(a) Marking scheme

1101模模模組組組06-12班班班 微微微積積積分分分2 期期期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. Consider F (x) =∫

x

1 x

sin(√ xt) dt.

(a) (3%) Show that F (x) = 1 x⋅ ∫

x² 1 sin(√

u) du.

(b) (7%) By using (a), find F^′(1).

Solution:

Marking Scheme for 1(a)

ˆ 2M for letting u = xt

ˆ 1M for completing the proof

Sample Solution for 1(a).

Let u = xt. (2M) Then du = xdt ⇒ 1

xdu = dt and

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

when t = 1 x, u = 1 when t = x, u = x²

(1M) Therefore,

F (x) =∫

x

1 x

sin(√

xt) dt =∫

x² 1

sin(√ u) ⋅1

xdt = 1 x⋅ ∫

x² 1

sin(√ u) du

Marking Scheme for 1(b)

ˆ 2M for the use of product/L’Hospital’s rule

ˆ 3M for differentiating the integral correctly (-2M for missing the term from chain rule)

ˆ 1M for F (1) = 0

ˆ 1M for answer Sample Solution I for 1(b).

By product rule (2M), F^′(x) = −1 x²⋅ ∫

x² 1

sin(√

u)du + 1

x⋅sin(x) ⋅ 2x

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(3M)

. Therefore,

F^′(1) = −1 ⋅ 0

®

(1M)

+sin(1) ⋅ 2 = 2 sin(1)

´¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶

(1M)

Sample Solution II for 1(b).

F^′(1)^def= lim

h→0

F (1 + h) − F (1)

h =lim

h→0

F (1 + h) − 0

h (1M) for F (1) = 0

=lim

h→0

∫

(1+h)² 1 sin(√

u) du h²+h

=lim

h→0

sin(1 + h) ⋅ 2(h + 1)

2h + 1 (2M) for L’H and (3M) for derivative

=2 sin(1) (1M) for answer

2. Evaluate the following integrals.

(a) (9%)∫ (x + 3)

√

1 − x²dx. (b) (9%)∫

1 0

ln(√

x + 2) dx.

Solution:

(a) Let x = sin θ with θ between −π/2 and π/2. So, dx = cos θdθ and

∫ (x + 3)

√

1 − x²dx

= ∫ (sin θ + 3) cos²θdθ (3%)

= − 1

3cos³θ + 3∫

1 + cos 2θ

2 dθ

= − 1

3cos³θ + 3

2(θ +sin 2θ

2 ) +C (3%)

= − 1

3(1 − x²)

3 2 +

3

2sin⁻¹x +3 2x

√

1 − x²+C. (3%)

(b) Let u =√

x. So, 2udu = dx and

∫

1 0

ln(√

x + 2)dx =∫

1 0

ln(u + 2)2udu (2%)

=ln(u + 2)u²∣

1 0− ∫

1 0 u² 1

u + 2du = ln 3 −∫

1 0

u²

u + 2du. (3%)

For the second term,

∫

1 0

u²

u + 2du =∫

1

0 u − 2 + 4

u + 2du (2%)

= u²

2 −2u + 4 ln ∣u + 2∣∣¹₀= − 3

2+4 ln 3 − 4 ln 2.

Therefore, the final answer is ln 3 +3

2−4 ln 3 + 4 ln 2 = 3

2−3 ln 3 + 4 ln 2. (2%)

3. (a) (8%) Evaluate ∫ cot³(x) dx and apply the result to show∫

π/2

0 cot³(x) dx is divergent.

(b) (4%) Use the Comparison Theorem for Improper Integral to determine whether

∫

∞ 1

1 e^x−2^xdx is convergent or divergent.

Solution:

(a) The following crucial steps must be shown clearly ⋯ 4%⋯ ∫ cot³(x) dx =∫ cot(x)(csc²(x) − 1) dx

= − ∫ cot(x) cot^′(x)dx −∫ cot(x) dx

= − cot²(x)

2 −ln ∣ sin(x)∣ + C

4%⋯ ∫

π/2 0

cot³(x) dx = lim

a→0⁺( cot²(a)

2 +ln ∣ sin(a)∣)

= lim

a→0⁺

1

2 sin²(a)(1 + sin²(a) ln ∣ sin(a)∣)

= lim

a→0⁺

1

2 sin²(a)(1 + 0) = ∞ (b) The following crucial steps must be shown clearly ⋯

1%⋯ e^x−2^x=e^x(1 −2^x

e^x) =e^x(1 − (2 e)^x) 1%⋯ lim

x→∞( 2

e)^x=0 ⇒ (2

e)^x<1/2 whenever x > b = ln(2) 1 − ln(2)

2%⋯ ∫

∞ 1

dx e^x−2^x = ∫

b 1

dx e^x−2^x+ ∫

∞ b

dx e^x−2^x

< ∫

b 1

dx e^x−2^x + ∫

∞ b

2dx

e^x =2e^−b+ ∫

b 1

dx

e^x−2^x < ∞ convergent

4. (a) (9%) Find the orthogonal trajectories of the family of curves y²=4 − Cx, where C is an arbitrary constant.

(b) (9%) Solve, for y = f (x), the equation 1 x⋅

dy

dx−2y = e^x2sin(x) cos(x) with f (π 3) =0.

Solution:

(a) Differentiating the given equation, we get

2ydy = −Cdx

or dy

dx= − C 2y. Solving the given equation for C ,

C =4 − y² x .

Plugging this in the differential equation to get rid of C from the expression of dy dx: dy

dx = y²−4

2xy . Another equivalent method to find dy

dx in terms of x, y is as follows: First, solve the given equation for C:

C =4 − y² x . Then differentiate both sides to get

0 = d dx(

4 − y² x ) =

−2y^dy_dxx − (4 − y²)

x² .

This is the differential equation whose solutions give the given family of curves. By simplification, we get dy

dx = y²−4

2xy .

The next step is to set up the differential equation for the orthogonal trajectories:

dy dx = −

1

y²−4 2xy

= − 2xy y²−4. Separating the variables, we have

y²−4

2y dy = −xdx.

Thus,

∫ y²−4

2y dy = −∫ xdx by which we get

y²

4 −2 ln ∣y∣ = −x² 2 +C.

(a) Marking scheme

ˆ 4 pts for finding the differential equation

2xydy

dx=y²−4

(or any equivalent form) of the given family of curves in x, y. Partial credits will be:

– 3 pts, if the student seems to try eliminating C, but incorrectly.

– 2 pts, if the differential equation is given correctly, but containing C.

– 1 pts, if the differential equation is given incorrectly, and containing C.

ˆ 2 pts for setting up the differential equation of the orthogonal trajectories dy

dx = − 2xy y²−4. Partial credits will be:

– 1 pt, if the student tries to set up the equation but incorrectly, such as dy

dx= 2xy y²−4 or dy

dx = −y²−4 2xy etc.

ˆ 3 pts for solving the differential equation of the orthogonal trajectories.

– 1 pt for separating the equation into the terms of x and those of y.

– 2 pts for writing down the final equation correctly.

* 1 pt is taken away if there are minor mistakes such as missing the absolute value sign of the logarithm.

– 1 pt for executing the integration correctly, but integrating incorrect integral caused by the previous mistakes.

(b) Making the given equation into the standard form, we have dy

dx−2xy = xe^x2sin x cos x.

Since

∫ (−2x)dx = −x

2+C

so we can choose the integration factor to be I(x) = e^−x2. Multiplying e^−x2 through, d

dx(e^−x2y) = x sin x cos x.

Integrating both sides, we get

e^−x2y =∫ x sin x cos xdx

= 1

2∫ x sin 2xdx

= 1

2x (−cos 2x 2 ) −

1 2∫ (−

cos 2x 2 )dx

= − 1

4x cos 2x +1

8sin 2x + C.

Hence,

y = e^x2(−

1

4x cos 2x +1

8sin 2x + C) . Using the initial condition

f (π 3) =0, we get

− 1

4x cos (2π 3 ) +

1

8sin (2π

3 ) +C = 0.

Hence

C = π 24−

√3 16.

(b) Marking scheme

ˆ 4 pts for finding the correct “integrable” form d

dx(e^−x2y) = x sin x cos x

– 3 pts for finding integration factor I(x) = e^−x2 correctly.

* 1 pt for rewriting the equation in the standard form.

ˆ 3 pts for executing the integration.

– 2 pts for the student tries integration by parts,

* 1 pt if the student just struggles, but gets to nowhere

ˆ 2 pts for determining the constant C using

f (π 3) =0.

It’s not necessary to write down the final equation, if the constant is determined correctly.

– 1 pts for setting up the equation for the constant.

If the student set up an incorrect equation, but execute the remaining calculations correctly, he/she will get 1 pt for the correct integration, 1 pt for setting up the equation for determining the constant (so at most 2 pts after setting up the integrable form).

5. (14%) Consider the region enclosed by the curve y = 5 x√

5 − x, the x-axis, the lines x = 1, x = 4 in the first quadrant.

Find the volume of the solid obtained by rotating this region about the x-axis.

Solution:

The volume is

V =∫

4 1

π ( 5 x√

5 − x)

2

dx = π∫

4 1

−25

x²(x − 5)dx (4%)

−25 x²(x − 5)=

A x +

B x²+

C x − 5 (3%) This shows us that

−25 = Ax(x − 5) + B(x − 5) + Cx²= (A + C)x²+ (B − 5A)x − 5B So we get

B = 5, A = 1, C = −1. (3%) Hence,

V = π (∫

4 1

1 xdx +∫

4 1

5 x²dx −∫

4 1

1 x − 5dx)

= π ( ln ∣x∣ − 5

x−ln ∣x − 5∣]

4

1

) = (4 ln 2 +15

4 )π (4%)

6. Consider a lemniscate C whose polar equation is given by r²=cos(2θ).

(a) (6%) Find all the points (in polar coordinates) of C at which the tangent to C is horizontal.

(b) (8%) The curve C is inscribed in a rectangle such that each side of the rectangle is tangent to C (see figure below). Find the area of the shaded region.

Solution:

Marking Scheme for 6(a)

ˆ 1M for realizing that dy dθ =0

ˆ 1M for writing y =√

cos(2θ) ⋅ sin θ

ˆ 1M for correct computation of dy dθ

ˆ 0.5M for each correct angular coordinates θ

ˆ 1M for the correct radical coordinate r Sample Solution for 6(a).

Since dy dx =

(dy/dθ)

(dx/dθ), a tangent is horizontal is equivalent to having dy

dθ =0. (1M)

Since y = r sin θ =

√

cos(2θ) ⋅ sin θ, (1M)

we have dy dθ =

−sin(2θ)

√ cos(2θ)

⋅sin θ +

√

cos(2θ) ⋅ cos θ. (1M)

So dy

dθ =0 implies tan(2θ) ⋅ tan θ = 1. Thus, we have 2 tan θ

1 − tan²θ ⋅tan θ = 1 ⇒ tan θ = ± 1

√

3 ⇒ θ = ±π 6, ±5π

6 . (2M) In all cases, r = 1

√2. (1M)

Therefore, the four required points are (r, θ) = ( 1

√ 2, ±π

6)and ( 1

√ 2, ±5π

6 ) Marking Scheme for 6(b)

ˆ 2M for the correct integrand for the area of lemniscate

ˆ 2M for the correct integration limits for the area of lemniscate

ˆ 1M for the correct area of lemniscate

ˆ 1M for the correct height of the rectangle OR y-coordinate of the highest point of lemniscate

ˆ 1M for the correct area of rectangle

ˆ 1M for the correct final answer Sample Solution for 6(b).

First we calculate the area enclosed by the lemniscate C : 4 ⋅∫

π 4

0

±

2M

1

2cos(2θ)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

2M

dθ = [sin(2θ)]

π 4

0 = 1

®

(1M)

The Cartesian coordinate of (r, θ) = ( 1

√ 2, ±π

6) is (x, y) =

⎛

⎜⎜

⎜⎜

⎜

⎝

√3 2√

2, 1 2√

2

²

(1M)

⎞

⎟⎟

⎟⎟

⎟

⎠

. Therefore, the area of the rectangle

equals

2 ⋅ 1

√ 2=

√ 2

°

(1M)

Hence, the area of the shaded region equals√

2 − 1 (1M).

7. For each real number p, the parametric equations x(t) = cos(3t)

t^p , y(t) =sin(3t)

t^p with 1 ≤ t < ∞ define an improper spiral.

(a) (12%) Find the arclength of the improper spiral when p = 3 (see figure).

(b) (2%) Find the range of values of p such that the arclength of the improper spiral is finite.

Solution:

(a) p = 3.

x(t) = cos(3t)

t³ , y(t) =sin(3t)

t³ with 1 ≤ t < ∞ x^′(t) =−3 sin(3t)

t³ −

3 cos(3t)

t⁴ , y^′(t) =3 cos(3t) t³ −

3 sin(3t) t⁴ [x^′(t)]²+ [y^′(t)]²=

9 t⁶ +

9 t⁸

∫

∞ 1

√

[x^′(t)]²+ [y^′(t)]²dt = lim

a→∞∫

a 1

3√ t²+1

t⁴ dt = lim

b→^π₂⁻∫

b π/4

3 sec³θ tan⁴θ dθ

= lim

b→^π₂⁻∫

b π/4

3 cos θ

sin⁴θ dθ = lim

b→^π₂⁻

[−(sin θ)⁻³]

b π/4=2√

2 − 1

(b)

x^′(t) =−3 sin(3t) t^p −

p cos(3t)

t^p+1 , y^′(t) =3 cos(3t) t^p −

p sin(3t) t^p+1 [x^′(t)]²+ [y^′(t)]²=

9 t^2p+

p² t^2p+2 The arc length is

∫

∞ 1

√ 9 t^2p +

p²

t^2p+2 dt =∫

∞ 1

1 t^p

√ 9 +p²

t² dt Since 3 <

√ 9 +p²

t² <

√

9 + p² bounded by constants, we can compare with∫

∞ 1

1 t^p dt.

Therefore the arc length is finite when p > 1.

Grading scheme:

(2 pts) for arc length formula.

(8 pts) for integration in (a). (4 pts) for trig-sub and (4 pts) for trig-integral.

(2 pts) for notation and answer in (a).

(1 pt) for setting up the improper integral with p in (b).

(1 pt) for range of p.