2011 hsc mx1 marathon (1 Viewer)

Wight · Oct 25, 2011

^ 10ln10 (sub in x=1)

Using the binomial theorem prove:
$C_k^{n+1} = C_k^n+C_{k-1}^n$

Sanical · Oct 25, 2011

^
Haha, I know how do it, but lol, I'm not gonna type it all out!

$\frac{d^2x}{dt^2}=-0.05v^3$
where x is the displacement in metres from the origin O and v metres per second is the velocity of the particle at time t seconds. When t = 0, the particle passes the origin with a velocity of 10m/s

Find the time for particle to travel 20 metres.

SImon05089 · Oct 25, 2011

Sorry Wight, i dont know how to use latex,

^ t = 7 seconds

barbernator · Oct 25, 2011

sqrt10s? not sure bout this one

barbernator · Oct 25, 2011

$dv/dt = -0.05v^{3}$
$dt/dv = 1/-0.05v^{3}$
$t=10/v^{2}+C=10/v^{2}-1$
$v^{2}=10/t+1$
$v=\sqrt{10}/\sqrt{t+1}$
$x=2\sqrt{10}t$
$t=\sqrt{10}$

left out integrating steps, it this correct?

Sanical · Oct 25, 2011

Haha, don't have the answers but I'm getting 9.9
EDIT: ERROR

Wight · Oct 25, 2011

I got 12 :S

$\frac{vdv}{dx}=-0.05v^3\newline \frac{-20dv}{v^2}=dx\newline x = \frac{20}{v}+C\newline v=10, x = 0, C=-2 \frac{20}{v}-2 = x\newline \frac{dx}{dt}=\frac{20}{x+2}\newline \frac{20dt}{dx}=x+2\newline 20t=0.5x^2+2x+C\newline x=0,t=0,C=0\newline 20t=0.5(20)^2+40\newline t = 12$

SImon05089 · Oct 25, 2011

barbernator,
i think your c should of been -0.1, if u were to sub the initial conditions

barbernator · Oct 25, 2011

sorry put in wrong c

Sanical · Oct 25, 2011

Damn I'm getting 8 now..

Sanical · Oct 25, 2011

nah it's 12

Wight · Oct 25, 2011

I used vdv/dx, so without using that (not sure if its in 3U course)
$\frac{dt}{dv}=\frac{-20}{v^3}\newline t = \frac{10}{v^2}+C\newline t = 0, v= 10,c = -0.1\newline v^2 = \frac{100}{10t+1}\newline \frac{dx}{dt}=\frac{10}{\sqrt{10t+1}}\newline u = 10t+1\newline du=10dt\newline x = \int \frac{1}{\sqrt{u}} \newline x=2u^{1/2}+C \newline t=0,x=0,c=-2\newline x=2 \sqrt{u}-2 \newline 11=\sqrt{u} \newline t = 12$

SImon05089 · Oct 25, 2011

i re-did it 3 times, i got 7, fml

Edit: found my error, should be 10t not 5t for v, yup t = 12 is right, gj boys

Sanical · Oct 25, 2011

SImon05089 said:
i re-did it 3 times, i got 7, fml

Well keep trying until you get 12.

Vidhya · Oct 25, 2011

barbernator said:
P(x) is an odd polynomial of degree 3. It has (x+4) as a factor and,
when it is divided by (x−3) , the remainder is 21. Find P(x).

What is the solution? Would you mind giving working out?

barbernator · Oct 25, 2011

Sanic - yay i got 12

barbernator · Oct 25, 2011

Vidhya said:
What is the solution? Would you mind giving working out?

as P(x) is odd, it has roots, Ax(x-4)(x+4)
P(-3)=-3A(-7)(-1)
= -21A=21
A=-1

P(x)= -x(x+4)(x-4)

Sanical · Oct 25, 2011

barbernator said:
Sanic - yay i got 12

Lol, nice work

A,B and C are three sequential points on a straight line on horizontal ground. A vertical flagpole PQ is situated close by the line (but its base P is not on the line).
The angles of elevation of the top of the flagpole from A, B and C are $\tan^{-1}\frac{1}{4}$ , $\tan^{-1}\frac{1}{2}$ and $\tan^{-1}\frac{1}{3}$ respectively. If AB = 90m and BC = 30m, find the height of the flagpole.

barbernator · Oct 25, 2011

Sanical said:
Lol, nice work

A,B and C are three sequential points on a straight line on horizontal ground. A vertical flagpole PQ is situated close by the line (but its base P is not on the line).
The angles of elevation of the top of the flagpole from A, B and C are $\tan^{-1}\frac{1}{4}$ , $\tan^{-1}\frac{1}{2}$ and $\tan^{-1}\frac{1}{3}$ respectively. If AB = 90m and BC = 30m, find the height of the flagpole.

jesus youve got some good questions

Sanical · Oct 25, 2011

barbernator said:
jesus youve got some good questions

Thanks, lol -_-

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