(a) Find the unit tangent and unit normal vectors T(t) and N(t). (b) Use Formula 9 to find the curvature.
r(1) = (t , 1/2t2, t2)

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

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