2003 WTW168 Semestertest 2 Answers

Vrae 1 tot 4 se antwoorde moet op die merkleesvorm op kant 1gemerk word. Gebruik ’n potlood.

The answers to Questions 1 to 4 must be marked on the optic readerform on side 1. Use a pencil.

Vraag 1 / Question 1

1.1 Bepaal / Evaluate

(i) ∫ cot2θ dθ

Kies een van / Choose one of

1(a) − 12

ln cos2θ + C

1(b) 12

ln|2sinθcosθ| + C

1(c) −2cosec22θ + C

1(d) ln|sinθ| + C

Ans: ∫ cot2θ dθ = 12 ∫

2cos 2θsin2θ

dθ = 12

ln sin2θ + C = 12

ln|2sinθcosθ| + C 1(b)

(ii) ∫ θsec2θ dθ

Kies een van / Choose one of

2(a) θ 2

2tanθ + C

2(b) θ tanθ − ln|cosθ | + C

2(c) θ tanθ + ln|cosθ | + C

2(d) geen van die / none of these

Ans: ∫ θsec2θdθ = θ tanθ − ∫ tanθdθ = θ tanθ − ∫ sinθcosθ dθ = θ tanθ + ln|cosθ | + C 2(c)

(iii) ∫ sin3θcos2θ dθ

Kies een van / Choose one of

3(a) 14

sin4θ + C

3(b) −13

cos3θ + C

3(c) 13

cos3θ − 15

cos5θ + C

3(d) − 13

cos3θ + 15

cos5θ + C

Ans: ∫ sin3θcos2θdθC = ∫ 1 − cos2θ cos2θsinθ dx

= ∫ − cos2θ − sinθ dx + ∫ cos4θ − sinθ dθ = − 13

cos3θ + 15

cos5θ + C 3(d)

(iv) ∫ 1x+1 x−2

dx

Kies een van / Choose one of

4(a) 13

ln x−2x+1

+ C

4(b) − ln 3x + 3 + ln 3x − 6 + C

wtw168 semester test 1 p 1 of 7

4(c) 13

ln x + 1 − 13

ln x − 2 + C

4(d) geen van die / none of these

Ans: 1x+1 x−2

=−13

x+1+

13

x−2

∴ ∫ 1x+1 x−2

dx = ∫−13

x+1+

13

x−2dx = − 1

3ln x + 1 + 1

3ln x − 2 + C = 1

3ln x−2

x+1+ C 4(a)

(v) ∫ xx−5

dx

Kies een van / Choose one of

5(a) ln x − 5 + C

5(b) x + 5ln x − 5 + C

5(c) −x + 5ln x − 5 + C

5(d) −x − 5ln x − 5 + C

Ans: ∫ xx−5

dx = ∫ x−5+5x−5

dx = ∫ 1 + 5x−5

dx = x + 5ln x − 5 + C 5(b)

1.2 Vir n ∈ Z, 0 ≤ n ≤ 3, word die parsiële breuke van x n

x−1 x−2 x+3 2 gegee deur

For n ∈ Z, 0 ≤ n ≤ 3, the partial fractions of x n

x−1 x−2 x+3 2 is given by

Kies een van / Choose one of

6(a) ax−1

+ bx−2

+ cx+3

6(b) ax−1

+ bx−2

+ cx+3 2

6(c) ax−1

+ bx−2

+ cx+3

+ dx+3 2

6(d) geen van die / none of these

Ans: graad(teller)=degree(numerator)< graad(noemer)=degree(denomenator)

x n

x−1 x−2 x+3 2 = ax−1

+ bx−2

+ cx+3

+ dx+3 2 6(c)

of / or ax−1

+ bx−2

+ Cx+Dx+3 2

waar / where Cx+Dx+3 2 =

C x+3−3 +D

x+3 2 =C x+3

x+3 2 + −3C+Dx+3 2 = c

x+3+ d

x+3 2

met / with c = C,d = −3C + D.

1.3 Die lengte van die kromme van y = x 2

3, 0 ≤ x ≤ 3 3

2word gegee deur

The length of the curve of y = x 2

3, 0 ≤ x ≤ 3 3

2is given by

Kies een van / Choose one of

7(a) ∫0

3 32 2π x 2

31 + 4x 2

9dx

7(b) ∫0

3 32 1 + x 2

9dx

7(c) ∫0

π3 3

2sec3u du

wtw168 semester test 1 p 2 of 7

7(d) ∫0

3 32 3

2sec2utanu du

Ans: L = ∫a

b 1 + dydx

2 dx

dydx = d

dxx 2

3= 2x

3∴ L = ∫

0

3 32 1 + 2x

32 dx

Laat / Let 2x3= tanu ∴ 2dx

3= sec2udu

x = 0 ⇒ tanu = 0 ⇒ u = 0,

x = 3 32

⇒ tanu =2 3 3

2

3= 3 ⇒ u = π

3

∴ L = ∫0

3 32 1 + 2x

32 dx = ∫

0

π3 1 + tanu 2 3

2sec2udu = ∫

0

π3 3

2sec3udu 7(c)

1.4 Laat / Let fx = 11+x 4 vir / for x ≥ 1.

Hoe sal jy die funksie gx kies, sódat met behulp van die Vergelykingstoets

vasgestel kan word dat ∫1

∞ fx dx konvergeer?

How will you choose the function gx in order to use the Comparison

Theorem to determine whether ∫1

∞ fx dx converges.?.

Kies een van / Choose one of

8(a) gx = 1x

8(b) gx = 11+x 4

8(c) gx = 1x 4

8(d) gx = 11+x 4

Ans: 1 + x4 ≥ x4 ≥ 0 ∴ 0 ≤ 11+x 4 ≤ 1

x 4 vir / for x ≥ 1.

maar / but ∫1

∞ 1x 4 dx konvergeer / converges

∴ ∫1

∞ 11+x 4 dx konvergeer ook / also converges ∴ Kies / Choose gx = 1

x 4 8(c)

1.5 Laat / Let fx = 21−x

.

Die derdegraadse Taylor polinoom vir fx rondom a = 2 is....

The third degreeTaylor polynomial for fx about a = 2 is...

Kies een van / Choose one of

9(a) T 3x = 2 − 2 x − 2 + 2 x − 2 2 − 2 x − 2 3

9(b) T 3x = −2 + 2 x − 2 − 2 x − 2 2 + 2 x − 2 3

9(c) T 3x = −2 + 2x − 2x2 + 2x3

9(d) T 3x = 2 − 2x + 2x2 − 2x3

Ans:

fx = 21−x

f2 = 21−2

= −2

f′x = d

dx2

1−x= 2

1−x 2 f′2 = 2

1−2 2 = 2

f′′x = d

dx2

1−x 2 = 41−x 3 f

′′2 = 4

1−2 3 = −4

f′′′x = d

dx4

1−x 3 = 121−x 4 f

′′′2 = 12

1−2 4 = 12

fx = 21−x

T 3x = −20!

x − 2 0 + 21!

x − 2 1 + −42!

x − 2 2 + 123!

x − 2 3

wtw168 semester test 1 p 3 of 7

= −2 + 2 x − 2 − 2 x − 2 2 + 2 x − 2 3 9(b)

1.6 Die grafiek van die kromme gegee deur x = 2cos t, y = sint met 0 ≤ t ≤ π, is

The graph of the curve given by x = 2cos t, y = sint with 0 ≤ t ≤ π, is

10(a)

2-2

2

2

10(b)

2 1

10(d)

2-2

10(c)

-1

-2-2 2

10(a)

2-2

2

2

10(b)

2 1

10(d)

2-2

10(c)

-1

-2-2 2

Ans: x2= cos t, y = sint maar / but cos t 2 + sint 2 = 1

∴ x2

2 + y 2 = 1 met /with −2 ≤ x ≤ 2, − 1 ≤ y ≤ 1 ellips / ellipse)

t x y x,y

0 2cos0 = 2 sin0 = 0 2,0

π2

2cos π2= 0 sin π

2= 1 0,1

π 2cosπ = −2 sinπ = 0 −2,0

10(d)

1.7 Die grafiek van die kromme gegee deur x = sin2θ, y = cos2θ, π ≤ θ ≤ 3π2

is

The graph of curve is given by x = sin2θ, y = cos2θ, π ≤ θ ≤ 3π2

is

11(d)11(c)11(a)

1

111(b)

1

1

-1

1

1-1

1

11(d)11(c)11(a)

1

111(b)

1

1

1

1

-1

1

-1

1

1-1

1

Ans: x = sin2θ, y = cos2θ maar / but cosθ 2 + sinθ 2 = 1

∴ x + y = 1 ∴ y = −x + 1 met /with 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

(reguit lyn / straight line)

θ x y x,y

π sin2π = 0 cos2π = −1 2 0,1 beginpunt / starting point

3π4

sin2 3π4

= 1

2

2= 1

2cos2 3π

4= −1

2

2= 1

212, 1

2

3π2

sin2 3π2

= −1 2 = 1 cos2 3π2

= 0 1,0 eindpunt / end point

, 11(b)

[11]

BEANTWOORD ALLE VERDERE VRAE OP HIERDIE VRAESTEL. /ANSWER ALL THE FOLLOWING QUESTIONS ON THE SCRIPT.

Vraag 2 / Question 2

wtw168 semester test 1 p 4 of 7

Laat / Let x = cos2t, y = sint, t ∈ 0,2π .

i Elimineer die parameter om die Cartesiese vergelyking van die kromme te kry.Eliminate the parameter to find the Cartesian equation of the curve.

Ans: x = cos2t, y = sint, maar / but cos t 2 + sint 2 = 1∴ x + y2 = 1 ∴ y2 = −x + 1 = 4 −1

4x − 1

met / with 0 ≤ x ≤ 1, − 1 ≤ y ≤ 1 (parabool / parabola)

ii Skets die kromme.Sketch the curve.

t x y x,y

0 cos20 = 1 sin0 = 0 1,0 beginpunt / starting point

π2

cos2 π2= 0 sin π

2= 1 0,1

π cos2π = 1 sinπ = 0 1,0

3π2

cos2 3π2

= 0 sin 3π2

= −1 0,−1 verander rigting / Change direction

2π cos22π = 1 sin2π = 0 1,0

, t ∈ 0,2π

y

x

(0,-1)

(0,1)

(1,0)

y

x

(0,-1)

(0,1)

(1,0)

3

wtw168 semester test 1 p 5 of 7

Vraag 3 / Question 3

Laat / Let fx = ln1 + x.

3.1 Bepaal die derde graadse Maclaurin polinoom (Taylor polinoom om a = 0).

Find the third degree Maclaurin polynomial (Taylor polynomial about a = 0).

Ans:

fx = ln1 + x f0 = ln1 + 0 = 0

f′x = d

dx ln1 + x = 11+x

f′0 = 1

1+0= 1

f′′x = d

dx1

1+x= −1

1+x 2 f′′0 = − 1

1+0 2 = −1

f′′′x = d

dx−1

1+x 2 = 21+x 3 f

′′′0 = 2

1+0 3 = 2

∴ fx = ln1 + x P3x = 00!

x0 + 11!

x1 + −12!

x2 + 23!

x3 = x − 12x2 + 1

3x3

3.2 Gebruik die antwoord in 3.1 om die waarde van ln1.1 te bepaal. Gee jou

antwoord tot 3 desimale syfers korrek.

Use the answer in 3.1 to find the value of ln1.1. Give your answer

correct to 3 decimals.

Ans: ln 1.1 = ln1 + 0.1 P30.1 = 0.1 − 12

0.1 2 + 13

0.1 3

= 0.09533 0.095(3 des / 3 dec)

4Vraag 4 / Question 4

Bepaal die oppervlakte van die omwentelingsliggaam gevorm wanneer y = x, 1 ≤ x ≤ 4

om die x − as wentel.

Find the area of the surface of revolution when y = x, 1 ≤ x ≤ 4 is rotated

about the x − axis.

Ans: S = ∫a

b 2πy 1 + f ′x 2 dx, f ′x = dydx = d

dx x = 12 x

∴ S = ∫a

b 2πy 1 + dydx

2 dx = ∫1

4 2π x 1 + 12 x

2 dx

= ∫1

4 2π x 4x+14x

dx = π ∫1

4 4x + 1dx = π6

4x + 1 32

1

4 = π6

17 32 − 5 3

2 30.846

4Vraag 5 / Question 5

’n Partikel beweeg een keer kloksgewys om die sirkel x2 + y − 1 2 = 4.

Die partikel begin by die punt 2,1 . Gee die parametriese vergelykings vir die beweging.

A particle moves around the circle x2 + y − 1 2 = 4 once and in a clockwise direction.

The particle starts at the point 2,1 . Give the parametric equations of the movement.

Ans: Laat / Let x = 2cos t, y − 1 = −2sint

wtw168 semester test 1 p 6 of 7

dan / then x = 2cos t, y = −2sint + 1, t ∈ 0,2π .

Toets antwoord: / Test answer

t x y x,y

0 2cos0 = 2 −2sin0 + 1 = 1 2,1 beginpunt / starting point

π2

2cos π2= 0 −2sin π

2+ 1 = −1 0,−1

π 2cosπ = 2 −1 = −2 −2sinπ + 1 = 1 −2,1

3π2

2cos 3π2

= 0 −2sin 3π2+ 1 = 3 0,3

2π 2cos 2π = 2 −2sin2π + 1 = 1 2,1 eindpunt / end point

Skets / Sketch 2cos t,−2sint + 1

Y

X

-1

0

1

3

-2 -1 1 2

Y

X

Y

X

-1

0

1

3

-2 -1 1 2

wtw168 semester test 1 p 7 of 7