Question

Compute the unit binormal vector and torsion of the curve. r(t) = <15t, 8 cos t, 8 sin t> A. B(t) = <8, -15 sin t, 15 cos t> / sqrt(289), tau = 15/289 B. B(t) = <0, 15 cos t, -15 sin t> / sqrt(289), tau = 1/289 C. B(t) = <0, -15 cos t, -15 sin t> / sqrt(289), tau = -15/289 D. B(t) = <8, 15 sin t, -15 cos t> / sqrt(289), tau = 15/289

          Compute the unit binormal vector and torsion of the curve.
r(t) = <15t, 8 cos t, 8 sin t>
A. B(t) = <8, -15 sin t, 15 cos t> / sqrt(289), tau = 15/289
B. B(t) = <0, 15 cos t, -15 sin t> / sqrt(289), tau = 1/289
C. B(t) = <0, -15 cos t, -15 sin t> / sqrt(289), tau = -15/289
D. B(t) = <8, 15 sin t, -15 cos t> / sqrt(289), tau = 15/289

Compute the unit binormal vector and torsion of the curve.
r(t) = <15t, 8 cos t, 8 sin t>
A. B(t) = <8, -15 sin t, 15 cos t> / sqrt(289), tau = 15/289
B. B(t) = <0, 15 cos t, -15 sin t> / sqrt(289), tau = 1/289
C. B(t) = <0, -15 cos t, -15 sin t> / sqrt(289), tau = -15/289
D. B(t) = <8, 15 sin t, -15 cos t> / sqrt(289), tau = 15/289

Added by Kelsey P.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sutara K

03/21/2022

Step 1

Step 1: Find the unit binormal vector \( \mathbf{B} \): Given the tangent vector \( \mathbf{T} = (15, -8\sin t, 8\cos t) \), we need to find the magnitude of \( \mathbf{T} \): \[ |\mathbf{T}| = \sqrt{15^2 + (-8\sin t)^2 + (8\cos t)^2} = \sqrt{225 + 64\sin^2 t + Show more…

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Compute the unit binormal vector and torsion of the curve. r(t) = <15t, 8 cos t, 8 sin t> A. B(t) = <8, -15 sin t, 15 cos t> / sqrt(289), tau = 15/289 B. B(t) = <0, 15 cos t, -15 sin t> / sqrt(289), tau = 1/289 C. B(t) = <0, -15 cos t, -15 sin t> / sqrt(289), tau = -15/289 D. B(t) = <8, 15 sin t, -15 cos t> / sqrt(289), tau = 15/289