Question

Given r(t)=sin(t)i+cos(t)j−ln(cos(t))k, find the unit normal vector N(t) evaluated at t=0,N(0).

Answer #1

Given r(t) = (et cos(t) )i + (et sin(t) )j
+ 2k. Find
(i) unit tangent vector T.
(ii) principal unit normal vector N.

Find the unit tangent vector T and the principle unit normal
vector N of ⃗r(t) = cos t⃗i + sin t⃗j + ln(cos t)⃗k at t = π .

Given that the acceleration vector is a(t)=(-9 cos(3t))i+(-9
sin(3t))j+(-5t)k, the initial velocity is v(0)=i+k, and the initial
position vector is r(0)=i+j+k, compute:
A. The velocity vector v(t)
B. The position vector r(t)

Find r(t) for the given
conditions.
r''(t) = −7
cos(t)j − 3
sin(t)k, r'(0)
=
3k, r(0)
= 7j

Consider the following vector function.
r(t) =
6t2, sin(t) − t cos(t), cos(t) + t sin(t)
, t > 0
(a) Find the unit tangent and unit normal vectors
T(t) and
N(t).
T(t)
=
N(t)
=
(b) Use this formula to find the curvature.
κ(t) =

Given that the acceleration vector is a ( t ) = (−9 cos( 3t ) )
i + ( −9 sin( 3t ) ) j + ( −5 t ) k, the initial velocity is v ( 0
) = i + k, and the initial position vector is r ( 0 ) = i +j + k,
compute: the velocity vector and position vector.

Find a unit tangent vector to the curve r = 3 cos 3t
i + 3 sin 2t j at t =
π/6 .

Find the vectors T and N and
the binormal vector B = T ⨯
N, for the vector-valued function
r(t) at the given value of
t.
r(t) = 6 cos(2t)i + 6
sin(2t)j +
tk, t0 =
pi/4
find:
T(pi/4)=
N(pi/4)=
B(pi/4)=

Q1 If r(t) = (2t2
- 5)i + (t - 2)j +
(4t + 10)k, find the curvature
k(t) at t = 1.
21733
3433
4173333
3433
Q2
Find the curvature k ( t ) for r ( t ) = 8 sin t i + 8 cos t
j
Group of answer choices
1
0
−sin2t+cos2t

8. Find r(t) given the following information.
r''(t)= 8 i + 12t k, r'(0)=6 j , r(0)= -4 i

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