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Given I = C: x=t y = cos(2t), z = sin ( 2t) > Sext Olt≤27. :: dx=1 dt ⇒ds= (x² + y² +2²) ds. 1: We know that ds = √ (d ² )² + (do ) ² + (dz) ² dt dx dy dz dt dt " dy वट dz = -2 sim (²t), d² = 2 COS(2t). dt i + 4 sin" (27) + 4 coś" (2+) dt = √5 dt [:: sin³0+ cos²³0=1] ,25 ⇒ I = √(t² + cos² (2 t) + sin(2t)) √5 dt 0 27 * I = √ ( 1² + 1) √5 dt = [#² ++] ² √5 = $T HET VE •√5 87³ +65 3 0

Given I = C: x=t y = cos(2t), z = sin ( 2t) > Sext Olt≤27. :: dx=1 dt ⇒ds= (x² + y² +2²) ds. 1: We know that ds = √ (d ² )² + (do ) ² + (dz) ² dt dx dy dz dt dt " dy वट dz = -2 sim (²t), d² = 2 COS(2t). dt i + 4 sin" (27) + 4 coś" (2+) dt = √5 dt [:: sin³0+ cos²³0=1] ,25 ⇒ I = √(t² + cos² (2 t) + sin(2t)) √5 dt 0 27 * I = √ ( 1² + 1) √5 dt = [#² ++] ² √5 = $T HET VE •√5 87³ +65 3 0

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Given I = S(x² + y² +2²³) ds. с C: x=t, y = cos(2t), z = sin(2t), olt≤27. ': We know that ds = √( @z j²₁+ (d ² ) ² + ( 1²₂ ) dz dx 12 dy dt dt. dx = 1, dy = -25'in (² t), d² = 2 cos(2t). dz (27) ⇒ds=√₁²+ 4 sin(²t) + 4c05² (2+) dt = √5 d+ [¹: sin³o+cos³²0=1] 2π ⇒ I = √(t² + cos² (2 t) + sin(2t)) √5 dt 0 → 27 • I = √ ( + ² + 1 ) √5 d t = [ + ² + + ] ² ²³√5 = 8x² + 6*. √5 t V5= 3 0 3
Transcribed Image Text:Given I = S(x² + y² +2²³) ds. с C: x=t, y = cos(2t), z = sin(2t), olt≤27. ': We know that ds = √( @z j²₁+ (d ² ) ² + ( 1²₂ ) dz dx 12 dy dt dt. dx = 1, dy = -25'in (² t), d² = 2 cos(2t). dz (27) ⇒ds=√₁²+ 4 sin(²t) + 4c05² (2+) dt = √5 d+ [¹: sin³o+cos³²0=1] 2π ⇒ I = √(t² + cos² (2 t) + sin(2t)) √5 dt 0 → 27 • I = √ ( + ² + 1 ) √5 d t = [ + ² + + ] ² ²³√5 = 8x² + 6*. √5 t V5= 3 0 3
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