Question

1) For this problem use the following space curve: F(t) =< t, 3 sin(t), 3 cos(t) > a) Determine the unit tangent vector: T. b) Determine the unit normal vector: Ñ. c) Determine the curvature of this space curve at the point: (0,0,3). d) Determine the arc length of the curve between t = 0 and t = 1.

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Answer 1

F(t) = <t, 3 sin (t) can be written as, rlt) t 3 cos(t) > 3 sin Lt) i + 3 cos (t) as given Unit tangent vector TE UTV IrtL Taking derivaltre of ret) rilta <1, 3.cost, -3 sint > 2 Ir'lt) a 12+ (3 cost) + (-3 sint) +9052t +9 slo't

Irltl [1+g Italcos²t + slot) JO T(t) < 1, 3 cost - sint Vio T(t) = L To I cost 3 cost, 3 sint -3 sint 77 b) Unit normal vector is given as UN V Ñ (t) = T'lt) IT'(+) Taking destrative of unit tangent vector T'(t) co -3 sint Jo -3 cost Vio T'lt) = + +-3 slot)i + (-3.cos the JTO a inttg lost 10 10 19

Ñ (t) = co, 3 sint, - 3 To cost) 3/To. Ñ (+) Co-sint - cost) 7 (Ans) Ñ (+) co-slot, - cost> cerve at co,0z) Curvature of space Curvature is diren dT dr dTldt dr Idt I di Idth I dr idth ko -3 sint y cost at Jio To JTO Hart 3 cost, s sint STO Curvature 3150 3 10 JIO

d) An length blw teo & tel Arc length stt) Ir't I dt a b 2 dt +(de at We found trilt) = 10 (already) Are length, To at to it) - STO (1-0) To (Ans)