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Edexcel A2 C4 Mathematics June 2016 - Official Thread Watch

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    (Original post by aymanzayedmannan)
    Are you sure that it's with respect to x?
    I meant with respect to y, sorry

    (Original post by Zacken)
    Assuming you mean \int 2\sin^2 y \cos y \, \mathrm{d}y then consider the substitution u = \sin y.
    I've written {dy}=\frac{1}{cosy} and put it back in to the equation, but I can't seem to get further than 2\int \frac{sin^2y}{cosy}.

    (Original post by BBeyond)
    You can do it by inspection if you mean dy: (2/3)sin^3(y) + c
    I try to avoid inspection if I can, nearly always end up getting it wrong It's the right equation though, but thank you anyway!
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    (Original post by blackdiamond97)
    I've written {dy}=\frac{1}{cosy} and put it back in to the equation, but I can't seem to get further than 2\int \frac{sin^2y}{cosy}.
    Surely that means you'd have \displaystyle \int \frac{2\sin^2 y \cos y}{\cos y} \, \mathrm{d}u?

    Since u = \sin y \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}y} = \cos y and dy = \frac{\mathrm{d}u}{\cos y}
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    (Original post by Zacken)
    Surely that means you'd have \displaystyle \int \frac{2\sin^2 y \cos y}{\cos y} \, \mathrm{d}u?

    Since u = \sin y \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}y} = \cos y and dy = \frac{\mathrm{d}u}{\cos y}
    C4 will be the death of me simply because I keep making silly mistakes :facepalm:I understand it from here, thank you
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    (Original post by blackdiamond97)
    C4 will be the death of me simply because I keep making silly mistakes :facepalm:I understand it from here, thank you
    You're welcome.
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    (Original post by blackdiamond97)
    I meant with respect to y, sorry
    I still recommend inspection as BBeyond said, but a sub will easily solve it too! You seem to be in good hands w/ Zacken, but if you want to do it by inspection, do consider \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sin^{3}x \right ) = 3\sin^2{x}\cos x.
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    Someone throw me a nice differential equation, quick
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    (Original post by Serine Soul)
    Someone throw me a nice differential equation, quick
    Terminal velocity equation? Mdv/dt =mg -kv . find velocity as a function of time
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    (Original post by Serine Soul)
    Someone throw me a nice differential equation, quick
    Use the substitution y = ux where u is a function of x to solve the DE:

    \displaystyle

\begin{equation*}\frac{\mathrm{d  }y}{\mathrm{d}x} = \frac{x}{y} + \frac{y}{x} \quad \quad (x>0, y>0)\end{equation*}

    that satisfied y=2 when x=1 in the form y = f(x) \quad (x > e^{-2}).
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    (Original post by samb1234)
    Terminal velocity equation? Mdv/dt =g -kv . find velocity as a function of time
    I don't do physics ;_;

    Am I still supposed to be able to do it? :lol:
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    (Original post by Serine Soul)
    I don't do physics ;_;

    Am I still supposed to be able to do it? :lol:
    Yeah its just that differential equation which is solvable via c4 methods. Assume that at time t=0, v=0.
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    (Original post by Serine Soul)
    I don't do physics ;_;

    Am I still supposed to be able to do it? :lol:
    Check the edit, missed out an m
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    (Original post by samb1234)
    Terminal velocity equation? Mdv/dt =mg -kv . find velocity as a function of time
    Surely you mean lowecase m in front of the dv/dt as well?
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    (Original post by Serine Soul)
    Someone throw me a nice differential equation, quick
    Idk, probably a really easy one as I made it up, but

    \displaystyle \frac{\mathrm{d} v}{\mathrm{d} t} = \frac{(v+4)\ln^{2}t}{v}
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    (Original post by Zacken)
    Surely you mean lowecase m in front of the dv/dt as well?
    Yeah im on my phone which corrected my grammar by making it a capital sorry
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    (Original post by BBeyond)
    You can do it by inspection if you mean dy: (2/3)sin^3(y) + c
    That was the first thing I thought aswell I.e integration by recognition when I saw that question. You get used to those type of questions.
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    (Original post by samb1234)
    Yeah im on my phone which corrected my grammar by making it a capital sorry
    Ah yeah that phone thing is always annoying :lol:
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    Cheers guys, I'm onto these now
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    (Original post by Serine Soul)
    Cheers guys, I'm onto these now
    I've got two follow ups to mine that builds on the concept introduced in this one.
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    (Original post by Zacken)
    I've got two follow ups to mine that builds on the concept introduced in this one.
    Y'all gonna have to wait a while

    I blame walking over 4 miles today. It's slowed me down
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    (Original post by Serine Soul)
    Y'all gonna have to wait a while

    I blame walking over 4 miles today. It's slowed me down
    Oh dear, take your time, savour the maths. No rush.
 
 
 
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