Contoh soal interpolasi kubik

1. Data hubungan antara temperatur dan kedalaman suatu danau :

Temperatur (oC) Kedalaman (m)

19.1 0

19.1 -1

19 -2

18.8 -3

18.7 -4

18.3 -5

18.2 -6

17.6 -7

11.7 -8

9.9 -9

9.1 -10

Tentukan temperatur danau pada kedalaman -7.5 meter dengan menggunakan Interpolasi Kubik

(Menggunakan Operasi Matrik) !

Penyelesaian :

Persamaan umum Metode Langsung (Direct Method) Interpolasi Kubik :

T (−9)=9. 9T (−8)=11. 7T (−7 )=17 . 6T (−6 )=18. 2

z=−9z=−8z=−7z=−6

T ( z )=a0+a1 z+a2 z2+a3 z

3

9 .9=a0−9a1+81a2−729a3 (1). .. .. . .. .. . .. .. .

T (−9)=a0+a1 (−9 )+a2(−9)2+a3 (−9)3

. .. .. . .. .. . .. .. . (2)

T (−8)=a0+a1 (−8 )+a2(−8)2+a3 (−8)3

11. 7=a0−8a1+64a2−512a3

9.9729819 3210 aaaa

7.11512648 3210 aaaa

6.17343497 3210 aaaa

2.18216366 3210 aaaa

)1(

)2(

)3(

)4(

T (−6 )=a0+a1 (−6 )+a2(−6 )2+a3 (−6 )3

(3 )

. .. .. . .. .. . .. .. .

. .. .. . .. .. . .. .. .

(4 )18 .2=a0−6 a1+36 a2−216a3

T (−7 )=a0+a1 (−7 )+a2(−7)2+a3 (−7 )3

17 . 6=a0−7a1+49a2−343a3

[1 −9 81 −7291 −8 64 −5121 −7 49 −3431 −6 36 −216

]×[a0

a1

a2

a3]=[ 9.9

11. 717 .618 .2

]

[a0

a1

a2

a3]=[ −615 .9

−262 .583−35 .55−1 .567

]

[a0

a1

a2

a3]=[1 −9 81 −729

1 −8 64 −5121 −7 49 −3431 −6 36 −216

]−1

×[ 9 .911.717 .618 .2

]

[a0

a1

a2

a3]=[ −56 189 −216 84

−24 .333 79 .5 −87 31.833−3.5 11 −11.5 4

−0.1667 0.5 −0 .5 0 .167]×[ 9 .9

11.717 .618 .2

]9 .9=a0−9a1+81a2−729a3

Dengan demikian, temperatur danau pada kedalaman -7.5 meter adalah :

[a0

a1

a2

a3]=[ −615 .9

−262 .583−35 .55−1 .567

]

a0=−615 .9a1=−262 . 583a2=−35 . 55

a3=−1. 5637

Jadi :

T (−7 . 5 )=a0+a1z+a2 z2+a3 z

3

T (−7 . 5 )=−615 . 9−262 .583 (−7 .5 )−35 .55 (−7 . 5)2−1 .567 (−7 . 5 )3

T (−7 . 5 )=−615 . 9+1969 .3725−1999. 6875+661 . 078125

T (−7 . 5 )=14 . 863

T (−7 . 5 )=−615 . 9−262 .583 (−7 .5 )−35 .55 (56 . 25 )−1 .567 (−421. 875 )