barbernator
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Re: HSC 2012 Marathon 
-1/32. Does anyone have an answer to the one above?
-1/32. Does anyone have an answer to the one above?
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HSC 2012 MX1 Marathon #2 (archive) (8 Viewers)
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Kingportable
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Re: HSC 2012 Marathon
Simple Harmonic Motion Fitzpatrick 3U Chapter 25(C) Q4(a)
The displacement x at time t of a point moving in a straight line is given by x=asin(nt+epsiton). Find the form wich this expression takes if initially:
b) x=0 and the velocity is negative
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Hey i'm trying to this question and currently working on question "a"
a) x'=0 and x=-5
Simple Harmonic Motion Fitzpatrick 3U Chapter 25(C) Q4(a)
The displacement x at time t of a point moving in a straight line is given by x=asin(nt+epsiton). Find the form wich this expression takes if initially:
b) x=0 and the velocity is negative
-------
Hey i'm trying to this question and currently working on question "a"
a) x'=0 and x=-5
Kingportable
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Re: HSC 2012 Marathon
Forget my previous question... Lol this one is an even more annoying one
This question's answer was completely different to mine.
Question 7 (CHAPTER 25C)
Solve the differential equation d^2x/dt^2 + 16x = 0 subject to the conditions x=3 and dx/dt=16 when t=0. Find the maximum displacement and the maximum speed if x metres is the displacement of a particle moving in a straight line at time t seconds.
what i did
i said that x=asin(nt+alpha)
and that d^2/dt^2 +16x=0
is the same as x''+16x=0
so x''=-16x <-- SHM where n=4
Since this is a sin function at t=0, x=0
so asin(alpha)=0
alpha=Sin^-1(0)
so, alpha=0 at t=0
since v=16 at t=0
i got that v=ancos(nt+alpha)=16
since at t=0, alpha =0
ancos(0)=16
an=16
since n=4
4a=16
a=4
so... x=4sin(4t)
however the back of John Fitzpatrick 3U says
x=4sin(4t) +3cos4t
the 4sin(4t) part makes total sense.... BUT WHERE THE HELL DID THAT 3cos4t COME FROM!!!!!!!!....WTTTFFFFF??????
So yeah im stuck on finding the displacement, is the axiliary methods in play in here?
Anyways the other answers are: 5m;20m/s
Forget my previous question... Lol this one is an even more annoying one
This question's answer was completely different to mine.
Question 7 (CHAPTER 25C)
Solve the differential equation d^2x/dt^2 + 16x = 0 subject to the conditions x=3 and dx/dt=16 when t=0. Find the maximum displacement and the maximum speed if x metres is the displacement of a particle moving in a straight line at time t seconds.
what i did
i said that x=asin(nt+alpha)
and that d^2/dt^2 +16x=0
is the same as x''+16x=0
so x''=-16x <-- SHM where n=4
Since this is a sin function at t=0, x=0
so asin(alpha)=0
alpha=Sin^-1(0)
so, alpha=0 at t=0
since v=16 at t=0
i got that v=ancos(nt+alpha)=16
since at t=0, alpha =0
ancos(0)=16
an=16
since n=4
4a=16
a=4
so... x=4sin(4t)
however the back of John Fitzpatrick 3U says
x=4sin(4t) +3cos4t
the 4sin(4t) part makes total sense.... BUT WHERE THE HELL DID THAT 3cos4t COME FROM!!!!!!!!....WTTTFFFFF??????
So yeah im stuck on finding the displacement, is the axiliary methods in play in here?
Anyways the other answers are: 5m;20m/s
Sy123
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Re: HSC 2012 Marathon
\frac{d}{dx}&=-16x \\ \frac{v^2}{2}&=-8x^2+C \\ \\ x=3 \Rightarrow v=16 \\ \frac{16^2}{2}&=-8(3)^2+C \\ C=200 \\ \frac{v^2}{2}&=200-8x^2 \\ v^2&=400-16x^2 \\ v&=+\sqrt{16(25-x^2)} \\ v&=+4\sqrt{25-x^2} \\ \\ \frac{dt}{dx}&=\frac{1}{4\sqrt{25-x^2}} \\ \\ t&=\frac{1}{4}\sin^{-1}{\frac{x}{5}}+C \end{align*})
 \\ \\ v_{max}=4\sqrt{25-0}=20m/s )
And if you notice, if you expand my x in terms of t, using the triangle to evaluate the inverse sine of 3/5
 )
+\cos{4t} \left (\frac{3}{5}\right) \right ) \\ \\ x&=4\sin{4t}+3\cos{4t} )
Which is the answer in the Fitzpatrick book. My way of solving this question is alot cleaner imo. (Also first time using alignment in latex, which turned out to be a semi failure since it was making my latexing invalid for some reason)
I will post my question up here very shortly
And if you notice, if you expand my x in terms of t, using the triangle to evaluate the inverse sine of 3/5
Which is the answer in the Fitzpatrick book. My way of solving this question is alot cleaner imo. (Also first time using alignment in latex, which turned out to be a semi failure since it was making my latexing invalid for some reason)
I will post my question up here very shortly
Last edited:
lolcakes52
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Re: HSC 2012 Marathon
no, can you add more terms. is the initial 1 part of the series.
OMGITzJustin
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Re: HSC 2012 Marathon
A particle moves with simple harmonic motion. At the extremities of the motion the absolute value of the acceleration is 1cm^-2, and when the particle is 3cm from the centre of motion the speed is 2*squareroot2 cm^-1.
Find period and amplitude
A particle moves with simple harmonic motion. At the extremities of the motion the absolute value of the acceleration is 1cm^-2, and when the particle is 3cm from the centre of motion the speed is 2*squareroot2 cm^-1.
Find period and amplitude
Sy123
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Re: HSC 2012 Marathon
 \\ \\ \therefore C=\frac{n^2 a^2}{2} \\ v^2=-n^2 x^2+n^2 a^2 )
^2=-9n^2+n^2 a^2 \\ \\ 8=-9n^2+n^2 a^2 \\ \\ \Abs{\ddot{x}}=1 \ \ \ when \ \ \ x=a \\ \\ if \ \ n>0 \ \ \ a>0 \therefore n^2 a=1 \\ \\ \therefore 8=-9n^2+a \\ n^2=\frac{1}{a} \\ \\ 8=\frac{-9}{a}+a \\ \\ a^2-8a-9=0 \\ a=9 \ \ a=-1 \ \ \ a>0 \\ \therefore amplitude=9 \\ \\ n^2 a=1 \\ \therefore n=\frac{1}{3} \\ \therefore Period=6\pi )
I have a feeling it might be wrong though but at least I attempted.
I have a feeling it might be wrong though but at least I attempted.
Last edited:
Sy123
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Re: HSC 2012 Marathon
Its a question entirely possible within 3u boundaries (depending on how you do it you will have to use 3u knowledge)
I made the question up after I proved that result. So enjoy :3
Its a question entirely possible within 3u boundaries (depending on how you do it you will have to use 3u knowledge)
I made the question up after I proved that result. So enjoy :3
Last edited:
Re: HSC 2012 Marathon
LHS = 1 + 2(r) + 3(r)^2 + 4(r)^3 + 5(r)^4 +....
r = 1/2
LHS = d/dr(r + r^2 + r^3 + r^4 + r^5 + ....)
=d/dr(r/(1-r)) --- > limiting sum of geometric series remember r + r^2 + r^3 + ... r^n = r-r^(n+1)/ (1-r) as n --> infinite, since |r| < 1 then r^(n+1) approaches zero.
= 1/(1-r)^2
therefore LHS = 1/(1-r)^2
r = 1/2
therefore LHS = 1/(1-1/2)^2 = 1/(1/2)^2 = 4
1 + 1 + 3/4 + 1/2 + 5/16 + 3/16 + .... approaches 4.
1 + 1 + 3/4 + 1/2 + 5/16 + 3/16 + ....find the limiting value of 1 + 1 + 3/4 + 1/2 + 5/16 + 3/16 + .... and i won't give you the general term because you want a challenge
LHS = 1 + 2(r) + 3(r)^2 + 4(r)^3 + 5(r)^4 +....
r = 1/2
LHS = d/dr(r + r^2 + r^3 + r^4 + r^5 + ....)
=d/dr(r/(1-r)) --- > limiting sum of geometric series remember r + r^2 + r^3 + ... r^n = r-r^(n+1)/ (1-r) as n --> infinite, since |r| < 1 then r^(n+1) approaches zero.
= 1/(1-r)^2
therefore LHS = 1/(1-r)^2
r = 1/2
therefore LHS = 1/(1-1/2)^2 = 1/(1/2)^2 = 4
1 + 1 + 3/4 + 1/2 + 5/16 + 3/16 + .... approaches 4.
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Re: HSC 2012 Marathon
taking derivatives of both sides
n(x)^(n-1) = 1/x
therefore x^n = 1/n
hence ln(x) = 1/n
x = (1/n)^(1/n)
ln((1/n)^(1/n)) = 1/n
(1/n)ln(1/n) = 1/n
ln(1/n) = 1
1/n = e
n = 1/e
but n > 0
it is not a constant.
Hence Proof by contradiction
x^n =/= ln(x)
lol i don't know what i did there but ye!
Assume x^n = lnx then
Its a question entirely possible within 3u boundaries (depending on how you do it you will have to use 3u knowledge)
I made the question up after I proved that result. So enjoy :3
taking derivatives of both sides
n(x)^(n-1) = 1/x
therefore x^n = 1/n
hence ln(x) = 1/n
x = (1/n)^(1/n)
ln((1/n)^(1/n)) = 1/n
(1/n)ln(1/n) = 1/n
ln(1/n) = 1
1/n = e
n = 1/e
but n > 0
it is not a constant.
Hence Proof by contradiction
x^n =/= ln(x)
lol i don't know what i did there but ye!
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Re: HSC 2012 Marathon
meh i was half asleep at the time i wrote this so yeah, sorry bro. don't know what i was doing. I'll work it out again!
ColdBoy...
becomes
if you are using log rules. Where did your x go?!
How did you get from
to
And alsoso your proof by contradiction is invalid. Also
is a constant.
meh i was half asleep at the time i wrote this so yeah, sorry bro. don't know what i was doing. I'll work it out again!
Re: HSC 2012 Marathon
Well really you can just prove this question graphically
ln(x) remains constant, x^n with increasing n, means it "goes up faster", hence if x =/= ln(x) (does not interesect this, then x^2 and x^3 and so on does not interesect)
that is graphical explanation of whats happening.
Let me write out the other proof neatly
Assume that x^(n) = ln(x)
then their derivatives must be equal so
n*x*(n-1) = 1/x
rearranging this gives x^(n) = 1/n
x = (1/n)^(1/n)
substituting this into x^(n) = ln(x) gives
[(1/n)^(1/n)]^n = ln((1/n)^(1/n)) giving
1/n = (1/n)*ln(1/n)
i.e. ln(1/n) = 1
n = 1/e by rearranging for n
now if n = 1/e, then x = e^e
substituting this into ln(x) = x^n
gives
e = [e^e]^n
i.e you obtain this expression
e = e^(e*n) --------- (1)
now for n > 0, assuming n is an integer, n = 1, 2, 3 ,4,5 .......
for n = 1
(1) tells us e=e^e --- (incorrect)
for n = 2
(1) tells us e = e^(2e) --- (incorrect)
.
.
.
.
so on
so by inspection e =/= e^(en) but we obtained e = e^(en) from assuming that ln(x) = x^n
so hence they must not be equal! QED. Shifty proof, but i'll take it! LOL!
ColdBoy...
How did you get from
And also
Well really you can just prove this question graphically
ln(x) remains constant, x^n with increasing n, means it "goes up faster", hence if x =/= ln(x) (does not interesect this, then x^2 and x^3 and so on does not interesect)
that is graphical explanation of whats happening.
Let me write out the other proof neatly
Assume that x^(n) = ln(x)
then their derivatives must be equal so
n*x*(n-1) = 1/x
rearranging this gives x^(n) = 1/n
x = (1/n)^(1/n)
substituting this into x^(n) = ln(x) gives
[(1/n)^(1/n)]^n = ln((1/n)^(1/n)) giving
1/n = (1/n)*ln(1/n)
i.e. ln(1/n) = 1
n = 1/e by rearranging for n
now if n = 1/e, then x = e^e
substituting this into ln(x) = x^n
gives
e = [e^e]^n
i.e you obtain this expression
e = e^(e*n) --------- (1)
now for n > 0, assuming n is an integer, n = 1, 2, 3 ,4,5 .......
for n = 1
(1) tells us e=e^e --- (incorrect)
for n = 2
(1) tells us e = e^(2e) --- (incorrect)
.
.
.
.
so on
so by inspection e =/= e^(en) but we obtained e = e^(en) from assuming that ln(x) = x^n
so hence they must not be equal! QED. Shifty proof, but i'll take it! LOL!
Carrotsticks
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Re: HSC 2012 Marathon
No need for all of this... just prove that f(x) = x^n - ln(x) > 0 for all x>0 and n>1....
No need for all of this... just prove that f(x) = x^n - ln(x) > 0 for all x>0 and n>1....
Carrotsticks
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Re: HSC 2012 Marathon
Give it a try! Show that the function is always positive for all x>0 (might have to take cases for when x E (0,1] and when x E (1,infinity) ).
There have been questions like this before in the HSC, proving that if say f(x) - g(x) > 0, then f(x) > g(x).
Give it a try! Show that the function is always positive for all x>0 (might have to take cases for when x E (0,1] and when x E (1,infinity) ).
There have been questions like this before in the HSC, proving that if say f(x) - g(x) > 0, then f(x) > g(x).
Re: HSC 2012 Marathon
f(x) = x^n - lnx
well you would obviously prove
f ' (x) > 0 (increasing for the domain (1,infinity))
f ' (x) < 0 (decreasing for domain (0, 1))
f'(x) = 0 for x = 1
AND
for domain (0,1)
f(x) > 0
at x = 1
f(x) = 1
for domain (1, infinity)
f(x) > 0 since f'(x) > 0 for x > 1 and f(1) "Turning point" = 1 > 0
hence whole f(x) greater than zero.
This is probably easier, but other way is more fun, doing random things LOL! Nice by spotting that by the way!
Meh i'm only doing this because i'm bored lol!
f(x) = x^n - lnx
well you would obviously prove
f ' (x) > 0 (increasing for the domain (1,infinity))
f ' (x) < 0 (decreasing for domain (0, 1))
f'(x) = 0 for x = 1
AND
for domain (0,1)
f(x) > 0
at x = 1
f(x) = 1
for domain (1, infinity)
f(x) > 0 since f'(x) > 0 for x > 1 and f(1) "Turning point" = 1 > 0
hence whole f(x) greater than zero.
This is probably easier, but other way is more fun, doing random things LOL! Nice by spotting that by the way!
Last edited:
Timske
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Re: HSC 2012 Marathon
<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{120} \textup{Use \~the ~substitution~ u} = \frac{1}{\sqrt{3}} \textup{tan}\textup{x} \textup{~to ~evaluate} \\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~ \int_{0}^{\frac{\pi}{4}} \frac{\textup{dx}}{3 - 2\textup{sin}^2\textup{x}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{120} \textup{Use \~the ~substitution~ u} = \frac{1}{\sqrt{3}} \textup{tan}\textup{x} \textup{~to ~evaluate} \\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~ \int_{0}^{\frac{\pi}{4}} \frac{\textup{dx}}{3 - 2\textup{sin}^2\textup{x}}" title="\dpi{120} \textup{Use \~the ~substitution~ u} = \frac{1}{\sqrt{3}} \textup{tan}\textup{x} \textup{~to ~evaluate} \\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~ \int_{0}^{\frac{\pi}{4}} \frac{\textup{dx}}{3 - 2\textup{sin}^2\textup{x}}" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{120} \textup{Use \~the ~substitution~ u} = \frac{1}{\sqrt{3}} \textup{tan}\textup{x} \textup{~to ~evaluate} \\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~ \int_{0}^{\frac{\pi}{4}} \frac{\textup{dx}}{3 - 2\textup{sin}^2\textup{x}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{120} \textup{Use \~the ~substitution~ u} = \frac{1}{\sqrt{3}} \textup{tan}\textup{x} \textup{~to ~evaluate} \\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~ \int_{0}^{\frac{\pi}{4}} \frac{\textup{dx}}{3 - 2\textup{sin}^2\textup{x}}" title="\dpi{120} \textup{Use \~the ~substitution~ u} = \frac{1}{\sqrt{3}} \textup{tan}\textup{x} \textup{~to ~evaluate} \\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~ \int_{0}^{\frac{\pi}{4}} \frac{\textup{dx}}{3 - 2\textup{sin}^2\textup{x}}" /></a>