Mathematics Marathon HSC 09 (2 Viewers)

xFusion · Oct 17, 2009

00iCon said:
so now... a not-general-maths question someone?

LOL like what? I seriously cant think of 2u questions.. So I use a exam question book

The harder questions are the graphing questions which I cbb posting

00iCon · Oct 17, 2009

xFusion said:
LOL like what? I seriously cant think of 2u questions.. So I use a exam question book The harder questions are the graphing questions which I cbb posting

Thread has been upgraded to Extension 2 Mathematics Marathon!
Q: The area bounded by y = 4 - x², x=2 and y=4 is rotated about the line x=4. Using the method of cylindrical shells:
i) Show that the volume of a cylindrical shell of thickness dx is
(pi)x²(8 - 2x)dx
ii) Find the volume of the solid generated

boxhunter91 · Oct 17, 2009

ooicon how bout posting in the 4U section of this site?

00iCon · Oct 17, 2009

boxhunter91 said:
ooicon how bout posting in the 4U section of this site?

but theres no actual marathon running, and if i get something wrong there, i will bring shame to my family for generations!
EDIT: made one: http://community.boredofstudies.org/showthread.php?p=4644070#post4644070

ktl_edwards · Oct 17, 2009

so confused...

how to do question 8a ) 2007 hsc advanced math paper

help!!

boxhunter91 · Oct 17, 2009

Thats a good question.
at t=1 N=Ae^k=1.6x10^9
at t=2 N=Ae^2k = 2.6x10^9

A=2.6x10^9/e^2k=1.6x10^9/e^k
.:e^k=2.6x10^0/1.6x10^9
.:k=ln(13/8)
.:A=1.6x10^9/e^k
.:A=12800/13x10^6

Part ii the answer is November 2010

raniaaa · Oct 17, 2009

ktl_edwards said:
so confused...

how to do question 8a ) 2007 hsc advanced math paper

help!!

a) i.

N= Ae^kt

when:
t=1 N=1600
t=2 N=2600

now:

1600=Ae^k .... (1)
2600=Ae^2k .... (2)

(2)/(1)

1.625= e^k
k= ln 1.625

sub k= ln 1.625 into (1)

Ae^ln 1.625 = 1600
1.625A= 1600
A= 984 8/13

Hornby · Oct 17, 2009

Q: If 8, p, q, 27 are the first four terms of a geometric series, find the values of p and q

boxhunter91 · Oct 17, 2009

Hornby said:
Q: If 8, p, q, 27 are the first four terms of a geometric series, find the values of p and q

EDIT scrape my answer have no idea what i just done.

hermand · Oct 17, 2009

Hornby said:
Q: If 8, p, q, 27 are the first four terms of a geometric series, find the values of p and q

$27=8r^3$
$\therefore r=\frac{3}{2}$
$\textup{since}\; p=8r,\; p=8\times \frac{3}{2}$
$\therefore p=12$
$\textup{since}\; q=8r^{2}=12r, \; p=12\times \frac{3}{2}$
$\therefore q=18$

Q.
When a valve is released, a chemical flows into a large tank that is initially empty. The volume, V litres, of chemical in the tank increases at the rate
$\frac{dV}{dt}=2e^{t}+2e^{-t}$
where t is measured in hours from the time the valve is released.
i. At what rate does the chemical initially enter the tank?
ii. Use integration to find an expression for V in terms of t.
iii. Show that $2e^{2t}-3e^{t}-2=0$ when V=3.
iv. Find t, to the nearest minute, when V=3.2

Hornby · Oct 17, 2009

At third line of working box, you forgot to multiply 216 by 8, working is still right you get p=12

then you get 8q = 144

q = 18

lychnobity · Oct 17, 2009

Use $\sum_{r=1}^{n}x^{2}$ to find the number of squares that can be found in a square measuring 8x8 (made up of unit squares)

Don't get what I'm asking? Look at the picture, and tell me how many squares can be made. Yes, they can overlap. Include all side lengths ie 1x1, 2x2, 3x3, 4x4, 5x5, 6x6, 7x7, 8x8

boxhunter91 · Oct 17, 2009

hermand said:
$27=8r^3$

Q.
When a valve is released, a chemical flows into a large tank that is initially empty. The volume, V litres, of chemical in the tank increases at the rate
$\frac{dV}{dt}=2e^{t}+2e^{-t}$
where t is measured in hours from the time the valve is released.
i. At what rate does the chemical initially enter the tank?
ii. Use integration to find an expression for V in terms of t.
iii. Show that $2e^{2t}-3e^{t}-2=0$ when V=3.
iv. Find t, to the nearest minute, when V=3.2

i. Dv/dt=4
ii.V=2e^t-2e^-t
iii.3=2e^t-2e^-t
3=2e^t-2/e^t
3et=2e^2t-2
.:2e^2t-2e^t-2=0
iv.Im seeing you changed the question from the one in the HSC. Makes it a bit harder

xFusion · Oct 17, 2009

boxhunter91 said:
i. Dv/dt=4
ii.V=2e^t-2e^-t
iii.3=2e^t-2e^-t
3=2e^t-2/e^t
3et=2e^2t-2
.:2e^2t-2e^t-2=0
iv.Im seeing you changed the question from the one in the HSC. Makes it a bit harder

I couldn't see the 3rd part when trying to do this XD
iv)22mins

Anyhow new question:When the temperature of a body is T degrees above the temperature(assumed constant) of its surroundings, its temperature falls at a rate proportional to T. This can be expressed using the formula T=Toe^(-kt), where To is the temperature that the body is initially above the surrounds, k is a constant and t denotes time(in minutes)
i)show by differentialtion that dT/dt =-kT
ii)if the body is initally at 85 degrees Celcius, the surroundings at 15 degress celcius, and the body cools to 65 degrees celcius after 10 mins, calculate the value of k correct to 3 sig figs.
iii)Calculate the temperature of the body after a further 10 mins

hermand · Oct 17, 2009

boxhunter91 said:
i. Dv/dt=4
ii.V=2e^t-2e^-t
iii.3=2e^t-2e^-t
3=2e^t-2/e^t
3et=2e^2t-2
.:2e^2t-2e^t-2=0
iv.Im seeing you changed the question from the one in the HSC. Makes it a bit harder

oh, oops. that was completely unintentional.

sorry! it was supposed to be V=3

hermand · Oct 17, 2009

lychnobity said:
Use $\sum_{r=1}^{n}x^{2}$ to find the number of squares that can be found in a square measuring 8x8 (made up of unit squares)

Don't get what I'm asking? Look at the picture, and tell me how many squares can be made. Yes, they can overlap. Include all side lengths ie 1x1, 2x2, 3x3, 4x4, 5x5, 6x6, 7x7, 8x8

so, are you saying where n=8, or 64?

lychnobity · Oct 17, 2009

hermand said:
so, are you saying where n=8, or 64?

8

tranny91 · Oct 18, 2009

xFusion said:
I couldn't see the 3rd part when trying to do this XD
iv)22mins

Anyhow new question:When the temperature of a body is T degrees above the temperature(assumed constant) of its surroundings, its temperature falls at a rate proportional to T. This can be expressed using the formula T=Toe^(-kt), where To is the temperature that the body is initially above the surrounds, k is a constant and t denotes time(in minutes)
i)show by differentialtion that dT/dt =-kT
ii)if the body is initally at 85 degrees Celcius, the surroundings at 15 degress celcius, and the body cools to 65 degrees celcius after 10 mins, calculate the value of k correct to 3 sig figs.
iii)Calculate the temperature of the body after a further 10 mins

i) Using the equation T = Toe^(-kt)

dT/dt = -k(Toe^(-kt))
but T = Toe^(-kt)

therefore dT/dt = -kT

ii- when t=0, To= 85,
when t=10, T=65
65=85e^(-10k)

65/80= e^(-10k)

log both sides

therefore k = 0.021 (3 sig fig)

ii- when t = 10
T = 85e^(-10k)

from calculation

T= 69.0625 degress

tranny91 · Oct 18, 2009

new question:
i) Prove the line y = x + 2 is a tangent to the parabola y = x^2 - 5x + 11
ii) Let Q be the point where the line y=x+2 touches the parabola x^2 - 5x +11. Show that the normal of the parabola at Q is y = -x + 8
iii) Find the area of the region enclosed between the parabola and the line y=-x+8

(last one i stuffed up xD)

boxhunter91 · Oct 18, 2009

LOL tranny this question came up in my half yearlies. I freaked out when i saw it and lost 8 marks like that

Now it is the easiest question lol.
i) x+2 = x^2 - 5x + 11
.:x^2-6x+9=0
.

x-3)^2=0
.:Since only one solution it is a tangent.
ii)At x=3 y=5
.:Mn=-1
.:y-5=-1(x-3)
.:y=-x +8
iii)Find the x-ordinates where they intersect then integrate -x + 8 - (x^2 - 5x + 11 ) dx.

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