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AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme]

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Original post by JRN-95
For 8b)i) isnt it dr/dt= (pi*r^2)/k


No because you were told that dr/dt is inversely proportional to the area (therefore 1/pi*r^2).

Also, I am pretty sure it told you in the question the constant was a. So the actual answer should have been dr/dt = a/pi*r^2.
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majka
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Original post by RavinderKlair97
how much ums would 55 marks get?


I think I got around 55 as well :// i hope very much for 80+
Original post by ForgottenApple
Probably your score


That's what a moron would say because it hasn't been marked yet
Original post by amyrah
coordinates for C were (-9, -2, 18) ?


I got this too! Does anyone know whether this is right/wrong with explanation?
(edited 8 years ago)
Do you remember the actual diff equation they asked you to solve?
Original post by datpr0
No because you were told that dr/dt is inversely proportional to the area (therefore 1/pi*r^2).

Also, I am pretty sure it told you in the question the constant was a. So the actual answer should have been dr/dt = a/pi*r^2.


****!!! i did dr/dt= pi*r^2/k damnnn!!! could i still get some marks and follow through marks for the next part
Does anyone remember the question for the implicit differentiation? Just want to go over it again as I don't remember if I got 11/8 or not
is there an agreed upon mark scheme anywhere?
Original post by HennersPD
Do you remember the actual diff equation they asked you to solve?


Yes.

It was: dx/dt = (4+5x)^1/2 / 5(1+t)^2
Anyone got a worked solution for the vectors question part b and c?
Original post by HennersPD
Do you remember the actual diff equation they asked you to solve?


if think it was dx/dt= (4+5x)^1/2 divided by 5(t+1)^2
Original post by jchap9776
Potentially, did anybody else show x tending to infinity = 2x-3 tending to infinity = 1/(2x-3) tends to 0? Therefore is decreasing. (Idk if it was 2x-3 but was something like that)


Thats what i literally done.
for 2)b)ii) I got

-1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

Therefore √29cos(x+1.19) = √29

cos(x+1.19) = 1
x + 1.19 = 0
x = -1.19

So why is it 5.09 where as it could be both...
For the last question, I forgot to divide 0.5 by 60. I wrote a quick comment, and some working-out to correct the value, but then I realise that I forgot to square root the 60. Now I'm pissed >:frown:

Also I forgot the +C in the differential equation. How many marks do you think I'll lose
Original post by Kennethm
for 2)b)ii) I got

-1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

Therefore √29cos(x+1.19) = √29

cos(x+1.19) = 1
x + 1.19 = 0
x = -1.19

So why is it 5.09 where as it could be both...


It said between 0 < x <2pi I think
question 2 needs sorting and for vectors I swear it was E(11,0,0) and E(23,4,-8)
Original post by Kennethm
for 2)b)ii) I got

-1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

Therefore √29cos(x+1.19) = √29

cos(x+1.19) = 1
x + 1.19 = 0
x = -1.19

So why is it 5.09 where as it could be both...


because 0<x<2pi
Original post by Kennethm
for 2)b)ii) I got

-1.19, they ask you for the MAXIMUM value from √29cos(x+1.19) which will be √29

Therefore √29cos(x+1.19) = √29

cos(x+1.19) = 1
x + 1.19 = 0
x = -1.19

So why is it 5.09 where as it could be both...


Because they told you to find the maximum value in the range 0 < x < 2pi :wink:!
Original post by jamal_m96
Nah love im pretty sure my core 3 exam had a question 9
There was deffo 8 questions, are you sure you're in the right thread?
Original post by JAW-97
There was deffo 8 questions, are you sure you're in the right thread?


He's trolling, ignore him.

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