...] A particle moves such that its position at time t is given by +++)i+ G + 5° – t - 5) i+ ( -* + 4t + 3) k. r(t) = The jerk of a curve is defined to be r (t), its third derivative with respect to time. Find all, if any, points where the jerk of the above curve is perpendicular to its velocity.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Needed to be solved all part's correctly in 30 minutes and get the thumbs up please show neat and clean work please provide correct solution
1. a)
...] A particle moves such that its position at time t is given by
r(t) =
+ t2 +1) i+
t3 + 5t2 – t – 5 )j+
-t3 – t² + 4t + 3 ) k.
The jerk of a curve is defined to be r (t), its third derivative with respect to
time.
Find all, if any, points where the jerk of the above curve is perpendicular to its
velocity.
] Let f(x, y) = g(u(x, y), v(x, y)) for suitable functions u, v and g.
af
b) [-
i) Write down an expression for
in terms of derivatives of q, u and v.
ii) Given that u(1,1) = 2 and v(1, 1) = -2, use appropriate entries in the
af
following table to find
(1, 1):
gv(1, 1) = 2,
V½(1, 1) = 11,
Ju (2, –2) = 3,
и, (2, — 2) — 13,
дь (2, —2) 3D 5,
Va (2, -2) = 17.
Ju(1, 1) = 1,
Uz (1,1) = 7,
1 Given that, for b > 0 we have
x2 +1
62 + 1
(x² + b²)²
dx = T
263
use differentiation under the integral sign to find
ONER
x2 +1
dx.
+ 4)3
(x²
Transcribed Image Text:1. a) ...] A particle moves such that its position at time t is given by r(t) = + t2 +1) i+ t3 + 5t2 – t – 5 )j+ -t3 – t² + 4t + 3 ) k. The jerk of a curve is defined to be r (t), its third derivative with respect to time. Find all, if any, points where the jerk of the above curve is perpendicular to its velocity. ] Let f(x, y) = g(u(x, y), v(x, y)) for suitable functions u, v and g. af b) [- i) Write down an expression for in terms of derivatives of q, u and v. ii) Given that u(1,1) = 2 and v(1, 1) = -2, use appropriate entries in the af following table to find (1, 1): gv(1, 1) = 2, V½(1, 1) = 11, Ju (2, –2) = 3, и, (2, — 2) — 13, дь (2, —2) 3D 5, Va (2, -2) = 17. Ju(1, 1) = 1, Uz (1,1) = 7, 1 Given that, for b > 0 we have x2 +1 62 + 1 (x² + b²)² dx = T 263 use differentiation under the integral sign to find ONER x2 +1 dx. + 4)3 (x²
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