BAB IV - Simponi MDP

BAB IV
FUNGSI VEKTOR DALAM
RUANG DIMENSI TIGA
4.1 FUNGSI VEKTOR

Fungsi Vektor dalam ruang dimensi tiga
ditentukan oleh
r(t) = f(t) i + g(t) j + h(t) k

Cara menggambar busur suatu persamaan
vektor
Substitusi nilai t dalam interval ke persamaan
vektor.
 Gambarkan
titik-titik tersebut dalam ruang
dimensi tiga.
 Hubungkan titik-titik tersebut.

4.2 KECEPATAN, PERCEPATAN,
DAN PANJANG BUSUR

Jika fungsi vektor r(t) = f(t) i + g(t) j + h(t) k
maka
kecepatan =
v(t) =
r’(t)
percepatan =
a(t) =
r”(t)
panjang busur
=
s
s

[f ' (t)]2  [f ' (t)]2  [f ' (t)]2 dt pada a  t  b
4.3 KELENGKUNGAN DAN
KOMPONEN VEKTOR

r(t) = f(t) i + g(t) j + h(t) k
di titik t = t1
| T ' (t ) |
κ=
| r ' (t ) |
r ' (t )
T (t ) =
| r ' (t ) |

Komponen Vektor

Vektor Singgung
aT

r '•r&quot;
=
| r '|
Vektor Normal
| r '&times;r&quot;|
aN =
| r '|
Contoh Soal Kelengkungan
Tentukan kelengkungan
r = (t2-1)i + (2t+3)j + (t2-4t)k
di t = 2
Solusi
r’ = 2ti + 2j + (2t-4)k
|r’| = 2√2t2-4t+5
r'
T=
| r '|
2[ti + j + (t - 2)k ]
T=
2
2 2t - 4t + 5
[ti + j + (t - 2)k ]
T=
2
2t - 4t + 5
T = [ti + j + (t-2)k] [ 2t2 - 4t + 5]-1/2
T = [t (2t2 - 4t + 5)-1/2 i] + [(2t2 - 4t + 5)-1/2 j] +
[(t-2) (2t2 - 4t + 5)-1/2 k ]
T’ = [(2t2- 4t+5)-1/2 + t(-1/2)(4t-4)(2t2 - 4t + 5)-3/2 ] i
+ [(-1/2)(4t-4)(2t2 - 4t + 5)-3/2] j
+ [(2t2- 4t+5)-1/2 + (t-2)(-1/2) (4t-4)
(2t2 - 4t + 5)-3/2] k
T’ = [(2t2- 4t+5)-1/2 + (-2t2+2t)(2t2 - 4t + 5)-3/2 ] i
+ [(-2t+2)(2t2 - 4t + 5)-3/2] j
+ [(2t2- 4t+5)-1/2+(-2t2+6t-4)(2t2 - 4t + 5)-3/2]
k
r’ = 2ti + 2j + (2t-4)k
r’(t=2) = 4i + 2j
|r’| = √20 = 2√5
T’(t=2)= [1/(5√5) i – 2/(5√5) j + 1/√5 k ]
|T’| = √1/125 + 4/125 + 1/5
|T’| = √30/125
| T '|
κ=
| r '|
30
125
κ=
2 5
κ=
30
1
125 2 5
30
κ=
50
=
30
5 52 5