2 2 − 1 = 1 =1 ⇒{ 1 2 = 1 , 1 √7 () = 2 −2 cos( ) 2 . 2 +2 + = 0 2 , () = 1 − + 2 − , { 1 = 2 −1 + 2 = 1 =2 ⇒{ 1 2 = 3 , () = 2 − + 3 − ℎ. 2 + 4 + 4 = 0 2 , () = 1 −2 + 2 −2 , 1 = 2 { −21 + 2 = 1 =2 ⇒{ 1 2 = 5 , () = 2 −2 + 5 −2 P7 () = − sin( + ) , , , , ℎ × sin() = 0 : , (0) = 0 → { ≠0 →=0 : , 0.4 ℎ 0.8 0.4 = → = 2.5 : , ℎ 4 →=4 : , ℎ ℎ 0.4 0.4 = 1 → = −2.52 2 ℎ, () = 4 −2.52× sin(2.5) P8 = 0− , (0− ) = 100 = 10 10Ω (0− ) = 0 = 0+ , ℎ ℎ , , = , + = 0 , 0+ +) (0 = (0 −) + ∫ () = (10 + 0) 0− = 0− 0+ 0+ (0 +) −) = (0 + ∫ () = (0 + 0) 0− = − 0− 0+ ℎ , + 1 (0 ) = (0+ ) = 0 / = ⇒ ℎ & , + 1 (0 ) = (0+ ) = 0 / ⇒ 2 1 + = 0 2 2 1 ⇒ + =0 2 ℎ , 2 + 1 =0 , 1 = √ − = { 2 √ , − () = 1 √ + 2 √ (0+ ) = 10 , | + = 0 / 0 1 + 2 = 10 = ∗ {√ (1 − 2 ) = 0 =5 ⇒{ 1 2 = 5 500 2 = () = 10 cos( ) = 10 cos( ) √ √5 P9 . , = 2 + , − = 2 = = & 1 ∫ () + + × 2 = 0 −∞ ⟹ 2 2 + + =0 2 . ℎ (), 2 + 2 1 + =0 ℎ (), + = 0 ⇒ 1 2 1 + ∫ () ⇒ + ( − (− ) × 2 ) = 0 −∞ ⇒ 2 2 + + =0 ℎ , 2 + 2 1 + =0 . = 0− , (0− ) = 10 (0− ) = 0 = 0+ , ℎ 1 → 2, 0+ +) (0 −) = (0 + ∫ () = (10 + 0) 0− 0− 0+ 0+ +) (0 −) = (0 + ∫ () = (0 + 0) 0− 0− 0+ ℎ & , 1 |0+ = ( (0+ ) − (0+ )2 ) { − = − = 1 1 |0+ = × (0+ ) = ( − (0+ )2 ) = 2.5 × 107 / ⇒ { (0+ ) |0+ = − = 0 / ℎ (), 2 + 2 1 + =0 , 2 = 4 × 107 4 = 2.5 × 1015 4 ⇒ 2 2 ( ) − = −9 × 1014 { √( 2 √ 2 2 4 7 7 + ( ) − −4 × 10 + 3 × 10 1 = = 2 2 2 4 − 2 − √( 2 ) − −4 × 107 − 3 × 107 = {1 = 2 2 − & 2 2 4 ) − = 3 × 107 ℎ, (), 7 7 7 () = −2×10 (1 1.5×10 + 2 −1.5×10 ) : (0+ ) = 10 , | + = 0 / 0 1 + 2 = 10 1 = 2∗ { −2 × 107 (1 + 2 ) + (1.5 × 107 1 − 1.5 × 107 2 ) = 0 20 3 ⇒{ 20 2 = 5 + 3 1 = 5 − 7 () = −2×10 ((5 − 20 20 7 7 ) × 1.5×10 + (5 + ) × −1.5×10 ) 3 3 4 7 = 10 × −2×10 × [cos(1.5 × 107 ) + sin(1.5 × 107 )] 3 = 10 × 5 7 × −2×10 × cos(1.5 × 107 + ) , ≥ 0 3 4 ℎ = arctan (− ) 3 . , ℎ ℎ () () − ℎ ℎ 20 + ∞ ℎ 20 , − , ′ () = ′ (+∞) + (′ (0) − ′ (+∞)) − 1 ℎ, ′ (+∞) = 10, ′ (0) = 0 ⇒ ′ () = 10 − 10 20 − 1 ⇒ ′ (20 ) = 10 − 10 −5×4 = 10(1 − −1 ) ℎ, ℎ (0− ) = ′ (20 ) = 10(1 − −1 ) 5 7 () = 10 × (1 − −1 ) × × −2×10 × cos(1.5 × 107 + ) , ≥ 0 3 4 ℎ = arctan (− ) 3 ℎ () ...
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