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AQA Core 4 Exam Discussion 14/06/2012 !Poll, paper and unofficial MS (first post)!

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Reply 980
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f1mad
13
Original post by Oromis263
And finally!
Question 8.

a) dhdt∝2−h⇒dhdt=k(2−h) \dfrac{dh}{dt} \propto 2-h \Rightarrow \dfrac{dh}{dt} = k(2-h)

bi) This is a lot of work, I used integration by parts, but integration by substitution was another viable method (or any other way that got you to the answer really!) :smile:

Final equation t=(3x+1)(2x−1)32−4 t = (3x+1)(2x-1)^{\dfrac{3}{2}} - 4

bii) Sub 2 in to above formula.

t=7×332−4=32.4 t = 7 \times 3^{\frac{3}{2}} -4 = 32.4 :smile: And that is it! :smile:


From what I recall, substitution wasn't possible?
Original post by f1mad
From what I recall, substitution wasn't possible?


I don't know, I didn't check it just went straight for parts, I just saw a few people say that's how they did it, perhaps u = 2x -1 or something.
Original post by Oromis263
It's actually the same thing :smile: I'll show you.

Unparseable latex formula:

t = 5x(2x-1)^{\dfrac{3}{2}} - (2x-1)^{\dfrac{5}{2}} - 4[br]\[br]Then\ factor\ out\ (2x-1)^{\dfrac{3}{2}} \Rightarrow (2x-1)^{\dfrac{3}{2}} (5x -(2x-1)) -4 = (2x-1)^{\dfrac{3}{2}}(3x+1) -4

:smile:


Ah fair enough - I was so happy to get an answer that I just left it completely unsimplified, i honestly couldn't believe it when 4 dropped out as the value for C.
Reply 983
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f1mad
13
Original post by Oromis263
I don't know, I didn't check it just went straight for parts, I just saw a few people say that's how they did it, perhaps u = 2x -1 or something.


Nah that doesn't work. IBP was the only viable method, hence why it was worth so many marks.
Original post by Oromis263
Question 6.

Unparseable latex formula:

9x^2 - 6xy + 4y^2 = 3[br]\[br]implicit\ differentiation[br]\[br]\18x - 6y -6x\dfrac{dy}{dx} + 8y\dfrac{dy}{dx} = 0[br]\[br]\Rightarrow \dfrac{dy}{dx} = \dfrac{6(3x-y)}{8y-6x}[br]\[br]\Stationary\ at \dfrac{dy}{dx} =0[br]\[br]3x-y = 0 \Rightarrow y = 3x[br]\[br]\Sub\ this\ into\ original\ equation[br]\[br]27x^2 = 3 \Rightarrow x = \dfrac{-1}{3}\ or \dfrac{1}{3} \Rightarrow y = 1\ or -1



I did this but i just somehow didnt manage to get the correct coordinates. How many marks do you think i will get for that?
(edited 11 years ago)
Original post by f1mad
Nah that doesn't work. IBP was the only viable method, hence why it was worth so many marks.


What? Sub worked it was (at least the easiest sub) u=(contents of square roots).
I did it this way.
Original post by Iepnauy
I know that now :tongue: but if I wrote my final answer up to x^4, will I have a mark deducted for not leaving it at x^2 do you reckon? :smile:


Oh, I thought you meant up to and including! Well, I would take one accuracy mark away for not putting the x^2 term if I was an examiner.

I made really silly mistakes with the last two questions! Thankfully I added up what I got right, judging from common answers in this thread and looks most likely that I would get my required B or higher in Maths after all. So no more Maths and now there's French and Economics left to go - where did all the time go?
Reply 987
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f1mad
13
Original post by In One Ear
What? Sub worked it was (at least the easiest sub) u=(contents of square roots).
I did it this way.


You're right :facepalm2: (pulling out the 15, makes it somewhat easier to see :tongue:).

Can't believe I used IBP on that :eek:.
Original post by Oromis263
Question 6.

Unparseable latex formula:

9x^2 - 6xy + 4y^2 = 3[br]\[br]implicit\ differentiation[br]\[br]\18x - 6y -6x\dfrac{dy}{dx} + 8y\dfrac{dy}{dx} = 0[br]\[br]\Rightarrow \dfrac{dy}{dx} = \dfrac{6(3x-y)}{8y-6x}[br]\[br]\Stationary\ at \dfrac{dy}{dx} =0[br]\[br]3x-y = 0 \Rightarrow y = 3x[br]\[br]\Sub\ this\ into\ original\ equation[br]\[br]27x^2 = 3 \Rightarrow x = \dfrac{-1}{3}\ or \dfrac{1}{3} \Rightarrow y = 1\ or -1



I did that and got the same coordinates as you but I took a slightly different approach in getting y = 3x
After I differentiated, the terms which are attached to the dy/dx terms cancel down to zero. You don't even have to rearrange the implicit function if you're finding stationary points.
According to the unofficial markscheme (thanks for that btw), I got everything right apart from the 6 mark vectors question which I left out entirely (left it till the end, ran out of time), and also the implicit differentiation question.
On that I got the x values of the stationary points, x=1/3 and x=-1/3, but then when I worked out the y-coordinates I got two co-ordinates for each x-value :s-smilie:

Hopefully I still get most of the marks out of 8 for getting up to that point, but do you think I'll drop more than 2 for that question?
(edited 11 years ago)
For the implicit I got to 18x-6y=0.

Therefore, for stationary points, it must equal 0.

So I got y = -3, x = -1. And y = 3, x = 1

How can it be anything else??
Reply 991
Avatar for mojopin1
mojopin1
If memory serves me correctly I think the cartesian equation was y2 = 9/4 x2(4-x2)

June 2007 question 6. b) is almost identical:

http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MPC4-W-QP-JUN07.PDF
Original post by JulietheCat
I did that and got the same coordinates as you but I took a slightly different approach in getting y = 3x
After I differentiated, the terms which are attached to the dy/dx terms cancel down to zero. You don't even have to rearrange the implicit function if you're finding stationary points.


Ah yeah, of course! ^_^ Thanks for pointing that out. Ah well, my method is just a wee bit more long winded then. :smile:
Reply 993
Avatar for mojopin1
mojopin1
Original post by Sagacious
For the implicit I got to 18x-6y=0.

Therefore, for stationary points, it must equal 0.

So I got y = -3, x = -1. And y = 3, x = 1

How can it be anything else??


6y = 18x
y = 3x

Pretty sure it was 1/3,1 and -1/3, -1
Original post by Sagacious
For the implicit I got to 18x-6y=0.

Therefore, for stationary points, it must equal 0.

So I got y = -3, x = -1. And y = 3, x = 1

How can it be anything else??


You have to sub a value for y back into the original equation to find the coordinates of x and y?
Original post by mojopin1
6y = 18x
y = 3x

Pretty sure it was 1/3,1 and -1/3, -1


Hmm, I wrote y=3x, how many marks do you think I'll lose?
Original post by f1mad
You're right :facepalm2: (pulling out the 15, makes it somewhat easier to see :tongue:).

Can't believe I used IBP on that :eek:.


Lol my first reaction was *use IBP!* but i must admit i looked at it and thought no ****ing way will it work- obviously i must have been having my own retarded moment given that it seems the favoured method by a long shot. Substitution did indeed make it a very easy integral (probs a 5 marker in c3 if they told you to use it) meaning 8 marks overall (including eliminating the constant of integration) seemed generous using a substitution.
Original post by sunnybacon
According to the unofficial markscheme (thanks for that btw), I got everything right apart from the 6 mark vectors question which I left out entirely (left it till the end, ran out of time), and also the implicit differentiation question.
On that I got the x values of the stationary points, x=1/3 and x=-1/3, but then when I worked out the y-coordinates I got two co-ordinates for each x-value :s-smilie:
Hopefully I still get most of the marks out of 8 for getting up to that point, but do you think I'll drop more than 2 for that question?


I think 6 marks will be just for the differentiation and then 2 answer marks for the co-ordinates. Possibly 1 mark for the x co-ordinates and one mark for the y co-ordinates as the two answer marks.
Reply 998
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f1mad
13
Original post by In One Ear
Lol my first reaction was *use IBP!* but i must admit i looked at it and thought no ****ing way will it work- obviously i must have been having my own retarded moment given that it seems the favoured method by a long shot. Substitution did indeed make it a very easy integral (probs a 5 marker in c3 if they told you to use it) meaning 8 marks overall (including eliminating the constant of integration) seemed generous using a substitution.


I used substitution at first, then thought **** it: IBP will definitely work anyway.

XD
Reply 999
Avatar for mojopin1
mojopin1
Original post by Sagacious
Hmm, I wrote y=3x, how many marks do you think I'll lose?


That's correct I think, then you substitute it back into the original equation to find x.

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