Calculo

1 -Das integrais duplas abaixo, calculeos seus respectivos limites?
A)0∫3 1∫2(1+8xy)dydx=
0∫3{1.dy+1∫28xy.dy}dx= 0∫3 {(y)21+8x(y2/2)21}dx=
0∫3 [(2-1)+8/2.x.(22-12)]dx= 0∫3 [1+4.x.(3)]dx= 0∫3 [1+12x]dx= 0∫3 1.dx+0∫312.x2/2= [(3-0)+12/2.(32-02)]= 3+6.9= 57
B) 1∫2 0∫3 x2 y dxdy=
1∫2 y{[x2]30 dy= 1∫2 [y/3.(33-03)]dy= 1∫2 [y/3.27]dy=
27/31∫2 y.dy= 27/3.[y2/2]21= 9/2.(22-12)4,5.3= 13,5
C) 1∫3 0∫1(1+4xy)dxdy=
1∫3 {[1.dx]10+4.y[x.dx]10}dy= 1∫3{[x]10+4.y[x2/2]10}dy=
1∫3 {(1-0)+4/2.y(12-02)}dy= 1∫3 1+2.y.(1)}dy= 1∫3 [1+2y]dy= 1∫3 1.dy
+2/2.[y2]=> [y]31+2/2.[y2]31= (3-1)+1.(32-12)= 2+8= 10
D) 1∫π/2 0∫π/2(senX.cosY)dydx=
1∫π/2senX1∫π/2{cosY.dy}dx= 1∫π/2sen X[senY]π/20 dx= 1∫π/2senX(sem 90°-sem 0)dx= 1∫π/2senX.(1-0)dx= 1∫π/2senX.dx1∫π/2-cosX dx= [-cosX]π/20= -(cos 90 – cos 0)= -(0-1)= -(-1)= 1
E) 1∫4 0∫2(x+√y)dxdy=
1∫4{0∫2x.dx+0∫2√ydx}dy=>1∫4{[x2/2]20+√y[x]20}dy=>
1∫4{[22/2-02/0]+√y[2-0]}dy=>1∫4{2+2√y]}dy=>1∫42.dy+1∫42√ydy=> 2[y]41+21∫2y1/2dy=> 2[4-1]+2[y3/2/3/2]41=> 6+2[√y3/3/2]41=> 6+2{1/2/3[√43-√13]}=> 6+2{2/3[8-1]}=> 6+2(0,67.7)=> 6+2.(4,69)=> 6+9,38= 15,38
F) 1∫4 1∫2(x/y+y/x)dydx=
1∫4[x.ln(y)+1/x.y2/2]21dx=>1∫4{x.(ln|2|-ln|1|]+1/2x[22-12]}dx=>1∫4{[x(0,69-0)]+1/2x.3}dx=>1∫40,69x.dx+3/21∫41/x.dx=> 0,69/2[x2]41+3/2[ln|x|]41=> 0,69/2(42-12)3/2(ln|4|-ln|1|)=> 0,69/2.15+3/2.1,38=> 5,18+2,07= 7,26
2 – Dada a função f(x,y)= X4-2XY-14, determine?
A) A função matemática do seu gradiente; f(x)= 4x3-2y e f(y)= -2x
Vf(2,1)=(4x3-2y, -2x)
B) O gradiente da função no ponto P(2,1);
Vf(2,1)= (4.(23)-2.(1), -2.(2)=> Vf(2,1)= 32-2, -4=>Vf(2,1)= (30,-4)
C) O módulo do gradiente desta função no ponto P(2,1)
//Vf//(2,1)=√a2+b2= √302+(-4)2= √916= 4.√229
3 – Dada a função: f(x,y)= X4-2XY3-14, determine a derivada direcional da função no ponto P(2,1), na direção do vetor V=(3,4)
Solução do 1º gradiente: df/dx(x,y)= 4x3-2y3 df/dy(x,y)= -6xy2
Vf(x,y)=(4x3-2y3. -6xy2)
Solução do 2º gradiente:
//√//= √a2+b2= √32+42= √25= 5 //√//=3/5,4/5)
df/dv=