FunCao Vetorial

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    Rn

    n = 2

    3

    (t)

    R2

    R3

    (t)

    (t) = (x(t), y(t))

    (t) = (x(t), y(t), z(t))

    I

    II

    x(t)

    y(t)

    z(t)

    t

    x y

    z

    t

    t

    t

    (t)

    t

    t

    (t)

    (t)

    x= x(t) y= y(t)

    z= z(t)

    t

    t

    : A R Rn n= 2 3 A

    (t) = (x(t), y(t), z(t))

    x(t)

    y(t)

    z(t)

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    (A) ={(t) Rn | tA}

    (t) = (2 +t, 1 +t2)

    t[1, 4]

    (t) = (2t + 1, 1 t)

    (3, 0)(R) (2, 1) /(R)

    (t) = (2 +t, 1 +t2)

    t[1, 4]

    (t) = (2t+ 1, 1 t)

    (t) = (2+t, 1+t2)

    t[1, 4]

    (t) = (2t+ 1, 1 t)

    (t) = (2 +t, 1 +t2)

    t[1, 4]

    (t) = (2t+ 1, 1 t)

    (t) = (2 t, 2t+ 1, 3t)

    (2, 1, 0)

    (0, 3, 3)

    (R)

    (t) = (2 t, 2t+ 1, 3t)

    (t) = (1 t)A + tB

    ([0, 1])

    A= (0, 1) B = (1, 3) A= (1, 1) B = (1, 3) A= (1, 0, 2) B = (2, 2, 3)

    (0) = (1,1)

    (1) = (2, 3)

    (t) = (2cos(t), 2sen(t))

    t R

    1

    1(t) = (cos(t), sen(t)) 2(t) = (cos(2t), sen(2t))

    3(t) = (cos(at+b), sen(at+b)), a= 0 4(t) = (sen(t), cos(t))

    (t) = (44t, 2t) (t) = (2 +4t, 12t)

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    (t) = (2 + cos(t), 1 + sen(t))

    (t) = (t3, t2)

    (t) = (t2, t)

    (t) = (t2, t4)

    (t) = (2+ cos(t), 1+sen(t))

    (0)

    (

    6)

    (4

    ) (

    3)

    (

    2)

    (t) = (t2, t)

    (0)

    (1)

    (1)

    (

    2)

    (2)

    (t) = (t3, t2)

    (0)

    (1)

    (1)

    ( 6

    4)

    ( 64)

    (t) = (t2, t4)

    (0)

    (1)

    (

    1)

    (

    2)

    (2)

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    (t) = (cos(t), 1, sen(t))

    (t) = (cos(2t), t, sen(2t))

    (t) = (1 + 3t, 1, 1 +t)

    (t) = (3 + 3t,3t, 3 + 2t)

    t= 0

    (t) = (cos(t), 1, sen(t))

    x 12

    =y+ 1

    3 =

    z 12

    x2 y2 = 1

    1669

    1690

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    1696

    XVII

    1697

    P

    C

    s

    C x=|OT||PQ|

    y=|CT| |CQ|

    s

    X

    C (0, r)

    P

    P = (x, y)

    C

    OT

    T P

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