HSC 2013 Maths Marathon (archive) (1 Viewer)

Smile12345 · Sep 27, 2013

Re: HSC 2013 2U Marathon

This is probably an 'easy' question but still good revision for 2U's and a question I'm not sure about... (Thanks in advance

Solve 1/ x^6 - 5 / x^2 + 4 = 0

Trust the above is clear!

Sy123 · Sep 27, 2013

Re: HSC 2013 2U Marathon

HSC2014 said:
Wolfram says no real solution; I couldn't do it

My bad I just tried to make a weird substitution quadratic

anthony_wheels · Sep 27, 2013

Re: HSC 2013 2U Marathon

Ahh, damn! I just did it for 'x^2-6x... = 1' before I saw Sy123's post. Would my answer be right if that was = 1?

anthony_wheels · Sep 27, 2013

Re: HSC 2013 2U Marathon

Wow... Also, Sorry for the blurry image :/ This one here is much better!

Sy123 · Sep 27, 2013

Re: HSC 2013 2U Marathon

$A particle is moving with a displacement$ \ \ x \ $respect to the origin, that moves such that$

$x= a\cos(n t) \ \ \ $for time$ \ \ t$

$$i) Show that$ \ \ \ddot{x} = -n^2 x$

$$ii) Find the displacement, velocity and acceleration when$ \ \ t = \frac{1}{2} a$

Menomaths · Sep 27, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$A particle is moving with a displacement$ \ \ x \ $respect to the origin, that moves such that$

$x= a\cos(n t) \ \ \ $for time$ \ \ t$

$$i) Show that$ \ \ \ddot{x} = -n^{2} x$

$$ii) Find the displacement, velocity and acceleration when$ \ \ t = \frac{1}{2} a$

$x=a\cos(n t)$

$\\\dot{x} = -an sin(nt)$

$\\\ddot{x}=-an^2cos(nt)$

$$Since x=a\cos(n t),\\\ddot{x}=-n^2x$

SpiralFlex · Sep 27, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
$x=a\cos(n t)$

$\\\dot{x} = -an sin(nt)$

$\\\ddot{x}= -an^_{2}$ $cos(nt)$

$$Since x=a\cos(n t), \\\ddot{x}= -n^_2\ x$

Omg I can't make it n^2 by itself...

$-n^2x$

Code:

-n^2x

Menomaths · Sep 27, 2013

Re: HSC 2013 2U Marathon

SpiralFlex said:
$-n^2x$

Code:

-n^2x

Thanks
Now I know what '^_' is for

HSC2014 · Sep 27, 2013

Re: HSC 2013 2U Marathon

what are these dots above x's

Menomaths · Sep 27, 2013

Re: HSC 2013 2U Marathon

HSC2014 said:
what are these dots above x's

${x} = displacement$

$\\\dot{x} = velocity$

$\\\ddot{x} = acceleration$

Makematics · Sep 27, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
$\\\dot{x} = displacement$

$\\\ddot{x} = velocity$

$\\\dddot{x} = acceleration$

not quite

Menomaths · Sep 27, 2013

Re: HSC 2013 2U Marathon

Makematics said:
not quite

.

Aysce · Sep 27, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
$x = displacement$

$\\\dot{x} = velocity$

$\\\ddot{x} = acceleration$

ftfy

HSC2014 · Sep 27, 2013

Re: HSC 2013 2U Marathon

How did those accents come about? I've never even seen them :'( Only like the different level derivatives: y', y"

Menomaths · Sep 27, 2013

Re: HSC 2013 2U Marathon

Guys don't change my comment OKAY? >:[

Sy123 · Sep 27, 2013

Re: HSC 2013 2U Marathon

$In an economy, in the short run the relationship between inflation and unemployment (both measured in percentages) can be modeled by$

$\frac{di}{du} = -k(i-c)^2 \ \ \ $where$ \ \ u \ $is unemployment$ \ \ i \ $is inflation$

$and$ \ \ k \ \ $and$ \ \ c \ \ $are constants$

$$i) Show that$ \ \ i = \frac{A}{u} + B \ \ \ $for constants$ \ \ A \ \ $and $ \ \ B \ \ $find these constants in terms of$ \ \ k \ \ $and$ \ \ c$

$$ii) It is known that when inflation is zero$ \ \ $unemployment is at$ \ \ 2\%$

$$after a policy by the government, inflation has risen to$ \ \ 2\% \ \ $and unemployment has decreased to$ \ \ 1.5\%$

$$Find the values of$ \ \ k \ \ $and$ \ \ c$

SpiralFlex · Sep 27, 2013

Re: HSC 2013 2U Marathon

X dotdotdot is the rate of change of acceleration also know as a jerk

Sy123 · Oct 5, 2013

Re: HSC 2013 2U Marathon

$$i) The point$ \ \ \left(p,q \right) \ \ $lies on the curve$ \ \ y = \frac{1}{x} \ \ \ $show that$ \ \ q = \frac{1}{p}$

$$ii) Show that the tangent to the curve at$ \ \ (p,q) \ \ $is given by$ \ \ x +p^2y = 2p$

$iii) This tangent cuts the$ \ \ x \ \ $and$ \ \ y \ \ $axes, at$ \ \ P \ \ $and$ \ \ Q \ \ $respectively$

$O \ \ $being the origin, show that the area of triangle$ \ \ OPQ \ \ $is constant$$

mahmoudali · Oct 7, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$i) The point$ \ \ \left(p,q \right) \ \ $lies on the curve$ \ \ y = \frac{1}{x} \ \ \ $show that$ \ \ q = \frac{1}{p}$

$$ii) Show that the tangent to the curve at$ \ \ (p,q) \ \ $is given by$ \ \ x +p^2y = 2p$

$iii) This tangent cuts the$ \ \ x \ \ $and$ \ \ y \ \ $axes, at$ \ \ P \ \ $and$ \ \ Q \ \ $respectively$

$O \ \ $being the origin, show that the area of triangle$ \ \ OPQ \ \ $is constant$$

i)

$$ point R(p,q) sub into$ \ \ y= \frac{1}{x} \ \ = q= \frac{1}{p}$
ii)

$\frac {dy}{dx} = \frac{-1}{x^2} \ \ $sub x= p to get gradient$ \ \ M=\frac{-1}{p^2}$

$$y-y_1= m (x-x_1)$ \ \ y-q= \frac{-1}{p^2}(x-p) \ \ y- \frac {1}{p} = \frac{-x}{p^2} +\frac{1}{p}$

$y= -\frac{x}{p^2} + {2}{p} \ \ $(times by p^2) $ \ \ \ \ yp^2 = -x+ 2p \ $(rearrange)$ \ yp^2 +x = 2p$

iii)
$when x=0$

$\frac{2}{p} \ \ $ at point Q when $ y=0 \ \ x=2p \ \ $at point P$$

$A of triangle OPQ$ = \frac{1}{2} \cdot \frac{2}{p} \cdot 2p = 2 \ \ \ $therefore constant$
$proof is complete [ ]$

Sy123 · Oct 7, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
i) point R(p,q) sub into y= 1/x ---> q=1/p
ii) dy/dx = -1/x^2 (sub x= p to get gradient) M=-1/p^2
y-y1= m(x-x1) ---> y-q= -1/p^2 (x-p) ---> y- 1/p = -x/p^2 +1/p
y= -x/p^2 +2/p (times by p^2 ) --> yp^2 = -x+ 2p ----> yp^2 +x = 2p
iii) when x = 0 y= 2/p at point Q when y=0 x = 2p at point P
A of triangle 0PQ= 1/2 x base x height = 1/2 x 2/p x 2p = 2 therefore constant
proof is complete [ ]
sy you need to teach me latex this is wasting too much time typing it this way put a harder question up fag

No latex 2/10

-----

$Find the number of solutions in$ \ \ x \ \ $to the equation$

$\ln(x+1) = mx$

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