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# Edexcel Maths C3 watch

1. Hi,

Can someone help me with C3 Jan 2009, Q8 part (C) and (D).

Thanks

http://dl.dropbox.com/0/view/4m89963...Jan%202009.pdf
2. (Original post by emdad16)
Hi,

Can someone help me with C3 Jan 2009, Q8 part (C) and (D).

Thanks

http://dl.dropbox.com/0/view/4m89963...Jan%202009.pdf
well,

from part b) the equation is 5 cos (x - 53)

the range of cos x is between -1 and 1

so the range of 5 cos x is between -5 and 5

max is 5, min is -5

so in c) when we add 10 to this the max is 15 and the min is 5

the min occurs when the cos graph is equal to -1 which occurs at 180 degrees

so this tells us that (x - 53) = 180 so x = 233

in c+d, x = 15t so 15t = 233 t = 15.5
3. (Original post by emdad16)
Hi,

Can someone help me with C3 Jan 2009, Q8 part (C) and (D).

Thanks

http://dl.dropbox.com/0/view/4m89963...Jan%202009.pdf
If you replace 15t with feta(pheta?) it should become clear.

You can then replace most of the equation with your answer to part a, and so the minimum temperature is 10 - 5 = 5.
4. also for x=4sin(2y+6) find dy/dx in terms of x i dont understand this arcsin thing
5. (Original post by emdad16)
Hi,

Can someone help me with C3 Jan 2009, Q8 part (C) and (D).

Thanks

http://dl.dropbox.com/0/view/4m89963...Jan%202009.pdf
ok here goes:
part c) the minimum on a cos graph is -1, so the minimum would be "10 + 5(-1)" which is 10 - 5 = 5

part d) put minimum=5 into equation so you get
5 = 10 + 5cos(15t - 53...)
-5 = 5cos(15t - 53..)
-1 = cos(15t - 53..)
therefore (15t - 53..) = 180
you can then work out "t" ...
hope that helped ...
6. (Original post by The Magnificent KoloToure)
If you replace 15t with feta(pheta?) it should become clear.

You can then replace most of the equation with your answer to part a, and so the minimum temperature is 10 - 5 = 5.
*theta.
7. why is X = 15t
8. (Original post by Loz17)
*theta.
Cheers.
9. oh my god the specimen papers a beast.
10. Should I do the specimen paper or Jan '10? Just started the latter.
11. (Original post by Narik)
Why though?
okay

you need to understand how composite functions work. It isn't as simple as sketching it and then looking at the possible y values on the graph for the range of the composite functions.

fg(x) means

2) put in into function g
3) take the output value g(x) and put into f
4) this gives us fg(x)

now any restrictions on the domains of f and g can have an impact on the composite function; also the ranges of f and g have an impact on the composite function; we have

f(x) = e^2x + 3 domain is anything
g(x) = ln(x-1) domain is RESTRICTED to x>1

so in step 1 our original values are restricted to just those >1

however any if we put in all values >1 into g ( ln(x-1) ) the range is every possible y value (sketch it!)

so any possible value enters f, the second function. if you sketch e^2x +3 for any possible x value, you will see it has an asymptote at y=3. It never reaches 3

Therefore the range is >3 and not >= 3

or think of it another way

what value of x do you put in to fg(x) get an answer of 3 out.

1.... but you can not put this in because we are told the domain of g is x>1....!!!!!!

the domain matters!
12. (Original post by emdad16)
why is X = 15t
that is what the question says it is.
13. Oh yh soz
and fanks BTW

much appreciated
14. (Original post by houseo2)
also for x=4sin(2y+6) find dy/dx in terms of x i dont understand this arcsin thing
dx/dy is very easy just differentiate with respect to y

dx/dy = 2 x 4cos(2y+6) = -8cos(2y+6)

so flip both sides

dy/dx = -1/(8cos(2y+6))

the annoying bit is making it in terms of x

now if x = 4sin(2y+6)

x/4 = sin(2y+6)

so we now need to move sin to the opposite side where it becomes its inverse function - this is called arc sin (sin^-1 on your calc)

arc sin (x/4) = 2y+6 now sub thi sback into dy/dx to get

dy/dx = -1/(8cos(arcsin(x/4)))
15. (Original post by Arsey)
dx/dy is very easy just differentiate with respect to y

dx/dy = 2 x 4cos(2y+6) = -8cos(2y+6)

so flip both sides

dy/dx = -1/(8cos(2y+6))

the annoying bit is making it in terms of x

now if x = 4sin(2y+6)

x/4 = sin(2y+6)

so we now need to move sin to the opposite side where it becomes its inverse function - this is called arc sin (sin^-1 on your calc)

arc sin (x/4) = 2y+6 now sub thi sback into dy/dx to get

dy/dx = -1/(8cos(arcsin(x/4)))
ohhh! thanks for explaining this!! ive always wanted to know what the heck arc sin is!
16. (Original post by Kaya_01)
imran that wouldnt be the case IF YOU WENT TO UR LESSONS :P

I do go to my lessons, i just dont learn anything in the lessons, oh well looks like i will be repeating c3 next year
17. (Original post by Arsey)
dx/dy is very easy just differentiate with respect to y

dx/dy = 2 x 4cos(2y+6) = -8cos(2y+6)

so flip both sides

dy/dx = -1/(8cos(2y+6))

the annoying bit is making it in terms of x

now if x = 4sin(2y+6)

x/4 = sin(2y+6)

so we now need to move sin to the opposite side where it becomes its inverse function - this is called arc sin (sin^-1 on your calc)

arc sin (x/4) = 2y+6 now sub thi sback into dy/dx to get

dy/dx = -1/(8cos(arcsin(x/4)))
Could you not also do implicit differntation so you get 1 = 8cos(2y+6)(dy/dx)

Just wondering where did you get the "-8" from?
18. (Original post by Arsey)
dx/dy is very easy just differentiate with respect to y

dx/dy = 2 x 4cos(2y+6) = -8cos(2y+6)

so flip both sides

dy/dx = -1/(8cos(2y+6))

the annoying bit is making it in terms of x

now if x = 4sin(2y+6)

x/4 = sin(2y+6)

so we now need to move sin to the opposite side where it becomes its inverse function - this is called arc sin (sin^-1 on your calc)

arc sin (x/4) = 2y+6 now sub thi sback into dy/dx to get

dy/dx = -1/(8cos(arcsin(x/4)))
thask but it then says 1/2root 16x^2-x^2
19. can someone answer my other questions as well please
20. also lol https://eiewebvip.edexcel.org.uk/Rep...e_20070118.pdf

question 8 ii

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