Video Example) Find the length of the arc of the circular helix with vector equation r(t) = 6 cos(t) i + 6 sin(t) j + tk from the point (6, 0, 0) to the point (6, 0, 27). Solution Since r'(t) = Ir'(t) = √(-6 sin(t))² + (6 cos(t))² + 1² = The arc from (6, 0, 0) to (6, 0, 27) is described by the parametric interval 0 ≤t≤ 2π, and so from the formula L -=[° \r' (t)\ dt, we have the following. 125 11 L= , we have |r'(t)\ dt = Nood Help? -1.³ (C ]) at dt
Video Example) Find the length of the arc of the circular helix with vector equation r(t) = 6 cos(t) i + 6 sin(t) j + tk from the point (6, 0, 0) to the point (6, 0, 27). Solution Since r'(t) = Ir'(t) = √(-6 sin(t))² + (6 cos(t))² + 1² = The arc from (6, 0, 0) to (6, 0, 27) is described by the parametric interval 0 ≤t≤ 2π, and so from the formula L -=[° \r' (t)\ dt, we have the following. 125 11 L= , we have |r'(t)\ dt = Nood Help? -1.³ (C ]) at dt
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Question
![Example
Video Example)
Find the length of the arc of the circular helix with vector equation r(t) = 6 cos(t) i + 6 sin(t) j + tk from the point (6, 0, 0) to the point (6, 0, 2π).
Solution
Since r'(t) =
Ir' (t) = √(-6 sin(t))² + (6 cos(t))² + 1²
2
L =
L =
The arc from (6, 0, 0) to (6, 0, 27) is described by the parametric interval 0 ≤ t ≤ 2π, and so from the formula
-SD 1²
Ja
we have the following.
-2π
1.²5
I
\r'(t)\ dt,
|r'(t)\ dt =
Need Help?
we have
2π
16² (1
JO
Read It
=
]) at
dt](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F078d0b9f-3892-49c8-9643-cdd840b9c9a9%2Fc2124d22-71c4-46c3-b415-c81adfc81d9a%2Famwspbp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example
Video Example)
Find the length of the arc of the circular helix with vector equation r(t) = 6 cos(t) i + 6 sin(t) j + tk from the point (6, 0, 0) to the point (6, 0, 2π).
Solution
Since r'(t) =
Ir' (t) = √(-6 sin(t))² + (6 cos(t))² + 1²
2
L =
L =
The arc from (6, 0, 0) to (6, 0, 27) is described by the parametric interval 0 ≤ t ≤ 2π, and so from the formula
-SD 1²
Ja
we have the following.
-2π
1.²5
I
\r'(t)\ dt,
|r'(t)\ dt =
Need Help?
we have
2π
16² (1
JO
Read It
=
]) at
dt
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