A
1.
• f (x
0
+ ∆x, y
0
+ ∆y) ≈ f (x
0
, y
0
) +
∂f
∂x
(x
0
, y
0
) · ∆x +
∂f
∂y
(x
0
, y
0
) · ∆y
• f (x
0
+ ∆x, y
0
+ ∆y) =
√
0.9 ln
1.2
0.9
• f (x, y) =
√
x
ln
y
x
• x
0
= 1, ∆x = −0.1, y
0
= 1, ∆y = 0.2
•
∂f
∂x
=
1
2
√
x
ln
y
x
−
1
√
x
,
∂f
∂y
=
√
x
y
,
•
∂f
∂x
(1, 1) = −1,
∂f
∂y
(1, 1) = 1, f (1, 1) = 0
•
√
0.9 ln
1.2
0.9
≈ 0 + (−1) · (−0.1) + 1 · 0.2 = 0.3
2.
• x + y + z = 27, z = 27
− x − y, x, y, z > 0
• I(x, y) = xy(27
− x − y), x > 0, y > 0, x + y < 27
•
∂I
∂x
= y(27 − 2x − y),
∂I
∂y
= x(27 − x − 2y)
•
∂I
∂x
= 0,
∂I
∂y
= 0
⇐⇒
y
(27 − 2x − y) = 0,
x
(27 − x − 2y) = 0
⇐⇒
x
= 9,
y
= 9
•
∂
2
I
∂x
2
= −2y,
∂
2
I
∂y
2
= −2x,
∂
2
I
∂x∂y
= 27 − 2x − 2y, W = det
−2y
27−2x−2y
27−2x−2y
−2x
• W (9, 9) =
−18
−9
−9 −18
= 243 > 0,
∂
2
I
∂x
2
(9, 9) = −18 < 0 – (9, 9) maksimum lokalne
właściwe
• uzasadnienie, że jest to maksimum globalne
• I = 729
3.
• f (x) =
x
2
3x
2
− 2
=
x
2
−2
1 −
3
2
x
2
• f (x) =
−
∞
X
n
=0
3
n
2
n
+1
x
2n+2
•
3
2
x
2
<
1 ⇐⇒ |x| <
r
2
3
, R =
r
2
3
• c
17
= 0 : f
(17)
(0) = 0
• c
18
= −
3
8
2
9
: f
(18)
(0) = −
3
8
2
9
18!.
1
4.
•
−1
1
2
y
x
•
ZZ
D
xy dxdy
=
1
Z
−1
dy
x
=2−y
2
Z
x
=y
2
xy dx
•
ZZ
D
xy dxdy
= ... =
1
Z
−1
y
"
x
2
2
#
x
=2−y
2
x
=y
2
dy
•
ZZ
D
xy dxdy
= ... = 2
1
Z
−1
y
1 − y
2
dy
•
ZZ
D
xy dxdy
= ... = 0
5.
• rysunek
•
|Σ| =
ZZ
D
s
∂z
∂x
2
+
∂z
∂y
2
+ 1 dxdy
•
∂z
∂x
=
−x
p
R
2
− x
2
− y
2
,
∂z
∂y
=
−y
p
R
2
− x
2
− y
2
•
|Σ| =
ZZ
D
R dxdy
p
R
2
− x
2
− y
2
, D : x
2
+ y
2
¬ r
2
•
|Σ| =
ZZ
∆
Rρ dϕdρ
p
R
2
− ρ
2
, ∆ = [0, 2π] × [0, r]
•
|Σ| =
2π
Z
0
dϕ
r
Z
0
Rρ dρ
p
R
2
− ρ
2
•
|Σ| = 2πR
R
−
p
R
2
− r
2
6.
• λ
2
+ 1 = 0, λ
1
= i, λ
2
= −i
• y(t) = C
1
cos t + C
2
sin t + ϕ(t)
• ϕ(t) = C
1
(t) cos t + C
2
(t) sin t
•
cos t
sin t
− sin t cos t
C
′
1
(t)
C
′
2
(t)
=
0
1
sin t
• C
′
1
(t) = −1, C
′
2
(t) =
cos t
sin t
• C
1
(t) = −t, C
2
(t) = ln sin t
• ϕ(t) =
−t cos t + sin t ln sin t
• y(t) = C
1
cos t + C
2
sin t + t cos t + sin t ln sin t
2
B
1.
• f (x
0
+ ∆x, y
0
+ ∆y) ≈ f (x
0
, y
0
) +
∂f
∂x
(x
0
, y
0
) · ∆x +
∂f
∂y
(x
0
, y
0
) · ∆y
• f (x
0
+ ∆x, y
0
+ ∆y) = (1.1)
0.9
√
0.9
• f (x, y) = x
y
√
y
• x
0
= 1, ∆x = 0.1, y
0
= 1, ∆y = −0.1
•
∂f
∂x
= yx
y−1
√
y
,
∂f
∂y
= x
y
ln x
√
y
+
1
2√y
!
•
∂f
∂x
(1, 1) = 1,
∂f
∂y
(1, 1) =
1
2
, f (1, 1) = 1
• (1.1)
0.9
√
0.9 ≈ 1 + 1 · 0.1 + 0.5 · (−0.1) = 1.05
2.
• C = (x, y, 2),
|CA| =
q
(x − 1)
2
+ y
2
+ 4, |CB| =
q
x
2
+ (y + 2)
2
+ 4,
• d(x, y) =
|CA|
2
+ |CB|
2
= 2x
2
− 2x + 2y
2
+ 4y + 13,
x, y
∈ R
•
∂d
∂x
= 4x − 2,
∂d
∂y
= 4y + 4
•
∂d
∂x
= 0,
∂d
∂y
= 0
⇐⇒
4x − 2 = 0,
4y + 4 = 0
⇐⇒
x
=
1
2
,
y
= −1
•
∂
2
d
∂x
2
= 4,
∂
2
d
∂y
2
= 4,
∂
2
d
∂x∂y
= 0, W = det
4 0
0 4
• W
1
2
,
−1
= 16 > 0,
∂
2
d
∂x
2
1
2
,
−1
= 4 > 0 –
1
2
,
−1
minimum lokalne właściwe
• uzasadnienie, że jest to minimum globalne
• C =
1
2
,
−1, 2
3.
• f (x) =
x
2x
2
+ 3
=
x
3
1 −
−
2
3
x
2
• f (x) =
∞
X
n
=0
(−1)
n
2
n
3
n
+1
x
2n+1
•
−
2
3
x
2
<
1 ⇐⇒ |x| <
r
3
2
, R =
r
3
2
• c
17
=
2
8
3
9
: f
(17)
(0) =
2
8
3
9
17!.
• c
18
= 0 : f
(18)
(0) = 0
3
C
1.
• f (x
0
+ ∆x, y
0
+ ∆y) ≈ f (x
0
, y
0
) +
∂f
∂x
(x
0
, y
0
) · ∆x +
∂f
∂y
(x
0
, y
0
) · ∆y
• f (x
0
+ ∆x, y
0
+ ∆y) =
√
1.1 arc tg
0.3
1.1
• f (x, y) =
√
x
arc tg
y
x
• x
0
= 1, ∆x = 0.1, y
0
= 0, ∆y = 0.3
•
∂f
∂x
=
1
2
√
x
arc tg
y
x
+
√
x
1
1 +
y
x
2
−
y
x
2
,
∂f
∂y
=
√
x
1
1 +
y
x
2
1
x
•
∂f
∂x
(1, 0) = 0,
∂f
∂y
(1, 0) = 1, f (1, 0) = 0
•
√
1.1 arc tg
0.3
1.1
≈ 0 + 0 · 0.1 + 1 · 0.3 = 0.3
2.
• C =
x, y,
1
xy
,
• d(x, y) = x
2
+ y
2
+
1
x
2
y
2
, xy 6= 0
•
∂d
∂x
= 2x −
2
x
3
y
2
,
∂d
∂y
= 2y −
2
x
2
y
3
, x, y ∈ R
•
∂d
∂x
= 0,
∂d
∂y
= 0
⇐⇒
2x−
2
x
3
y
2
= 0,
2y−
2
x
2
y
3
= 0
⇐⇒
x
1
= 1,
y
1
= −1
∨
x
2
= 1,
y
2
= 1
∨
x
3
= −1,
y
3
= 1
∨
x
4
= −1,
y
4
= −1
•
∂
2
d
∂x
2
= 2 +
6
x
4
y
2
,
∂
2
d
∂y
2
= 2 +
6
x
2
y
4
,
∂
2
d
∂x∂y
=
4
x
3
y
3
, W = det
2+
6
x
4
y
2
4
x
3
y
3
4
x
3
y
3
2+
6
x
2
y
4
• W (1,
−1) = W (1, 1) = W (−1, −1) = W (−1, 1) = 48 > 0,
∂
2
d
∂x
2
(1, −1) =
∂
2
d
∂x
2
(1, 1) =
∂
2
d
∂x
2
(−1, −1) =
∂
2
d
∂x
2
(−1, 1) = 8 > 0 – (1, −1), (1, 1), (−1, −1), (−1, 1) minima lokalne
właściwe
• uzasadnienie, że są to minima globalne
• C
1
= (1, −1, −1), C
2
= (1, 1, 1), C
3
= (−1, −1, 1), C
4
= (−1, 1, −1)
3.
• f (x) =
x
2
4 − 2x
2
=
x
2
4
1 −
x
2
2
!
• f (x) =
∞
X
n
=0
x
2n+2
2
n
+2
•
x
2
2
<
1 ⇐⇒ |x| <
√
2, R =
√
2
• c
17
= 0 : f
(17)
(0) = 0
• c
18
=
1
2
10
: f
(18)
(0) =
18!
2
10
.
5
4.
•
1
2
y
x
•
ZZ
D
x
2
y
2
dxdy
=
2
Z
1
dy
x
=y
Z
x
=
1
y
x
2
y
2
dx
•
ZZ
D
x
2
y
2
dxdy
= ... =
2
Z
1
"
x
3
3y
2
#
x
=y
x
=
1
y
dy
•
ZZ
D
x
2
y
2
dxdy
= ... =
1
3
2
Z
1
y
−
1
y
5
dy
•
ZZ
D
x
2
y
2
dxdy
= ... =
27
64
5.
• rysunek
•
|Σ| =
ZZ
D
s
∂z
∂x
2
+
∂z
∂y
2
+ 1 dxdy
•
∂z
∂x
= x,
∂z
∂y
= y
•
|Σ| =
ZZ
D
q
x
2
+ y
2
+ 1, D : r
2
¬ x
2
+ y
2
¬ R
2
•
|Σ| =
ZZ
∆
q
ρ
2
+ 1ρ dϕdρ, ∆ = [0, 2π] × [r, R]
•
|Σ| =
2π
Z
0
dϕ
R
Z
r
q
ρ
2
+ 1ρ dρ
•
|Σ| =
2
3
π
1 + R
2
p
1 + R
2
−
1 + r
2
p
1 + r
2
6.
• λ
2
+ 1 = 0, λ
1
= i, λ
2
= −i
• y(t) = C
1
cos t + C
2
sin t + ϕ(t)
• ϕ(t) = C
1
(t) cos t + C
2
(t) sin t
•
cos t
sin t
− sin t cos t
C
′
1
(t)
C
′
2
(t)
=
0
1
cos t
• C
′
1
(t) = −
sin t
cos t
, C
′
2
(t) = 1
• C
1
(t) = ln cos t, C
2
(t) = t
• ϕ(t) = cos t ln cos t + t sin t
• y(t) = C
1
cos t + C
2
sin t + cos t ln cos t + t sin t
6
D
1.
• f (x
0
+ ∆x, y
0
+ ∆y) ≈ f (x
0
, y
0
) +
∂f
∂x
(x
0
, y
0
) · ∆x +
∂f
∂y
(x
0
, y
0
) · ∆y
• f (x
0
+ ∆x, y
0
+ ∆y) = ln
√
1.1 +
4
√
0.8 − 1
• f (x, y) = ln
√
x
+
4
√
y
− 1
• x
0
= 1, ∆x = 0.1, y
0
= 1, ∆y = −0.2
•
∂f
∂x
=
1
√
x
+
4
√
y
− 1
1
2
√
x
,
∂f
∂y
=
1
√
x
+
4
√
y
− 1
1
4
4
p
y
3
•
∂f
∂x
(1, 1) =
1
2
,
∂f
∂y
(1, 1) =
1
4
, f (1, 1) = 0
• ln
√
1.1 +
4
√
0.8 − 1
≈ 0 + 0.5 · 0.1 + 0.25 · (−0.2) = 0
2.
• C = (x, y,
−x − y), |CA| =
q
(x − 1)
2
+ (y − 2)
2
+ (x + y + 3)
2
• d(x, y) =
|CA|
2
= (x − 1)
2
+ (y − 2)
2
+ (x + y + 3)
2
, x, y ∈ R
•
∂d
∂x
= 2(x − 1) + 2(x + y + 3),
∂d
∂y
= 2(y − 2) + 2(x + y + 3)
•
∂d
∂x
= 0,
∂d
∂y
= 0
⇐⇒
2(x − 1) + 2(x + y + 3) = 0,
2(y − 2) + 2(x + y + 3) = 0
⇐⇒
x
= −1,
y
= 0
•
∂
2
d
∂x
2
= 4,
∂
2
d
∂y
2
= 4,
∂
2
d
∂x∂y
= 2, W = det
4 2
2 4
• W (
−1, 0) = 12 > 0,
∂
2
d
∂x
2
(−1, 0) = 4 > 0 – (−1, 0) minimum lokalne właściwe
• uzasadnienie, że jest to minimum globalne
• C = (
−1, 0, 1)
3.
• f (x) =
x
3 + 4x
2
=
x
3
1 −
−
4
3
x
2
• f (x) =
∞
X
n
=0
(−1)
n
4
n
3
n
+1
x
2n+1
•
−
4
3
x
2
<
1 ⇐⇒ |x| <
√
3
2
, R =
√
3
2
• c
17
=
4
8
3
9
: f
(17)
(0) =
4
8
3
9
17!.
• c
18
= 0 : f
(18)
(0) = 0
7
4.
•
1
e
2
e
1
y
x
•
ZZ
D
y
x
dxdy
=
2
Z
1
dy
x
=e
2
Z
x
=e
y
y
x
dx
•
ZZ
D
y
x
dxdy
= ... =
2
Z
1
[y ln |x|]
x
=e
2
x
=e
y
dy
•
ZZ
D
y
x
dxdy
= ... =
2
Z
1
y
(2 − y) dy
•
ZZ
D
y
x
dxdy
= ... =
2
3
5.
• rysunek
•
|Σ| =
ZZ
D
s
∂z
∂x
2
+
∂z
∂y
2
+ 1 dxdy
•
∂z
∂x
= 2x,
∂z
∂y
= 2y
•
|Σ| =
ZZ
D
q
4x
2
+ 4y
2
+ 1, D : x
2
+ y
2
¬ 1
•
|Σ| =
ZZ
∆
q
4ρ
2
+ 1ρ dϕdρ, ∆ = [0, 2π] × [0, 1]
•
|Σ| =
2π
Z
0
dϕ
1
Z
0
q
4ρ
2
+ 1ρ dρ
•
|Σ| =
π
6
5
√
5 − 1
6.
• λ
2
+ 1 = 0, λ
1
= i, λ
2
= −i
• y(t) = C
1
cos t + C
2
sin t + ϕ(t)
• ϕ(t) = C
1
(t) cos t + C
2
(t) sin t
•
cos t
sin t
− sin t cos t
C
′
1
(t)
C
′
2
(t)
=
0
sin
2
t
• C
′
1
(t) = − sin
3
t
, C
′
2
(t) = cos t sin
2
t
• C
1
(t) = cos t
1 −
1
3
cos
2
t
, C
2
(t) =
1
3
sin
3
t
,
• ϕ(t) =
1 −
1
3
cos
2
t
cos
2
t
+
1
3
sin
4
t
=
1
3
1 + cos
2
t
• y(t) = C
1
cos t + C
2
sin t +
1
3
1 + cos
2
t
8
