General Thoughts: Mathematics Extension 1 (1 Viewer)

HyperComplexxx · Oct 26, 2011

lpodnano said:
I got that. How do you do the next part?

if it lost 1/8 of volume therefore there be 7/8V left
Intially, V = h^3pi/3
So make the 7/8V = 7h^3pi/28 = pi(h^3 -L^3)/3
Then equate L = h/2 ?
then i got -5pi/2

Wight · Oct 26, 2011

lpodnano said:
I got that. How do you do the next part?

lol I skipped the next part cause I wanted to do the binomial...and had 5 minutes left (got up to 7)b)iii) -.- ). I'll solve it now but.

alstah · Oct 26, 2011

It was hard!

But on a brighter note, maths - never again in my life! (besides basic arithmetic)

damo_g · Oct 26, 2011

Wight said:
$V=\frac{\pi }{3}(h^3-l^3)\newline dV=\frac{\pi }{3}(-3l^2)dl\newline \frac{dV}{dl}\cdot \frac{dl}{dt} = -10\pi l^2\newline l=2\newline \frac{dV}{dt}=-40\pi$

someone confirm that just in case.

yes! That's 2 more, much needed, marks.

tranx · Oct 26, 2011

Re: holy fucking shit i might as well just die now.

Think i got 83/84 in that exam, lost it cos i made a silly..

brachester · Oct 26, 2011

damo_g said:
yes! That's 2 more, much needed, marks.

this is so annoying, i can"t remember if i did square the l or not!!!!!! Ahrggggggg

apollo1 · Oct 26, 2011

Re: holy fucking shit i might as well just die now.

fuk those 1 markers totally put me off. i thought 1 markers meant just questions with straight out one line answers.

clementc · Oct 26, 2011

*sigh* i did every little bit, all the differentiion and shit, and accidentally wrote (dl/dt)^2 in the formula instead of (dl/dt). Thus all my answers are out of whack fml ='( ANd its a formula so probably no carry forward *crawls into a ball and cries*

bobthebuilder94 · Oct 26, 2011

wow this is so akward.... wrote down -40pi^2 ... then evaluated it to 378 for some reason -___- there goes another mark

b00m · Oct 26, 2011

Re: holy fucking shit i might as well just die now.

was the test similar to last year? i remember there being a lot of shitty 1 markers that took so much time

Wight · Oct 26, 2011

okay my solution to the next part...wish I thought like this under exam conditions -.-
$l = 0, V\rightarrow full\newline \therefore V = \frac{\pi }{3}h^3\newline \frac{7}{8}\cdot\frac{\pi}{3}h^3 = \frac{\pi}{3}(h^3-l^3)\newline l = \frac{h}{2}\newline \therefore \textrm{as } \frac{dV}{dt} = -10\pi l^2\newline \frac{dV}{dt}=\frac{-10\pi h^2}{4}$

clementc · Oct 26, 2011

Re: holy fucking shit i might as well just die now.

Shitty 1 marker AND 2 markers GRRRR. Ironically the 3 markers were nicest because they actually deserved the marks. Not like the half a page for 1 markers >='(

Also, for cos^-1(-1/2), did anyone write the general solution? I was being a smartarse and wrote it, but now it seems so stupid, as arccos has restricted domain =\

Implying · Oct 26, 2011

Wight said:
okay my solution to the next part...wish I thought like this under exam conditions -.-
$l = 0, V\rightarrow full\newline \therefore V = \frac{\pi }{3}h^3\newline \frac{7}{8}\cdot\frac{\pi}{3}h^3 = \frac{\pi}{3}(h^3-l^3)\newline l = \frac{h}{2}\newline \therefore \textrm{as } \frac{dV}{dt} = -10\pi l^2\newline \frac{dV}{dt}=\frac{-10\pi h^2}{4}$

yeah i screwed that bit up hugely

CombustionManX · Oct 26, 2011

HyperComplexxx said:
pi squared ? i got pi

I can't remember what I got exactly

Deep Blue · Oct 26, 2011

Re: holy fucking shit i might as well just die now.

A hard paper is better than an easy paper. Your marks will probably align to the same as they would have been anyway.

hup · Oct 26, 2011

is the restriction y =/= 1/4

cssftw · Oct 26, 2011

lpodnano said:
I got that. How do you do the next part?

Q7(a) PART (iii)
$\textup{For volume to have lost}\; \frac{1}{8}\: \textup{of its water, it means there is}\: \frac{7}{8}\: \textup{part\s remaining from the full volume}\\\\\textup{When the lower cone is full,}\: V_{max}=\frac{\pi }{3}h^3\; cm^3\; \textup{from part (i) as }\; V_{max}\; \textup{occurs when}\;\l=0\\\\\textup{Now}\: \frac{7}{8}ths\: \textup{of}\; V =\; \frac{7\pi }{24}(h^3)\: cm^3\\\\\textup{But from part (i) we know that}\: V=\frac{\pi}{3}(h^3 - l^3)\\\\ \textup{In order to find}\; \frac{dV}{dt}\; \textup{we need an expression for}\: l\: \textup{as}\; \frac{dV}{dt}\: =\: -10\pi(l^2)\\\\\textup{Thus we equate these two expressions for}\: V\:.......\\\\\frac{7\pi }{24}(h^3)\; =\; \frac{pi}{3}(h^3 - l^3)\\\\\therefore \; \frac{7}{8}(h^3)\; =\: h^3 - l^3\\\\\therefore 7h^3 = 8h^3 - 8l^3\\\\\therefore 8l^3 = h^3\\\\\therefore l=\frac{h}{2}\\\\\textup{Subsitute}\; l=\frac{h}{2}\; \textup{into the expression for}\;\frac{dV}{dt}\;........\\\\\therefore \frac{dV}{dt}=-10\pi(\frac{h}{2})^2$

$\therefore \frac{dV}{dt}\; =\; \frac{-5}{2}\pi(h^2)$

Latex took fucking forever....hope I...answered your question....pheeeeeeeeeeeeeeeeeeeeeeeeeeww

CombustionManX · Oct 26, 2011

Wight said:
$V=\frac{\pi }{3}(h^3-l^3)\newline dV=\frac{\pi }{3}(-3l^2)dl\newline \frac{dV}{dl}\cdot \frac{dl}{dt} = -10\pi l^2\newline l=2\newline \frac{dV}{dt}=-40\pi$

someone confirm that just in case.

Did it again, yes that is what I actually got.
dv/dt= -10pi(l^2)
when l=2, dv/dt= -40pi

Sent from my Nexus S using Tapatalk

Wentworth · Oct 26, 2011

hup said:
is the restriction y =/= 1/4

I got y<1/4 not too sure though :/

rawrence · Oct 26, 2011

I thought it was okay.. already lost about 4-5 marks but on top of stupid mistakes, more to come!

A lot harder than I expected I think.

Also what was the total distance the particle travelled? And also what did everyone get as 'p'?

General Thoughts: Mathematics Extension 1 (1 Viewer)

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