Calculus Final Exam

Calculus Final Exam

1. (a) (4 points) Find the gradient of f (x, y) = ln(x²+ y²) for (x, y) 6= (0, 0).

Solution: ∇f = 1 x²+ y²

¡2x, 2y¢

(b) (4 points) In what direction does f (x, y) = ln(x²+ y²) increase most rapidly at (x, y) 6= (0, 0)? Find the largest directional derivative of f .

[Note: a direction is a unit vector.]

Solution: In the same direction of the gradient of f , i.e. 1 p

x²+ y²

¡x, y¢

, the function f increase most rapidly, and its directional derivative is∇f · (x, y)

p

x²+ y² = 2 p

x²+ y².

(c) (4 points) Find an equation of the tangent plane to the surface z = ln(x²+ y²) at p = (1, 1, ln2).

Solution: Let g(x, y, z) = ln(x²+ y²) − z. Then ∇g(p) =¡

1, 1, −1¢

is pointing to the normal direction to the surface z = ln(x²+ y²), at p. Hence, (x − 1) + (y − 1) − (z − ln2) = 0 is an equation of the tangent plane to the surface z = ln(x²+ y²) at p = (1, 1, ln2).

2. (12 points) Use Lagrange multipliers to find the maxima and minima of f (x, y) = xy under the constraint x²+ y²= 4.

Solution: Let g(x, y) = x²+y². To find the candidate point on g = 4 at which f achieves an extremum, we solve the equations ∇f =λ∇g, i.e. Solving

·y

x

¸

=

·2λ^x 2λ^y

¸

. Substituting the second equation into the first one, we get (1 − 4λ²)y = 0 which implies that either y = 0, orλ ^{= ±1}

2.

If y = 0, then we have x = 0 by using the second equation. Since g(0, 0) 6= 4, (0, 0) is not a point on the curve x²+ y²= 4.

Ifλ^{= ±1}

2, then we have y = ±x by using the first equation. Since g = 4, we get 2x²= 4, and (x, y) = (√

2,√

2), or (−√ 2, −√

2), or (√ 2, −√

2), or (−√ 2,√

2). Hence f (±√ 2, ±√

2) = 2 is the maximum value of f on the curve x²+ y²= 4, while f (√

2, −√

2) = f (−√ 2,√

2) = −2 is the minimum value of f on the curve x²+ y²= 4.

3. For each of the following systems i. − iii.

(a) Find the eigenvalues and eigenvectors.

(b) Classify the equilibrium (0, 0), and analyze the stability of the equilibrium.

(c) Sketch several trajectories of solution curves in the phase plane i.e. x₁x₂−plane.

i. (12 points) dX dt =

·3 −2 2 −2

¸ X

Calculus Final Exam June 25, 2007

Solution: (a) λ1 = −1,ξ⁽¹⁾ =

·1 2

¸

; λ2 = 2,ξ⁽²⁾ =

·2 1

¸

. (b) saddle point, unstable. (c) The general solution is X = C₁e^−t

·1 2

¸

+C₂e^2t

·2 1

¸

=

·C₁e^−t+ 2C₂e^2t

2C₁e^−t+C₂e^2t

¸ .

x y

1 -0.5

0.5

0.5 1

-1

-1 -0.5 0

0

ii. (12 points) dX dt =

·5 −1

3 1

¸ X

Solution: (a)λ1= 2, ξ^{(1) =}

·1 3

¸

; λ2 = 4, ξ⁽²⁾⁼

·1 1

¸

. (b) node or source, unstable. (c) The general solution is X = C1e^2t

·1 3

¸

+C2e^4t

·1 1

¸

=

·C1e^2t+C2e^4t

3C1e^2t+C2e^4t

¸ .

0

-0.5 1

x y

0.5

0 -1

-1 -0.5 1

0.5

iii. (12 points) dX dt =

·1 −5 1 −3

¸ X

Solution: (a)λ1,λ2= −1 ± i;ξ^(1),ξ⁽²⁾⁼

·2 ± i 1

¸

(b) spiral point, stable.(c) Two linearly inde-

pendent solutions are X = e^(−1+i)t

·2 + i 1

¸

and X = e^(−1−i)t

·2 − i 1

¸

. The general solution can be

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Calculus Final Exam June 25, 2007

written as X_h = C1ReX + C2ImX = C1e^−t

·2 cost − sint cost

¸

+ C2e^−t

·cost + 2 sint sint

¸

, where ReX, ImX denote the real and imaginary parts of X , respectively.

2 1.5 1 0.5 0 1.5

1

0.5

-0.5 0 -1

-0.5

-1

4. Consider the system d dt

·x y

¸

=

· x(1 − 0.5y) y(−0.75 + 0.25x)

¸

=

· x − 0.5xy

−0.75y + 0.25xy

¸

(a) (8 points) Find all the equilibria of the system.

Solution: (x(t), y(t)) ≡ (0, 0) or (3, 2) for all t ∈ R are equilibria.

(b) (6 points) Find the linearized system at each equilibrium.

Solution: At (0, 0), the linearized system is d dt

·x

y

¸

=

·1 0 0 −0.75

¸ ·x

y

¸

At (3, 2), the linearized system is d dt

·x

y

¸

=

· 0 −1.5

0.5 0

¸ ·x − 3

y − 2

¸

(c) (6 points) Classify and analyze the (local) stability of each equilibrium.

Solution: The eigenvalues of the linearized system at (0, 0) areλ1= 1;λ2= −0.75, the origin is a saddle point of both the linearized system and of the nonlinear system, and therefore is unstable.

The eigenvalues of the linearized system at (3, 2) areλ1= 1,λ2= ±

√3i

2 , the equilibrium (3, 2) is a center of both the linearized system and is a stable equilibrium of the nonlinear system.

5. (20 points) Find the general solution of the system dX dt =

·2 −1 3 −2

¸ X +

·e^−t 0

¸

= AX + g(t), where X =

·x1

x₂

¸ .

Solution: The eigenvalues and eigenvectors areλ1= −1,ξ⁽¹⁾⁼

·1 3

¸

;λ2= 1,ξ⁽²⁾⁼

·1 1

¸ . The homogeneous solution is X_h= C₁e^−t

·1 3

¸ +C₂e^t

·1 1

¸

=

· C₁e^−t+C₂e^t

3C₁e^−t+C₂e^−t

¸

where C₁,C₂ are arbi- trary constants.

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Calculus Final Exam June 25, 2007

LetΨ=

·1 1 3 1

¸

, thenΨ⁻¹= −1 2

· 1 −1

−3 1

¸ . By setting X =ΨY, we have Y⁰=Ψ⁻¹AΨY +Ψ⁻¹g.

Hence, we have y⁰₁+ y₁= −1

2e^−t, and y⁰₂− y₂= 3

2e^−t which are equivalent to

¡e^ty₁¢₀

= −1 2, and¡

e^−ty₂¢₀

=3 2e^−2t.

We get

·y₁

y₂

¸

=



−1 2te^−t

−3 4e^−t



 , and a particular solution is Xp=ΨY =



−¡3 4+1

2t¢ e^−t

−¡3 4+3

2t¢ e^−t



 .

The general solution is X = X_h+ X_p=



−¡3 4+1

2t¢

e^−t+C₁e^−t+C₂e^t

−¡3 4+3

2t¢

e^−t+ 3C₁e^−t+C₂e^t



 .

Note that if one find the general solution of Y by adding constants C₁, and C₂in integrating equations for y₁, and y₂, one will also get the general solution X =ΨY .

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