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7/21/2019 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)
7/21/2019 FunCao Vetorial
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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)
7/21/2019 FunCao Vetorial
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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)
7/21/2019 FunCao Vetorial
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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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