shit things for exam (1 Viewer)

hup · Oct 13, 2011

are these necessary

pencil for diagram
compass ruler for circles and graphs

what proof for cylindrical shells?
do you have to do the whole r +dx etc and neglect dx^2
or can you just draw the rectangle thing and show V = 2pirh dx

i will add more if i come across

artosis · Oct 14, 2011

i wanna know about the compass thing too.

if u look at the bos sample answers for the 2009 ext 2 paper, they didnt use a compass :S

Official · Oct 14, 2011

hup said:
are these necessary

pencil for diagram
compass ruler for circles and graphs

what proof for cylindrical shells?
do you have to do the whole r +dx etc and neglect dx^2
or can you just draw the rectangle thing and show V = 2pirh dx

i will add more if i come across

For cylindrical shell I might do the whole r+dx, etc, especially if it is a 4 marker, it's too risky not to imo

blackops23 · Oct 14, 2011

hup said:
do you have to do the whole r +dx etc and neglect dx^2

.... what is the point of making one of the most simplest things in 4U (a formula that is only 2pirh) into a complicated thing? I don't think I've ever learnt r+dx but im guessing it takes into the width of the thickness of the cylinder?

AWARE me if i'm wrong, but i thought in 4 u they didnt take marks off for insufficient working - and that only occurred in 2u, and to a lesser extent in 3u

timting95 · Oct 14, 2011

For the cylindrical shells i would include it, it doesn't take much time extra maybe 5 more seconds?? But if you have a look at the MX2 syllabus it actually says that you need to be able to come to V = 2pirh etc... So i'd say do it, unless you don't know how to - then don't do it

Riproot · Oct 14, 2011

timting95 said:
For the cylindrical shells i would include it, it doesn't take much time extra maybe 5 more seconds?? But if you have a look at the MX2 syllabus it actually says that you need to be able to come to V = 2pirh etc... So i'd say do it, unless you don't know how to - then don't do it

Yeah, I have no idea what you're talking about...

AAEldar · Oct 14, 2011

In the MANSW past paper + solution books they include the sum and exclusion of $(\Delta x)^2$ and all that in their answers and my teacher/head teacher have both advised us to include it.

Pencil for diagram? No thank you!

Omnipotence · Oct 14, 2011

blackops23 said:
what is the point of making one of the most simplest things in 4U (a formula that is only 2pirh) into a complicated thing? I don't think I've ever learnt r+dx but im guessing it takes into the width of the thickness of the cylinder?

AWARE me if i'm wrong, but i thought in 4 u they didnt take marks off for insufficient working - and that only occurred in 2u, and to a lesser extent in 3u

LOL I've always did it r + dx

Riproot · Oct 14, 2011

Omnipotence said:
LOL I've always did it r + dx

Wtf is r+dx?
Someone please awares me!

D94 · Oct 14, 2011

blackops23 said:
oh fuck that shit.... what is the point of making one of the most simplest things in 4U (a formula that is only 2pirh) into a complicated thing? I don't think I've ever learnt r+dx but im guessing it takes into the width of the thickness of the cylinder?

AWARE me if i'm wrong, but i thought in 4 u they didnt take marks off for insufficient working - and that only occurred in 2u, and to a lesser extent in 3u

The 2(pi)xy or 2(pi)rh method skips steps which are evident in the setup for the r + dx or x+ dx method. Essentially, the r + dx method leads onto a 2(pi)rh method.

Riproot said:
Wtf is r+dx?
Someone please awares me!

You take a cylindrical shell with larger radius as x + dx and then a smaller radius as x and then you apply the volume of a cylinder formula. So you get:

$\\\pi (x+dx)^{2}y - \pi (x)^{2}y \\=\pi y (x+dx)^{2} - \pi (x)^{2} \\=\pi y (x+dx+x)(x+dx-x) \\=\pi y (2x+dx)(dx) \\=\pi y (2xdx+dx^{2}) \\=\pi y (2xdx) \\=2\pi xy dx$

Then you substitute in the height ( $y$ ) and integrate it.

$\Delta V = \sum_{a}^{b} 2\pi xy\ dx \\V = 2\pi\int_{a}^{b} xy\ dx$

It's the same thing, but the setup which I just showed 'could' be worth a mark.

Official · Oct 14, 2011

D94 said:
The 2(pi)xy or 2(pi)rh method skips steps which are evident in the setup for the r + dx or x+ dx method. Essentially, the r + dx method leads onto a 2(pi)rh method.

You take a cylindrical shell with larger radius as x + dx and then a smaller radius as x and then you apply the volume of a cylinder formula. So you get:

$\\\pi (x+dx)^{2}y - \pi (x)^{2}y \\=\pi y (x+dx)^{2} - \pi (x)^{2} \\=\pi y (x+dx+x)(x+dx-x) \\=\pi y (2x+dx)(dx) \\=\pi y (2xdx+dx^{2}) \\=\pi y (2xdx) \\=2\pi xy dx$

Then you substitute in the height ( $y$ ) and integrate it.

$\Delta V = \sum_{a}^{b} 2\pi xy\ dx \\V = 2\pi\int_{a}^{b} xy\ dx$

It's the same thing, but the setup which I just showed 'could' be worth a mark.

<3

Riproot · Oct 14, 2011

D94 said:
You take a cylindrical shell with larger radius as x + dx and then a smaller radius as x and then you apply the volume of a cylinder formula. So you get:

$\\\pi (x+dx)^{2}y - \pi (x)^{2}y \\=\pi y (x+dx)^{2} - \pi (x)^{2} \\=\pi y (x+dx+x)(x+dx-x) \\=\pi y (2x+dx)(dx) \\=\pi y (2xdx+dx^{2}) \\=\pi y (2xdx) \\=2\pi xy dx$

Then you substitute in the height ( $y$ ) and integrate it.

$\Delta V = \sum_{a}^{b} 2\pi xy\ dx \\V = 2\pi\int_{a}^{b} xy\ dx$

It's the same thing, but the setup which I just showed 'could' be worth a mark.

My teacher never mentioned that.

I'm such a stressball right now.

shit things for exam (1 Viewer)

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