HSC 2016 MX1 Marathon (archive) (4 Viewers)

leehuan · Dec 22, 2015

Re: HSC 2016 3U Marathon

Here's an informal and potentially inaccurate answer (refer to final part):

$\\ $The famous result is derived using the following means: \\ $ \ddot { x } =\frac { dv }{ dt } \\ \ddot { x } =\frac { dv }{ dt } \frac { dx }{ dx } \\ \ddot { x } =\frac { dx }{ dt } \frac { dv }{ dx } \\ \ddot { x } =v\frac { dv }{ dx }$

$\\ $The important factor lies in step 2. We introduce the term $\frac {dx}{dx}$ into our equation, so we just made the assumption that $dx \neq 0$. By consequence, $\frac {dx}{dt} \neq 0 \Rightarrow v \neq 0$. If we restate that $v=0$ then our equation will break down due to the fact that $\frac {0}{0}$ is an indeterminate form which we are multiplying by, and the resulting equation is consequently not necessarily true.$$

$\\ $One may question, in this case, why is $dt \neq 0$? This can be argued by the fact that the derivative is a ratio: $\lim _{ \Delta t\rightarrow 0 }{ \frac { \Delta x }{ \Delta t } } =\frac { dx }{ dt } $ where $\frac { \Delta x }{ \Delta t } $ represents average velocity. The fact that the derivative expresses change in time limiting towards 0, however never reaching it, assumes that the quantity will never equal to 0.$$

$\\ $Incidentally, if $dt=0$, then we would have $v \rightarrow \infty$

kawaiipotato · Dec 28, 2015

Re: HSC 2016 3U Marathon

$Question 1$
$When the area in square units of an expanding circle is increasing twice as fast as its radius in linear units, what is the radius?$

$Question 2$
$\lim_{t \rightarrow \infty} \left(4t^{2}\left(\sin \left(\frac{2}{t}\right)\right)^2 \right)$

Paradoxica · Dec 28, 2015

Re: HSC 2016 3U Marathon

kawaiipotato said:
$Question 1$
$When the area in square units of an expanding circle is increasing twice as fast as its radius in linear units, what is the radius?$

$Question 2$
$\lim_{t \rightarrow \infty} \left(4t^{2}\left(\sin \left(\frac{2}{t}\right)\right)^2 \right)$

$Question 1$

$$\noindent Suppose the radius is increasing at a rate (in dimensionless units) of $ \rho s^{-1} $, i.e. $\frac{\mathrm d r}{\mathrm d t} = \rho$. Then the area of the circle has a rate of increase of $\frac{\mathrm d A}{\mathrm d t}$. By the area of a circle formula, we have $\frac{\mathrm d A}{\mathrm d t} = \frac{\mathrm d (\pi r^2)}{\mathrm d t}= 2 \pi r \frac{ \mathrm d r}{\mathrm d t} = 2\pi r \rho $. From the given scenario, at some arbitrary time $ t$, $2\pi r \rho = 2\rho $. So $r = \frac{1}{\pi}$, since the circle is expanding, so the rate of increase is non-zero.$

$Question 2$

$$\noindent Substitute $t = \frac{2}{u}$, swap limits to yield $ \lim_{u \rightarrow 0} \frac{16}{u^2}\sin^2{u} = \lim_{u \rightarrow 0} 16 \left ( \frac{\sin{u}}{u} \right )^2 = 16$

InteGrand · Jan 1, 2016

Re: HSC 2016 3U Marathon

$\noindent \textbf{NEW QUESTION}$

$\noindent Evaluate $\int _{-3} ^{1} \frac{x}{\sqrt{1+x^2}}\text{ d}x$ by using the substitution $u = 1+x^2$.$

DatAtarLyfe · Jan 1, 2016

HSC 2016 3U Marathon

Soz for sideways

InteGrand · Jan 1, 2016

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
Soz for sideways

$When we do the evaluation at the end, since we changed the limits already, we evaluate them in terms of $u$, i.e. $\Big{[} \sqrt{u}\Big{]}_{10} ^{2} = \sqrt{2} -\sqrt{10}$. We don't need to convert back to $x$ in the end (this is the purpose of changing the limits of the integral in the first place). If we convert back to $x$ in the end, we'd need to change limits back to what they were for $x$ initially.$

InteGrand · Jan 1, 2016

Re: HSC 2016 3U Marathon

$\noindent \textbf{NEW QUESTION}$

$\noindent Find $\int _0 ^1 x \sqrt{1-x^2} \text{ d}x$ using the substitution $u = 1-x^2$.$

DatAtarLyfe · Jan 1, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$When we do the evaluation at the end, since we changed the limits already, we evaluate them in terms of $u$, i.e. $\Big{[} \sqrt{u}\Big{]}_{10} ^{2} = \sqrt{2} -\sqrt{10}$. We don't need to convert back to $x$ in the end (this is the purpose of changing the limits of the integral in the first place). If we convert back to $x$ in the end, we'd need to change limits back to what they were for $x$ initially.$

Shit yeh, what an amateur mistake. This is why i hate definite integrals, frickin so much easier just to chuck in the c at the end.
Thanks for that

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leehuan · Jan 1, 2016

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
Shit yeh, what an amateur mistake. This is why i hate definite integrals, frickin so much easier just to chuck in the c at the end.
Thanks for that

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Lol the only thing annoying about definite integrals is pulling out the calculator really...

$u=1-{ x }^{ 2 }\Rightarrow du=-2x \, dx\\ \int _{ 0 }^{ 1 }{ x\sqrt { 1-{ x }^{ 2 } } dx } \\ =\int _{ 1 }^{ 0 }{ \frac { -1 }{ 2 } \sqrt { u } du } \\ ={ \left[ \frac { { u }^{ \frac { 3 }{ 2 } } }{ 3 } \right] }_{ 0 }^{ 1 }\\ =\frac { 1 }{ 3 }$

$\\ $ \bf {NEW QUESTION }$$
$\\ \int _{ 1 }^{2 }{ { x }^{ 2 }\sqrt { x-1 } \, dx } $ using the substitutions:$ \\ $a) $ u=x-1\\ $b) $ { u }^{ 2 }=x-1$

DatAtarLyfe · Jan 1, 2016

Re: HSC 2016 3U Marathon

Wow, another definite

Can i make a request for the next question for either IG or lee? Can you guys give a harder 3u, almost 4u level indefinite?

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Drsoccerball · Jan 1, 2016

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
Can i make a request for the next question for either IG or lee? Can you guys give a harder 3u, almost 4u level indefinite?

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I am neither of them but here's one from the BOS trials.
$$ Solve using the substitution $ x=\frac{1-u}{1+u}$

$\int{\frac{ln(1+x)}{1+x^2}}dx$

InteGrand · Jan 1, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
I am neither of them but here's one from the BOS trials.
$$ Solve using the substitution $ x=\frac{1-u}{1+u}$

$\int{\frac{ln(1+x)}{1+x^2}}dx$

$\noindent Haha this should have limits from 0 to 1.$

Drsoccerball · Jan 1, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$\noindent Haha this should have limits from 0 to 1.$

NO. She asked for no limits.

InteGrand · Jan 1, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
NO. She asked for no limits.

Requires non-elementary functions then.

Drsoccerball · Jan 1, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
Requires non-elementary functions then.

unlucks

leehuan · Jan 1, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
NO. She asked for no limits.

This is ridiculous for even the 4U level though when you get rid of the boundaries.
__________________________________________

@ DAL

$\int { \tan ^{ 5 }{ x } \sec ^{ 3 }{ x } dx }$

Also, it is ok to go from u=x-1 to du=dx as I have found. And with u^2=x-1, just go straight to 2u du = dx

(Note: 4U don't get given a substitution)

Paradoxica · Jan 1, 2016

Re: HSC 2016 3U Marathon

leehuan said:
This is ridiculous for even the 4U level though when you get rid of the boundaries.
__________________________________________

@ DAL

$\int { \tan ^{ 5 }{ x } \sec ^{ 3 }{ x } dx }$

Also, it is ok to go from u=x-1 to du=dx as I have found. And with u^2=x-1, just go straight to 2u du = dx

(Note: 4U don't get given a substitution)

Polylogs. heh.

Of course it's ok to do that. It would be the same as differentiating with respect to a third variable, then removing that third variable from the equation. In fact, it is identical.

DatAtarLyfe · Jan 1, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
I am neither of them but here's one from the BOS trials.
$$ Solve using the substitution $ x=\frac{1-u}{1+u}$

$\int{\frac{ln(1+x)}{1+x^2}}dx$

Its k, you're just as adept

InteGrand said:
$\noindent Haha this should have limits from 0 to 1.$

InteGrand said:
Requires non-elementary functions then.

wait so should i do it with the limits or without?

leehuan said:
This is ridiculous for even the 4U level though when you get rid of the boundaries.
__________________________________________

@ DAL

$\int { \tan ^{ 5 }{ x } \sec ^{ 3 }{ x } dx }$

Also, it is ok to go from u=x-1 to du=dx as I have found. And with u^2=x-1, just go straight to 2u du = dx

(Note: 4U don't get given a substitution)

Paradoxica said:
Polylogs. heh.

Of course it's ok to do that. It would be the same as differentiating with respect to a third variable, then removing that third variable from the equation. In fact, it is identical.

K cool, the more you know

InteGrand · Jan 1, 2016

Re: HSC 2016 3U Marathon

DatAtarLyfe said:
Its k, you're just as adept

wait so should i do it with the limits or without?

K cool, the more you know

Must use limits from 0 to 1. The indefinite integral (primitive) can't be expressed in terms of elementary functions (turns out it requires polylogs; recall that in general it is likely that an arbitrary function will not have an anti-derivative in terms of elementary functions). More info about non-elementary primitives: https://en.wikipedia.org/wiki/Nonelementary_integral .

Drsoccerball · Jan 12, 2016

Re: HSC 2016 3U Marathon

$$ Use the substitution $ u= \sin{4x} $ to evaluate : $ \int{\frac{\sin{8x}}{9+\sin ^4(4x)}}dx$

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