Mathematics Marathon HSC 09 (1 Viewer)

addikaye03 · Oct 16, 2009

khorne said:
Here's a big cup of harden the hell up:

Find the volume of the ellipsoidal solid given by 2x^2 + 4y^2 + z^2 = 16

First ever maths troll

lol now that's just silly khorne

untouchablecuz · Oct 16, 2009

yer khorne

random-1005 · Oct 16, 2009

khorne said:
Here's a big cup of harden the hell up:

Find the volume of the ellipsoidal solid given by 2x^2 + 4y^2 + z^2 = 16

First ever maths troll

whoa your mad, you shouldn't be up past your bed time

00iCon · Oct 16, 2009

khorne said:
Here's a big cup of harden the hell up:

Find the volume of the ellipsoidal solid given by 2x^2 + 4y^2 + z^2 = 16

First ever maths troll

64sqroot(2)(pi)/3
=94.7815...
DID IT IN ONE STEP!

khorne · Oct 16, 2009

00iCon said:
64sqroot(2)(pi)/3
=94.7815...
DID IT IN ONE STEP!

Wrong bitch, I believe you will find it's 64*pi/3

00iCon · Oct 16, 2009

khorne said:
Wrong bitch, I believe you will find it's 64*pi/3

Wolframalpha is never wrong!

ninetypercent · Oct 16, 2009

New question:

If dx/dt = 1/(x+4) and x =0 when t=0, find t when x =2

hermand · Oct 16, 2009

ninetypercent said:
New question:

If dx/dt = 1/(x+4) and x =0 when t=0, find t when x =2

$\frac{dx}{dt}=\frac{1}{x+4}$
$\therefore \frac{dt}{dx}=x+4$
$\therefore t=\frac{x^{2}}{2}+4x+c$
$\textup{since}\: \: x=0 \: \textup{when}\: t=0, \: c=0$
$\therefore t=\frac{x^{2}}{2}+4x$
$\textup{when}\; x=0,\: t=\frac{2^{2}}{2}+4(2)$
$\therefore t=10\; \textup{at}\; x=2$

Q.
a. Find the limiting sum of the series $1+\cos^{2}x+\cos^{4}x+...$ in simplest form.
b. Why does this series have a limiting sum?

Timothy.Siu · Oct 16, 2009

hermand said:
$\frac{dx}{dt}=\frac{1}{x+4}$
$\therefore \frac{dt}{dx}=x+4$
$\therefore t=\frac{x^{2}}{2}+4x+c$
$\textup{since}\: \: x=0 \: \textup{when}\: t=0, \: c=0$
$\therefore t=\frac{x^{2}}{2}+4x$
$\textup{when}\; x=0,\: t=\frac{2^{2}}{2}+4(2)$
$\therefore t=10\; \textup{at}\; x=2$

Q.
a. Find the limiting sum of the series $1+\cos^{2}x+\cos^{4}x+...$ in simplest form.
b. Why does this series have a limiting sum?

it only does if cos x =/=1 or it'd just keep getting bigger.

hermand · Oct 16, 2009

Timothy.Siu said:
it only does if cos x =/=1 or it'd just keep getting bigger.

it's not a number.

*taps nose knowingly*.

haha.

s2Vicki · Oct 16, 2009

Question:

Show that the maximum value of <a href="http://www.codecogs.com/eqnedit.php?latex=\frac{\ln x}{x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{\ln x}{x}" title="\frac{\ln x}{x}" /></a> is <a href="http://www.codecogs.com/eqnedit.php?latex=\frac{1}{e}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{1}{e}" title="\frac{1}{e}" /></a>

xFusion · Oct 16, 2009

$\frac{\partial \frac{lnx}{x}}{\partial x} = \frac{x\times\frac{1}{x}-lnx\times 1 }{x^{2}}$
$when \frac{\partial \frac{lnx}{x}}{\partial x} = 0$
$\frac{1-lnx}{x^{2}}=0 \therefore x=e$
Test for maximum in dy/dx
$test \frac{1}{e}>0$
$test e=0$
$test e^{2}<0$
$\therefore x=e$ is maximum

p.s are you sure your question is correct? or did i do something wrong =/
Question:The rate of sales, R dollars per day of a certain australian company is given by R=2t +120, where t=time in days.
a)what is the rate of sales at the end of 10 days?
b)calculate total amount of sales in first 8 days
c) on what day will accumulated sales exceed $10000?

s2Vicki · Oct 16, 2009

xFusion said:
$\frac{\partial \frac{lnx}{x}}{\partial x} = \frac{x\times\frac{1}{x}-lnx\times 1 }{x^{2}}$
$when \frac{\partial \frac{lnx}{x}}{\partial x} = 0$
$\frac{1-lnx}{x^{2}}=0 \therefore x=e$
Test for maximum in dy/dx
$test \frac{1}{e}>0$
$test e=0$
$test e^{2}<0$
$\therefore x=e$ is maximum

p.s are you sure your question is correct? or did i do something wrong =/

i don't know...i got it off a sheet my teacher wrote.

xFusion · Oct 16, 2009

s2Vicki said:
i don't know...i got it off a sheet my teacher wrote.

lol alright, i'm quite sure i did it right unless someone can correct me :uhhuh:

elmoateme · Oct 17, 2009

You found the x value for which lnx/x is a maximum. To find that actual maximum value you need to sub x=e into the original equation. Doing that you get 1/e as required. =)

So yes you did it right, but you just didn't finish the question. Make sure you do in the HSC.

ktl_edwards · Oct 17, 2009

put a loan repayment question in! Love it!!

xFusion · Oct 17, 2009

elmoateme said:
You found the x value for which lnx/x is a maximum. To find that actual maximum value you need to sub x=e into the original equation. Doing that you get 1/e as required. =)

So yes you did it right, but you just didn't finish the question. Make sure you do in the HSC.

OMG genius

I couldn't think how to get 1/e. Im gonna rep you for that

anyhow loan repayment question..: Keirren borrows $150,000 on a 7.5% p/a interest, calculated at the end of each month after Keirren repays an amount.
a)How much is each monthly repayment if Keirren pays off the loan in 10 years.
b)How much would Keirren save if he pays off the loan in 5 years as opposed to 10 years.

addikaye03 · Oct 17, 2009

xFusion said:
OMG genius I couldn't think how to get 1/e. Im gonna rep you for that
anyhow loan repayment question..: Keirren borrows $150,000 on a 7.5% p/a interest, calculated at the end of each month after Keirren repays an amount.
a)How much is each monthly repayment if Keirren pays off the loan in 10 years.
b)How much would Keirren save if he pays off the loan in 5 years as opposed to 10 years.

P=$150,000, I=0.075/12=1/160 (monthly interest), n=120 monthly repayments

i) A_1=P+I-M

=150000+150000(1/160)-M

=150000(1+1/160)-M

A_2=A1+A1(I)-M

=A1(1+1/160)-M

=[150000(1+1/160)-M](1+1/160)-M

=150000(1+1/160)^2-M(1+1/160)-M

=150000(1+1/160)^2-M[1+(1+1/160)]

A_3=A2(1+1/160)-M

=[150000(1+1/160)^2-M[1+(1+1/160)](1+1/160)-M

=150000(1+1/160)^3-M[1+(1+1/160)+(1+1/160)^2]

...

A_10=150000(1+1/160)^10-M[1+(1+1/160)+(1+1/160)^2+...+(1+1/160)^9]

After 10 years, A_10=0; a=1, r=(1+1/160)

M[(1+1/160)^9-1]/(1/160)=150000(1+1/160)^10

M=[1875(1+1/160)^10]/ 2[(1+1/160)^9 -1]

Don't have a calculator here atm, but that's the answer

ii) Where ever n=10, insert n=5. For the (n-1)term (ie. the 9) insert 4

Then minus it from i) to get difference

Enjoy, Addison

ktl_edwards · Oct 17, 2009

a)
r=7.5%/12 n=12X10
=0.625% = 120
=1.00625

A1=150000X(1.00625)- M
A2= 150000.(1.00625)^2 - M(1+1.00625)
A120=150000.(1.00625)^120 - M (1+1.00625+....+1.00625^119)

S119=1.(1.00625^119 -1/0.00625)
= 175.8314

so M= 150000.(1.00625)^120 / 175.8314
= $1801.78

Total Payment = M x 120
=$216231.6

b) r=1.00625 n=60

A60=150000X 1.00625^60 - M (1+1.00625+...+1.00625^59)

S59= 1.( 1.00625^59 -1 /0.00625)

M= 150000.(1.00625)^60 / S59
= $3066.79

Total payment = M x 60
=$184005.73

so Kierren saved $216213.6 - $184005.73
= $32207.87

00iCon · Oct 17, 2009

so now... a not-general-maths question someone?

Mathematics Marathon HSC 09 (1 Viewer)

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