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Higher Maths 2011 Paper

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    Can anybody help me on the waves functions question, It's Q6 a) and b) on paper 2. I tried checking the Mark scheme but it didn't make any sense to me.
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    a) It's simply the normal way of the wave funtion. Compare with double angle, work out quadrants, square and add, divide.
    b) Have you done intergration of trig between limits yet?
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    (Original post by aroy45)
    a) It's simply the normal way of the wave funtion. Compare with double angle, work out quadrants, square and add, divide.
    b) Have you done intergration of trig between limits yet?
    I'm pretty sure I can do the 1st part. I go:

    Root(34) Sin(x+301) for the 1st equation. 301 is the answer in degrees, and 5.3 is something is that answer in radians, but 301 is easier to use to work through in b)

    But I'm not sure how to do b). The only thing left to do in the course is the end of Logarithms and Exponentials but I'm still not sure what to do.
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    The formula sheet tells you how to integrate trig functions as I remember, have a look there first and try integrating it. You'll need to use the result from the first part after that.
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    Solution for 6 a.)

    Wave function

    3sinx - 5cosx = Rsin(x+a)
    = R(sinxcosa+cosxsina)
    =Rsinxcosa + Rcosxsina
    Rcosa = 3
    Rsina = -5

    => R = root 34

    Tana = 5/3 (Remember to put your calculator in radians!)
    a = 1.03 RADS
    => a = 5.253

    Hope that helps
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    (Original post by INT2 Warrior)
    Solution for 6 a.)

    Wave function

    3sinx - 5cosx = Rsin(x+a)
    = R(sinxcosa+cosxsina)
    =Rsinxcosa + Rcosxsina
    Rcosa = 3
    Rsina = -5

    => R = root 34

    Tana = 5/3 (Remember to put your calculator in radians!)
    a = 1.03 RADS
    => a = 5.253

    Hope that helps
    It's the second part I'm not sure on. :/
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    You need to be in radians for calculus.

    Basically integrate, then substitute your answer for part a) into it. Then basically substitue t and 0 and as the whole thing equals 3, work through it like any trig equation.
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    (Original post by (:emily.)
    You need to be in radians for calculus.

    Basically integrate, then substitute your answer for part a) into it. Then basically substitue t and 0 and as the whole thing equals 3, work through it like any trig equation.
    Also how would I go about differentiating and integrating:

    sin(2x+2)^5

    We have done ones like: sin(2x+2) or (2x+2), but what happens if you have both a power for the bracket and a trig function in front of the bracket?
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    (Original post by JaggySnake95)
    Also how would I go about differentiating and integrating:

    sin(2x+2)^5

    We have done ones like: sin(2x+2) or (2x+2), but what happens if you have both a power for the bracket and a trig function in front of the bracket?
    Use the chain rule. Have you learnt dy/dx = dy/du * du/dx ?

    If not, just remember differentiate round the bracket and then in the bracket, with the 'bracket' being sin(x) etc.

    So if f(x) = sin^5(x)
    f'(x) = 5sin^4(x) * 2cos(x)

    If there is anything other than x in the bracket you need to use the chain rule on that part as well, but don't know if you'll need to do that.

    As far as I know, integrating it is more complicated and you also don't need to know that for higher? (Someone correct me on that if I'm wrong...)
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    (Original post by (:emily.)
    Use the chain rule. Have you learnt dy/dx = dy/du * du/dx ?

    If not, just remember differentiate round the bracket and then in the bracket, with the 'bracket' being sin(x) etc.

    So if f(x) = sin^5(x)
    f'(x) = 5sin^4(x) * 2cos(x)

    If there is anything other than x in the bracket you need to use the chain rule on that part as well, but don't know if you'll need to do that.

    As far as I know, integrating it is more complicated and you also don't need to know that for higher? (Someone correct me on that if I'm wrong...)
    don't you multiple just by cos(x) rather than 2cos(x) ?
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    (Original post by JaggySnake95)
    don't you multiple just by cos(x) rather than 2cos(x) ?
    Yeah,

    \frac{d}{dx}(sin^5x) = 5sin^4x~cosx
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    Here's my solution to the question. I sat the higher paper last year so...

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    I hope that makes some sense. I wrote that like 10/11 months ago so I don't remember the question well. If you have an questions about the solution then ask and I'll try and answer them =)
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    (Original post by mimx)
    Yeah,

    \frac{d}{dx}(sin^5x) = 5sin^4x~cosx
    How would you do: sin(2x+2)^5

    and also

    Whenever there is a bracket do you always differentiate it, or do you only do it if there is a power? OR can you drop the power of 1 from a bracket to 0 to get rid of the bracket?
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    (Original post by laughylolly)
    Here's my solution to the question. I sat the higher paper last year so...

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    I hope that makes some sense. I wrote that like 10/11 months ago so I don't remember the question well. If you have an questions about the solution then ask and I'll try and answer them =)
    I would have done it differenty and subsituted the wave equation in a few lines later, but I'm curious how you do the wave equation minus the other wave euations after you substituted in t and 0.
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    (Original post by JaggySnake95)
    don't you multiple just by cos(x) rather than 2cos(x) ?
    Oops sorry, I originally differentiated sin(2x) then realised it was different so changed it
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    (Original post by (:emily.)
    Oops sorry, I originally differentiated sin(2x) then realised it was different so changed it
    Thanks.

    Have you any idea about the questions I asked a few posts ago?
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    (Original post by JaggySnake95)
    I would have done it differenty and subsituted the wave equation in a few lines later, but I'm curious how you do the wave equation minus the other wave euations after you substituted in t and 0.
    You mean this line?


    \sqrt{34}sin(t+5.25...)-\sqrt{34}sin(0+5.25...)=3


    Well \sqrt{34}sin(0+5.25...)=-5 (if you use the exact value from part a). Then since it's -\sqrt{34}sin(0+5.25...) it will become positive 5 and so on.
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    (Original post by laughylolly)
    You mean this line?


    \sqrt{34}sin(t+5.25...)-\sqrt{34}sin(0+5.25...)=3


    Well \sqrt{34}sin(0+5.25...)=-5 (if you use the exact value from part a). Then since it's -\sqrt{34}sin(0+5.25...) it will become positive 5 and so on.
    See I'm thinkig because both sides are basically the same, they would cancel each other out?

    I would personally do it differently I'm just wondering how you do that subtraction of both wave equations.
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    (Original post by JaggySnake95)
    See I'm thinkig because both sides are basically the same, they would cancel each other out?

    I would personally do it differently I'm just wondering how you do that subtraction of both wave equations.
    Not quite following you there... They are the same but one has a t in it and the other doesn't and that makes all the difference.

    For me, that was the most straight forward way to find a solution for t.
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    (Original post by JaggySnake95)
    How would you do: sin(2x+2)^5

    and also

    Whenever there is a bracket do you always differentiate it, or do you only do it if there is a power? OR can you drop the power of 1 from a bracket to 0 to get rid of the bracket?
    Like I said, doubt you'd need to do it for higher, but it would be:

     \frac{d}{dx}sin^5(2x+2)) = 5sin^4(2x+2) * \frac{d}{dx}sin(2x+2)

= 5sin^4(2x+2) * 2cos(2x+2)

= 10sin^4(2x+2)cos(2x+2)

    And I don't really understand the last question... but technically you could put a bracket around anything and then it would be to the power of 0, but remember that that equals 1 and then you need to multiply by the derivative of the bit in the brackets so you would get the same answer anyway?

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