DMO’L.St Thomas More C4: Starters Revise formulae and develop problem solving skills. 123456789...

DMO’L.St Thomas More

C4: Starters

Revise formulae and develop problem solving skills.

1 2 3 4 5 6 7 8 9

10

11

12

13

14

15

16

17

18

19

20 21

22 23 24 25 26 27

28 29 30 31


Starter 1

Express in partial fractions.

Hence find

)3)(2(

26

xx

x

dxxx

x

)3)(2(

26


Starter 1

Express in partial fractions.)3)(2(

26

xx

x

32)3)(2(

26

x

B

x

A

xx

x

)2()3(26 xBxAx

Bx 5203

4B

Ax 5102

2A

Hence

3

4

2

2

)3)(2(

26

xxxx

x


Starter 1

dxxx

dxxx

x

3

4

2

2

)3)(2(

26

cxx 3ln42ln2

Back


Starter 2


Hence find

42

2

x

x

dxx

x 42

2


Starter 2


221

4

41

4 22

2

x

B

x

A

xx

x

)2()2(4 xBxA

Bx 442

1B

Ax 442

1A

Hence

2

1

2

11

42

2

xxx

x

42

2

x

x


Starter 2

dxxx

dxx

x

2

1

2

11

42

2

cxxx 2ln2ln

Back

4ln 2 xkx


Starter 3

2tx

Find the cartesian equation of the curve given by the parametric equations

2ty


Starter 32tx

Find a way to eliminate t

2ty

2yt22 )2( ytx

xy 2)2(Back


Starter 4

Find the cartesian equation of the curve given by the parametric equations

1

1

tx

2

1

ty


Starter 4

1

1

tx

2

1

ty

Find a way to eliminate t

xt

11 3

12

xt

3

11

x

y

x

xy

31Back


Starter 5

Find the cartesian equation the curve given by the parametric equations

tx sin2

ty cos3


Starter 5

tx sin2 ty cos3Find a way to eliminate t

tx 2

2

sin2

ty 2

2

cos3

ttyx 22

22

cossin32

Back132

22

yx


Starter 6

Find the coordinates of the points where the following curves meet the x,y axes

tx 5 ty 6

050 tx 115 yt )11,0(

060 ty 116 xt )0,11(

Back


Starter 7

Find the coordinates of the points where the following curves meet the x,y axes

ttx 1

2 9ty

020 tx 90 yt )9,0(

090 ty 10189 xt )0,8.1(

Back


Starter 8

Find dy/dx leaving your answer in terms of t. tx 2 tty 42

2dt

dx

Back

42 tdt

dy

2

42

t

dx

dt

dt

dy

dx

dy

2tdx

dy


Starter 9

Find dy/dx leaving your answer in terms of t. tx sec ty tan

ttdt

dxtansec

Back

tdt

dy 2sec

tt

t

dx

dt

dt

dy

dx

dy

tansec

sec2

tdx

dy

sin

1


Starter 10Find the equation of the tangent to the curve defined by the following parametric equations at the point P where t = /2

ttx sin46 ttty cos2 tdt

dx cos46 ttttdtdy sincos2

43 xAt P t = /2 so that

4

2y 12dx

dy

)43(124

2 xygiving 412 xy

Back


Starter 11

Evaluate

xdx4

0

2cos

xdx4

0

2cos

1cos22cos 2 xx

Back

)2cos1(cos 212 xx dxx)2cos1(4

021

4

022sin

21

xx

41

8


Starter 12

Complete the table:

ydxdy

x

x

n

a

e

x

x

x

x

x

kx

ln

sec

tan

cos

sin

aa

e

xx

x

x

x

nkx

x

x

x

n

ln

tansec

sec

sin

cos

1

2

1

Back


Starter 13

Complete the table:

ydxdy

x

x

n

e

x

x

x

x

x

x

5

6

3

2

3

6ln

7sec

4tan

cos

2sin

)13(

3ln)3(5

6

7tan7sec7

4sec4

sincos3

2cos2

)13(6

5

6

1

2

2

12

x

x

x

n

e

xx

x

xx

x

xnx

Back


Starter 14

Complete the table:

y dxy

xe

e

xx

x

xx

x

x

x

x

x

cos

tansec

4sec

sincos

2sin

13

sin

6

321

2

3

2

x

x

x

x

x

e

e

x

x

cxx

sin

661

21

44tan

4cos

22cos

3

32ln

sec

4

Back


Starter 15

Evaluate

xdx4

0

2tan

xdx4

0

2tan

xx 22 sectan1

Back

1sectan 22 xxdxx )1(sec4

0

2

40tan

xx

41


Starter 16

Evaluate

dxxx

x

4

1 222

2

dxxx

x

4

1 222

2 xxu 22

Back

22 xdudx

24

322

4

1

22udu

xdu

x

x ux

243ln u

3ln24ln

8ln


Starter 17

In each case finddxdy

1022 yx

Back

in terms of x and y

32 23 xxyy

yx xeye

022 dxdyyx y

xdxdy

23226 xxyy dxdy

dxdy xy

yxdxdy

2623 2

dxdyyy

dxdyxx xeeeye xx

xy

xee

yeedxdy


Starter 18

Find xdxx ln2

xdxx ln2 xu ln

Back

2xdxdv

dxxxx 231

3 ln3

cx xx 93

33

ln

xdxdu 1 3

3xv


Starter 19

Find xdxln

xdxln xu ln

Back

1dxdv

dxxx 1ln

cxxx ln

xdxdu 1 xv


Starter 20

Find dxex x2

dxex x2

2xu

Back

xdxdv e

dxxeex xx2

xdxdu 2 xev

xu 2 xdxdv e

2dxdu xev

dxexeex xxx 222

cexeex xxx 222


Starter 21Use the trapezium rule with 6 strips to estimate

dxx 3

0

2 )1ln( x 1st/last others

0 0

0.5 0.2231

1 0.6931

1.5 1.1787

2 1.6094

2.5 1.9810

3 2.3026

2.3026 5.6854



x 1st/last others

0 0

0.5 0.2231

1 0.6931

1.5 1.1787

2 1.6094

2.5 1.9810

3 2.3026

2.3026 5.6854

dxx 3

0

2 )1ln(



x 1st/last others

0 0

0.5 0.2231

1 0.6931

1.5 1.1787

2 1.6094

2.5 1.9810

3 2.3026

2.3026 5.6854

)6854.523026.2(25.0 42.3 To 3 sig.

fig.

Back

dxx 3

0

2 )1ln(



dxx 3

0tan1

x 1st/last others

0 1 /12 1.1260/6 1.2559/4 1.4142/3 1.6529

2.6529 3.7962



dxx 3

0tan1

x 1st/last others

0 1 /12 1.1260/6 1.2559/4 1.4142/3 1.6529

2.6529 3.7962

)7962.326529.2(24

34.1 To 3 sig. fig.

Back


Region A is bounded by the curve with equation , the lines x = 1, x = 0 and the x-axis.

The region A is rotated through 360o about the x-axis

Find the volume generated.

Starter 23

32 xy

1

0

2dxyVolume

1

0

24 )96( dxxx

1

0

35 925

xxx 5

56

Back


Points A and B have position vectors i + j + k and 2i - 3j + 2k respectively.

Find the vector equation of the straight line through A and B.

Starter 24

AB = (2i - 3j + 2k) – (i + j + k)


Points A and B have position vectors i + j + k and 2i - 3j + 2k respectively.

Find the vector equation of the straight line through A and B.

Starter 24

AB = (2i - 3j + 2k) – (i + j + k)

= i – 4j + k Hence, a vector equation is;

r = i + j + k + (i – 4j + k) Back

DMO’L.St Thomas Moreangle

Find the acute angle between the two lines with vector equations

r = 2i + j + k +t(3i – 5j – k)

and r = 7i + 4j + k +s(2i + j – 9k)

Starter 25

Consider the angle between their direction vectors; a = (3i – 5j – k) and b = (2i + j – 9k)

Cosine of angle bab.a

863510 1823.0

o5.79 Back


Starter 26

The direction vector of the line is

a = i + j +k

A line has vector equation

r = 3i + 5j - k +t(i + j +k)

Find the position vector of the point P, on the line, such that OP is perpendicular to the line.

When t = OP a


Starter 26

The direction vector of the line is

a = i + j +k

A line has vector equation

r = 3i + 5j - k +t(i + j +k)

Find the position vector of the point P, on the line, such that OP is perpendicular to the line.

When t = OP a

OP . a = 0


Starter 26

When t = OP a

OP . a = 0

0 k) j k).(i)1(j)5(i)3((

0 )1()5()3(

0 73

3-7

So P has position vector

OP = 3i + 5j - k -7/3(i + j +k)

Back


Starter 27Find the of the tangent to the given curve at the point (1,0).

yxyx 23)(

Differentiate;

dxdy

dxdy xyx 2)1()(3 2

At (1,0) dx

dydxdy 2)1(3

21dx

dy

Hence tangent is12 xy

Back


Starter 28A curve has parametric equations x = 4cos and y =

8sin

(a)Find the gradient of the curve at P, the point where = /4

(b)Find the equation of the tangent to the curve at P.

(c) Find the coordinates of the point R where the tangent meets the x-axis.

(d)Find the area of the region bounded by the curve, the tangent and the x-axis.



8sin

(a)Find the gradient of the curve at P, the point where = /4 dx

dtdtdy

dxdy

sin41cos8 dx

dy

cot48dx

dy

cot48dx

dy

At P = /4;

2dxdygradient



8sin

(b) Find the equation of the tangent to the curve at P.At P = /4; 22cos4

24

4 x

2gradient

24sin82

84 y

Equation of tangent; 282 xy



8sin

(c) Find the coordinates of the point R where the tangent meets the x-axis.

At R y=0 282 x

24 x

)0,24( R



8sin

(d) Find the area of the region bounded by the curve, the tangent and the x-axis.

0

4

ydxAreaArea

d

d

dxy

4

02

2422

d 4

0)sin4(sin88

d 4

0

2sin328

d 4

02cos168 4

022sin168

4

Back


Starter 29Find the general solution of each differential

equation: xydxdy tantan

10dxdyyxe

xdxdy

y tantan1

dxxydy tancot

cxy seclnsinln

xky secsin

10 dxdyyxee

dxedye xy 10

cee xy

Back


The region R is bounded by the curve C, the x-axis and the lines x = -8 and x = 8.

The parametric equations for C are x = t3 and y = t2

Find the area of R.

Area under curve

8

8ydx

8

8

x

xdt

dt

dxy

23tdt

dx

2

2

43t

tdtt

2

2

5

5

3

t4.38

Starter 30


The region R is bounded by the curve C, the x-axis and the lines x = -8 and x = 8.

The parametric equations for C are x = t3 and y = t2

The region R is rotated about the x-axis, find the volume generated.

Volume 8

8

2dxy

8

8

2x

xdt

dt

dxy 23t

dt

dx

2

2

63t

tdtt

2

2

7

7

3

t 7

768

Starter 30

Back


A curve has equation

Find the coordinates of the points on the curve where

Differentiate w.r.t. x

01632 22 yxyx

06222 dxdy

dxdy yxyx

Starter 31

0dxdy

0dxdy 00022 yx

xy

What’s this?

Sub. back 01632 222 xxx

2x

Back2;2 yx

DMO’L.St Thomas More C4: Starters Revise formulae and develop problem solving skills. 123456789...

Documents

Transcript of DMO’L.St Thomas More C4: Starters Revise formulae and develop problem solving skills. 123456789...