HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

[barbernator]

Re: HSC 2012 Marathon

Can you explain what I bolded. Does that mean, since the temperature goes from 0 to 20, it becomes 0.2cm^3/20degrees/20cm^3 of metal laid?

dv/dtemp=0.01 at any point. Technically my statement up there isn't correct, because as 1cm^3 is expanding, the extra volume makes extra volume which makes extra volume infinitessimally, so the change in volume would be more like 0.010101010101... or something. But at any point in time, dv/dtemp=0.01 is the correct definition

 

[RealiseNothing]

Re: HSC 2012 Marathon

dv/dtemp=0.01 at any point. Technically my statement up there isn't correct, because as 1cm^3 is expanding, the extra volume makes extra volume which makes extra volume infinitessimally, so the change in volume would be more like 0.010101010101... or something. But at any point in time, dv/dtemp=0.01 is the correct definition

Ok good, that's where I was getting confused since it didn't make sense as it would keep expanding etc lol.

 

[barbernator]

Re: HSC 2012 Marathon

anyway, new question. Differentiate <a href="http://www.codecogs.com/eqnedit.php?latex=ln(\frac{x^3@plus;x^2@plus;1}{5x^4-e^{2x}-4})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?ln(\frac{x^3+x^2+1}{5x^4-e^{2x}-4})" title="ln(\frac{x^3+x^2+1}{5x^4-e^{2x}-4})" /></a>

 

[J-Wang]

Re: HSC 2012 Marathon

dont forget that there is a hole at x=0, thats why i posted it

is there another way of graphing this other than plotting points and graphing them? (u r still able to determine there is a hole at x=0)

 

[Timske]

Re: HSC 2012 Marathon

anyway, new question. Differentiate <a href="http://www.codecogs.com/eqnedit.php?latex=ln(\frac{x^3@plus;x^2@plus;1}{5x^4-e^{2x}-4})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?ln(\frac{x^3+x^2+1}{5x^4-e^{2x}-4})" title="ln(\frac{x^3+x^2+1}{5x^4-e^{2x}-4})" /></a>

there is an easier way of doing this right?

 

[fortune_cookie]

Re: HSC 2012 Marathon

there is an easier way of doing this right?

use log laws to simplify it first. Then it becomes trivial

= [(3x^2 +2x)/(x^3+x^2+1) ] - [ (20x^3 -2e^(2x) ) /(5x^4 -e^(2x) -4)]

 

[Timske]

Re: HSC 2012 Marathon

16 i think if my idea of anticlockwise/clockwise is correct.

care to share your working out

 

[barbernator]

Re: HSC 2012 Marathon

care to share your working out

9 must be locked next to 2 and 1.

8 can only go next to 1,2 or 3, but seeing as 1 and 2 are next to 9, 8 must definitely be next to 3, and either 1 or 2.

7 can be next to 1,2,3 or 4. As 1,2,3 are next to 8 and 9, 7 must be next to 4 and either 1,2,3.

From these variables, we can note that once 291 is locked, there is the comb 38291 or 29183.
Taking both of these, we can place the 74 group. 3829174 or 4738291 or 4729183 or 2918374.
5 and 6 can be placed in between either way.
and then we have both anticlockwise and clockwise.

So from those 4 seven digit strings, we then multiply by 2 for the perm of 5 and 6, and then multiply by 2 for anti/clockwise. resulting in 16

 

[Timske]

Re: HSC 2012 Marathon

9 must be locked next to 2 and 1.

8 can only go next to 1,2 or 3, but seeing as 1 and 2 are next to 9, 8 must definitely be next to 3, and either 1 or 2.

7 can be next to 1,2,3 or 4. As 1,2,3 are next to 8 and 9, 7 must be next to 4 and either 1,2,3.

From these variables, we can note that once 291 is locked, there is the comb 38291 or 29183.
Taking both of these, we can place the 74 group. 3829174 or 4738291 or 4729183 or 2918374.
5 and 6 can be placed in between either way.
and then we have both anticlockwise and clockwise.

So from those 4 seven digit strings, we then multiply by 2 for the perm of 5 and 6, and then multiply by 2 for anti/clockwise. resulting in 16

Thanks

 

[Timske]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{100} \int \frac{1}{3@plus;\sqrt{x}} ~\textup{dx} ~~,~~ \textup{given x} = (u-3)^2" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{100} \int \frac{1}{3+\sqrt{x}} ~\textup{dx} ~~,~~ \textup{given x} = (u-3)^2" title="\dpi{100} \int \frac{1}{3+\sqrt{x}} ~\textup{dx} ~~,~~ \textup{given x} = (u-3)^2" /></a>

 

[Sy123]

Re: HSC 2012 Marathon

Will post question

 

[Aesytic]

Re: HSC 2012 Marathon

x = (u-3)^2
u-3 = rootx
u = rootx + 3
dx/du = 2(u-3)
dx = 2(u-3) du
.'. integral of 1/(3+rootx) dx = integral (2(u-3))/(3+u-3) du
= integral (2(u-3))/u du
= integral (2 - 6/u) du
= 2u - 6ln(u)
= 2(rootx + 3) - 6ln(rootx + 3) + C

 

[Sy123]

Re: HSC 2012 Marathon

A particle is launched in projectile motion from the origin, at an initial velocity of 20 m/s. At an angle 45 degrees.
Given that:



Find the time when the particle is decreasing in height AND is at an angle of 15 degrees to the horizontal.
Give your answer to 2 decimal places.

(I dont know how to do dots on top of the x :/)

 

[bleakarcher]

Re: HSC 2012 Marathon

I got 10.35 seconds.

 

[Sy123]

Re: HSC 2012 Marathon

I keep getting the same answer of 1.79 seconds

EDIT: It is supposed to say 15 degrees to the horizontal, man Im sorry

 

[bleakarcher]

Re: HSC 2012 Marathon

lol I just looked at my working again after you said that and I saw my time in terms of g:

t=[20sqrt(2)/g][1-tan(15)]

and when I substituted g=10 I put 10sqrt(2)[1-tan(15)] for some retarded reason lol.

 

[bleakarcher]

Re: HSC 2012 Marathon

But I now get 2.07 seconds..

 

[Sy123]

Re: HSC 2012 Marathon

Did you use tan(-15) or tan(15) because that was the trick I was aiming at when I said while the particle is decreasing in height

Ive rechecked my solution and I am pretty sure its correct.

I substituted this into my x in terms of t.

x=-20(tan(-15)-1)

From

tan(-15)=1-x/20

 

[nightweaver066]

Re: HSC 2012 Marathon

But I now get 2.07 seconds..

What i got but i'm doubting it. Will need to double check.

Edit: Actually yeah it's the answer. Made sure by subbing in to x and y equations, finding the angle and making sure vertical velocity was negative.

 

[Nooblet94]

Re: HSC 2012 Marathon

Use \dot{} or \ddot{} to do the dots.

Also, I don't like the wording of the question. "At an angle of 15 degrees to the horizontal" is pretty ambiguous.