Flashcards in C4 Deck (54):

1

## what is the remainder when p(x) is divided by (ax-b)

###
p(b/a)

solution of ax-b=0

2

## how can (5x+1)/(x-1)(2x+1)(x-5) be expressed in partial fraction form

### A/(x-1) + B/(2x+1) + C/(x-5)

3

## how can (5x+1)/(x-1)(2x+1)^2 be expressed in partial fractions

### A/(x-1) + B/(2x+1) + C/(2x+1)^2

4

## how to convert a pair of parametric equations into a single cartesian equation

###
elimiate the parameter

e.g x=t+4 and y=1-t^2

using x=t+4 t=x-4

so t^2=x^2-8x+16

put into the y equation to make y=1-(x^2-8x+16)

5

## curve x=rcosϴ and y=rsinϴ

### circle with radius r and centre is the origin

6

## curve x=rcosϴ+p and y=rsinϴ+q

### circle with radius r and centre (p,q)

7

## curve x=acosϴ and y=bsinϴ

###
ellipse

centre is the origin

width is 2a and height is 2b (because a and b are only half the measurement as to the radius)

8

##
curve is x=2cosϴ+3 and y=sinϴ -1

find cartesian equation of curve

###
isolate cos and sin

cosϴ=(x-3)/2

sinϴ=y+1

use the identity sin^2ϴ+cos^2ϴ=1

(y+1)^2 + (x-3/2)^2 = 1

9

##
translation (a)

(b)

### add a to x function and b to y function

10

## stretch in x direction

### multiply the x function by the required factor

11

## stretch in y direction

### multiply the y function by the required factor

12

## reflection in y axis

### multiply the x function by -1

13

## reflection in x axis

### multiply the y function by -1

14

## formula for binomial expansion of (1+ax)^n

### 1+ nax + n(n-1)/2! (ax)^2 + n(n-1)(n-2)/3! (ax)^3 + ...

15

## to do binomial expansion of (x+5)/(3-x)(1+3x)

### split expression into partial fractions then expand

16

##
sin(A+B)=

sin(A-B)=

###
sinAcosB + sinBcosA

sinAcosB - sinBcosA

17

##
cos(A+B)=

cos(A-B)=

###
cosAcosB - sinAsinB

cosAcosB + sinAsinB

18

##
tan(A+B)=

tan(A-B)=

###
tanA+tanB/(1-tanAtanB)

tanA-tanB/(1+tanAtanB)

19

## sin2A=

### 2sinAcosA

20

## cos2A=

###
cos^2A-sin^2A

2cos^2A-1

1-2sin^2A

21

## tan2A=

### 2tanA/(1-tan^2A)

22

## write 3sinx + 4cosx in the form Rcos(x-a)

###
make them equal each other

3sinx+4cosx = Rcos(x-∝)

sub in ange formula for cos(x-∝)

3sinx+4cosx=R(cosxcos∝+sinxsin∝)

group together the constants

=Rcos∝cosx + Rsin∝sinx

left with 3=Rsin∝ and 4=Rcos∝ as cosx and sinx cancel as on both sides

find R by squaring both equations and adding them together

3^2+4^2=R^2sin^2∝+R^2cos^2∝

25=R^2(sin^2∝+cos^2∝)

25=R^2

R=+or- 5

we will choose R=5 and sub into one of the equations

3=5sin∝

sin∝=3/5

∝=0.643501

put value into other equation

4=5cos0.643501

4=4 so correct value

(if it doesn't work find another value on cos graph)

5cos(x-0.644)

23

## find the maximum value of 3sinx+4cosx

###
y value

5cos(x-0.644)

translation (0.644,0)

stretch s.f 5 in y axis so maximum value 5

24

## fidn the smallest value of x for which the maximum value occurs

###
5cos(x-0.644)=5 (max value)

cos(x-0.644)=1

let A = x-0.644

cosA=1

A=cos^(-1)(1)

A=0

so x-0.644=0

x=0.644

25

##
solve this differential equation

dy/dx=2x(y+4)

###
separate the variables so you have all xs on one side and all ys on the other

dy/(y+4)=2xdx

integrate both sides

∫1/(y+4)dy = ∫2xdx

ln(y+4) = x^2 + c

26

## what equation represents exponential growth

### y=ae^bt

27

## what equation represents exponential decay

### y=ae^-bt

28

## rate of change inversely proportional to cube of time t

###
dx/dt ∝ 1/t^3

dx/dt = k/t^3

29

## rate of decrease directly proportional to square root of x

### dx/dt ∝ -(x)^1/2

30

##
P(t) is population at any time t

Po is intial population at time t=0

population grows exponentially

###
dP/dt ∝ P

dP/dt = kP

separate the variables

1/P dP = kdt

intergrate

∫1/P dP = ∫k dt

ln|P| = kt + c

|P| = e^(kt+c)

|P|=e^c x e^kt

when t=0, P=Po

|Po|=e^c x e^kx0

Po=e^c

put back into original equation

|P|=Poe^kt

P=Poe^kt

31

## differentiate xy^2 with respect to x

###
using the product rule

d/dx (xy^2) = d/dx(x)y^2+xd/dxy^2

1xy^2 + x2ydy/dx

y^2 + 2xy dy/dx

32

## ∫(x+9)/(x-3)(x+1) dx

###
split up into partial fractions

x+9=A(x-3) +B(x+1)

when x+3, 12=4B so B=3

when x=-1, 8=-4A so A = -2

3/(x-3) - 2/(x+1)

integrate separately

=∫3/(x-3)dx - ∫2/(x+1)

then make top equal differential of bottom (in this case both 1) so you can apply the ln rule

3∫1/(x-3)dx-2∫1/(x+1)dx

=3ln|x-3|- 2ln|x+1| +c

33

## ∫sinxcosxdx

###
use double angle formula of sin2x=2sinxcosx

so sinxcosx=1/2sin2x

∫1/2sin2x = -1/4cos2x + c

34

## the modulus of a vector

###
is its magnitude

|a|

35

##
A = 2

B = 3

find magnitude of AB

###
do pythagorus

AB = 2^2 + 3^2

= (13)^1/2

36

## what is the direction of a vector

### the angle measured anticlockwise from vector i (x axis)

37

##
find magnitude and hence direction of

(6)

(5)

###
magnitude = 6^2 + 5^2 = (61)^1/2

angle we are trying to calculate is between 6 and (61)^1/2

opposite angle = 5

adjacent angle = 6

hypotenuse= (61)^1/2

use Toa so tanϴ=5/6

ϴ=tan^-1(5/6)

ϴ=39.8 degrees

38

## coordinates of -2i-3j

###
(-2,-3)

i is parallel to x axis

j is parallel to y axis

k is parallel to z axis

39

## if a vector is parallel to a

###
it is a multiple of a

e.g a = kb

40

## -a and a

### same magnitude but opposite direction

41

## when are vectors equal

### if they have the same magnitude and direction

42

## base vector component form

### e.g 4i-2j

43

## column vector form

###
e.g (7)

(2)

44

## unit vector

### vector with magnitude of 1

45

## position vector

###
position of a point in vector form

e.g OP (1)

(2)

(3)

46

## displacement vector

###
how to get from one point to another

e.g P(1,2,3) and Q(-1,2,-3)

so PQ(-2)

(0)

(-6)

47

## collinear

###
if they lie on the same straight line

proven by showing AB = kAC

48

## how to find AB when given OA and OB

### AB = OB - OA

49

## vector equation

###
r = a + tb

r is position vector of a general point on the line

a is position vector of a particular point on the line

b is a vector in the direction of the line

50

##
find the vector equation of the line through (5)& (7)

(6) (9)

###
pick either vector for a

b is the vector between them so

(x) = (5) + t(2)

(y) (6) (3)

51

## dot product rule

### a.b = |a| |b|cosϴ

52

##
find angle between a(2) and b (1)

(3) (-1)

###
|a| is (2^2 + 3^2)^1/2 = (13^1/2)

|b| is (1^2+(-1)^2))^1/2 = (2)^1/2

(2).(1)=13^1/2x2^1/2cosϴ

(3) (-1)

(2).(1) = (2x1) + (3x-1)= -1

(3) (-1)

-1= (13)^1/2 x (2)^1/2 x cosϴ

ϴ = cos^-1(-1/(13^1/2 x 2^1/2)

ϴ = 101.3

53

## How to show 2 vector line equations intersect

###
Combine the line equation to include the t and u values

Make the 2 new equations equal to each other

Find u and find t

Put values back into new equations to find point of intersection

54