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    (Original post by MathsNerd1)
    Indeed but I'll get an hours break between the two which should hopefully be enough, otherwise my brain will just melt from all the intense Maths that I'll be doing whilst all my friends go to someone's BBQ during the day, oh how I hate that I can't go to it but I'd rather get into Warwick instead!
    4 30ish it's all over on june 25th 2013!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!
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    (Original post by MathsNerd1)
    It is for all my ramblings to understand a question Also yeah it's going to be a tough day, although on the 25th will be both STEP and AEA which I think might be hard and then I've got to attend awards evening on the same night
    Oh, you're taking both STEP I and Il, then?
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    (Original post by ACBLISS)
    4 30ish it's all over on june 25th 2013!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!
    Indeed!
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    (Original post by justinawe)
    Oh, you're taking both STEP I and Il, then?
    Yeah, not too sure why though as I suck at both of them pretty much
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    So I've done the stupid thing of leaving this module right to the last minute. I have Integration and Vectors left to tackle. Whats the best way to tackle these? Should I just keep up with what I am currently doing and following the guides on examsolutions.com and then just do every question?
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    (Original post by MathsNerd1)
    Yeah, not too sure why though as I suck at both of them pretty much
    I very much doubt that
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    (Original post by F1Addict)
    You're told that the rate the temperature decreases is proportional to the difference in temperature between the water and the room.

    Temperature rate is given by \frac{d\theta}{dt}, and the difference in temperature is given by (\theta - \theta_R), where \theta_R is the room temperature.

    Can you set up the differential equation now?

    Spoiler:
    Show
    This is what you should get:
    \frac{d\theta}{dt} = -(\theta - \theta_R)

    \Rightarrow \frac{d\theta}{dt} = -(\theta - 20)

    Or alternatively,
    \frac{d\theta}{dt} = (20 - \theta)

    They are both the same (assuming that \theta is greater than or equal to 20)

    I have done this , but I don't know where the T comes in . It says to connect t and theta.

    Is the T part the differential? Ie- d(theta)/ dt?
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    (Original post by justinawe)
    I very much doubt that
    Trust me, I'll be lucky to get a 3 right now
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    Edexcel, You have failed this city!
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    (Original post by MathsNerd1)
    Trust me, I'll be lucky to get a 3 right now
    inb4 you smash them both and get S/S :mmm:
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    (Original post by bloomingblossoms)
    4. During a chemical reaction, a compound is being made from two other substances.
    At time t hours after the start of the reaction, x g of the compound has been produced.
    Assuming that x = 0 initially, and that
    dx/dt
    = 2(x − 6)(x − 3),
    (a) show that it takes approximately 7 minutes to produce 2 g of the compound. (10)
    (b) Explain why it is not possible to produce 3 g of the compound. (2)

    Can anyone explain why you need to integrate and why the inverse of the derivative is equal to the integration of 2?

    The mark scheme is here: http://www.school-portal.co.uk/Group...urceId=3992046

    Thanks in advance!
    I'm trying to do this question but I'm stuck because after I integrated , I can't put x=0 in the equation since I have ln|x-6| and that would give me an error. Could you maybe change the marking scheme link because i can't access it, or just give me the paper and the year.
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    (Original post by Scienceisgood)
    EVERY SINGLE PAPER minus C1 and C2 was replaced I think.
    Wow, I didn't realise so many papers have been replaced! Fingers crossed for a good paper
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    (Original post by justinawe)
    inb4 you smash them both and get S/S :mmm:
    inb4? And I'd be over the moon if I actually got something like that!! Even a 1 would be great to see
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    (Original post by Tanomc)
    I have done this , but I don't know where the T comes in . It says to connect t and theta.

    Is the T part the differential? Ie- d(theta)/ dt?
    When the question talks about rate, its with respect to time, t. The rate the temperature decreases means how much does the temperature decrease in a given amount of time. So yes, the connection between t and \theta is made by the differential part \frac{d\theta}{dt}, and the -(\theta -20) part says how t and \theta are related.

    Btw, there is no upper case 'T'. I take it thats a typo and you meant 't'.

    Also, I forgot to include the constant of proportionality in my earlier post. So it should be \frac{d\theta}{dt} = -k(\theta -20), where k is some constant.
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    Dear Edexcel

    If this C4 paper is horrible, then I will track you down and haunt you.

    :mob:
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    (Original post by MathsNerd1)
    inb4? And I'd be over the moon if I actually got something like that!! Even a 1 would be great to see
    http://lmgtfy.com/?q=inb4

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    (Original post by F1Addict)
    When the question talks about rate, its with respect to time, t. The rate the temperature decreases means how much does the temperature decrease in a given amount of time. So yes, the connection between t and \theta is made by the differential part.

    Btw, there is no upper case 'T'. I take it thats a typo and you meant 't'.
    Typo. Ok, thanks a lot. If rate is always with respect to time then that cleared a lot of confusion and sorry for the long time to reply, I had to get something to eat .
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    Can some one do the last question no 8 on june 2007
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    Anyone else finding C4 so difficult....
    Need 100ums in this exam...
    No confidence after last thursdays C3 exam...

    Urgh.
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    So ****ing pumped for C4 after hearing how the C3 exam went...

    BRING IT EDEXCEL!!!!!


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