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

4.

•

1

√

3

y

x

•

ZZ

D



1

x

2

+

1

y

2



dxdy

=

√

3

Z

1

dy

x

=y

2

+1

Z

x

=y



1

x

2

+

1

y

2



dx

•

ZZ

D



1

x

2

+

1

y

2



dxdy

= ... =

√

3

Z

1

y



−

1

x

+

x

y

2



x

=y

2

+1

x

=y

dy

•

ZZ

D



1

x

2

+

1

y

2



dxdy

= ... =

√

3

Z

1



1 +

1

y

2

−

1

y

2

+ 1



dy

•

ZZ

D



1

x

2

+

1

y

2



dxdy

= ... =

2

√

3

−

π

12

5.

• rysunek

•

|Σ| =

ZZ

D

s



∂z

∂x



2

+



∂z
∂y



2

+ 1 dxdy

•

∂z

∂x

=

−x

p

4 − x

2

− y

2

,

∂z
∂y

=

−y

p

4 − x

2

− y

2

•

|Σ| =

ZZ

D

2 dxdy

p

4 − x

2

− y

2

, D : x

2

+ y

2

¬ 3

•

|Σ| =

ZZ

∆

2ρ dϕdρ

p

4 − ρ

2

, ∆ = [0, 2π] ×

h

0,

√

3

i

•

|Σ| =

2π

Z

0

dϕ

√

3

Z

0

2ρ dρ

p

4 − ρ

2

•

|Σ| = 4π

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

cos

2

t





• C

′

1

(t) = − cos

2

t

sin t, C

′

2

(t) = cos

3

t

• C

1

(t) =

1
3

cos

3

t

, C

2

(t) = sin t



1 −

1
3

sin

2

t



• ϕ(t) =

1
3

cos

4

t

+



1 −

1
3

sin

2

t



sin

2

t

=

1
3



1 + sin

2

t



• y(t) = C

1

cos t + C

2

sin t +

1
3



1 + sin

2

t



4

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