ExEdsonAmanda26-01-2021, R

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# Gráfico em 3 dimensões

# Curva
# Superfície z=f(x,y)

x <- seq(-5,2, by=0.05)
y <- seq(-2,2,by=0.05)
mf<-function(s,r){
x=s
y=r
f1=y*x^3+y^4+2*x*y+x+y+1 
f2=y*x^2+2*x*y+x+5
f1^2+f2^2 # |f(x,y)|^2; auxiliar para gráfico
}

require(grDevices) # for trans3d
z <- outer(x, y, mf)
z[is.na(z)] <- 4
op <- par(bg = "white")
persp(x, y, z, theta = 60, phi = 20, expand = 1)
contour(x,y,z,levels=c(0.5,1,5,10))

#
persp(x, y, z, theta = 60, phi = 20, expand = 1)->res

# Método de Euler oara resolver a equação u'(t)=-grad f(u(t)), e(0)=(2,-2)

gradmf<-function(u){ # gradiente de f(x,y)
s=u[1]
r=u[2]
h=10^(-5)
dnfx= mf(s+h,r)-mf(s-h,r)
dnfy=mf(s,r+h)-mf(s,r-h)
c(dnfx,dnfy)/(2*h)
}

t0=0  # tempo inicial
tf=1  # t final 
e0=c(2,-2) # condição inicial 
mf(e0[1],e0[2]) # teste da escolha
n=1000
h=(tf-t0)/n  # Tamanho do passo

tt=seq(t0,tf,by=h)

Y=matrix(0,2,length(tt))
Y[,1]=e0

for ( i in 1:(length(tt)-1)){
Y[,i+1]=Y[,i]-h*gradmf(Y[,i])
}

print("Aproximação para o ponto de máximo"); Y[,length(tt)]
print("Aproximação para o valor máximo"); mf(Y[1,length(tt)],Y[2,length(tt)])

lines(trans3d(Y[1,],Y[2,],mf(Y[1,],Y[2,]), res), col = "blue", lwd = 2) # Inclui curvas

plot(Y[1,],Y[2,],col="blue")

plot(tt[1:25],mf(Y[1,1:25],Y[2,1:25]), col = "blue")

[ + ] Show input
Absolute running time: 2.28 sec, cpu time: 1.25 sec, memory peak: 56 Mb, absolute service time: 2,53 sec