Honours Maths Further Calculus Option

sully88

Can any one help me with 1995 paper 2 question 8 part c? Its the first part. I think its the 2d diagram thats throwing me off, i tried to draw a 3d one using the same measurements but cant get my head around it. Any help would be greatly appreciated

mp3guy

Those old ones are terrible, and to think you'd have had to do that without a calculator. Personally I don't go back past 1997, because the syllabus changed then.

spartacus93

You have to get the tan of the angle in the bottom left...

Tan A = 4 / 4-r

or tan A = 11 / 4

depending on if you take the big triangle or the little one...

then just leave them equal each other. 11 / 4 = h / 4-r

4h = 44-11r
h = 11-11r/4

JC 2K3

Firstly, for the start of the question forget you're looking at a cone and a cylinder, you're looking at a TRIANGLE and a SQUARE.

Taking "x" as the distance between the bottom corner of the square and the bottom corner of the triange:
2r + 2x = 8
2x = 8 - 2r
x = 4 - r

4*11 = 44 = Area of triangle
2rh = Area of square
44 - 2rh = Area of triangle - area of square

Now the area of the tringle - area of the square is also equal to the three smaller triangles you see, hence:
r(11 - h) + hx = r(11 - h) + h(4-r) = Area of triangle - square

Putting the two equations together:
44 - 2rh = 11r - rh + 4h - rh
44 = 11r + 4h
h = 11 - 11r/4

Now max volume... i'm getting stuck here myself....
Volume of cylinder = V
V = (pi)r^2(h) = (pi)r^2(11 - 11r/4)
= (pi)11r^2 - (11(pi)r^3)/4
dV/dr = 22(pi)r - (33(pi)r^2)/4
0 = 22(pi)r - (33(pi)r^2)/4
(33(pi)r^2)/4 = 22(pi)r
(33r)/4 = 22
r = 88/33 = 8/3

V = (pi)11r^2 - (11(pi)r^3)/4
= 11((pi)(64/9) - (pi)(512/108))
(pi)704/9 - (pi)5632/108
= (pi)2816/108......

Answer says 216(pi), what have I done wrong?

MathsManiac

mp3guy wrote:

Those old ones are terrible, and to think you'd have had to do that without a calculator. Personally I don't go back past 1997, because the syllabus changed then.

Actually, the syllabus changed in 1994. (i.e. changed in 1992 and examined first in 1994).

mp3guy

http://www.examinations.ie/index.php?l=en&mc=en&sc=sy

Looks an awful lot like 1997 to me

Naikon

mp3guy wrote:

Those old ones are terrible, and to think you'd have had to do that without a calculator. Personally I don't go back past 1997, because the syllabus changed then.

Nor do I, mind you I am in Ordinary maths but they changed both levels in the same year.

md99

BOLLOCKS!!!!!! I've been practicing the years before 1997, no wonder I'm getting my ass kicked :mad: :mad:

Anyone try the 1997 paper? Now, that was just lovely, 3 theorems 20 marks each!

So when did the course change, 1992, 1994 or 1997? I thought they wouldn't have the 'old' syllabus in our exam papers??

JC 2K3

^Same.

I find nothing is different on the 1994-96 papers besides the fact that they're a little bit more difficult, which doesn't indicate a change in syllabus.

Actually, I'm gonna call bollocks on this one

Typo on Examinations.ie tbh.

boger

come on guys these questions are easy

JC 2K3

^Point out to me where I've gone wrong above then.

mp3guy

JC 2K3 wrote:

^Same.

I find nothing is different on the 1994-96 papers besides the fact that they're a little bit more difficult, which doesn't indicate a change in syllabus.

Actually, I'm gonna call bollocks on this one

Typo on Examinations.ie tbh.

Or you could just be wrong. Claiming a typo on a site is a bit far just to keep yourself in the knowledge you're right. I'll believe the site over you thanks.

Donald-Duck

JC 2K3 wrote:

^Same.

I find nothing is different on the 1994-96 papers besides the fact that they're a little bit more difficult, which doesn't indicate a change in syllabus.

Actually, I'm gonna call bollocks on this one

Typo on Examinations.ie tbh.

Now I may be completly wrong on this, but they really look like scanned files so its not an examinations.ie typo

Crumbs

As MathsManiac said, the syllabus changed for the 1994 LC. The only change I'm aware of for 1997 was that the "Alternative Ordinary Level" became "Foundation Level".

JC 2K3 wrote:

^Point out to me where I've gone wrong above then.

Your answer seems correct JC. It can be simplified to 704(pi)/27. I've seen a couple of wrong answers given for that question and 216(pi) is definitely wrong as that's greater than the volume of the cone itself.

Ishmael

I'd Assume the solution is correct as the answer is 26.1(pi by you calculations)

so its probably a typo.

Gangsta

Lads- Similar Triangles!!! thats the easiest way to do it.

JC 2K3

mp3guy wrote:

Or you could just be wrong. Claiming a typo on a site is a bit far just to keep yourself in the knowledge you're right. I'll believe the site over you thanks.

Donald-Duck wrote:

Now I may be completly wrong on this, but they really look like scanned files so its not an examinations.ie typo

I scanned my own exam papers from 1994 and highlighted the area where it says marks will be taken off if it is not shown where calculators are used. Therefore calculators were allowed and since calculators weren't allowed on the old syllabus it must have changed in or before 1994.

Also:

The Ordinary level and Higher level syllabuses were introduced in September 1992 and first examined in 1994.

http://www.curriculumonline.ie/index.asp?docID=788

Need any mroe proof, mp3guy?

JC 2K3

Crumbs wrote:

Your answer seems correct JC. It can be simplified to 704(pi)/27. I've seen a couple of wrong answers given for that question and 216(pi) is definitely wrong as that's greater than the volume of the cone itself.

Ah, good point, lol.

God.Looking back.Leaving cert maths wan not that bad.