Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    0
    ReputationRep:
    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
    Offline

    7
    (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
    Offline

    0
    ReputationRep:
    (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.
    Offline

    0
    ReputationRep:
    also for x=4sin(2y+6) find dy/dx in terms of x i dont understand this arcsin thing
    Offline

    0
    ReputationRep:
    (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 ...
    Offline

    3
    (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.
    Offline

    0
    ReputationRep:
    why is X = 15t
    Offline

    0
    ReputationRep:
    (Original post by Loz17)
    *theta.
    Cheers.
    Offline

    0
    ReputationRep:
    oh my god the specimen papers a beast.
    Offline

    0
    ReputationRep:
    Should I do the specimen paper or Jan '10? Just started the latter.
    Offline

    7
    (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

    1) start with input value x
    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!
    Offline

    7
    (Original post by emdad16)
    why is X = 15t
    that is what the question says it is.
    Offline

    0
    ReputationRep:
    Oh yh soz
    and fanks BTW

    much appreciated
    Offline

    7
    (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)))
    Offline

    0
    ReputationRep:
    (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!
    Offline

    2
    ReputationRep:
    (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
    Offline

    0
    ReputationRep:
    (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?
    Offline

    0
    ReputationRep:
    (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
    Offline

    0
    ReputationRep:
    can someone answer my other questions as well please
    Offline

    0
    ReputationRep:
    also lol https://eiewebvip.edexcel.org.uk/Rep...e_20070118.pdf

    question 8 ii
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.