# CCEA C4 Mathematics - 22nd May 2014Watch

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4 years ago
#21
(Original post by stephen_mcgarry)
Hi! I sure that's right for the point, but wasn't t out of the range to get these coordinates?
I used t=5π/4 which is smaller than 3π/2
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4 years ago
#22
(Original post by Mr Tall)
how did you do the last part of the vector question
Set the vector equal to the vector equation of the line, and showed that λ was equal to 2 for both i and j values.
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4 years ago
#23
(Original post by TheWiseSalmon)
I used t=5π/4 which is smaller than 3π/2
Oooh yeah! That's OK! Well done

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4 years ago
#24
Just out of curiosity does anyone remember their answer for the volume of revolution question (the volume of the bowl)?
I thought my answer looked a bit odd, something along the lines of pi( 1/2e^2 + e + ?)
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4 years ago
#25
(Original post by Rebecca-h)
Just out of curiosity does anyone remember their answer for the volume of revolution question (the volume of the bowl)?
I thought my answer looked a bit odd, something along the lines of pi( 1/2e^2 + e + ?)
Can't remember my exact answer, but it was 24.0 to 3sf

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4 years ago
#26
(Original post by Mr Tall)
but what if there was 2 points with gradient root2 in that domain? did you just have to pick one?
I really not sure what happens, I'm going to do it again now

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4 years ago
#27
(Original post by Rebecca-h)
Just out of curiosity does anyone remember their answer for the volume of revolution question (the volume of the bowl)?
I thought my answer looked a bit odd, something along the lines of pi( 1/2e^2 + e + ?)
I think I got something like that, but it didn't say exact value so I just gave 24.0 cubic units
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4 years ago
#28
(Original post by TheWiseSalmon)
I think I got something like that, but it didn't say exact value so I just gave 24.0 cubic units
Yay same answer!

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4 years ago
#29
Just done the last question again and it seems that there are two sets of coordinates, I cannot see how there can only be a "point"
The two points I got are (2pi,-4) and ((5pi-2)/2,-2sqrt2)
If anybody can correct me, or explain why it may be one point feel free, I would really appreciate it

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4 years ago
#30
Can I post the full paper up as somebody is asking for the questions? But I can't send them in a private message? Will I be allowed to post the paper up?

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4 years ago
#31
(Original post by stephen_mcgarry)
Can I post the full paper up as somebody is asking for the questions? But I can't send them in a private message? Will I be allowed to post the paper up?

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yep you can, it's done in one sitting!
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4 years ago
#32
The co-ordinate (2pi, -4) does not work, if you put your value for T which is equal to pi, back into your gradient equation, you'll see that it comes out as 0 and not root 2.
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4 years ago
#33
(Original post by 1MacG)
The co-ordinate (2pi, -4) does not work, if you put your value for T which is equal to pi, back into your gradient equation, you'll see that it comes out as 0 and not root 2.
Doesn't that happen for 5pi/4 as well?

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4 years ago
#34
(Original post by stephen_mcgarry)
Doesn't that happen for 5pi/4 as well?

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lol no

dy/dx = -4cos(5pi/4) / 2 - 2cos(5pi/2

= 2root2 / 2 = root2
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4 years ago
#35
Not 100% sure it even matters, but I actually think its 24.0units^2 . Think its due to an area being multiplied by pi rather than a distance. Don't quote me on it but this is from another mark scheme.

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4 years ago
#36
(Original post by 1MacG)
lol no

dy/dx = -4cos(5pi/4) / 2 - 2cos(5pi/2

= 2root2 / 2 = root2
Hmm, isn't dy/dx = -4sint/(2-2cos2t)

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4 years ago
#37
OK so here is the full paper

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4 years ago
#38
(Original post by stephen_mcgarry)
OK so here is the full paper

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In case you can't see, question 3 is the integral between 3 and 2

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Thread starter 4 years ago
#39
(Original post by stephen_mcgarry)
Hmm, isn't dy/dx = -4sint/(2-2cos2t)

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Hey, could you post the paper please?

Edit: I see your post now, thanks
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4 years ago
#40
(Original post by GingerCodeMan)
Hey, could you post the paper please?

Edit: I see your post now, thanks
Haha that's OK, spent ages trying to send it in pm, gave up and put it up here

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