Matdid 998907 - esercitazione di matematica PDF

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esercitazione di matematica...

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A Giulia con la speranza che almeno nella matematica non assomigli al papà ,

✍

α>0

y(t)  2  y′ = y t y2 + 4  y(0) = 2 y(t) tα

t → +∞

Z  4 y− = y

y2 + 4 y2



dy =

Z

t dt + C

  Z  Z  2 y +4 t2 4 dy = 1 + 2 dy = +C 2 y y2

C=0 y2 − 4 − y

t2 y 2

= 0. y

2

y(t) =

t /2 ∓

q

t4 4

+ 16

2

=

t2 ∓

√

t4 + 64 . 4 y(0) = −2

y(t) = √ α=2

t4 + 64 ∼ t2

t2 +

√

t4 + 64 . 4

t → +∞

y(t) ∼

x2 2

α

y(t)

✍

t2 + t 2e2y + 6ey  y(0) = 0  

2y

y

e + 6e =

Z

2y

y′ =

y

(2e + 6e ) dy = C=7 e2y + 6ey −

Z

(t2 + t) dt =

t3 t2 − − 7 = 0. 2 3 y

(z + 3)2 = z 2 + 6z + 9 = 9 +

e = z = −3 ∓

r

y(t) = log −3 +

r

y

16 +

t3 t2 + + C. 2 3

z = ey

t3 t2 + + 7. 2 3

t3 t2 + . 2 3

t3 t2 16 + + 3 2

!

.

y(t)

✍

(

y ′ = 3ex − y 2 y(0) = 1

y(t) 2 2 2 2

x0 = 1 y ′ (0) = 3 − 0 = 3 > 0 y ′ (0) > 0 x0 = 1 y ′′ = 3ex − 2y y ′ y ′′(0) = 3 − 2y (0) y ′ (0) = 3 − 2 · 1 · 3 = −3 < 0. x0 = 1

✍

y(t) √ e−x y + 1 y = −x e +1  y(0) = 1  

′

✒  2 √ 1 1 −x y = − log(e + 1) + 2 + log 2 − 1. 2 2

y(t)

✍ (

y ′ = (e−3y + 1)(2x − 1) y(0) = −1

✒

y(t) =

h i 1 2 log (1 + e−3 )e3x −3x − 1 . 3

y(t)

✍ (

y ′ = (3 + 27y 2 ) (xe3x − 2x2 ) y(0) = 0

✒

y(t) =

✍ 2xe−3y .

  1 tan 3t e3t − e3t − 6t3 + 1 . 3

y′ =

y(t)

✍

(

y t+1 y(0) = 1

y′ =

y(1) = 2 e 2 √ 2 2 √ 2 e y(t)

✍

    y ′ − (t + 2) 1 + 1 = 0 y  y(0) = 1

y(t) 2 2 2 2 ✒ y(t)

✍

( y(t) 2 2 2 2 ✒

x0 = 0

y ′ = 3 sin t + y 2 y(0) = π

y(t)

✍

 ′′ ′   y − 6y + 9y = 3t + 2 y(0) = −1   ′ y (0) = 2

r 2 − 6r + 9 = 0 r=3 z(t) = c1 e3t + c2 t e3t c1 , c2 ∈ R

y ′ (t) = A

y(t) = At + B

0 − 6A + 9(At + B) = 3t + 2

1 A= , 3

4 B= . 9

4 1 y(t) = z(t) + y(t) = c1 e3t + c2 t e3t + t + , 3 9 c1 , c2 ∈ R 4 −1 = y(0) = c1 + ; 9

y ′ (t) = 3 c1 e3t + 3 c2 t e3t + c2 e3t +

1 3

y ′′(t) = 0

1 2 = y ′ (0) = 3c1 + c2 + . 3 c1

c1 = −

13 , 9

1 5 13 c2 = 2 − − 3c1 = + = 6. 3 3 3

y(t) = −

13 3t 1 4 e + 6 t e3t + t + . 9 9 3

y(t)

✍

 ′′ ′   y + 2y − 3y = 0 y(0) = 0   ′ y (0) = 1.

limt→+∞ y(t) = 2 2 2 +∞ 2 −∞ y ′′ + 2y ′ − 3 r=1

r 2 + 2r − 3 = 0

r = −3 y(t) = c1 et + c2 e−3t ,

c1 , c2 ∈ R 0 = y(0) = c1 + c2 y ′ (t) = c1 et − 3c2 e−3t 1 = y ′ (0) = c1 − 3c2 1 1 c1 = ; c2 = − . 4 4

1 1 y(t) = et − e−3t . 4 4 lim y(t) = +∞

t→+∞

y(t)  ′′ ′   y − y − 2y = cos(2t) y(0) = 1   ′ y (0) = 0.

✍

y ′′ − y ′ − 2y = cos(2t)

r2 − r − 2 = 0 r = −1

r=2 z(t) = c1 e−t + c2 e2t ,

c1 , c2 ∈ R y(t) = α sin(2t) + β cos(2t).

y ′ (t) = 2α cos(2t) − 2β sin(2t);

y ′′ (t) = −4α sin(2t) − 4β cos(2t).

−4α sin(2t) − 4β cos(2t) − 2α cos(2t) + 2β sin(2t) − 2α sin(2t) − 2β cos(2t) = cos(2t)

(−6α + 2β) sin(2t) + (−2α − 6β) cos(2t) = cos(2t)

(

 1  α = − −6α + 2β = 0 20 ⇔  −2α − 6β β = − 3 . 20

y(t) = −

1 3 cos(2t). sin(2t) − 20 20

y(t) = c1 e−t + c2 e2t −

y(t) =

3 1 sin(2t) − cos(2t). 20 20

11 −t 5 2t 1 3 cos(2t). e + e − sin(2t) − 20 15 12 20

y(t)

✍

 ′′ ′ −2t   y − 4y + 8y = e y(0) = −1   ′ y (0) = 0.

✒

r 2 − 4r + 8 = 0

r = 2 ∓ 2i

y(t) = c1 e2t sin(2t) + c2 e2t cos(2t). y(t) = A e−2t A = 1/20

y(t) = c1 e2t sin(2t) + c2 e2t cos(2t) +

y(t) =

1 −2t e . 20

21 2t 1 −2t 11 2t e sin(2t) − e cos(2t) + e . 10 20 20

y(t)

✍

✒

 ′′ ′   y − y − 2y = sin(2t) y(0) = 0   ′ y (0) = 1. 7 5 3 1 y(t) = − e−t + e2t − cos(2t). sin(2t) + 20 15 12 20

y ′′ − 4y ′ +

✍ 13y = 4x.

✍ 2y ′′ + 3y ′ + 4y = 0.

✍ y ′′ + 6y ′ + 8y = e4t + t2 ,

y (1) = 2,

y ′ (1) = 3.

✍ y ′ + 2(tan t)y = t,

✍

y (−1) = 4.

y(t)  ′′ ′ x   y + y − 2y = −e y(0) = 0   ′ y (0) = 0.

✒ 1 1 1 y(t) = et − e−2t − t et 9 9 3

✍ y ′ + 3(cot t)y = t,

y(−1) = −π.

✍ y ′′ + 6y ′ + 8y = e2t +

√

πt2 ,

y (1) = 0,

✍ y ′ − 6(cot t)y = t,

y(2) = π.

✍ y ′ + (tan t)y = 2t2 ,

y (1) = 3.

✍ y (3) − 2y ′′ + 5y ′ = 0. y(t) lim y(t) = π.

t→−∞

y (3) − 2y ′′ + 5y ′ = 3tet .

y ′ (−1) = 2.

✍ y ′′ − 6y ′ + 9y = 0. √ y ′′ − 6y ′ + 9y = cos( 2t).

✍ y ′′ − 4y ′ + 13y = 0.

y ′′ − 4y ′ + 13y = 1 + e2t .

✍ y ′′′ + y ′ = 0.

y ′′′ + y ′ = 1 + et .

✍ y ′′ − 2y ′ + 17y = 0.

y ′′ − 2y ′ + 17y = sin(2t).

✍ ρ(ϑ) = ϑ2 + 1,

γ 0 ≤ ϑ ≤ 2π.

γ, γ

  x(ϑ) = (ϑ2 + 1) cos ϑ  y(ϑ) = (ϑ2 + 1) sin ϑ

γ(π)

0 ≤ ϑ ≤ 2π

  x′ (ϑ) = 2ϑ cos ϑ − (ϑ2 + 1) sin ϑ  y ′ (ϑ) = 2ϑ sin ϑ + (ϑ2 + 1) cos ϑ ϕ′ (π) = (−2π, −(π 2 + 1)).

T (π) =



−2π π2 + 1 √ , −√ π 4 + 6π 2 + 1 π 4 + 6π 2 + 1



 N (π) = −√

 2π π2 + 1 ,√ . π 4 + 6π 2 + 1 π 4 + 6π 2 + 1

  x = −(π 2 + 1) − 2πt  y = −t(π 2 + 1) y=

t∈R

π2 + 1 (π 2 + 1)2 x+ . 2π 2π γ f (x) = (1/3)(2x − 1)3/2

✍ x≤1

f

1/2 ≤

γ

γ1 =

  x=t

 y = 1(2t − 1)3/2 3  x = 3 − 1t 2 2 γ2 = 1  y = (2 − t) 3 γ1′ (t) = (1,

√

1 2

≤ t ≤ 1;

1 ≤ t ≤ 2.

2t − 1). s

 2  2 1 1 1 + (2t − 1) dt + L(γ) = L(γ1 ) + L(γ2 ) = + 2 3 1/2 √ √ √ √   Z 1 √ √ 2 3/2 1 13 2 2 1 13 13 2t dt + = 2 t = = + − + 3 6 6 3 6 3 1/2 1/2 Z

p

ρ = 2θ 2

γ

✍ −π ≤ θ ≤ π,

1

γ; θ=ε

ε → 0+

Φ(t) = (ρ(t) cos t, ρ(t) sin t) = (2t2 cos t, 2t2 sin t) −π ≤ t ≤ π, Φ′ (t) = (4t cos t − 2t2 sin t, 4t sin t + 2t2 cos t) |Φ′ (t)|2 = 16t2 + 4t4

L=

Z

π

−π

√

2|t| 4 +

t2

dt = 2

Z

|Φ′ (t)| =

π√

t2

0

p 

√ 4t2 (4 + t2 ) = 2|t| 4 + t2 ,

4 2 (t + 4)3/2 + 4 2t dt = 3

π 0

32 4 = (π 2 + 4)3/2 − . 3 3

(4ε cos ε − 2ε2 sin ε, 4ε sin ε + 2ε2 cos ε) Φ′ (ε) = √ |Φ′ (ε)| 2|ε| 4 + ε2 ε > 0, τ (ε) =



2 cos ε ε sin ε 2 sin ε ε cos ε √ −√ ,√ +√ 2 2 2 4 + ε2 4+ε 4+ε 4+ε

2π,

→ (1, 0)

(et cos t, et sin t)

γ

✍



γ;

−2π ≤ t ≤

t = 0. Φ(t) = (et cos t, et sin t) Φ′ (t) = (et (cos t − sin t), et (sin t + cos t), p √ |Φ′ (t)| = et cos2 t + sin2 t − 2 sin t cos t + sin2 t + cos2 t + 2 sin t cos t = et 2. L(γ) =

Z

2π

−2π

t=0

′

√ √ et 2 dt = 2(e2π − e−2π ).

Φ(0) = (1, 0) Φ (0) = (1, 1),   x= 1+t y=t

y = x − 1.

✍

ρ = 2θ,

π θ = 2

  x(θ ) = ρ(θ ) cos θ = 2θ cos θ

 y(θ) = ρ(θ) sin θ = 2θ sin θ.

θ

  x′ (θ) = 2 cos θ − 2θ sin θ

 y ′ (θ) = 2 sin θ + 2θ cos θ.

θ=

π 2

   π  = −π  x′ 2 π   y′ = 2. 2 θ0 θ0 =

π , 2

y ′ (θ0 )(x − x(θ0 )) = x′ (θ)(y − y (θ0 )) 2(x − 0) = −π(y − π)

lγ

✍ γ(t) =



 2 + 3t , 2t − 1, ln(t) , 8t r

π

2x + πy = π 2 .

1 ≤ t ≤ 2. 2 γ(1)

R3

✍ x = t e2t ,

y = 3 t e2t ,

z = 2 t e2t

t ∈ [0, 1].

t = 1/2.

✍ γ(t) := (x(t), y(t)) = A = γ(0), √ 5( 8 − 1) . 3



 5t3 5t2 , −1 , 3 2

t ≥ 0.

B = γ(s),

AB

γ

B.

✍ γ(t) := (x(t), y(t), z(t)) = (et cos t, et sin t, et t0

√

3),

t ∈ (−∞, +∞).

γ(t0 )

t1

γ(t0 ) γ(t)

√ γ(t1 ) 2 5. γ(t1 ).

✍   2 γ(t) = cos 2t + 2t sin 2t, sin 2t − 2t cos 2t, t3 3

t ∈ [0, +∞).

t¯ > 0

γ(0) σ

γ

π  2

.

γ( ¯t)

38 . 3

ex f (x, y) = yex ;

✍ x = 1;

γ

γ

x=0

γ.

γ γ=

  x(t) = t

t ∈ [0, 1]

 y(t) = et ; γ ′ (t) = (1, et ).

Z

f (x, y)ds = γ

Z

1

t t

e e

√

1+

e2t

0

1 = 2

L(γ) =



(1 + e2t )3/2 3/2

Z 1√

1

0

=

1 [(1 + e2 )3/2 − 1]. 3

1 + e2t dt.

0

Z Z

√

√

√

1 + e2t dt

1+

e2 t

dt =

L(γ) =

Z

Z

z z 2 dz = z −1

1√

1 + e2t dt =

0

√ 1 = 1 + e2 + log 2

Z  1+

"

√

1 2 z −1



1 1 + e2t + log 2

2t = log(z 2 −1).

1 + e2t =: z

  z−1 1 dz = z + log + C. 2 z+1 √

1 + e2t − 1 √ 1 + e2t + 1

! √ √ 1 1 + e2 − 1 √ − 2 − log 2 1 + e2 + 1

!#1

0

! √ 2−1 √ . 2+1

γ

✍ p

f (x, y) =

Φ(t) = (t cos 2t, −t sin 2t), t = 0 −1 ≤ t ≤ 1.

x2 + y 2

Φ′ (t) = (cos 2t − 2t sin 2t, − sin 2t − 2t cos 2t), (0, 0) (1, 0),

Φ′ (0) = (1, 0);

Φ(0) = (0, 0) x.

|Φ′ (t)|2 = cos2 2t + 4t2 sin2 2t − 4t sin 2t cos 2t + sin2 2t + 4t2 cos2 2t + 4t = 1 + 4t2

|Φ′ (t)| =

Z

1

−1

=2

p

t2

Z 1√

√

1 + 4t2 .

cos2

2t +

t2

Z Z 1√ √ 2 2 t 1 + 4t dt = sin 2t |Φ (t)| dt = 2

−1

1 4

1 + 4t2 t dt =

0

Z

5

√

s ds =

1

γ γ

γ

γ sin x ′ ′ γ (π/2) = (0, 1, 0) γ (π) = (−1, 1, 0).

Z

xyz γ

p

1+

cos2

y ds =

Z

2

sin t t (1 + cos t) dt =

0

Z

0

sin t]02π +

(sin t, t, 1) 0 ≤ t ≤ 2π; t = 0, t = π2 , t = p xyz 1 + cos2 y. γ ′ (t) = (cos t, 1, 0) γ ′ (0) = (1, 1, 0)

z = 1.

2π

=[−t cos t +

−1

√ |t| 1 + 4t2 dt

1 2 √ 5 1 √ [s s]1 = (5 5 − 1). 4 3 6

γ

✍ π,

1

′

2π

t sin t dt +

Z

2π

t sin t cos2 t dt 0

2π  1 sin3 t 8 cos3 t + sin t − = − π. −t 3 9 0 3 3

✍ f (x, y) = √ γ

∂E E=

Z



x2 (x, y) : x ≥ 0, x + y ≥ 1, 0 ≤ y ≤ 1 − 4   3 1, . γ 4 2

f (x, y) = γ

Z

f (x, y) + γ1

γ1 =

γ2 =

f (x, y) = γ2

Z

π/2

0

y=0

f (x, y) =

f (x, y) + γ2

Z



f (x, y) γ3

1 ≤ t ≤ 2;

f (x, y) = 0

γ1

  x = cos t

0 ≤ t ≤ 2π;

√ √ √ sin t cos t π/2 √ dt = [− 4 + cos2 t]0 = [−2 + 5] = 5 − 2 4 + cos2 t

p

γ3

Z

 y = sin t

γ3 =

Z

2

 x=t Z

Z

xy 4 + x2

Z

0

  x=t

2   y = 1− t 4 r

(x′ )2 + (y ′ )2 =

2

2

t(1 − t4 ) √ · 4 + t2 Z

√

0 ≤ t ≤ 2; t2 1+ = 4

4 + t2 1 dt = 2 2

f (x, y) = γ

√

√

3 5− . 2

Z

4 + t2 . 2 2

0

  t3 1 t− dt = . 4 2

  3 1, 4

  x=t

x′ = 1, y ′ = −

t 2



1, 34



2   y = 1− t 4

0 ≤ t ≤ 2;

t=1

  x = 1+t

3 1  y = − t 4 2

t ∈ R;

1 5 y = − x+ . 4 2 f (x, y) =

✍ xy 

 x2 y 2 =1 (x, y) ∈ R : + 4 9 2

s = sin t . . . γ

γ(θ) =

  x = 3 cos θ  y = 2 sin θ

γ ′ (θ) =

|γ ′ (θ)| = Z

f ds = γ

Z

 y = 2 cos θ

p

9 sin2 θ + 4 cos2 θ =

π/2

f (γ (θ)) |γ ′ (θ)| ds =

0

3 = 5

  x = −3 sin θ

Z

π/2

10 sin θ cos θ 0

p

Z

h πi θ ∈ 0, 2 h πi θ ∈ 0, 2 p

5 sin2 θ + 4.

π/2

6 sin θ cos θ

0

3 5 sin θ + 4 dθ = 5 2



p

5 sin2 θ + 4 dθ

(5 sin2 θ + 4)3/2 3/2

π/2 0

=

38 . 5

✍ α α

α(t) = (t cos t, t sin t, t), t ∈ [0, 2π]. (−π, 0, π)

α(t) := (α1 (t), α2 (t), α3 (t)) = (t cos t, t sin t, t)

R

αz

ds,

t ∈ [0, 2π]

α′ (t) = (cos t − t sin t, sin t + t cos t, 1) |α′ (t)| =

Z

p

cos2 t + t2 sin2 t − 2 t sin t cos t + sin2 t + t2 cos2 t + 2 t sin t cos t + 1 =

t2 + 2.

f (x, y, z ) = z

z ds = α

√

Z

2π 0

f (α1 (t), α2 (t), α3 (t)) |α′ (t)| dt =

Z

2π

t

0

√

2π 2 3/2 1 (t + 2)   t2 + 2 dt =  3/2 2 0

1 = [(4π 2 + 2)3/2 − 23/2 ]. 3

α : R → R3

(x0 , y0 , z0 )

(x − x0 ) α′1 (t0 ) + (y − y0 ) α2′(t0 ) + (z − z0 ) α3′ (t0 ) = 0 (x + π) α1′ (π) + (y − 0) α2′ (π ) + (z − π ) α3′ (π) = 0 z − x − y π = 2 π. ✍ R p α

x2 + y 2 ds,

α α

α(t) = (t sin t, t cos t, 2 t), t ∈ [0, 2π]. (π/2, 0, π)

α(t) := (α1 (t), α2 (t), α3 (t)) = (t sin t, t cos t, 2 t)

t ∈ [0, 2π]

α′ (t) = (sin t + t cos t, cos t − t sin t, 2)

|α′ (t)| =

Z p

p

sin2 t + t2 cos2 t + 2 t sin t cos t + cos2 t + t2 sin2 t − 2 t sin t cos t + 4 = f (x, y, z) =

x2 + y 2 ds =

α

Z

2π ′

0

f (α1 (t), α2 (t), α3 (t)) |α (t)| dt =

2π

Z

t

0

√

p

√

t2 + 5.

x2 + y 2

2 π 2 3/2  (t + 5) 1   t2 + 5 dt =  3/2 2 0

1 = [(4π 2 + 5)3/2 − 53/2 ]. 3

α : R → R3

(x0 , y0 , z0 )

(x − x0 ) α′1 (t0 ) + (y − y0 ) α2′(t0 ) + (z − z0 ) α3′ (t0 ) = 0 

x−

π   π π  ′  π α1 + (y − 0) α′2 + (z − π) α3′ =0 2 2 2 2 

x−

 π π 1 + (y − 0) − + 2 (z − π) = 0 2 2 2x − π y + 4 z = 5 π.

✍

R √ α

2

α

α(t) = (t , cos t, sin t), t ∈ [0, 2π]. (π , −1, 0) 2

α

α(t) := (α1 (t), α2 (t), α3 (t)) = (t2 , cos t, sin t)

t ∈ [0, 2π]

α′ (t) = (2t, − sin t, cos t) |α′ (t)| =

p

4t2 + sin2 t + cos2 t =

√

4t2 + 1.

x ds,

f (x, y, z ) = Z

√

x ds =

α

=

Z

2π 0

f (α1 (t), α2 (t), α3 (t)) |α′ (t)| dt =

Z

2π

0

1 [(16π 2 + 1)3/2 − 1]. 12

t

√

(x − x0 ) α′1 (t0 ) + (y − y0 ) α2′(t0 ) + (z − z0 ) α3′ (t0 ) = 0

(x − π 2 ) α1′ (π) + (y + 1) α2′ (π) + (z − 0) α3′ (π) = 0

z = 2 π x − 2 π 3.

x

2 π 2 3/2 (4 t + 1) 1   4 t2 + 1 dt =  3/2 8

α : R → R3

(x − π 2 ) 2 π − z = 0

√

0

(x0 , y0 , z0 )

✍  x y 1)f (x, y) = 3 1 − − 2 4 2 2 2)f (x, y) = x − y x−y 3)f (x, y) = x+y f  x y 3 1− − =C 2 4 C x y + = 1− 3 2 4

0≤C≤3

x=y C

x2 − y 2 = C x = −y

C=0

✍ f (x, y) = arcsin(xy − y − 2x) arcsin x

x ∈ [−1, 1]

f

D = {(x, y) ∈ R2 : −1 ≤ xy − y − 2x ≤ 1}. x=1 (x, y) ∈ D x>1

(x, y) ∈ D

D.

2x − 1 1 + 2x ≤y≤ x−1 x−1 2x + 1 2x − 1 ≤y≤ x−1 x−1

f

f

x0 2−x   −x2 + 2x ≤ y ≤ x2 − 2x     1 3 , y = 6 x = 6  2 2     −1 < x < 2.

f

✍ f (x, y) = c f

1 + xy x2

c ∈ R

z = f (x, y)

c,

Ec = {(x, y ) ∈ R2 : f (x, y ) = c}. c

f

Ec

Ec

1 + xy = c, x2

c∈R

1 y = cx − , x

c ∈ R.

f

f (x, y) =

✍ √ 1) + xey − yex

√

x − 1+ ln(y −

D = {(x, y) ∈ R2 : x ≥ 1} ∩ {(x, y ) ∈ R2 : y > 1} ∩ {(x, y ) ∈ R2 : xey − yex ≥ 0}.

xey − yex ≥ 0 ⇔ xey ≥ yex ⇔ xe−x ≥ ye−y . f (x) = xe−x . limx→+∞ f (x) = 0; f ′ (x) = e−x [1 − x] f.

R, f(0) = 0, limx→−∞ f (x) = −∞ x=1 1/e.

f

3

2 y 1

–3

–2

–1

0

1

2

3