ᵬ08-13I ᦟʟὃY

微甲08-13班 統一教學期中考解答

1. (12%) (a) Use logarithmic differentiation to find the derivative of the function y = x^1x, x > 0.

(b) Find the tangent line of the function y = tan⁻¹(e^x) at x = 0.

Sol:

(a)

∵ ln y = 1 xln x

=⇒ 1 y · dy

dx =

1

xx − ln x x²

=⇒ dy

dx = x^x1 · 1 − ln x x² . (b) When x = 0, y(0) = tan⁻¹1 = π

4,

=⇒ dy

dx = e^x 1 + e^2x

=⇒ dy

dx |^x=0= 1 2, the tangent line:

y − π 4 = x

2.

2. (10%) Find the highest and the lowest points of the curve given by x²+ xy + 2y² = 28.

Sol:

First note that the curve is an elliptic (for example, by examining the discriminant = 1−4·1·2 <

0), so it does have an unique highest and an unique lowest points.

i) Implicit differentiation with respect to x ⇒ 2x + y + xdy

dx + 4ydy dx = 0.

ii) At the highest and the lowest points, dy

dx = 0, so 2x + y = 0.

iii) By ii) and the equation of the elliptic, we get x = ±2, and then y = ∓4.

So the highest point is (−2, 4), and the lowest point is (2, −4).

3. (14%) Evaluate the limits.

(a) lim

x→∞(1 + 3 x+ 5

x²)^x, (b) lim

x→0

R¹

cos x 2

tdt − x²

x⁴ .

Sol:

(a) Taking logarithm.

log

 1 + 3

x + 5 x²

^x

= x log

 1 + 3

x + 5 x²



= log

 1 + 3

x + 5 x²

  1 x



To find the limit above. Consider the following limit log

 1 + 3

x + 5 x²

⁰

 1 x

⁰

=



− 3 x² − 15

x³

 

1 + 3 x + 5

x²

 

− 1 x²

 

→ 3 as x → ∞. Hence by l’Hospital Rule, we conclude that

x→∞lim log

 1 + 3

x + 5 x²

^x

= 3 This is equivalent to

x→∞lim

 1 + 3

x + 5 x²

^x

= e³

(b) Since this limit is of the form 0/0. So we consider the following limit

Z ¹

cos x

2

tdt − x²

⁰

. x^4
0

= 2 tan x − 2x 4x³

The last equality holds by the Fundamental Theorem of Calculus. Again this limit is of the form 0/0, we consider the limit

(2 tan x − 2x)⁰

(4x³)⁰ = 2 sec²x − 2 12x²

This is still a indeterminated form. Hence we differentiate again. We arrived that 4 sec²x tan x

24x We meet 0/0, so differentiate it again

4 sec⁴x + 8 sec²x tan²x

24 → 1

6 as x → 0 Therefore, by l’Hospital Rule, we conclude that

x→0lim

Z ¹

cos x

2

tdt − x²



 x⁴



= 1 6 2

4. (10%) See the figure below. A box with cover is to be constructed from a square piece of cardboard, 30cm wide, by cutting out a square or a rectangle (shaded region) from each of the four corners and bending up the remaining cardboard (unshaded region) along the dotted lines. What is the largest volume that such a box can have? Justify that the volume you obtain actually is the maximum volume.

Sol:

2y + 2x = 30 ⇒ y = 15 − x

Volume= V (x) = x(15 − x)(30 − 2x) = 2x³− 60x²+ 450x on [0, 15]

V⁰(x) = 6x²− 120x + 450 = 6(x − 15)(x − 5) 發生最大值的位置可能點有端點和 critical point 0, 5, 15 V (0) = 0 = V (15), V (5) = 5 · 10 · 20 = 1000.

Then the largest volume is 1000.

5. (20%) Study the function y = f (x) = 20x³ − 3x⁵ on R, answer the following questions, and sketch the graph of this function. Write down all necessary calculation.

(a) Is f (x) an odd function? .

Is f (x) an even function? .

(b) x-intercept(s) is(are) , y-intercept is .

(c) f⁰(x) =

f is increasing on interval(s) .

f is decreasing on interval(s) .

the coordinate(s) of local maximum point(s) is(are) (x, y) = . the coordinate(s) of local minimum point(s) is(are) (x, y) = .

(d) f⁰⁰(x) = .

f is concave upward on interval(s) .

f is concave downward on interval(s) .

the coordinate(s) of inflection point(s) is(are) (x, y) = . (e) The equation(s) of asymptote(s) of y = f (x) is(are) .

(Answer none if there is no asymptote.)

(f) Sketch the graph of y = f (x).

Sol:

(a) f (−x) = 20(−x)³− (−x)⁵ = −(20x³ − 3x⁵) = −f(x), hence f (x) is odd not even.

(b) 0 = 20x³− 3x⁵ ⇒ x-intercepts are 0, r 20

3 , −r 20 3 . f (0) = 0 ⇒ y-intercept is 0.

(c) Since f⁰(x) = 60x²− 15x⁴ = 15x²(2 − x)(2 + x), hence we have f⁰(x) > 0 on (−2, 0) , (0, 2) ⇒ f(x) is increasing on (−2, 0) , (0, 2).

f⁰(x) < 0 on (−∞, −2) , (2, ∞) ⇒ f(x) is decreasing on (−∞, −2) , (2, ∞).

And it is easy to get f (x) has local maximum at (x, y) = (2, 64),

4

and local minimum at (x, y) = (−2, −64).

(d) f⁰(x) = 120x − 60x² = 60x(√

2 − x)(√

2 + x), hence f⁰⁰(x) > 0 on (−∞, −√

2)), (0,√

2) ⇒ f(x) is concave upward on (−∞, −√ 2)), (0,√

2).

f⁰⁰(x) < 0 on (0, ∞), (−√

2, 0) ⇒ f(x) is concave downward on (0, ∞), (−√ 2, 0).

hence the inflection points are (0, 0), (√ 2, 28√

2), (−√

2, −28√ 2).

Claim: f (x) has no asymptote.

Proof: lim

x→∞f (x) − (ax + b) = ∞, and lim_x→−∞f (x) − (ax + b) = −∞ , for all a, b ∈ R.

(e) As follows.

-4 -3.2 -2.4 -1.6 -0.8 0 0.8 1.6 2.4 3.2 4

-50 -25 25 50

6. (10%) A particle moves along the curve with equation x³+ y⁴+ xy = 3. Let (x(t), y(t)) be the position coordinate, measured in meters, of the particle at time t, measured in seconds, and S(t) = px(t)²+ y(t)² be the distance between the particle and the origin at time t. Suppose that at t = 1 the particle is located at (x, y) = (1, 1) with dS

dt  

_t=1 =√

2 m/sec.

(a) Find x⁰(1) and y⁰(1).

(b) Find the speed, px⁰(t)²+ y⁰(t)², of the particle at t = 1.

Sol:

(a)

s² = x²+ y² 2sds

dt = 2xdx

dt + 2ydy dt

input t = 1, x(1) = 1, y(1) = 1 ⇒ s =√ 2,ds

dt =√ 2 2√

2 ·√

2 = 2dx

dt + 2dy

dt ⇒ 2 = dx dt + dy

dt (1)

x³+ y⁴+ xy = 3

3x²˙x + 4y³˙y + ˙xy + x ˙y = 0

inputt = 1 ⇒ 3 ˙x + 4 ˙y + ˙x + ˙y = 0 ⇒ 4 ˙x + 5 ˙y = 0 (2)

solve (1) and (2), then we obtain ˙y = −8, ˙x = 10.

(b) √

8²+ 10² = 2√ 41

6

7. (10%) Consider the limit lim

n→∞n²

n

X

k=1

k n⁴+ k⁴.

(a) Explain carefully why the limit exists. Express the limit as a definite integral.

(b) Evaluate this definite integral.

Sol:

(a)

n→∞lim n²

n

X

k=1

k

n⁴+ k⁴ = lim

n→∞

n

X

k=1

1

n · k · n⁻¹ 1 + (k · n⁻¹)⁴ This is exact the Riemann Sum of the function f (x) = x

1 + x⁴ on [0, 1]. Note that f (x) is continuous on [0, 1]. Hence integrable. We conclude the limit exists and is equal to the integral of f (x) on [0, 1], i.e.,

n→∞lim n²

n

X

k=1

k n⁴+ k⁴ =

Z ¹

0

x 1 + x⁴dx

(b) Evaluate the integral. Let u = x². du = 2xdx and when x varies from 0 to 1, u varies from 0 to 1. Hence

Z ¹

0

f (x)dx = Z ¹

0

x

1 + x⁴dx = 1 2

Z ¹

0

1

1 + u²du = 1

2tan⁻¹u    

1

0

= π 8

8. (14%) Evaluate the following integrals.

(a)

Z e^{sin x}

sec xdx, (b) Z ⁶⁴

1

4√

x + 7√³ x

√6

x dx.

Sol:

(a) Let u = sin x, then

Z e^{sin x} sec xdx =

Z

e^udu = e^u+ C = e^{sin x}+ C

(b) Z ⁶⁴

1

4x¹² + 7x¹³ x¹⁶ dx =

Z ⁶⁴

1

4x¹³ + 7x¹⁶ = 3x⁴³ + 6x⁷⁶|⁶⁴¹ = (768 + 768) − (3 + 6) = 1527

8