1052ᵬᵬᵬ06-10IIIʟʟʟὃὃὃYYYȤȤȤᑖᑖᑖ 1. (15% total) (a) (5%) Derive the MacLaurin series of tan

1052微微微甲甲甲06-10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% total)

(a) (5%) Derive the MacLaurin series of tan⁻¹x.

(b) (5%) Find the value of a ∈ R such that the limit lim_x→0sin(ax) − sin x − tan⁻¹x

x³ is finite.

(c) (5%) Evaluate the above limit.

Solution:

(a) (tan⁻¹x)^′= 1

1 + x² (1%) tan⁻¹x =∫

1

1 + x²dx =∫

∞

j=0∑(−1)^jx^2jdx =

∞

j=0∑∫ (−1)

jx^2jdx =

∞

j=0∑(−1)^jx^2j+1

2j + 1+c (3%)

(P.S. tan⁻¹x = x −x³ 3 +

x⁵

5 +... by calculate f^′(0), ..., f⁽⁵⁾(0) (2%)).

Let x = 0, c = tan⁻¹0 = 0. Its radius of convergence is 1. ∣x∣ < 1 (1%)

(b) sin x = x −x³ 3! +

x⁵

5! −... (3%) (P.S. sin x = x − x³

3! +... (2%)) sin(ax) − sin x − tan⁻¹x

x³ = (a − 2)1 x²+ {

1 − a³ 3! +

1

3} +O(x²).

Thus, a = 2. (2%) (c) The limit is 1

6(1 − 8) +1 3 =

−5 6 (5%)

Page 1 of 8

2. (10% total) Consider the power series

∞

∑

n=1

(−1)ⁿxⁿ (n + 1) ln(n + 1). (a) (5%) Determine its radius of convergence.

(b) (5%) Determine its interval of convergence.

Solution:

(a) Let cn=

(−1)ⁿ

(n + 1) ln(n + 1). Use Ratio Test we have

n→∞lim

∣c_n+1xⁿ⁺¹∣

∣c_nxⁿ∣

= ∣x∣ lim

n→∞

n + 1 n + 2

ln(n + 1)

ln(n + 2) = ∣x∣ ⋅ 1 < 1 Where lim

n→∞

ln(n + 1) ln(n + 2)

L

= lim

n→∞

n + 2 n + 1 =1.

Thus the radius of convergence R = 1.

(b) Let bn= ∣cn∣.

Try x = 1. (i) bn clearly decreasing (ii) lim

n→∞bn=0 obviously.

Thus by Alternating Series Test

∞ n=1∑

(−1)ⁿb_n converges.

Try x = −1. (i) bn positive (ii) bn decreasing (iii) f (n) = bn continuous Then by Integral Test,∫

∞ 1

f (x)dx = ln ln(x + 1)∣^∞₁ diverges implies

∞

∑

1

bn diverges.

Therefore the interval of convergence is (−1, 1].

Grading

(a) (2 pts) State correct test.

(2 pts) Correct calculation.

(1 pt) Correct answer

(b) (1 pt) Case x = 1, state correct test.

(1 pt) Case x = 1, correct calculation (1 pt) Case x = −1, state correct test.

(1 pt) Case x = −1, correct calculation (1 pt) Correct answer

Remarks

ˆ If lim_n→∞an

bn

=0, then ∑ bn converges implies ∑ an converges, but ∑ bn diverges means nothing.

3. (20% total) Consider the space curve r(t) = ti + t²j +2t³ 3 k.

(a) (5%) Find the arc length of the curve from t = 0 to t = a.

(b) (5%) Find the curvature κ(0) at t = 0.

(c) (5%) Find the unit tangent T(0) at t = 0.

(d) (5%) Find the unit normal N(0) at t = 0.

Solution:

(a)

r^′(t) = (1, 2t, 2t²) (2%)

∣r^′(t)∣ =

√

1 + 4t²+4t⁴=2t²+1 (1%) if a ≥ 0

s =∫

a 0

∣r^′(t)∣dt =∫

a 0

(2t²+1)dt = 2

3a³+a (2%) if a < 0

s =∫

0 a

∣r^′(t)∣dt =∫

0 a

(2t²+1)dt = −2 3a³−a (b)

(Method I)

dT

ds =κN ⇒ κ = ∣^dT_dt∣

∣r^′(t)∣ (1%) T (t) = r^′(t)

∣r^′(t)∣ = ( 1

2t²+1, 2t

2t²+1, 1 − 1 2t²+1) T^′(t) = ( −4t

(2t²+1)², −4t²+2

(2t²+1)², 4t

(2t²+1)²) (2%) κ(0) =∣T^′(0)∣

∣r^′(0)∣ = ∣(0, 2, 0)∣

∣(1, 0, 0)∣=2 (2%) (Method II)

κ = ∣r^′×r^′′∣

∣r^′∣³

(1%) r^′(t) = (1, 2t, 2t²) r^′(0) = (1, 0, 0)

r^′′(t) = (0, 2, 4t) r^′′(0) = (0, 2, 0)

∣r^′(0) × r^′′(0)∣ = ∣(0, 0, 2)∣ = 2 (2%)

∣r^′(0)∣³= ∣(1, 0, 0)∣³=1 κ(0) = ∣r^′(0) × r^′′(0)∣

∣r^′(0)∣³ =2 (2%) 沒有代入t=0扣2%

(c)

T (0) = r^′(0)

∣r^′(0)∣ = (1, 0, 0) (5%) 若因前面r^′(t), ∣r^′(t)∣算錯而算錯T (0)會酌量扣分

沒有代入t=0扣2%

(d)

N (0) ∥ T^′(0) = (0, 2, 0) (3%)

Page 3 of 8

若因前面r^′(t), ∣r^′(t)∣, T^′(t)算錯而算錯T^′(0)會酌量扣分 N (0) = (0, 2, 0)

∣(0, 2, 0)∣ = (0, 1, 0) (2%) 沒有代入t=0扣2%

4. (11%) Let z = f (x, y) and x = r cos θ, y = r sin θ.

(a) (6%) Express ∂z

∂x in terms of r, θ and partial derivatives with respect r, θ.

(b) (5%) Express ∂²z

∂x² in terms of r, θ and partial derivatives with respect r, θ.

Solution:

(a). Note that r =

√

x²+y² and θ = tan⁻¹y x.

∂z

∂x

=

∂z

∂r

∂r

∂x+

∂z

∂θ

∂θ

∂x (♣ 3%)

=

∂z

∂r⋅ x

√

x²+y² +

∂z

∂θ⋅

−y

x²+y² (♣ 2%)

=

∂z

∂rcos θ +∂z

∂θ⋅

−sin θ

r (♣ 1%) (b). From above, we know ∂

∂x=cos θ ∂

∂r− sin θ

r

∂

∂θ.

∂²z

∂x²

= (cos θ ∂

∂r− sin θ

r

∂

∂θ) (

∂z

∂rcos θ +∂z

∂θ⋅

−sin θ

r )(♣ 3%)

=

∂²z

∂r²cos²θ − ∂²z

∂r∂θ

sin θ cos θ

r +

∂z

∂θ

sin θ cos θ r² +

sin θ r (−

∂²z

∂r∂θcos θ +∂z

∂rsin θ +∂²z

∂θ² sin θ

r +

∂z

∂θ cos θ

r )

=

∂²z

∂r²cos²θ − ∂²z

∂r∂θ

2 sin θ cos θ

r +

∂²z

∂θ² sin²θ

r² +

∂z

∂r sin²θ

r +

∂z

∂θ

2 sin θ cos θ

r² (♣ 2%)

Page 5 of 8

5. (20% total) Let f (x, y, z) = (x²+z²)sinπxy

2 +yz² and a point p = (1, 1, −1). Answer the following:

(a) (5%) Find the gradient of f at p.

(b) (5%) Find the approximate value of f (0.98, 1.02, −0.97).

(c) (5%) Find the plane tangent to the level surface through p defined by f (x, y, z) = f (p) = 3.

(d) (5%) If a bird flies through p directly to the point (2, −1, 1) with speed 5, what is the rate of change of f as seen by the bird at p?

Solution:

(a)

∂f

∂x RR RR RR RR RR R_(1,1,−1)

= [2x sinπxy

2 + (x²+z²) (cosπxy 2 )

πy 2 ]

RR RR RR RR RR R_(1,1,−1)

=2

∂f

∂y RR RR RR RR RR R_(1,1,−1)

= [(x²+z²) (cosπxy 2 )

πx 2 +z²]

RR RR RR RR RR R_(1,1,−1)

=1

∂f

∂z RR RR RR RR RR R_(1,1,−1)

= [2z sinπxy 2 +2yz]

RR RR RR RR RR R_(1,1,−1)

= −4

∴∇f (1, 1, −1) = [∂f

∂xi +∂f

∂yj +∂f

∂zk]

RR RR RR RR RR R_(1,1,−1)

=2i + j − 4k Remark: (4%) for the partial derivatives; (1%) for evaluation of the gradient (b) The linear approximation L(x, y, z) of f (x, y, z) at the point p = (1, 1, −1) is

L(x, y, z) = f (1, 1, −1) +∂f

∂x RR RR RR RR RR R(1,1,−1)

(x − 1) +∂f

∂y RR RR RR RR RR R(1,1,−1)

(y − 1) +∂f

∂z RR RR RR RR RR R(1,1,−1)

(z + 1)

=3 + 2(x − 1) + (y − 1) − 4(z + 1) (3%)

∴f (0.98, 1.02, −0.97) ≈ L(0.98, 1.02, −0.97) = 3 + 2(−0.02) + 0.02 − 4(0.03) = 2.86 (2%)

(c) Notice that f (p) = f (1, 1, −1) = 3, which means that the plane tangent to the level surface is the tangent plane of f (x, y, z) at the point p. Therefore, the tangent plane equation is

2(x − 1) + (y − 1) − 4(z + 1) = 0 (5%) (d) The unit vector u from point p = (1, 1, −1) to point q = (2, −1, 1) is

u = q − p

∣q − p∣= (1 3, −2

3,2

3) (1%) Then, the directional derivative

Duf (p) = ∇f (p) ⋅ u = (2, 1, −4) ⋅ (1 3, −2

3,2 3) = −

8

3 (3%)

The rate of change of f as seen by the bird at p with speed v = 5 is D_uf (p) × v = −40

3 (1%)

6. (12%) Find the local extreme values and saddle points of f (x, y) = x²y − xy²+xy − y².

Solution:

{ fx= 2xy − y²+y = y(2x − y + 1) = 0 ⋯(∗) fy= x²−2xy + x − 2y = (x + 1)(x − 2y) = 0 ⋯(„) For (∗):

i. If y = 0, then x(x + 1) = 0 from („) ⇒ x = 0 or −1.

Hence, we have (0, 0), (−1, 0).

ii. If y = 2x + 1, then 3x²+5x + 2 = 0 from („) ⇒ x = −1 or −2 3. Hence, we have (−1, −1), (−2

3, −1 3).

Note that we can get the same result by considering („). Therefore, the critical points are (0, 0), (−1, 0), (−1, −1), (−²₃, −¹₃). (1% for each point)

Since

f_xx=2y, f_xy=f_yx=2x − 2y + 1, f_yy= −2x − 2, we now have

D(x, y) = f_xxf_yy−f_xy² = −4(x + 1)y − (2x − 2y + 1)².

i. D(0, 0) = −1 < 0 ⇒ (0, 0) is a saddle point. (2%)

ii. D(−1, 0) = −1 < 0 ⇒ (−1, 0) is a saddle point. (2%)

iii. D(−1, −1) = −1 < 0 ⇒ (−1, −1) is a saddle point. (2%)

iv. D(−2 3, −1

3) = 1

3 >0, fxx(−

2 3, −1

3) = − 2 3 <0

⇒ f (x, y) has a local maximum at(−²₃, −¹₃). (1%)

And the local maximum value at (−2 3, −1

3)is f (−2 3, −1

3) = ₂₇¹ . (1%)

Page 7 of 8

7. (12%) Find the maximum and the minimum of the function f (x, y) = 3x²−2y² on the curve 2x²−2xy + y²=1.

Solution:

Let g(x, y) = 2x²−2xy + y²−1 = 0 By applying the method of Lagrange multipliers, we need to solve

∇f = λ∇g [2 points] and g(x, y) = 0

or

⎧⎪

⎪⎪

⎪

⎨

⎪⎪

⎪⎪

⎩

6x = λ(4x − 2y) (1)

−4y = λ(−2x + 2y) (2) 2x²−2xy + y²−1 = 0 (3)

[6 points]

(1 points per coefficient in (1)(2))

Clearly, x ≠ 0, y ≠ 0, λ ≠ 0, or g(x, y) fails to be 0. So, dividing (1) by (2) gives 3x

−2y = 2x − y

−x + y ⇒ 3x²−7xy + 2y²=0 (3x − y)(x − 2y) = 0 ∴ 3x = y or x = 2y

ˆ Case 1: 3x = y. Plug this into (3) can get x²=1/5, y²=9/5 ∴ f (x, y) = 3x²−2y²=

3 5−

18 5 = −3

ˆ Case 2: x = 2y. Plug this into (3) can get y²=1/5, x²=4/5 ∴ f (x, y) = 3x²−2y²=

12 5 −

2 5 =2

Since the extreme value must exist, 2 is the absolute maximum [2 points]

and − 3 is the absolute minimum [2 points]