thus the radius of convergence of A(x) is 1

1012微微微甲甲甲01-04班班班期期期中中中考考考解解解答答答和和和評評評分分分標標標準準準

1. (15 points) Let sk =

k

X

j=1

1

j, k = 1, 2, · · · , and A(x) =

∞

X

k=1

skx^k. (a) Find the interval of convergence of A(x).

(b) Express A(x) in terms of elementary functions by comparing A(x) and xA(x).

Solution:

(a)

Using ratio test, (knowing how to use ratio test or root test in the correct way earn 1 point) sk+1

sk

= 1 +(_k+1¹ ) sk

→ 1 as k → ∞,

thus the radius of convergence of A(x) is 1. (having computed the radius of convergence earn 3 points) (The radius of convergence can also be calculated using root test.

√k

sk = ^k r1

1 +1

2 + · · · + 1 k ≤√^k

1 + 1 + · · · + 1 = ^k

√ k

√k

sk = ^k r1

1 +1

2 + · · · + 1 k ≥ ^k

r1 k+1

k+ · · · + 1 k = 1 Since

√k

k = e^{ln kk} and

lim

t→∞

ln t t = lim

t→∞

(¹_t)

1 = 0 by l’hospital rule.

Thus

lim

k→∞

√k

k = 1 and by squeezing

lim

k→∞

√k

sk = 1.

)

Note that

|sk(±1)^k| = sk → ∞ 6= 0 as k → ∞,

hence A(x) diverges at x = ±1. The interval of convergence of A(x) is (−1, 1). (obtaining the endpoints behavior earn 1 point)

(b)

For x ∈ (−1, 1),

A(x) =

∞

X

k=1

skx^k = x +

∞

X

k=2

skx^k

xA(x) =

∞

X

k=1

s_kx^k+1=

∞

X

k=2

s_k−1x^k.

Subtracting them, we obtain

(1 − x)A(x) = x +

∞

X

k=2

(s_k− sk−1)x^k = x +

∞

X

k=2

1 kx^k=

∞

X

k=1

1

kx^k. (5 points)

But ∞

X

k=1

1

kx^k = − ln(1 − x), So

A(x) = −ln(1 − x)

1 − x . (5 points)

2. (15 points)

(a) Expand f (x) = (x − 1) ln(1 + 3x) in powers of x − 1.

(b) For what values of x is the above expansion valid?

(c) Find the sum

∞

X

k=1

1 k

 3 4

^k .

Solution:

(a) f (x) = (x−1) ln(1+3x) = (x−1) ln[3(x−1)+4] = (x−1) ln[4(1+3

4(x−1))] = (x−1)[ln 4+ln[1+3

4(x−1)]] = (x − 1)[ln 4 + Σ^∞_k=1(−1)^k+1

k [3

4(x − 1)]^k] (b) −1 < 3

4(x − 1) ≤ 1, which is say if −1

3 < x ≤ 7 3 (c) f (0) = 0 = − ln 4 + Σ^∞_k=11

k(3 4)^k.

⇒ Σ^∞_k=11 k(3

4)^k= ln 4.

3. (15 points) Let C be the curve given by r(t) = 2

3(1 + t)³²i +2

3(1 − t)³²j + atk, t ∈ (−1, 1), a ∈ R \ {0}.

(a) Find the length s(b) of the curve from t = 0 to t = b ∈ (0, 1).

(b) Find the unit tangent, the principal normal, and the osculating plane of C at r(t).

(c) Find the curvature κ(t) of C at r(t).

Solution:

γ⁰(t) = ((1 + t)¹², −(1 − t)¹², a)

|T⁰(t)| =p a²+ 2 s(b) =

ˆ b 0

|T⁰(t)|dt = bp

a²+ 2...(3pts) T(t) = γ⁰(t)

|γ⁰(t)| = 1

√a²+ 2((1 + t)¹², −(1 − t)¹², a)...(3pts) T⁰(t) = 1

√

a²+ 2(1

2(1 + t)⁻¹² ,1

2(1 − t)⁻¹² , 0) = 1 2√

a²+ 2(1 + t)⁻¹² , (1 − t)⁻¹² , 0)

|T⁰(t)| = 1

p2(a²+ 2)(1 − t²) N(t) = T⁰(t)

|T⁰(t)| = 1

√

2((1 − t)¹², (1 + t)¹², 0)...(3pts) B(t) = T(t) × N(t) = 1

p2(a²+ 2)(−a(1 + t)¹², a(1 − t)¹², 2) Osculating plane atγ(t) : (x − 2

3(1 + t)²³, y − 2

3(1 − t)²³, z − at) · B(t) = 0...(3pts) κ(t) = |T⁰(t)|

|γ⁰(t)| = 1

√2(a²+ 2)√

1 − t², t ∈ (−1, 1)...(3pts)

4. (15 points) Let f (x, y) =





 x²y

x²+ y², (x, y) 6= (0, 0);

0, (x, y) = (0, 0).

(a) Compute fx(0, 0) and fy(0, 0).

(b) Calculate fx(x, y) and fy(x, y) for (x, y) 6= (0, 0).

(c) Are f_x and f_y continuous at (0, 0)?

(d) Determine f_xy(0, 0) and f_yx(0, 0) if they exist. If they do not exist, explain why.

(e) Is f (x, y) differentiable at (0, 0)?

Solution:

(a) by definition ,

∂f

∂x|(0,0)= lim

h→0

f (h, 0) − f (0, 0)

h = 0

similaritly ,

∂f

∂y|_(0,0)= lim

h→0

f (0, h) − f (0, 0)

h = 0 (each 1 pts.) (b) when f on (x, y) 6= (0, 0)

∂f

∂x = 2xy(x²+ y²) − x²y(2x)

(x²+ y²)² = 2xy³ (x²+ y²)²

∂f

∂y = x²(x²+ y²) − x²y(2y)

(x²+ y²)² = x²(x²− y²)

(x²+ y²)² (each 2pts.) (c) observe f along y = mx , m arbitrary

lim

(x,y)→(0,0)

f_x(x, y) = lim

(x,mx)→(0,0)

2x(mx)³

(x²+ (mx)²)² = 2m³ (1 + m²)² 6= fx(0, 0) = 0

lim

(x,y)→(0,0)fy(x, y) = lim

(x,mx)→(0,0)

x²(x²− (mx)²)

(x²+ (mx)²)² = 1 − m² (1 + m²)² 6= fy(0, 0) = 0

⇒ limit doesn’t exist at (0, 0)

⇒ fx, fy not conti. at (0, 0) (3 pts.) (d) fxy= ∂fx

∂y (0, 0) = lim

h→0

fx(0, h) − fx(0, 0)

h = 0

fyx= ∂fy

∂x(0, 0) = lim

h→0

fy(h, 0) − fy(0, 0)

h = lim

h→0

1

h −→ ∞ (3 pts.) (e) sol.1 If f diff. at (0, 0) , then_√ lim

h²+k²→0

f (h, k) − f (0, 0) − ∇f (0, 0) · (h, k)

√h²+ k² = 0 (1 pts)

⇒_√ lim

h²+k²→0

h²k

(h²+ k²)³² = 0 But along h = k

√ lim

h²+k²→0

h²k

(h²+ k²)³² = lim

h→0

h³

2³²h³ = 2⁻³² 6= 0 so f not diff. at(0, 0) (3 pts.) sol.2 set −→u = ( 1

√

m²+ 1, m

√

m²+ 1) fu(0, 0) = lim

h→0

f (^{√ h}

m²+1,^√mh

m²+1) − f (0, 0)

h = m

(1 + m²)³² But ∇f (0, 0) · −→u =< fx(0, 0), fy(0, 0) > ·−→u = 0

a contradiction , so f not diff. at (0, 0) (3 pts.)

5. (15 points) Let f (x, y, z) = e^xyln z. Find the directional derivatives of f at P (1, 0, e) in the following directions.

(a) In the direction in which f increases most rapidly at P .

(b) In the directions parallel to the line in which the planes x + y − z = 2 and 4x − y − z = 1 intersect.

(c) In the direction of increasing t along the path r(t) =p

1 + t²i + tan tj + e^2t+1k

Solution:

f (x, y, z) = e^xyln z

⇒ ∇f (x, y, z) = (ye^xyln z) i + (xe^xyln z) j + (1

ze^xy) k (2 pts)

⇒ ∇f (1, 0, e) = j +1

e k (1 pt)

(Get the correct expression of ∇f (x, y, z) but the wrong value of ∇f (1, 0, e): 2-point deduction for whole question.)

(a)

The desired directional derivative is k∇f (1, 0, e)k =

√ 1 + e²

e . (4 pts)

(b)

The directions of this line are v = (1, 1, −1) × (4, −1, −1) = (−2, −3, −5) and −v = (2, 3, 5).

⇒ The unit vectors are u = v kvk = 1

√

38(−2, −3, −5) and −u = 1

√

38(2, 3, 5) . (2 pts)

⇒ The directional derivative in u is ∇f (1, 0, e) · u = −1

√38(3 + 5e⁻¹), and directional derivative the in −u is

√1

38(3 + 5e⁻¹). (2 pts)

(If you only write one of the two derivatives, you get at most 3 pts.)

(c)

r⁰(t) = t

√1 + t² i + sec²t j + 2e^2t+1 k.

⇒ r⁰(0) = j + 2e k. (2 pts)

⇒ The desired directional derivative is ∇f (r(0)) · r⁰(0)

kr⁰(0)k = 3

√1 + 4e². (2 pts)

(Calculation error: 1-point deduction for each error.)

(Correct formula but with wrong answer form: 1-point deduction for each error.) (Did not use unit vectors: 1-point deduction for each error.)

6. (15 points) Suppose f (x, y) = x²+ cxy + 2y²where c is a constant.

(a) Find all values of c such that (0, 0) is a stationary point of f . (b) Find all values of c such that (0, 0) is a saddle point of f . (c) Find all values of c such that f has a local minimum at (0, 0).

(d) Find all values of c and all (x0, y0) 6= (0, 0) such that f has a local minimum at (x0, y0).

Solution:

(a) (3 %) solution:

For Of = (2x + cy)ˆi+ (cx + 4y)ˆj, we have a point (x, y) is a stationary point if Of (x, y) = 0, that is 2x + cy = 0 and cx + 4y = 0. So, for (0, 0) to be a stationary point of f , it is clear that c can be any real number, i.e. c ∈ R.

(b) (4 %) solution:

For Of = (2x + cy)ˆi + (cx + 4y)ˆj = ∂f

∂xˆi + ∂f

∂yˆj, we have ∂²f

∂x²(x, y) = 2, ∂²f

∂y²(x, y) = 4, and ∂²f

∂x∂y(x, y) =

∂²f

∂y∂x(x, y) = c for all (x, y) ∈ R². Thus, at (0, 0), A = ∂²f

∂x²(0, 0) = 2 ,C = ∂²f

∂y²(0, 0) = 4, and B = ∂²f

∂x∂y(0, 0) =

∂²f

∂y∂x(0, 0) = c.

The discriminant is D = AC − B²= 8 − c².

By second partials test, for (0, 0) to be a saddle point, we must have D < 0, that is 8 − c²< 0, so c > 2√ 2 and c < −2√

2.

If you do this problem only until here, you can get 4 points ,but the check for the case D = 0 will be 2 points in next problem (c).

When D = 0, we have c = ±2√

2, so if c = 2√

2, we have f (x, y) = x²+ 2√

2xy + 2y² = (x +√

2y)² ≥ 0 for all (x, y) ∈ R², thus (0, 0) is a local minimum if c = 2√

2. Similarly, when c = −2√

2, we have f (x, y) = x²− 2√

2xy + 2y²= (x −√

2y)²≥ 0 for all (x, y) ∈ R², thus (0, 0) is a local minimum if c = −2√ 2.

Therefore, the point (0, 0) si a saddle point only when c > 2√

2 and c < −2√ 2.

(c) (4 %) solution:

The discriminant is D = AC − B²= 8 − c².

By second partials test, for (0, 0) to be a local minimum, we must have D > 0 and A > 0, but A = ∂²f

∂x²(0, 0) = 2 > 0, which is clear. So we only need to consider D > 0, that is 8 − c²> 0, so −2√

2 < c < 2√ 2 . If you do this problem only until here, you can get 2 points.

By the argument in the problem (b), we know that when c = ±2√

2, (0, 0) is a locl minimum. Thus (0, 0) si a locl minimum only when −2√

2 ≤ c ≤ 2√ 2 (d) (4%)

solution:

Note that if a point (x0, y0) is a locl minimum of f , we must have the point (x0, y0) satisfies Of (x⁰.y0) = 0, that is 2x0+ cy0= 0 and cx0+ 4y0 = 0. But to have the point (x0, y0) 6= (0, 0), we need the above system of equations (2x0+ cy0= 0 & cx0+ 4y0= 0) have solutions other than (0, 0) this is equivalent to 8 − c²= 0 (which is the determinant of the matrix of the coefficints of the above system of equations ).

So c = ±2√

2, when c = 2√

2, we have f (x, y) = x²+ 2√

2xy + 2y²= (x +√

2y)²≥ 0 for all (x, y) ∈ R², thus we have all the points on the line x + 2√

2y = 0 are local minimum of f . Similarly, when c = −2√

2,we have f (x, y) = x²− 2√

2xy + 2y² = (x −√

2y)² ≥ 0 for all (x, y) ∈ R², thus we have all the points on the line x − 2√

2y = 0 are local minimum of f . Therefore, the value of c are ±2√

2, and the corresponding (x0, y0) 6= (0, 0) are the set {(x, y) 6= (0, 0) : x +√ 2y = 0} and {(x, y) 6= (0, 0) : x −√

2y = 0}, respectively.

7. (15 points) A rectangular box has three of its faces on the coordinate planes and one vertex in the first octant on the paraboloid z = 4 − 5x²− 6y². Determine the maximum volume of the box.

Solution:

We want to find the maximum of xyz with side condition z = 4 − 5x²− 6y². So putting f (x, y, z) = xyz and g(x, y, z) = 5x²+ 6y²+ z, and using Lagrang’s method by setting ∇f = λ∇g, we have







yz = 10λx xz = 12λy xy = λ.

Substituting xy = λ to the first and the second equation, we have

 yz = 10x²y xz = 12λxy². Hence, we get z = 10x²= 12y², since z = 4 − 5x²− 6y², we get x²= 1

5, y²= 1

6, and z = 2 (also we get λ = 1

√30) when xyz attains extrema. We then deduce the maximum should be

r2 15.

評分標準：

(a) 算出∇f及∇g，並列出∇f = λ∇g以明示使用Lagrange方法，得2分

(b) 滿足前述條件且列出yz = 10λx, xz = 12λy, xy = λ之明顯的等價敘述，得3分 (c) 滿足前述條件且列出10x²= 12y²之明顯的等價敘述，得3分

(d) 滿足前述條件且列出z = 10x²= 12y²之明顯的等價敘述，得2分

(e) 滿足前述條件且得到正確答案，得5分。但若滿足前述條件且得到達最大值之座標、比值或λ卻計算出錯誤

答案，得3分。

(f) 使用其他方法（例如：第二偏導數判定法、算幾不等式等等）斟酌給分。