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    (Original post by drandy76)
    Zaken once tried to integrate to find the area of his ****, only to stop in horror upon finding out that the upper limit was epsilon


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    U sayin it's 2D????
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    (Original post by Student403)
    U sayin it's 2D????
    It can be considered to have 0 dimensions, he traded penis size for maths intuition


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    (Original post by drandy76)
    It can be considered to have 0 dimensions, he traded penis size for maths intuition


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    This is so funny
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    physicsmaths and Zacken


    https://www.youtube.com/watch?v=Zvm8SjCNKgE

    From 6:24 and onwards, is the way he is doing it correct? Shouldn't he be using the Quotient Rule? I have no idea what he's done there.
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    (Original post by Infinity999)
    physicsmaths and Zacken


    https://www.youtube.com/watch?v=Zvm8SjCNKgE

    From 6:24 and onwards, is the way he is doing it correct? Shouldn't he be using the Quotient Rule? I have no idea what he's done there.
    What the **** is this guy doing? His "simplification" at 6:23 is all wrong, he shouldn't be allowed to do maths...
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    (Original post by Infinity999)
    physicsmaths and Zacken


    https://www.youtube.com/watch?v=Zvm8SjCNKgE

    From 6:24 and onwards, is the way he is doing it correct? Shouldn't he be using the Quotient Rule? I have no idea what he's done there.
    WTF!!
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    (Original post by Infinity999)
    physicsmaths and Zacken


    https://www.youtube.com/watch?v=Zvm8SjCNKgE

    From 6:24 and onwards, is the way he is doing it correct? Shouldn't he be using the Quotient Rule? I have no idea what he's done there.
    If this is the Max. Power Transfer Theorem, there is a step where you use the quotient rule. What on earth is he doing there?
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    (Original post by Zacken)
    What the **** is this guy doing? His "simplification" at 6:23 is all wrong, he shouldn't be allowed to do maths...
    Laughing so hard right now
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    (Original post by EricPiphany)
    Laughing so hard right now
    I was hoping it'd be a joke video or something. :erm:
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    (Original post by Zacken)
    I was hoping it'd be a joke video or something. :erm:
    Unfortunately, this guys teaches at my school and said I couldn't get into Oxbridge.
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    (Original post by Infinity999)
    Unfortunately, this guys teaches at my school and said I couldn't get into Oxbridge.
    Bloody hell... anyway:

    \displaystyle P = \frac{\epsilon^2 R}{(R+r)^2} \Rightarrow \frac{ \mathrm{d}P }{ \mathrm{d}R } = \frac{(R+r)^2\frac{ \mathrm{d} }{ \mathrm{d}R } \left(\epsilon^2 R\right) - \epsilon^2 R\frac{\mathrm{d}}{\mathrm{d}R } \left((R+r)^2\right)}{(R+r)^4}
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    (Original post by Marxist)
    If this is the Max. Power Transfer Theorem, there is a step where you use the quotient rule. What on earth is he doing there?
    It is wrong and my maths teacher said the same. This is the max power theorem, yes. I am ahead of the class so I asked him to do it.
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    (Original post by Infinity999)
    It is wrong and my maths teacher said the same. This is the max power theorem, yes. I am ahead of the class so I asked him to do it.
    go slap his **** tommorow.
    :rofl:
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    (Original post by Infinity999)
    It is wrong and my maths teacher said the same. This is the max power theorem, yes. I am ahead of the class so I asked him to do it.
    Okay so you're looking for the maximum; after I did the quotient rule, my final results turns out to be R=r after setting dP/dR=0. If you get this, then there should be no problem. And you can meet your conclusion.
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    (Original post by Marxist)
    Okay so you're looking for the maximum; after I did the quotient rule, my final results turns out to be R=r after setting dP/dR=0. If you get this, then there should be no problem. And you can meet your conclusion.
    Right, that makes sense now. I got the same result as Zacken did for the derivative. And would I need to do the second derivative to show that it is a maximum?
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    (Original post by Infinity999)
    And would I need to do the second derivative to show that it is a maximum?
    Yeah, that'd be a good idea.
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    (Original post by Infinity999)
    Right, that makes sense now. I got the same result as Zacken did for the derivative. And would I need to do the second derivative to show that it is a maximum?
    Use R sub s = R sub l instead, makes life easier!
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    (Original post by Zacken)
    Yeah, that'd be a good idea.
    Thanks!
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    (Original post by Marxist)
    Use R sub s = R sub l instead, makes life easier!
    Thanks! Oh, I see what you mean. One for internal and one for sum?
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    (Original post by Infinity999)
    Thanks!
    He's replied to my comment. :lol:
 
 
 
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