Answer:
a. i. (i + tj + 2tk)/√(1 + 5t²)
ii. (-5ti + j + 2k)/√[25t² + 5]
b. √5/[√(1 + 5t²)]³
Step-by-step explanation:
a. The unit tangent
The unit tangent T(t) = r'(t)/|r'(t)| where |r'(t)| = magnitude of r'(t)
r(t) = (t, t²/2, t²)
r'(t) = dr(t)/dt = d(t, t²/2, t²)/dt = (1, t, 2t)
|r'(t)| = √[1² + t² + (2t)²] = √[1² + t² + 4t²] = √(1 + 5t²)
So, T(t) = r'(t)/|r'(t)| = (1, t, 2t)/√(1 + 5t²) = (i + tj + 2tk)/√(1 + 5t²)
ii. The unit normal
The unit normal N(t) = T'(t)/|T'(t)|
T'(t) = dT(t)/dt = d[ (i + tj + 2tk)/√(1 + 5t²)]/dt
= -5ti/√(1 + 5t²)⁻³ + [-5t²j/√(1 + 5t²)⁻³] + [-10tk/√(1 + 5t²)⁻³]
= -5ti/√(1 + 5t²)⁻³ + [-5t²j/√(1 + 5t²)⁻³] + j/√(1 + 5t²)+ [-10t²k/√(1 + 5t²)⁻³] + 2k/√(1 + 5t²)
= -5ti/√(1 + 5t²)⁻³ - 5t²j/[√(1 + 5t²)]⁻³ + j/√(1 + 5t²) - 10t²k/[√(1 + 5t²)]⁻³ + 2k/√(1 + 5t²)
= -5ti/√(1 + 5t²)⁻³ - 5t²j/[√(1 + 5t²)]⁻³ - 10t²k/[√(1 + 5t²)]⁻³ + j/√(1 + 5t²) + 2k/√(1 + 5t²)
= -(i + tj + 2tk)5t/[√(1 + 5t²)]⁻³ + (j + 2k)/√(1 + 5t²)
We multiply by the L.C.M [√(1 + 5t²)]³ to simplify it further
= [√(1 + 5t²)]³ × -(i + tj + 2tk)5t/[√(1 + 5t²)]⁻³ + [√(1 + 5t²)]³ × (j + 2k)/√(1 + 5t²)
= -(i + tj + 2tk)5t + (j + 2k)(1 + 5t²)
= -5ti - 5²tj - 10t²k + j + 5t²j + 2k + 10t²k
= -5ti + j + 2k
So, the magnitude of T'(t) = |T'(t)| = √[(-5t)² + 1² + 2²] = √[25t² + 1 + 4] = √[25t² + 5]
So, the normal vector N(t) = T'(t)/|T'(t)| = (-5ti + j + 2k)/√[25t² + 5]
(b) Use Formula 9 to find the curvature.
The curvature κ = |r'(t) × r"(t)|/|r'(t)|³
since r'(t) = (1, t, 2t), r"(t) = dr'/dt = d(1, t, 2t)/dt = (0, 1, 2)
r'(t) = i + tj + 2tk and r"(t) = j + 2k
r'(t) × r"(t) = (i + tj + 2tk) × (j + 2k)
= i × j + i × 2k + tj × j + tj × 2k + 2tk × j + 2tk × k
= k - 2j + 0 + 2ti - 2ti + 0
= -2j + k
So magnitude r'(t) × r"(t) = |r'(t) × r"(t)| = √[(-2)² + 1²] = √(4 + 1) = √5
magnitude of r'(t) = |r'(t)| = √(1 + 5t²)
|r'(t)|³ = [√(1 + 5t²)]³
κ = |r'(t) × r"(t)|/|r'(t)|³ = √5/[√(1 + 5t²)]³
