Question

Question 7ii and iii please! http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w07_qp_3.pdf

Answer 1

from part (i)
\[Log_e {\frac{N}{125}} = 50 K \sin (0.002t)\]OR\[N = 125 e^{50Ksin(0.002t)}\]

Answer 2

use either of these equations; just plug in N = 166 and t = 30, you''ll get something like \[\log_e {\frac{166}{125}} = 50 K \sin(0.002\times 30)\]

Answer 3

solve for k

Answer 4

Yeah I can't seem to be getting the a answr

Answer 5

# correction its sin(0.02t) and not sin(0.002t)

Answer 6

Yeah yeah I did my own working and all alrd can't get the answer.

Answer 7

\[k = \frac{\log(\frac{166}{125})}{50\sin (0.6)} =0.28367\]

Answer 8

Thats not the answer. It's 0.01

Answer 9

you are right, it should be 0.0104 . I mixed up the calculation. Solve the expression up there for k and you'll get 0.0104

Answer 10

set you calculator to radians

Answer 11

Yessss finally.

Answer 12

What about iii

Answer 13

Thank You @IrishBoy123
when we plug this value of k in the expression for N, we get\[N = 125 e^{0.5\sin(0.02t)}\]
Since N varies exponentially with t. t is time here so it cannot take negative values thus the smallest vale of N will be when t = 0
at t = 0, \[N = 125e^{0.5sin0 } = 125 \times e^0 = 125 \]

Answer 14

Youmean ln 125??

Answer 15

just 125. on day 0 according to this model we had 125 insects

Answer 16

ln125 ~ 4.828? is this a possible number of insects?

Answer 17

if have used the second equation here; given in my first post (on top)

Answer 18

I thought we must sett sin 0.02t = -1?

Answer 19

i think you need to differentiate and checkout the stationary points

Answer 20

yes

Answer 21

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w07_ms_3.pdf Taken a look at the answer scheme.

Answer 22

Why sin 0.02t =-1 tho

Answer 23

Take *

Answer 24

dN/dt = 125 (0.5) cos (t/50) (1/50) exp (0.5 sin (t/50))

that will be a max or min at dN/dt = 0, ie cos t/50 = 0 so check out t/50 = π/2, 3π/2, 5π/2, easy on a claculator

Answer 25

that's when cos t/50 = 0 as per the derivative

Answer 26

I don't get it

Answer 27

crap i messed up. @IrishBoy123 is right. the derivative test will fail here as at (t = 25pi) the second derivative will also be zero, and will give a saddle point.
Instead just think about the value of N being least. For this the exponential function must be least. which will happen when 0.5sin(0.02t) takes the lest value possible.
Sin (0.02t) = -1 because that the least value of the sine function and for it s least value the exponent will give is the least value N = 125 e^(-0.5) ~ 75.81

Answer 28

I see thanks! Could help me with question 5ii ? http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w06_qp_3.pdf

Answer 29

when you rationalize the denominator, you'll get\[\frac{1}{\sqrt{1+x}+\sqrt{1-x}}\times \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\]

Answer 30

\[\frac{\sqrt{1+x}-\sqrt{1-x}}{(\sqrt{1+x})^2-(\sqrt{1-x})^2}= \frac{\sqrt{1+x}-\sqrt{1-x}}{1+x-1+x}= \frac{\sqrt{1+x}-\sqrt{1-x}}{2x}\]

Answer 31

This is what I did for i

Answer 32

I'm asking help for ii not I

Answer 33

now it can be expanded using Taylor's series

Answer 34

binomial

Answer 35

or you could just use the binomial theorem. are you familiar with either?

Answer 36

Yes I'm using the (1+x)^n binomial formula, I still can't get my answer

Answer 37

make sure you apply the second binomial as (1 + (-x)) and not as (1-x).

trite comment but common error.

Answer 38

\[\frac{\sqrt{1+x}}{x}= \frac{1+(1/2)x-(1/4)x^2+(3/8)x^3 }{x}\]

Answer 39

missing the factorials

Answer 40

I still can't get it. Been trying for a few days tho. 😭