MedVision ad

Mathematics Marathon HSC 09 (2 Viewers)

luv2luvurmama

Member
Joined
Mar 6, 2009
Messages
312
Gender
Male
HSC
2009
ò <SUP>(Penis)3 Dp
</SUP>Find volume generated through rotating the area between Y=Penis<SUP> </SUP>and the axis around the y axis.
Tip: if having difficulty sketching graph, google penis.
 

xFusion

Member
Joined
Sep 28, 2008
Messages
161
Gender
Female
HSC
2009
i) find ∫tan(x)dx by writing the integral in the form f'(x)/f(x)

ii) a(t)=t3+sec2t

with initial conditions x(0)=6 m and v(0)=4 ms-1

using i), find x as a function of t
i)∫tan(x)dx =∫sinx/cosx dx
let u=cosx
du/dx=-sinx
---> -∫du/u =-ln(u) =-ln(cosx) +C
ii)a(t)=t3+sec2t
v(t)=t^4/4 + tan(t) +C
whent=0, v=4, C=4
v(t)=(t^4)/4 + tan(t) +4
x(t)=(t^5)/20 -ln(cos(t)) +4t +C
when x=6, t=0, C=6
hence x=(t^5)/20 -ln(cos(t)) +4t +6
 

max powers

New Member
Joined
Jun 14, 2008
Messages
9
Gender
Male
HSC
2009
i) find ∫tan(x)dx by writing the integral in the form f'(x)/f(x)

ii) a(t)=t3+sec2t

with initial conditions x(0)=6 m and v(0)=4 ms-1

using i), find x as a function of t

part i.
you let tanx = sinx/cosx
= - ln cosx ?

part ii.
a= t^3 + Sec^2 t
v= t^4/3 + tan t + C
let v=4 and t=0
therefore C=4
v= t^4/3 + tan t +4
x= t^5/15 - ln cosx + 4t +C
let x=6 and t=0
x= t^5/15 - ln cosx + 4t +6

...I'm no good at typing it out lol.
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
39th Question (guesstimate.)

This is a higher-order question. Note that it requires English language rather than mathematical notation; which is not always necessary for the skilled mathematician.

Question: Explain why a circular function is equal to infinity. (And yes, there IS an answer...)

:)

NOW no taking the piss!! Serious answer only pls .....
Wow....what a shit question.
 

xFusion

Member
Joined
Sep 28, 2008
Messages
161
Gender
Female
HSC
2009
3 sin(^2)x cos x

Question:

Find the value of n for which the equation (n+2)x^2 + 3x - 5 = 0 has one root triple the other.
let roots be @, 3@
-b/a = -3/(n+2) =@ +3@ =4@
@= -3/[4(n+2)] -------1
c/a= -5/(n+2) =3@^2 --------2
sub 1 into 2
n= -2.3375
lol nice answer

Question: Jodi takes out a $50,000 loan. The interest is 6% P/A (calculated at the end of each quarter before she pays). She pays this back in quarterly installments. How much would she have to pay each quarter to pay it off in 5 years?
 

jet

Banned
Joined
Jan 4, 2007
Messages
3,148
Gender
Male
HSC
2009


I'm gonna make this question up and see how it goes.

To cut a plank of timber, builders cut a rectangular prism out of a perfectly circular log of timber. If a cross section were viewed, it would be a rectangle inscribed into a circle. Consider a log with cross sectional radius 'r'. What is the maximum area that the cross section of the plank can have? i.e. the maximum area of the rectangle.
 

untouchablecuz

Active Member
Joined
Mar 25, 2008
Messages
1,693
Gender
Male
HSC
2009
let A be the area of the rectangle and [x, y] be the sides of this rectangle

A=xy => A2=x2y2 [1]

x2+y2=(2r)2=4r2 => y2=4r2-x2 [2]

sub [2] in [1]

A2=x2y2=x2(4r2-x2)=4r2x2-x4

maximising A2 is equivalent to maximising A, since dA2/dx=(dA2/dA)(dA/dx) = 2A(dA/dx)

dA2/dx=8r2x-4x3=4x(2r2-x2)=4x(√2r-x)(√2r+x)

let dA2/dx=2A(dA/dx)=0

x=√2r, since x>0

d2A2/dx2=8r2-12x2

when x=√2r, d2A2/dx2=8r2-24r2<0

.'. x=√2r is a maximum turning point

sub in [2], y=√2r

sub x=√2r and y=√2r into [1]

Amax=(√2r)2=2r2
 
Last edited:

untouchablecuz

Active Member
Joined
Mar 25, 2008
Messages
1,693
Gender
Male
HSC
2009
you can also do it like this:

consider rectangle WXYZ, where XY=2r is a diameter

let angle WXZ=k, WZ=a and WX=b

Arectangle=ab [1]

in ∆WXZ,

sin(k)=a/(2r) => a=2rsin(k) [2]

cos(k)=b/(2r) => b=2rcos(k) [3]

sub [2] and [3] in [1]

Arectangle=(2r2)(2sinkcosk)=(2r2)sin(2k)

Arectangle is maximum when sin(2k) is maximum

i.e. when sin(2k)=1

.'. Amax=2r2
 

random-1005

Banned
Joined
Dec 15, 2008
Messages
609
Gender
Male
HSC
2009
Q.It is estimated that 85% of students in Australia own a mobile phone.
(i) Two students are selected at random. What is the probability that neither
of them owns a mobile phone?

0.15 x 0.15 = 0.0225 => 2.25%

(ii) Based on a recent survey, 20% of the students who own a mobile phone
have used their mobile phone during class time. A student is selected at
random. What is the probability that the student owns a mobile phone
and has used it during class time?

0.85 x 0.2 = 0.17 => 17%





Solve tan2x = cot40

cot 40= tan50 (complementary angles)

therefore tan2x=tan(50+360n) {angles of any magnitude} for n=0,1,2,3.....

2x=50 +360n
x= (25 +180n) degress

god knows what the hell untouchable did
 
Last edited:

random-1005

Banned
Joined
Dec 15, 2008
Messages
609
Gender
Male
HSC
2009
tan2x=cot40=tan(5π/18)


.'. 2x=πk+(5π/18) => x=0.5(πk+(5π/18)) i.e. x=(π/36)(18k+5)

where the hell did this come from ? so this 5n/18 just magically appears.

im assuming this is general solution of tan in the second line, where did the k come from??

its n(pi)+Tan^-1 (B)
 
Last edited:

untouchablecuz

Active Member
Joined
Mar 25, 2008
Messages
1,693
Gender
Male
HSC
2009
where the hell did this come from ? so this 5n/18 just magically appears.

im assuming this is general solution of tan in the second line, where did the k come from??

its n(pi)+Tan^-1 (B)
cot(x)=tan(90-x)

when x=40,

cot(40)=tan(90-40)=tan(50)=tan(5π/18)

now, tan2x=cot(40)=tan(5π/18)

let tan(5π/18)=y

tan2x=y

2x=πk+tan-1y=πk+tan-1(tan(5π/18))=πk+5π/18

x=(π/36)(18k+5)
 

random-1005

Banned
Joined
Dec 15, 2008
Messages
609
Gender
Male
HSC
2009
cot(x)=tan(90-x)

when x=40,

cot(40)=tan(90-40)=tan(50)=tan(5π/18)

now, tan2x=cot(40)=tan(5π/18)

let tan(5π/18)=y

tan2x=y

2x=πk+tan-1y=πk+tan-1(tan(5π/18))=πk+5π/18

x=(π/36)(18k+5)

yes, that makes perfect sense!

i love how the people doing 4 unit always have to make something appear so much harder than it actually is.

how about the soln is x=25 +180n n=0,1,2,3.....

not your rubbish with ks and shit
 
Last edited:

untouchablecuz

Active Member
Joined
Mar 25, 2008
Messages
1,693
Gender
Male
HSC
2009
yes, that makes perfect sense!

i love how the people doing 4 unit always have to make something appear so much harder than it actually is.

how about the soln is x=25 +180n n=0,1,2,3.....

not your rubbish with ks and shit
k is just a dummy variable; it can be interchanged with x, y, z, m, Φ, ω, θ

point being, k (or n) is simply representative integer values

also, your solution is wrong

the solution is x=(π/36)(18k+5)=90k+25=90n+25=90ω+25=90θ+25 where k or n or ω or θ is integral
 

aero135

Super Ultimate User!
Joined
Oct 23, 2007
Messages
242
Gender
Male
HSC
2008
Hey guys, anyone got any question 10 type questions. Ive done the WHOLE success one boook. Any one got any other good ones?
 

hiddimp

New Member
Joined
Mar 15, 2008
Messages
4
Gender
Undisclosed
HSC
N/A
given that d/dx(b^x)=b^x.loge.b

evaluate: integrate pi^x dx where x=pi and x=0
im not sure, let me noe if im wrong:

let u = pi^x
du/dx = pi^x.loge(pi)
du/loge(pi) = pi^x.dx
if x= pi, u = pi^pi
if x=0, u = 1

Integrating pi^x where u = pi^pi and u = 1

= 1/loge(pi) from u = pi^pi to 1
= 1/loge(pi). (pi^pi -1)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top