Consider the following curve. x = sin(6t), y = -cos(6t), z = 24t Using the given parametric equations, give the corresponding vector equation r(t). r(t) = (sin (6t), - cos (6t),241) Find r'(t) and Ir'(t)\. r'(t) = |r'(t) = √612 (6 cos (61),6 sin (6t),24) Find the equation of the normal plane of the given curve at the point (0, 1, 4x). -6x+12z-48=0 x Now consider the osculating plane of the given curve at the point (0, 1, 4x). Determine each of the following. T(t) = T'(t) IT'() = N(t) = -sin (61)j + cos(61)k x Find the equation of the osculating plane of the given curve at the point (0, 1, 4x). X

Algebra and Trigonometry (MindTap Course List)
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Author:James Stewart, Lothar Redlin, Saleem Watson
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Chapter8: Polar Coordinates And Parametric Equations
Section8.CT: Chapter Test
Problem 8CT
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Consider the following curve.
x = sin(6t), y = -cos(6t), z = 24t
Using the given parametric equations, give the corresponding vector equation r(t).
r(t) = (sin (6t), - cos (6t),241)
Find r'(t) and Ir'(t)\.
r'(t) =
Ir'(t) =
=
(6 cos (61),6 sin (6t),24)
Find the equation of the normal plane of the given curve at the point (0, 1, 4x).
-6x + 12z-48m=0 x
T'(t)
√612
Now consider the osculating plane of the given curve at the point (0, 1, 4x). Determine each of the following.
T(t) =
IT'() =
N(t) = -sin (61)j + cos(61)k x
Find the equation of the osculating plane of the given curve at the point (0, 1, 4x).
Xx
Transcribed Image Text:Consider the following curve. x = sin(6t), y = -cos(6t), z = 24t Using the given parametric equations, give the corresponding vector equation r(t). r(t) = (sin (6t), - cos (6t),241) Find r'(t) and Ir'(t)\. r'(t) = Ir'(t) = = (6 cos (61),6 sin (6t),24) Find the equation of the normal plane of the given curve at the point (0, 1, 4x). -6x + 12z-48m=0 x T'(t) √612 Now consider the osculating plane of the given curve at the point (0, 1, 4x). Determine each of the following. T(t) = IT'() = N(t) = -sin (61)j + cos(61)k x Find the equation of the osculating plane of the given curve at the point (0, 1, 4x). Xx
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