SOLUTION: A vector equation of the circular helix is R(t) = a cos ti + a sin tj + tk So D,R(t) =-a sin ti + a cos tj + k and |D,R(t)|= Va +1. From (3) we get T(t) = Vat + (-a sin ti + a cos tj + k) NSIONAL SPACE AND SOLID ANALYTIC GEOMETRY So D,T(t) = VI (-a cos ti – a sin tj) Applying (8), we obtain K(t) a² +1 (-a cos ti -- a sin tj) The curvature, then, is given by K(t) = |K(t)| = and so the curvature of the circular helix is constant. From (11) we get N(t) =-cos ti – sin tj Applying (12), we have B() = VT (-a sin ti + a cos tj + k) × (-cos ti – sin tj) Va? +1 1 Va + 1 (sin ti – cos tj + ak) - COS
SOLUTION: A vector equation of the circular helix is R(t) = a cos ti + a sin tj + tk So D,R(t) =-a sin ti + a cos tj + k and |D,R(t)|= Va +1. From (3) we get T(t) = Vat + (-a sin ti + a cos tj + k) NSIONAL SPACE AND SOLID ANALYTIC GEOMETRY So D,T(t) = VI (-a cos ti – a sin tj) Applying (8), we obtain K(t) a² +1 (-a cos ti -- a sin tj) The curvature, then, is given by K(t) = |K(t)| = and so the curvature of the circular helix is constant. From (11) we get N(t) =-cos ti – sin tj Applying (12), we have B() = VT (-a sin ti + a cos tj + k) × (-cos ti – sin tj) Va? +1 1 Va + 1 (sin ti – cos tj + ak) - COS
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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Transcribed Image Text:SOLUTION: A vector equation of the circular helix is
R(t) = a cos ti + a sin tj + tk
So D,R(t) =-a sin ti + a cos tj + k and |D,R(t)|= Va +1. From (3) we
get
T(t) =
Vat +
(-a sin ti + a cos tj + k)
NSIONAL SPACE AND SOLID ANALYTIC GEOMETRY
So
D,T(t) = VI
(-a cos ti – a sin tj)
Applying (8), we obtain
K(t)
a² +1
(-a cos ti -- a sin tj)
The curvature, then, is given by
K(t) = |K(t)| =
and so the curvature of the circular helix is constant. From (11) we get
N(t) =-cos ti – sin tj
Applying (12), we have
B() = VT
(-a sin ti + a cos tj + k) × (-cos ti – sin tj)
Va? +1
1
Va + 1
(sin ti – cos tj + ak)
- COS
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