Lista 4 (1)

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Universidade Federal do Paraná - UFPR
Centro Politécnico
Departamento de Matemática
Disciplina: Cálculo 2 Código: CM312 Semestre: Semestre 2024/2
Lista 4
1. Determine as derivadas parciais indicadas.
(a) f(x, y) = x3y5; fx(3,−1) (b) f(x, y) =
√
2x+ 3y; fy(2, 4)
(c) f(x, y) = xe−y + 3y;
∂f
∂y
(1, 0) (d) f(x, y) = sen(y − x);
∂f
∂y
(3, 3)
(e) z =
x3 + y3
x2 + y2
,
∂z
∂x
,
∂z
∂y
(f) z = x
√
y − y√
x
;
∂z
∂x
,
∂z
∂y
(g) z =
x
y
+
y
x
;
∂z
∂x
(h) z = (3xy2 − x4 + 1)4;
∂z
∂x
,
∂z
∂y
(i) u = xy sec(xy);
∂u
∂x
(j) u =
x
x+ t
;
∂u
∂x
,
∂u
∂t
(k) f(x, y, z) = xyz; fy(0, 1, 2) (l) f(x, y, z) =
√
x2 + y2 + z2; fz(0, 3, 4)
(m) u = xy + yz + zx; ux, uy, uz (n) u = x2y3t4; ux, uy, ut
2. Determine as derivadas parciais de primeira ordem da função.
(a) f(x, y) = x3y5 − 2x2y + x (b) f(x, y) = x2y2(x4 + y4)
(c) f(x, y) = ex tg(x− y) (d) f(s, t) = s/
√
s2 + t2
(e) g(x, y) = y tg(x2y3) (f) g(x, y) = ln(x+ ln y)
(g) f(x, y) = exy cosx sen y (h) f(s, t) =
√
2− 3s2 − 5t2
(i) f(u, v) = arctg(u/v) (j) f(x, t) = esen(t/x)
(k) z = ln
(
x+
√
x2 + y2
)
(l) z = xx
y
(m) f(x, y) =
∫ y
x e
t2dt (n) f(x, y) =
∫ x
y e
t/tdt
(o) f(x, y, z) = xyz (p) f(x, y, z) = xey + yez + zex
(q) u = z sen y
x+z (r) u = xy2z3 ln(x+ 2y + 3z)
(s) u = xy
z
(t) f(x, y, z, t) = x−y
z−t
(u) f(x, y, z, t) = xy2z3t4
3. Use a derivação impĺıcita para encontrar ∂z/∂z e ∂z/∂y.
(a) xy + yz = xz (b) xyz = cos(x+ y + z)
(c) x2 + y2 − z2 = 2x(y + z) (d) xy2z3 + x3y2z = x+ y + z
4. Determine ∂z/∂x e ∂z/∂y se z = f(ax+ by).
5. Determine todas as derivadas parciais de segunda ordem .
1
(a) f(x, y) = x2y + x
√
y (b) f(x, y) = sen(x+ y) + cos(x− y)
(c) z = (x2 + y2)3/2 (d) z = cos2(5x+ 2y)
(e) z = t arcsen
√
x (f) z = xln t
6. Verifique que a conclusão do Teorema de Clairaut é válida, isto é, uxy = uyx.
(a) u = x5y4 − 3x2y3 + 2x2 (b) u = sen2 x cos y
(c) u = arcsen(xy2) (d) u = x2y3z4
7. Determine as derivadas parciais indicadas.
(a) f(x, y) = x2y3 − 2x4y; fxxx (b) f(x, y) = exy
2
; fxxy
(c) f(x, y, z) = x5 + x4y4z3 + yz2; fxyz (d) f(x, y, z) = exyz; fyzy
(e) z = x sen y; ∂z
∂y2∂x
(f) z = ln sen(x− y); ∂z
∂y∂x2
(g) u = ln
(
x+ 2y2 + 3z3
)
; ∂3u
∂x∂y∂z
8. Se f e g são funções duas vezes deriváveis de uma única variável, mostre que a função
u(x, y) = xf(x+ y) + yg(x+ y)
satisfaz a equação uxx − 2uxy + uyy = 0.
9. Mostre que a função
f(x, · · · , xn) = (x21 + · · ·xn)(2−n)/2
satisfaz a equação
∂2f
∂x21
+ · · ·+ ∂2f
∂x2n
= 0
.
10. Determine uma equação do plano tangente à superf́ıcie no ponto especificado.
(a) z = x2 + 4y2, (2, 1, 8) (b) z = x2 − y2, (3,−2, 5)
(c) z = 5 + (x− 1)2 + (y + 2)2, (2, 0, 10) (d) z = xy, (−1, 2,−2)
(e) z =
√
x− y, (5, 1, 2) (f) z = y2 − x2, (−4, 5, 9)
(g) z = sen(x+ y), (1,−1, 0) (h) z = ln(2x+ y), (−1, 3, 0)
(i) z = ex ln y, (3, 1, 0)
11. Desenhe a superf́ıcie e o plano tangente no ponto dado. ( Escolha o domı́nio e o ponto de vista de modo a ver
tanto a superf́ıcie quanto o plano tangente.) Em seguida, dê zoom até que a superf́ıcie e o plano tangente se
tornem indistingúıveis.
(a) z = xy, (−1, 2,−2) (b) z =
√
x− y, (5, 1, 2)
12. Explique por que a função é diferenciável no ponto dado. Em seguida, encontre a linearização L(x, y) da função
naquele ponto.
(a) f(x, y) = y lnx, (2, 1) (b) f(x, y) =
√
1 + x2y2, (0, 2)
13. Determine a diferencial da função.
2
(a) z = x2y3 (b) v = ln(2x− 3y)
(c) w = x sen(yz) (d) z = x4 − 5x2y + 6xy3 + 10
(e) z =
1
x2 + y2
(f) z = yexy
(g) z = ex cos(xy) (h) w = x2y + y2z
(i) w =
x+ y
y + z
14. Use diferenciais para aproximar o valor de f em um dado ponto.
(a) f(x, y) =
√
20− x2 − 7y2, (1, 95; 1, 08) (b) f(x, y) = ln(x− 3y), (6, 9; 2, 06)
(c) f(x, y, z) = x2y3z4, (1, 05; 0, 9; 3, 01) (d) f(x, y, z) = xy2 sen(πz), (3, 99; 4, 98; 4, 03)
15. Use diferenciais para aproximar o número.
(a) 8, 94
√
9, 99− (1, 01)3 (b) (
√
99 + 3
√
124)4
(c)
√
0, 99e0,02 (d)
√
(3, 02)2 + (1, 97)2 + (5, 99)2
Respostas:
1. (a) −27
(b) 3
8
(c) 2
(d) −1
(e) x4+3x2y2−2xy3
(x2+y2)2
, 3x2y2+y4−2yx3
(x2+y2)2
(f)
√
y + y
2x3/2 , x
2
√
y −
1√
x
(g) 1
y −
y
x2
(h) 4(3xy2−x4 + 1)3(3y2− 4x3), 24xy(3xy2−x4 + 1)3
(i) y sec(xy)[1 + xy tg(xy)]
(j) t
(x+t)2
, − x
(x+t)2
(k) 0
(l) 4
5
(m) y + z, x+ z, y + x
(n) 2xy3t4, 3x2y2t4,4x2y3t3
2. (a) fx(x, y) = 3x2y5 − 4xy + 1, fy(x, y) = 5x3y4 − 2x2
(b) fx(x, y) = 6x5y2 + 2xy6, fy(x, y) = 6y5x2 + 2yx6
(c) ex[tg(x− y) + sec2(x− y)], fy(x, y) = −ex sec2(x− y)
(d) fs(s, t) = t2
(t2+s2)3/2
, ft(s, t) = − st
(t2+s2)3/2
(e) gx(x, y) = 2xy4 sec2(x2y3), gy(x, y) = tg(x2y3) + 3x2y3 sec2(x2y3)
(f) gx(x, y) = 1
x+ln y , gy(x, y) = 1
y(x+ln y)
(g) fx(x, y) = exy sen y(y cosx− senx), fy(x, y) = exy cosx(x sen y − cos y)
(h) fs(s, t) = − 3s√
2−3s2−5t2 , ft(s, t) = − 5t√
2−3s2−5t2
(i) fu(u, v) = v
u2+v2
, fv(u, v) = − u
u2+v2
(j) fx(x, t) = −t cos
(
t
x
)
esen(
t
x )
x2 , ft(x, t) = esen(
t
x )
x cos
(
t
x
)
(k) ∂z
∂x = 1√
x2+y2
, ∂z
∂y = y
x
√
x2+y2+x2+y2
(l) ∂z
∂x = xy−1xx
y
(1 + y lnx), ∂z
∂y = xx
y+y( lnx)2
(m) fx(x, y) = −ex2
, fy(x, y) = ey
2
(n) fx(x, y) = ex
x , fy(x, y) = − ey
y
(o) fx(x, y, z) = yzxyz−1, fy(x, y, z) = zxyz lnx, fz(x, y, z) = yxyz lnx
3
(p) fx(x, y, z) = ey + zex, fy(x, y, z) = xey + ez, fz(x, y, z) = yez + ex
(q) ∂u
∂x = −yz
(x+z)2
cos
(
y
x+z
)
, ∂u
∂y = z
x+z cos
(
y
x+z
)
, ∂u
∂z = sen( y
x+z )− yz
(x+z)2
cos
(
y
x+z
)
(r) ux = y2z3[ln(x+ 2y + 3z)+ x
x+2y+3z ], uy = 2xyz3[ln(x+ 2y + 3z)+ y
x+2y+3z ], uz = 3xy2z2[ln(x+ 2y + 3z)+
z
x+2y+3z ]
(s) ux = yzxy
z−1, uy = xy
z
yz−1z lnx, uz = xy
z
yz lnx ln y
(t) fx(x, y, z, t) = 1
z−t , fy(x, y, z, t) = − 1
z−t , fz(x, y, z, t) = y−x
(z−t)2 , ft(x, y, z, t) = x−y
(z−t)2
(u) fx(x, y, z, t) = y2z3t4, fy(x, y, z, t) = 2xy2z3t4, fz(x, y, z, t) = 3xy2z2t4, ft(x, y, z, t) = 4xy2z3t3
3. (a) z−y
y−x , x+z
y−x
(b) − yz+sen(x+y+z)
xy+sen(x+y+z) , −
xz+sen(x+y+z)
xy+sen(x+y+z)
(c) x−y−z
x+z ,y−xx+z
(d) 1−y2z3−3x2y2z
3xy2z2+x3y2−1 , 1−2xyz3−2x3yz
3xy2z2+x3y2−1
4. af ′(ax+ by), bf ′(ax+ by)
5. (a) fxx = 2y, fxy = 2x+ 1
2
√
y , fyx = 2x+ 1
2
√
y , fyy = − x
4y3/2
(b) fxx = − sen(x + y) − cos(x− y), fxy = − sen(x + y) + cos(x− y), fyx = − sen(x + y) + cos(x− y),
fyy = − sen(x+ y)− cos(x− y)
(c) zxx = 3(2x2+y2)√
x2+y2
, zxy = 3xy√
x2+y2
, zyx = 3xy√
x2+y2
, zyy = 3(x2+2y2)√
x2+y2
(d) zxx = 50[sen2(5x+ 2y)− cos2(5x+ 2y)], zxy = 20[sen2(5x+ 2y)− cos2(5x+ 2y)], zyx = 20[sen2(5x+ 2y)−
cos2(5x+ 2y)], zyy = 8[sen2(5x+ 2y)− cos2(5x+ 2y)]
(e) zxx = t(2x−1)
4(x−x2)3/2
, zxt = 1
2
√
x−x2
, ztx = 1
2
√
x−x2
, ztt = 0
(f) zxx = (ln t)[(ln t)− 1]x(ln t)−2, zxt = x(ln t)−1 1+ln t lnx
t , ztx = x(ln t)−1 1+ln t lnx
t , ztt = xln t lnx (lnx)−1
t2
6. (a)
(b)
(c)
(d)
7. (a) −48xy
(b) 2y3exy
2
(2 + xy2)
(c) fxyz = 48x3y3z2
(d) x2z(2 + xyz)exyz
(e) − sen y
(d[f]) −2 cossec2(x− y) cotg(x− y)
(g) 72yz2
(x+2y2+3z3)3
8.
9.
10. (a) 4x+ 8y − z = 8
(b) 6x+ 4y − z = 5
(c) 2x+ 4y − z = −6
(d) 2x− y − z = −2
(e) x− y − 4z = −4
(f) z = 8x+ 10y − 9
(g) z = x+ y
(h) z = 2x+ y − 1
(i) z = e3y − e3
11. (a) (b)
12. (a) 1
2x+ (ln 2)y − 1 (b) 1
13.
4
(a) 2xy3dx + 3x2y2dy
(b) 1
2x−3y (2dx− 3dy)
(c) (sen yz)dx + (xz cos yz)dy + (xy cos yz)dz
(d) (4x3 − 10xy + 6y3)dx + (−5x2 + 18xy2)dy
(e) − 2
(x2+y2)2
(xdx + ydy)
(f) y2exydx + exy(1 + xy)dy
(g) ex(cosxy − y senxy)dx− (xex senxy)dy
(h) 2xydx + (x2 + 2yz)dy + y2dz
(i) (y+z)dx+(z−x)dy−(x+y)dz
(y+z)2
14. (a) 2, 846
(b) −0, 28
(c) 65, 88
(d) 3π
15. (a) 26, 76
(b) 49, 770
(c) 1, 015
(d) 6, 9914
5