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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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 =λ2v2 = 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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