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PartII: Problem-SolvingProblems( Showallwork ) PartI: , ,

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Calculus Student ID number:

Part I:

單選

,

填充

,

是非題

1. (10 pts) Match the surfaces to the functions?

x2+ y2: B , log(x2+ y2): D , xye−(x2+y2): E , sin(x + y): A , sin(x) sin(y): C .

(A) (B) (C)

(D) (E)

Part II: Problem-Solving Problems (

計算與證明題

Show all work)

2. Evaluate the given integrals (a) (5 pts)

1

x−4/3dx

1

x−4/3dx = lim

R→∞

R

1

x−4/3dx = lim

R→∞−3x−1/3|R1 = 3 (b) (5 pts)

0

x−4/3dx

0

x−4/3dx =

1

0

x−4/3dx +

1

x−4/3dx

1

0

x−4/3dx = lim

R→0+

1

R

x−4/3dx = lim

R→0+−3x−1/3|1R DIV

0

x−4/3dx DIV

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Name: Student ID number:

3. (5 pts) Compute the Taylor polynomial of degree 3 about a = 0 for f (x) = e2x. f (x) = e2x, f(1)(x) = 2e2x, f(2)(x) = 4e2x, f(3)(x) = 8e2x. f (0) = 1, f(1)(0) = 2, f(2)(0) = 4, f(3)(0) = 8.

⇒ P3(x) = 1 + 2x + 4

2!x2+ 8

3!x3 = 1 + 2x + 2x2+4 3x3.

4. (10 pts) Solve dy

dx = x− 1

y , with y(0) =−2, i.e. y = −2 when x = 0.

y dy =

(x− 1) dx ⇒ 1

2y2 = 1

2x2− x + C ⇒ y = ±√

x2− 2x + C

∵ y(0) = −2 ∴ y = −√

x2− 2x + 4

5. Suppose that dy

dx = y(y− 1)(y − 2)

(a) (5 pts) Find the equilibria of the differential equation.

y(y− 1)(y − 2) = 0 ⇒ y = 0, 1, 2 (b) (5 pts) Discuss the stability of the equilibria.

Let g(y) = y(y− 1)(y − 2) = y3− 3y2+ 2y, g0(y) = 3y2− 6y + 2.

g0(0) = 2 > 0⇒ y = 0 is unstable g0(1) =−1 < 0 ⇒ y = 0 is locally stable

g0(2) = 2 > 0⇒ y = 0 is unstable

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Name: Student ID number:

6. Given a matrix A =

[ 5 4

−2 −1 ]

(a) (10 pts) Find the eigenvalues and eigenvectors of A

0 = det (A− λI) = λ2− 4λ + 3 = (λ − 1)(λ − 3) ⇒ λ1 = 1, λ2 = 3.

λ1 = 1 : Av1 = λ1v1

[ 5x + 4y

−2x − y ]

= [ x

y ]

⇒ y = −x ⇒ Take v1 = [ 1

−1 ]

λ2 = 3 : Av2 = λ2v2

[ 5x + 4y

−2x − y ]

= [ 3x

3y ]

⇒ x = −2y ⇒ Take v2 = [ −2

1 ]

(b) (10 pts) Let v1, v2 be two eigenvectors found in part (6a). Find α, β so that

[ −3 2

]

= αv1+βv2. Compute A10 [ −3

2 ]

without using a calculator.

[ −3 2

]

= α [ 1

−1 ]

+ β [ −2

1 ]

⇒ α = −1, β = 1.

⇒ A10 [ −3

2 ]

= A10(−v1+ v2) = −A10v1+ A10v2 =−λ101 v1+ λ102 v2

= [ 1

−1 ]

+ 310 [ −2

1 ]

=

[ −1 − 2 · 310 1 + 310

]

7. (5 pts) Let A be a matrix and v1 and v2 be two eigenvectors of A, with eigen values λ1 and λ2, respectively (i.e. Av1 = λ1v1 and Av2 = λ2v2). Assume that λ1 6= λ2, prove that there is no constant c satisfying that v2 = cv1

Suppose that v2 = cv1 for some c. Then

Av2 2v2 = Acv1 = cAv1 = cλ1v1 = λ1v2.

⇒(λ2− λ1)v2 = 0⇒ λ1 = λ2, (∵ v2 6= −→

0 )⇒ contradiction

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Name: Student ID number:

8. (a) (5 pts) Compute lim

(x,y)→(1,2)

2x2y x4+ y2 lim

(x,y)→(1,2)

2x2y

x4+ y2 = lim(x,y)→(1,2)2x2y lim(x,y)→(1,2)x4 + y2 = 4

5. (b) (5 pts) Show that lim

(x,y)→(0,0)

2x2y

x4+ y2 does not exist.

• Along path x = 0, lim

(x,y)→(0,0)

2x2y

x4+ y2 = lim

y→0

0

04+ y2 = 0

• Along path y = x2, lim

(x,y)→(0,0)

2x2y

x4+ y2 = lim

x→0

2x2x2

x4+ x4 = lim

x→0

2x4 2x4 = 1

⇒ The limit does not exist.

9. (10 pts) Given f (x) = exy x

y, find fx, fy, fxx, fxy and fyy.

fx = yexy 1

y, fy = xexy+ x y2, fxx = y2exy, fxy = (1 + xy)exy + 1

y2, fyy = x2exy 2x y3

10. (10 pts) Find the equation of the tangent plane to the surface at the given point.

z = x2 − y2+ 1 at (1, 2,−2) Let f (x, y) = x2− y2+ 1

fx(x, y) = 2x, fy(x, y) = −2y, fx(1, 2) = 2, fy(1, 2) =−4, The equation of the tangent plane:

z = L(x, y) = f (1, 2) + fx(1, 2)(x−1)+fy(1, 2)(y−2) = −2+2(x−1)−4(y −2).

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