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The first three derivatives of the position vector r(t) As we mentioned above, we will indicate derivation with respect to t by prime. According to equations we have dr == \r\T = s'T ⇒ r" = s″T + s'T' = s'T + (s)2 = 8" + (s')²³ KN dt ds Similarly find "" using further differentiations with respect to t. T' 5. Show that we have and, Txr" = (s')³ KB ⇒ K = (r² x r").p"" = (s')® K²T ⇒ T = |²x| |²|3 (²x gol).!!! |r² x r" |2

The first three derivatives of the position vector r(t) As we mentioned above, we will indicate derivation with respect to t by prime. According to equations we have dr == \r\T = s'T ⇒ r" = s″T + s'T' = s'T + (s)2 = 8" + (s')²³ KN dt ds Similarly find "" using further differentiations with respect to t. T' 5. Show that we have and, Txr" = (s')³ KB ⇒ K = (r² x r").p"" = (s')® K²T ⇒ T = |²x| |²|3 (²x gol).!!! |r² x r" |2

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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The first three derivatives of the position vector r(t) As we mentioned above, we will indicate derivation with respect to t by prime. According to equations we have dr dt Similarly find "" using further differentiations with respect to t. p² and, =\r'\T = s'T ⇒ r" = 8″T + s'T' = s"T + (s')2 = 8″T + (s')² kN ds 5. Show that we have r' xr" = (s')³ KB ⇒ (r' xr").r" = (s')³K²T ⇒ K = T = |r² x r"| |r1|3 (² x ²).p!!! |r' xr" |2
Transcribed Image Text:The first three derivatives of the position vector r(t) As we mentioned above, we will indicate derivation with respect to t by prime. According to equations we have dr dt Similarly find "" using further differentiations with respect to t. p² and, =\r'\T = s'T ⇒ r" = 8″T + s'T' = s"T + (s')2 = 8″T + (s')² kN ds 5. Show that we have r' xr" = (s')³ KB ⇒ (r' xr").r" = (s')³K²T ⇒ K = T = |r² x r"| |r1|3 (² x ²).p!!! |r' xr" |2
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