Question

More than one option could be correct.

Question 3: (4 pts) Fill in the circle next to each vector-valued function that parameterizes the ellipse 4.x2 + y2 = 4 in the xy-plane. A Fi(t)= (t, V4 – 412) B r2(t) = (2 cos(t), sin(t)) © F3(t) = (cos(t), 2 sin(t)) D F4(t) = (cos(t), 2 sin(2t))

Answer 1

3 (t) = 444) Ĉ + y(t) {(t) = < Kitt), y t)) Ellipse 4x+y+= 4 . LHS = 4x++y2 R15 = 4 LMS means Left hand side RHS means Right hand side Vector valued function will parameterise when

X(t), yt) of {(t)= < htt), yt) makes LHS = RHS ® {(t) = <t, J4-47²) 1 + = + . g(t) = J4-4+2 . Putting these in LHS LHS= 48²42 Lus= 4+'+(04-44)

'Lus= 4*+ 4-44 LHS= 4 LHS=KHS so; option A is correct (B) 32(t)=(2cost, seint> 744)= 2lost y(t) = sint peeting these in Las

LHS= 4x²+ y2 215 = 4(2cost)? + Esentj2 Ius = 414 cos't) + sin til LH S=16 cos'tt sinit Lus=15 cost + cos²tt sin't LHS=15cost til

(senza . tcostA=1] so, LKS & RHS. so, option B is corong e az(t) = 5 cost, 2 sint) ut) = cost y(t) = sint putting these in LHS LHS= 4x²+42

2 LHS = 4(cost)+(2 sin t) LUS= 4 cost + 4 sin't LHS = 4 (cost tsinnt) LKS= 4 tsin' At Costa -17 LHS= RHS. so, option c is correct correct () Tact) = < cosct), 2 sin(2t)) a(t) = cost

ЧЕ) = 2 (26) рина Илих лh Lux LHS = 4x+y2 Lus = 4 (ct) + 2 даn(21)] LHS= 4 cos²t + 4 sin 2t

LHS = 4 cost t 4 sin2t LHs= 4 [cost + sen 22t] Lms=4[cost + [sen 2t)) L95 = 4 [ cos't + (Aseint cost)) LHs = 4 [ coo*t + 4 sein't cooty LH6 = 4 205²t [ + 4 sent] LASERHS: so, option () is wrong. uro

so, option (A) and option() are correct. VA) Rit)= <t,d4-46) ver) R35+) =< cost, 2sist>