Pengerjaan 23 Oktober 2019

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Revision as of 17:53, 23 October 2019 by Alesdaniel (talk | contribs) (Created page with " ==1A== w4 = 4 t1 = 1 t2 = 2 t3 = 3 a = [[w4,-t2,-t3], \ [w4,-t1,-t3], \ [w4,-t1,-t2]] b = [[t1], \ [t2], \ [t3]] def gaussElimin(a,b): n = len...")
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1A

w4 = 4

t1 = 1

t2 = 2

t3 = 3

a = [[w4,-t2,-t3], \

    [w4,-t1,-t3], \
    [w4,-t1,-t2]]

b = [[t1], \

    [t2], \
    [t3]]


def gaussElimin(a,b):

   n = len(b)
   for i in range(0,n-1):
       for k in range(i+1,n,i+1):
           if a[i][k] != (0.0):
               a[i][k] = a[i][k]/a[i-1][k]
               lam = len(b)
   for i in range(n,k):
       b = a[i][k] - lam*(a[i][k]/a[i+1][k])
       return b

print(gaussElimin(a,b))

Hasil yang didapatkan adalah none; kesalahan dikarenakan kesalahan pada algoritma yang sudah dilakukan.


2A

h = 0.1

v = 0

a = 2

Cd = 0.01

m = 100

def diff_t2(t2,v):

 return((v2-v)*((m)/((Cd*(v2)**3)**0.5))

def rungeKutta(t2,v,v0,h):

 v = v0
 k1 = h*diff_t2(v0+h*t2)
 k2 = h*diff_t2(v0+0.5*k1*h+v2)
 k3 = h*diff_t2(v0+0.5*k2*h+v2)
 k4 = h*diff_t2(v0+k3*h+v2)
 t2 = t0 + (1/6)*(k1+2*k2+2*k3+k4)

return(t2) print('waktu yang diperlukan', rungeKutta(v2,v0,t2,h))

Hasilnya adalah syntax error, dikarenakan masih ada kesalahan input algoritma yang dikerjakan, terutama pada penulisan define persamaan rungeKutta(t2,v,v0,h)