Question

Help me with 6ii and 8i please.
http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w05_qp_3.pdf

Answer 1

@ganeshie8 @perl dont mind lending a hand?

Answer 2

6ii try sub x = cos^2 theta.

Answer 3

I don't get you

Answer 4

ooop sorry, they gave you the equivalent sub in 6 i. OK, now you just need to integrate 2 sin^2 (theta) and plug in the limits. is that the problem?

easiest way to integrate a sin^2 or cos^ is to use double angle formula.

so cos 2ø = 2 cos^2 ø - 1 =1 - 2 sin^ ø

manipulate that and the integration is easy. does that help?

Answer 5

I did that. Idk how to integrate it.

Answer 6

sub \[\sin^{2}\theta = \frac{ 1-\cos2\theta }{ 2 }\]

Answer 7

from the double angle formula given above:

∫2sin^2 ø = ∫ 2(1 - cos2ø) /2 = ∫ (1 - cos2ø)

yes?

Answer 8

Yess,

Answer 9

and change the limits too. you used x = sin^ theta so lims are now ?? <= theta =< ??

Answer 10

then for the \[\cos 2\theta\] , use u=2theta, du=2dx

Answer 11

I realized the limit is 1/6 theta.how do yo you change that?

Answer 12

So is it ( theta - sin theta/2) ?

Answer 13

mmm.

you used x = sin^2 theta as the sub, ie theta = arcsin (sqrt(x))

thus, the limits are arcsin (sqrt(0)) and arcsin (sqrt(1/4)).

yes?

Answer 14

I don't get you

Answer 15

""So is it ( theta - sin theta/2) ?""

y.

Answer 16

Could please type out the steps for me ? I guess, it'd help more that way.

Answer 17

is this what you have now?

Answer 18

I dont get the limits. hoW do change 1/4 to theta/6??

Answer 19

Heyy I got it alrd. thanks!

Answer 20

Could you pls help me out with question 8i)?

Answer 21

ok, in the integration you let x = sin^ theta. where 0 =< x =< 1/4. so at the upper limit , x = sin^2 theta = 1/4
thus sin theta = 1/2, so theta = arcsin (1/2)

OK?

after that, you just use your experience - as sin theta is 1/2, you have opposite = 1, hypoteneuse = 2, and adjacent = sqrt(3). and you should recognise this as a 60deg, 30 deg right angled triangle.

Answer 22

Yes I got it alrd. Help me wih wuestion 8 Pls?

Answer 23

sure. have you done part i OK? and did you get something like x(t) = 100 exp(-kt^2 /2)

Answer 24

Im asking for i

Answer 25

ok, sorry.

you hve dx/dt = -kxt

thus dx/x = -kt dt

or ∫1/x dx = ∫ -kt dt

that ok?

Answer 26

Uh huh

Answer 27

this has a fancy name -- separation of variables

you are given dx/dt = - kxt

x is something that varies with both t and itself...

but you need to separate the x's and t's -- ie divide both sides by x to get x to the LHS keeping t on the RHS. you get:
1/x dx/dt = - kt

this bit i really do not like but it is how it is commonly taught nowadays-- multiply both sides by dt to get dt from LHS to RHS

1/x dx = -kt dt

you can then integrate from here.

you have ∫ 1/x dx = ∫ -kt dt

integrate LHS and RHS separately. but keeping the equals sign!!

Answer 28

Ln x = -1/2 kt^2 + c

Answer 29

Right?

Answer 30

excellent

as an *aside*, the k/2 could be replaced by another constant as the /2 bit will just get in the way, but maybe it would be more instructional to keep it as k/2 rather than say P , q or whatever, for now.

so, now for the limits.

x(0) = 100. what is c?

Answer 31

So ln x = -k/2 t^2 + c
x = e^0 . e^c
Right?

Answer 32

e^c = 100?

Answer 33

Is c = ln100??

Answer 34

you can keep it as ln(100) = c, or c = ln(100) obviously.

that gives you:
ln x = -k/2 t^2 + ln(100)

then you can bring the ln(100) to the LHS and combine it with ln(x) using your knowledge of logs.

that makes it neater. does that make any sense?

Answer 35

sorry, crossed messages. my broadband is very slow and i have this problem all the time.

in the last message, i suggest combining ln(x) and ln(100). is that OK?

Answer 36

It's okay. I feel you lol.
The answer scheme says just leave it as it is. :) what about ii?

Answer 37

ii has got some messy algebra in it, but you see that you are given x(20) = 90 and you have to find T where x(T) = 50. [i made T up, obviously, call it t* or whatever.....]

the first one is given so you can calc k, then using that k you can what T is.

Answer 38

ln 90 = -k/2 20^2 + ln 100
ln 50 = -k/2 T^2 + ln100

agreed?

Answer 39

if you agree with the equations, is there a smart way to do this that avoids the mess and skips having to solve for k?!?!

Answer 40

t 2 = 400(ln 100 – ln 50)/(ln 100 – ln 90)

Answer 41

bingotastic!!!

of course ln100 - ln 50 = ln 100/50 yes?