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OCR MEI C4 [June 14th] Discussion

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Reply 260
same :smile: gl all
Original post by Amy-Rose
Just put the x in || on one side, number on other side and simplify. As is modulus, change any - in the || to a plus. :smile:

|-x^2| < 4
|x^2| < 4
|x| < 2


How do you know if its less than or less than or equal to?
Original post by Amy-Rose
Just put the x in || on one side, number on other side and simplify. As is modulus, change any - in the || to a plus. :smile:

|-x^2| < 4
|x^2| < 4
|x| < 2


How'd you know if its modulus or otherwise because sometimes it's -1/2<x<1/2 for example

is there any set of way of knowing
Original post by getenoughsuarez
How'd you know if its modulus or otherwise because sometimes it's -1/2<x<1/2 for example

is there any set of way of knowing


|x| < 1/2 is the same as saying -1/2<x<1/2
Original post by JumpingFrog
Heres a scan of what I did about 30 minutes ago. You've made a slip early on by the looks of it.
http://i.imgur.com/wvbe6.jpg

It tells you D is 15 metres from B. You already worked out AE and that was 25. So you know the length of the direction vector (which is AE) is 25. Since you only need to move 15 you work this out as a fraction of the direction vector's length which is 3/5 giving you your lambda. It's all to do with the lines being parallel.


I still don't quite understand :s-smilie: but thanks for helping :smile:
You know in the June 11 paper, the vectors question about the horizontal plane... can someone please explain to me why it was that?
also what are the other vectors for other planes, like the vertical one

please help! :smile:
Reply 266
Original post by helpme12345
You know in the June 11 paper, the vectors question about the horizontal plane... can someone please explain to me why it was that?
also what are the other vectors for other planes, like the vertical one

please help! :smile:


That's because its a horizontal plane... the point for horizontal and vertical will always be (0,0,1)
can anyone help me with 8iii on jan 2012 please?
Reply 268
Original post by coolcurrypuff
http://www.mei.org.uk/files/papers/c406ja_ssi4c0.pdf

can someone please explain 7(ii)... am i supposed to manipulate what they ask me to show? or i dont even understand the answer key ): oh dear....


You basically have to differentiate the original equation on both sides using the quotient rule to show the answer:

tanθ = 6y / (160 + y^2)

sinθ/cosθ = 6y / (160 +y^2)

The rest is straight forward, hope this helps. :smile:
Reply 269
Original post by nasira372
can anyone help me with 8iii on jan 2012 please?


You need to find the value of t and plot it in the x equation..
So t=y/4 plot it in x equation u get 2(y/4)^2 which will give u x=y^2/8 take the 8 to the other side which gives you 8x=y^2

To find the volume of revolution, since its rotating about the x axis u will use the formula pie (integrate) y^2
we know that y^2 is 8x so all u do is integrate 8x
The limit would be 0 and 4 because in the question they have told you that t is between rt -2 and rt 2 by plotting rt 2 in x equation 2(rt2)^2 which is 4
The answer is 64 pie
(edited 11 years ago)
Original post by Palak16
That's because its a horizontal plane... the point for horizontal and vertical will always be (0,0,1)


thanks! :smile:

for part 7 iii june 11 (same question)
it says CD = 1/2AB

but the ratio is CD:AB = 1:2
:confused:
I thought it would be the other way around?
Reply 271
Original post by Hirarara
You basically have to differentiate the original equation on both sides using the quotient rule to show the answer:

tanθ = 6y / (160 + y^2)

sinθ/cosθ = 6y / (160 +y^2)

The rest is straight forward, hope this helps. :smile:


I've just looked at that question and on the last bit how did they move the sec^2 to cos^2?

Edit: Duh, I see it now. sec^2(theta) = 1/cos^2(theta), so just multiplied though :biggrin:
(edited 11 years ago)
Original post by Palak16
You need to find the value of t and plot it in the x equation..
So t=y/4 plot it in x equation u get 2(y/4)^2 which will give u x=y^2/8 take the 8 to the other side which gives you 8x=y^2

To find the volume of revolution, since its rotating about the x axis u will use the formula pie (integrate) y^2
we know that y^2 is 8x so all u do is integrate 8x
The limit would be 0 and 4 because in the question they have told you that t is between rt -2 and rt 2 by plotting rt 2 in x equation 2(rt2)^2 which is 4
The answer is 64 pie


If it was rotated about the y axis would you sub rt 2 into y?

ty for this
Goodnight, and goodluck everyone :smile: x
Reply 274
Original post by nasira372
If it was rotated about the y axis would you sub rt 2 into y?

ty for this


Not really it depends on the equation.. It would be a good idea to get the value of t which is the easiest... when it says rotate about the y axis the formula will be pie (integrate) x^2 and the limits will be the values on the y axis...
Reply 275
Out of all my exams this is probably the one i'm most prepared for we've done every past paper since like 2006 ._.

That said I still can't properly do vectors and that'll probably be 1 of the 2 for section B...
Reply 276
Original post by helpme12345
thanks! :smile:

for part 7 iii june 11 (same question)
it says CD = 1/2AB

but the ratio is CD:AB = 1:2
:confused:
I thought it would be the other way around?


Erm even i am confused on that one..
Well good everyone. Hope paper B is nice tomorrow. !! :biggrin:


This was posted from The Student Room's iPhone/iPad App
Original post by Palak16
Erm even i am confused on that one..


thanks anyway :smile:
can someone please help me with this? :smile:

for part 7 iii june 11
it says CD = 1/2AB

but the ratio is CD:AB = 1:2

I thought it would be the other way around?

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