Author Topic: Mathematics Question Thread (Read 1301725 times) Tweet Share

RuiAce · « **Reply #570 on:** October 17, 2016, 10:01:39 pm »

Quote from: nibblez16 on October 17, 2016, 09:59:50 pm

Hello, can you please explain the last question... I always get confused on those types... Thanks

Can we please have a link to the paper or the answers to all the previous parts? For a question like this that might be relevant.

nibblez16 · « **Reply #571 on:** October 17, 2016, 10:11:38 pm »

Quote from: RuiAce on October 17, 2016, 10:01:39 pm

Can we please have a link to the paper or the answers to all the previous parts? For a question like this that might be relevant.

RuiAce · « **Reply #572 on:** October 18, 2016, 12:43:31 am »

Quote from: nibblez16 on October 17, 2016, 10:11:38 pm

$\text{Observe that }P\text{ and }r\text{ remain, however }\theta\text{ vanishes in what we need to prove.}$
$\text{The intuition to this question is not nice.}\\ \text{Notice that we are proving a restriction}\\ \text{(Or in particular, a restriction on }r)$
$\text{Since }\theta\text{ vanishes, we must identify if there are any restrictions on }\theta\\ \text{This is now targetting the very upper end of 2U capabilities.}$
$\text{Fundamentally, }\boxed{0 < \theta < 2\pi}\\ \text{Hence }2 < \theta+2 < 2\pi+2\\ \text{So }2r < r(\theta+2) < 2r(\pi + 1)$
$\text{Thus }2r < P < 2r(\pi +1)$
$\text{Split the inequality up at this point:}\\ 2r < P \implies r < \frac{P}{2}\\ P < 2r(\pi + 1) \implies r > \frac{P}{2(\pi+1)}\\ \text{Recombine and done.}$
$\textit{The main question was where was the inspiration in all of this?}\\ \text{Once again, notice how there was only a restriction on }\theta\\ \text{Since }\theta\text{ vanishes, starting on it is probably a good idea.}$
$\text{You may have chosen to work backwards in pencil.}\\ \text{For some algebraic proofs, working backwards gives you a better sense of direction.}$
$\text{Working backwards, }P < 2r(\pi + 1) < P(\pi + 1)\\ \text{But we know from part (i) that }P=r(\theta + 2)\\ \text{It becomes harder to see what's going on here.}\\ \text{This last bit just assesses your analytical skills.}\\ \text{Ultimately you have to watch out for small details. That's how to get 100 percent in 2U.}$

FallonXay · « **Reply #573 on:** October 18, 2016, 09:04:14 am »

Hello,

with this question, why does the minimum of D occur at the same time as the minimum of D^2?

Also, why do you use the negative root in the volume question?

And, how do you solve c part ii (Using the graph... find the values of m... one solution)

Thanks!

kimmie · « **Reply #574 on:** October 18, 2016, 09:49:54 am »

Hey can you please help me with this question. It's from the 2013 hsc paper. I did part (i) but I don't understand how to do part (ii)

RuiAce · « **Reply #575 on:** October 18, 2016, 09:49:57 am »

Quote from: FallonXay on October 18, 2016, 09:04:14 am

Hello,

with this question, why does the minimum of D occur at the same time as the minimum of D^2?

Also, why do you use the negative root in the volume question?

And, how do you solve c part ii (Using the graph... find the values of m... one solution)

Thanks!

$\text{In Extension 2, a trick known as implicit differentiation is taught.}\\ \frac{d(D^2)}{dt}=2D\frac{dD}{dt}\\ \text{So seeting }\frac{d(D^2)}{dt}=0\text{ has the same effect as setting }\frac{dD}{dt}=0$
$\text{This can also be interpreted graphically.}\\ \text{Open up GeoGebra. Sketch some functions and their squares. Compare where the stationary points are.}\\ \text{e.g. }(x-3)(x-2)(x-1)\text{ and }[(x-3)(x-2)(x-1)]^2, \sin x\text{ and }\sin^2 x$
$\text{To prove the result that }D^2\text{ has a S.P. when }D\text{ has a S.P.}\\ \text{Let }y=[f(x)]^2\\\text{so that }y\text{ is just the square of }f(x) \\\text{Then }\frac{dy}{dx}=2f(x)f^\prime(x)$
$\text{For a stationary point, }\frac{dy}{dx}=0\\ \text{So }f(x)=0 \text{ or }f^\prime(x)=0$
$\text{The }f(x)=0\text{ case tells us something extra, but all we are interested is that}\\ \frac{dy}{dx}=0\text{ when }f^\prime(x)=0\\ \text{i.e. the derivative of the square is 0, when the derivative of the original thing is 0.}$

RuiAce · « **Reply #576 on:** October 18, 2016, 09:52:08 am »

Quote from: FallonXay on October 18, 2016, 09:04:14 am

Hello,

with this question, why does the minimum of D occur at the same time as the minimum of D^2?

Also, why do you use the negative root in the volume question?

And, how do you solve c part ii (Using the graph... find the values of m... one solution)

Thanks!

$\text{For the volumes question, consider what is going on.}$
$\text{You have the equation of the parabola }y=(x-2)^2\\ \text{Taking roots, }x-2 = \pm \sqrt{y} \implies x = 2\pm \sqrt{y}$
$\text{But what is the difference? This is an important feature.}\\ \text{The positive root corresponds to the }\textbf{right branch}\text{ of the parabola}\\ \text{The negative root corresponds to the }\textbf{left branch}\text{ of the parabola}$
$\text{Again, you should graph }x-2=\sqrt{y}\text{ and }x-2=-\sqrt{y}\text{ separately on GeoGebra or Desmos.}$

RuiAce · « **Reply #577 on:** October 18, 2016, 09:54:31 am »

Quote from: kimmie on October 18, 2016, 09:49:54 am

Hey can you please help me with this question. It's from the 2013 hsc paper. I did part (i) but I don't understand how to do part (ii)

Already done in post #454

FallonXay · « **Reply #578 on:** October 18, 2016, 10:00:47 am »

Quote from: RuiAce on October 18, 2016, 09:52:08 am

$\text{For the volumes question, consider what is going on.}$
$\text{You have the equation of the parabola }y=(x-2)^2\\ \text{Taking roots, }x-2 = \pm \sqrt{y} \implies x = 2\pm \sqrt{y}$
$\text{But what is the difference? This is an important feature.}\\ \text{The positive root corresponds to the }\textbf{right branch}\text{ of the parabola}\\ \text{The negative root corresponds to the }\textbf{left branch}\text{ of the parabola}$
$\text{Again, you should graph }x-2=\sqrt{y}\text{ and }x-2=-\sqrt{y}\text{ separately on GeoGebra or Desmos.}$

Why do we specifically need to use the left branch though? Sorry - I still don't understand

RuiAce · « **Reply #579 on:** October 18, 2016, 10:08:26 am »

Quote from: FallonXay on October 18, 2016, 10:00:47 am

Why do we specifically need to use the left branch though? Sorry - I still don't understand

$\text{Sketch it firstly so that you can convince yourself.}\\ \text{If you need to, also draw up a table of values.}$
$\text{Geometrically, why this occurs is}\\ \text{Recall that }y=(x-2)^2\text{ is saying that }y\text{ is a function of }x\\ \text{But }x=-2\pm \sqrt{y}\text{ fails a 'horizontal line test'.}\\ \text{So }x\text{ is not a function of }y$
$\text{Indeed, }x=-2+\sqrt{y}\text{ and }x=-2-\sqrt{y}\text{ split up are functions}\\ \text{In both cases (split up), }x\text{ is a function of }y.$
$\textit{Take care that the analysis slightly differs now that we're considering }x\textit{ in terms of }y\\ \textit{Not the usual vice-versa}$
$x=\sqrt{y}\text{ and }x=-\sqrt{y}\text{ are reflections of each other.}\\ \text{Note that a square root curve is just a parabola, with half of it chopped off.}\\ \text{Recall that }x^2=y\text{ is just }y=\sqrt{x}\text{ and its reflection }y=-\sqrt{x}.\text{ This works similarly here.}$
$\text{The reason why }y=\sqrt{x}\text{ and }y=-\sqrt{x}\text{ are reflections is because}\\ \textbf{they are the negative of each other.}\\ \text{Recall that the graph of the negative is just the same graph flipped about the }x\text{-axis.}$
$\text{Instead, here, we shift our analysis to the }y\text{-axis instead.}\\ x=\sqrt{y}\text{ and }x=-\sqrt{y}\text{ are reflections about the }y\text{-axis.}\\ x=\sqrt{y}\text{ represents what's to the right of the }y\text{-axis}\\ \text{whereas }x=-\sqrt{y}\text{ represents wha's to the left of the }y\text{-axis}$
$\text{The only difference is that we have }x=-2+\sqrt{y}\text{ and }x=-2-\sqrt{y}\\ \text{i.e. an extra -2 floats around.}\\ \text{This just means that we use this analogy along }x=2\text{ and not }x=0(y\text{-axis})$

FallonXay · « **Reply #580 on:** October 18, 2016, 10:20:43 am »

Quote from: RuiAce on October 18, 2016, 10:08:26 am

$\text{Sketch it firstly so that you can convince yourself.}\\ \text{If you need to, also draw up a table of values.}$
$\text{Geometrically, why this occurs is}\\ \text{Recall that }y=(x-2)^2\text{ is saying that }y\text{ is a function of }x\\ \text{But }x=-2\pm \sqrt{y}\text{ fails a 'horizontal line test'.}\\ \text{So }x\text{ is not a function of }y$
$\text{Indeed, }x=-2+\sqrt{y}\text{ and }x=-2-\sqrt{y}\text{ split up are functions}\\ \text{In both cases (split up), }x\text{ is a function of }y.$
$\textit{Take care that the analysis slightly differs now that we're considering }x\textit{ in terms of }y\\ \textit{Not the usual vice-versa}$
$x=\sqrt{y}\text{ and }x=-\sqrt{y}\text{ are reflections of each other.}\\ \text{Note that a square root curve is just a parabola, with half of it chopped off.}\\ \text{Recall that }x^2=y\text{ is just }y=\sqrt{x}\text{ and its reflection }y=-\sqrt{x}.\text{ This works similarly here.}$
$\text{The reason why }y=\sqrt{x}\text{ and }y=-\sqrt{x}\text{ are reflections is because}\\ \textbf{they are the negative of each other.}\\ \text{Recall that the graph of the negative is just the same graph flipped about the }x\text{-axis.}$
$\text{Instead, here, we shift our analysis to the }y\text{-axis instead.}\\ x=\sqrt{y}\text{ and }x=-\sqrt{y}\text{ are reflections about the }y\text{-axis.}\\ x=\sqrt{y}\text{ represents what's to the right of the }y\text{-axis}\\ \text{whereas }x=-\sqrt{y}\text{ represents wha's to the left of the }y\text{-axis}$
$\text{The only difference is that we have }x=-2+\sqrt{y}\text{ and }x=-2-\sqrt{y}\\ \text{i.e. an extra -2 floats around.}\\ \text{This just means that we use this analogy along }x=2\text{ and not }x=0(y\text{-axis})$

Ahh k, I see. thx

kimmie · « **Reply #581 on:** October 18, 2016, 10:28:28 am »

Quote from: RuiAce on October 18, 2016, 09:54:31 am

Already done in post #454

I checked it and I'm still confused why would we use 2 for the gradient thing?

RuiAce · « **Reply #582 on:** October 18, 2016, 10:33:07 am »

Quote from: kimmie on October 18, 2016, 10:28:28 am

I checked it and I'm still confused why would we use 2 for the gradient thing?

$\text{Like mentioned, 2 is a crucial number here.}\\ \text{2 and -2 are the gradients of }y=|2x-3|$
$\text{This implies that 2 and -2 are where the gradients become equal}\\ \text{This is critical because we are analysing when the lines are parallel.}\\\text{The instant lines become parallel and stop being parallel tells us a lot about the points of intersection.}$
$\text{Hence 2 and -2 become the basis for comparison,}\text{Utilise the GeoGebra simulation to have a better idea}$

kimmie · « **Reply #583 on:** October 18, 2016, 10:46:13 am »

Oh I get it now thank you so much

Quote from: RuiAce on October 18, 2016, 10:33:07 am

$\text{Like mentioned, 2 is a crucial number here.}\\ \text{2 and -2 are the gradients of }y=|2x-3|$
$\text{This implies that 2 and -2 are where the gradients become equal}\\ \text{This is critical because we are analysing when the lines are parallel.}\\\text{The instant lines become parallel and stop being parallel tells us a lot about the points of intersection.}$
$\text{Hence 2 and -2 become the basis for comparison,}\text{Utilise the GeoGebra simulation to have a better idea}$

kimmie · « **Reply #584 on:** October 18, 2016, 10:48:01 am »

Can you help me with part (iv) and (v) please

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