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Getting to Cambridge: STEP by STEP!

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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
@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.
Reply 1162
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...
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!!
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?
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
Reply 1166
Original post by EricPiphany
Laughing so hard right now


I was hoping it'd be a joke video or something. :erm:
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.
Reply 1168
Original post by Infinity999
Unfortunately, this guys teaches at my school and said I couldn't get into Oxbridge.


Bloody hell... anyway:

P=ϵ2R(R+r)2dPdR=(R+r)2ddR(ϵ2R)ϵ2RddR((R+r)2)(R+r)4\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}
(edited 8 years ago)
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.
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 dick tommorow.
:rofl:
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.
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?
Reply 1173
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.
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!
Original post by Zacken
Yeah, that'd be a good idea.


Thanks!
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?
Reply 1177
Original post by Infinity999
Thanks!


He's replied to my comment. :lol:
Original post by Infinity999
Thanks! Oh, I see what you mean. One for internal and one for sum?


No, 'the source resistance' and 'to a load with resistance' is what they're for.
Original post by Marxist
No, 'the source resistance' and 'to a load with resistance' is what they're for.


Oh... my bad. I think I get it now. :s-smilie:

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