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fun_jirachi

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Re: 4U Maths Question Thread
« Reply #2310 on: June 15, 2019, 02:48:27 pm »
+3
Hey there!

Consider on a cartesian plane the semicircle

Then, as you're taking slices as it rotates from front to back, the slices will be semicircles that have radius x for 0<=x<=r. They'll go from zero, reach a radius of r, then go back down to zero. ie

Then, since the semicircle has radius r, and when you rotate it into a hemisphere, it will have depth 2r (from r to -r), your integral just becomes


Hope this helps :)
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luckystrike826

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Re: 4U Maths Question Thread
« Reply #2311 on: June 15, 2019, 06:51:29 pm »
0
Hello! I have a question from class when doing the harder 3u section.

Given that d/dx (lnx) = 1/x,

Prove that:

Thanks a lot!
« Last Edit: June 15, 2019, 06:53:46 pm by luckystrike826 »

goodluck

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Re: 4U Maths Question Thread
« Reply #2312 on: June 15, 2019, 11:05:11 pm »
0
That's basically the double recurrence relation method actually. It's not the binomial theorem.

(Double recurrences are outside the scope of MX2 but as seen in that example, they're handled the same way as usual recurrence formulas. i.e. with integration by parts. But I feel as though the question is misleading to mention the binomial theorem in that case.)

Yeah, I think it's like some of those recurrence formulas were, which is probably why they got away with it. would it be something that we could expect?

I'm stuck on some other problems (Is it just me or is complex numbers a more complicated area that everyone seems to treat it as? I've heard a lot of people saying it's super easy, which makes me feel awful)

1. Let z= (i+ix)/(1-ix). Show that:
                                                 if x is less than or equal to 1, arg(z)= pi/2+2arctan(x)
                                                 if x is greater than 1, arg(z)= -3pi/2+2arctan(x)
Challenge q: Are there interesting patterns you can spot as x approaches plus and minus infinity

2. Find the real and imaginary part of (1+cos2theta+isin2theta)/(1+cos2theta-isin2theta). I tried to use the 2cos^2theta-1 for it but it didn't come out nicely at all

kaustubh.patel

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Re: 4U Maths Question Thread
« Reply #2313 on: June 16, 2019, 08:52:39 am »
0
I'm struggling with a volumes question help would be great thanks.

A hemisphere has radius r. By considering cross-sections parallel to the base of the hemisphere, show the volume is given by V=2/3(pi)r^3.


Umm what i did was imagine a QUATER of a semi circle and found the volume of that, you can simply multiply that by 2 to get the volume of the semi circle.

RuiAce

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Re: 4U Maths Question Thread
« Reply #2314 on: June 16, 2019, 09:50:22 am »
+1
Yeah, I think it's like some of those recurrence formulas were, which is probably why they got away with it. would it be something that we could expect?

I'm stuck on some other problems (Is it just me or is complex numbers a more complicated area that everyone seems to treat it as? I've heard a lot of people saying it's super easy, which makes me feel awful)

1. Let z= (i+ix)/(1-ix). Show that:
                                                 if x is less than or equal to 1, arg(z)= pi/2+2arctan(x)
                                                 if x is greater than 1, arg(z)= -3pi/2+2arctan(x)
Challenge q: Are there interesting patterns you can spot as x approaches plus and minus infinity

2. Find the real and imaginary part of (1+cos2theta+isin2theta)/(1+cos2theta-isin2theta). I tried to use the 2cos^2theta-1 for it but it didn't come out nicely at all
Nah, double recurrence relations are actually not examinable in the HSC. Can't remember where I was first told of this but it is an official statement.
________________________________________________________________________

Just before I jump into 1, I want to confirm that you meant \( z = \frac{i+ix}{1-ix} \) as opposed to \( z = \frac{1+ix}{1-ix} \)? Haven't tried it yet, just wanted to double check first.
\[ \text{Double angle formulas should be fine for the second one.}\\ \text{It's just that even more work is necessary after you use it.} \\ \begin{align*} \frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta} &= \frac{2\cos^2\theta+2i\sin\theta\cos\theta}{2\cos^2\theta-2i\sin\theta\cos\theta}\\ &= \frac{\cos\theta+i\sin\theta}{\cos\theta-i\sin\theta}\\ &= \frac{\cos\theta+i\sin\theta}{\cos(-\theta) + i\sin(-\theta)}\\ &= \frac{\operatorname{cis}\theta}{\operatorname{cis}(-\theta)}\\ &= \operatorname{cis}2\theta\\ &= \cos2\theta+i\sin2\theta \end{align*} \]
Note that the complex numbers questions you have given are on the upper end of the spectrum - they tend to rock up in Q14-16 instead of Q11-13 of the exam. But at the E4 range, recognising things like \( \operatorname{cis}(-\theta) = \cos\theta - i\sin\theta \) becomes quite common, as you're expected to have a full mastery of trigonometric identities. And of course, the fact that division of complex numbers in mod-arg form involves subtracting their arguments.

goodluck

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Re: 4U Maths Question Thread
« Reply #2315 on: June 16, 2019, 10:27:19 am »
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Nah, double recurrence relations are actually not examinable in the HSC. Can't remember where I was first told of this but it is an official statement.
________________________________________________________________________

Just before I jump into 1, I want to confirm that you meant \( z = \frac{i+ix}{1-ix} \) as opposed to \( z = \frac{1+ix}{1-ix} \)? Haven't tried it yet, just wanted to double check first.
\[ \text{Double angle formulas should be fine for the second one.}\\ \text{It's just that even more work is necessary after you use it.} \\ \begin{align*} \frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta} &= \frac{2\cos^2\theta+2i\sin\theta\cos\theta}{2\cos^2\theta-2i\sin\theta\cos\theta}\\ &= \frac{\cos\theta+i\sin\theta}{\cos\theta-i\sin\theta}\\ &= \frac{\cos\theta+i\sin\theta}{\cos(-\theta) + i\sin(-\theta)}\\ &= \frac{\operatorname{cis}\theta}{\operatorname{cis}(-\theta)}\\ &= \operatorname{cis}2\theta\\ &= \cos2\theta+i\sin2\theta \end{align*} \]
Note that the complex numbers questions you have given are on the upper end of the spectrum - they tend to rock up in Q14-16 instead of Q11-13 of the exam. But at the E4 range, recognising things like \( \operatorname{cis}(-\theta) = \cos\theta - i\sin\theta \) becomes quite common, as you're expected to have a full mastery of trigonometric identities. And of course, the fact that division of complex numbers in mod-arg form involves subtracting their arguments.

Whoops, I made a typo!: the original statement was i(1+ix)/(1-ix) so I just multiplied it in. So it should equal (i-x)/(1-ix). Sorry about that!

Also for the previous question, when you divided the fraction by costheta, do we need to say smth like costheta can't equal 0? or is that not required?
« Last Edit: June 16, 2019, 10:31:19 am by goodluck »

RuiAce

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Re: 4U Maths Question Thread
« Reply #2316 on: June 16, 2019, 04:05:09 pm »
+2
Whoops, I made a typo!: the original statement was i(1+ix)/(1-ix) so I just multiplied it in. So it should equal (i-x)/(1-ix). Sorry about that!

Also for the previous question, when you divided the fraction by costheta, do we need to say smth like costheta can't equal 0? or is that not required?
With the double angle formula, technically speaking you should, and then check the solutions for \(\cos x = 0\), i.e. \( x=2n\pi \pm \frac\pi2\) separately. There is another method that completely avoids this issue, but it's a bit harder to see. You can request to see that method if you wish in a later post.
\[ \text{Looking at that question now, the question would be split into multiple parts}\\ \text{should it appear in an exam.}\\ \text{The nature of the question looks very assignment-like.} \]
\[ \text{An assumption is made that }x\text{ is real.} \]
\[ \text{The initial expression given is not very manageable, so we're tempted to simplify it.}\\ \text{This should be done in the classic way of division by complex numbers.}\\ \begin{align*} \frac{i(1+ix)}{1-ix} &= \frac{i(1+ix)^2}{(1+ix)(1-ix)}\\ &= \frac{i(1+2ix-x^2)}{1^2+x^2}\\ &= \frac{i(1+2ix-x^2)}{1+x^2}\\ &= \frac{-2x}{1+x^2} + \frac{1-x^2}{1+x^2}. \end{align*} \]
\[ \text{Personally I feel there should be more cases - the case }x\leq 1\text{ looks a bit too powerful}\\ \text{at first glance. But I could be wrong.}\\ \text{For now, here's the solution to the second case:}\]
It might be fine if it is replaced by \( 0 < x \leq 1\) though, but then the bit about \(-\infty\) would be bad.
\[ \text{Suppose now that }x > 1. \\ \text{Then, note that }-\frac{2x}{1+x^2} < 0\\ \textbf{and }\frac{1-x^2}{1+x^2} < 0.\\ \text{Therefore, the complex number will be in the third quadrant.}\]
\[ \text{Upon drawing the diagram, we see that the related angle }\theta\text{ satisfies}\\ \tan \theta = \frac{x^2-1}{2x}.\\ \text{Therefore the argument will be}\\ \arg z = -\pi + \tan^{-1} \left(  \frac{x^2-1}{2x}\right) \]
___________________________________
\[ \text{So we've boiled the problem down to proving that}\\ \boxed{-\pi + \tan^{-1} \frac{x^2-1}{2x} = -\frac{3\pi}{2} + 2\tan^{-1}x}. \]
Of course, this may not be the recommended approach. But if the source of your question hasn't given any hints, anything goes. Here, I will now treat this problem similarly to a harder 3U question. On a separate sheet of paper, I've done some working backwards to make sure that this proof works. But I only present my answer with the working in a forwards, logical manner. (The intuition is that proving this statement is equivalent to proving that \( \frac\pi2 - 2\tan^{-1}x = \tan^{-1} \frac{1-x^2}{2x} \) for \(x > 1\).)
\[ \text{Observe that}\\\begin{align*} \tan \left( \frac\pi2 - 2\tan^{-1}x \right) &= \cot \left(2\tan^{-1}x \right)\\ &= \frac{1}{\tan (2\tan^{-1}x)}\\ &= \frac{1}{\left( \frac{2\tan(\tan^{-1}x)}{1-\tan^2(\tan^{-1}x)} \right)}\\ &= \frac{1}{\left( \frac{2x}{1-x^2} \right)}\\ &= \frac{1-x^2}{2x}\end{align*}\]
\[ \text{Hence taking inverse tan on both sides,}\\ \begin{align*} \frac\pi2 - 2\tan^{-1}x &= \tan^{-1} \frac{1-x^2}{2x}\\ \therefore \frac\pi2 - \tan^{-1} \frac{1-x^2}{2x} &= 2\tan^{-1}x\\ \frac\pi2 + \tan^{-1} \frac{x^2-1}{2x} &= 2\tan^{-1}x\\ \therefore -\pi + \tan^{-1} \frac{x^2-1}{2x} &= -\frac{3\pi}{2} + 2\tan^{-1}x \end{align*} \]
There's one huge subtlety here - I did not justify why we could cancel out the tan-inverse with the tan on the LHS. In general, \( \tan^{-1}(\tan x) = x \) only when \( -\frac\pi2 < x < \frac\pi2\).

Indeed, this situation holds here. Note that if \(x > 1\), then \( \frac\pi4 < \tan^{-1}x < \frac\pi2\). This can then be rearranged to give \( -\frac\pi2 < \frac\pi2 - 2\tan^{-1}x < 0 \), which satisfies the criteria that \( -\frac\pi2 < \frac\pi2 - 2\tan^{-1}x < \frac\pi2 \) regardless, so the cancellation is justified here.

Ollierobb1

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Re: 4U Maths Question Thread
« Reply #2317 on: June 19, 2019, 12:28:17 pm »
0
Four letters are chosen from the letters of the word TELEGRAPH.
These four letters are then placed alongside one another to form a four letter arrangement. Find the number of distinct four letter arrangements which are possible, considering all choices. (2 marks)

Help would be much appreciated

RuiAce

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Re: 4U Maths Question Thread
« Reply #2318 on: June 19, 2019, 01:08:34 pm »
+2
Four letters are chosen from the letters of the word TELEGRAPH.
These four letters are then placed alongside one another to form a four letter arrangement. Find the number of distinct four letter arrangements which are possible, considering all choices. (2 marks)

Help would be much appreciated
Hint: You will need to consider three separate cases:
- 2 E's among the four letters
- 1 E among the four letters
- 0 E's among the four letters

This is because E is the only repeated letter present. Note that the problem quickly becomes catastrophic when we have more repeated letters.

You should post any progress you have if you require further help.

DrDusk

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Re: 4U Maths Question Thread
« Reply #2319 on: June 19, 2019, 06:50:06 pm »
0
Hello! I have a question from class when doing the harder 3u section.

Given that d/dx (lnx) = 1/x,

Prove that:

Thanks a lot!
« Last Edit: June 19, 2019, 07:38:18 pm by DrDusk »

RuiAce

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Re: 4U Maths Question Thread
« Reply #2320 on: June 19, 2019, 07:34:18 pm »
+3
Hello! I have a question from class when doing the harder 3u section.

Given that d/dx (lnx) = 1/x,

Prove that:

Thanks a lot!
With this question, I will say that I heavily dislike this question as it has been provided without any context whatsoever, especially since the supposed result to be proved is commonly taken as a definition. Without the source of the problem, or any extra parts that were provided, I fail to see how this is appropriate even at the MX2 level.

In saying that, here is another approach. The intuition relies on the fact that the derivative and the limit are more or less related by the first principles formula.
\begin{align*} \frac{1}{y} &= \frac{d}{dy}\ln y\\ &= \lim_{h\to 0}\frac{\ln (y+h) - \ln y}{h}\\ &= \lim_{h \to 0} \ln \frac1h \left(1 - \frac{h}{y} \right) \tag{difference log law}\\ &= \lim_{h\to 0}\ln \left( 1-\frac{h}{y}\right)^{1/h}\\ &= \lim_{n\to \infty}\ln \left( 1 - \frac{1}{ny} \right)^n \end{align*}
\[ \text{where we perform the substitution }n=\frac1h.\\ \text{Furthermore, we sub }x=\frac1y\text{ to obtain}\\ x = \lim_{n\to \infty} \ln \left( 1 - \frac{x}{n} \right)^{n}. \]
\[ \text{Due to the continuity of }\ln(\cdot), \text{ the limit can be moved inside the logarithm to obtain}\\ x = \ln \left( \lim_{n\to \infty} 1-\frac{x}{n} \right).\\ \text{Which immediately rearranges to }\boxed{e^x = \lim_{n\to \infty} \left( 1-\frac{x}{n}\right)^n} \]

joelg

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Re: 4U Maths Question Thread
« Reply #2321 on: June 20, 2019, 04:39:18 pm »
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Hey guys, I'm struggling with part iv. (see photo) help would be greatly appreciated

Thanks!

RuiAce

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Re: 4U Maths Question Thread
« Reply #2322 on: June 20, 2019, 04:49:15 pm »
+4
Hey guys, I'm struggling with part iv. (see photo) help would be greatly appreciated

Thanks!
Hint: You can try to show that the interval \(QR\) subtends angles of equal size at \( P\) and \(S\), which are both on the same side of \(QR\). That'll help you prove that the points are concyclic.

To do so, you may need \(b^2=a^2(1-e^2)\), but it's likely you'll also need the angle between two lines formula to find \( \tan \angle QPR \) and \(\tan \angle QSR\). I haven't tried the problem yet, but I wouldn't be surprised by some nasty algebra.

Ollierobb1

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Re: 4U Maths Question Thread
« Reply #2323 on: June 21, 2019, 02:00:40 pm »
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Hint: You can try to show that the interval \(QR\) subtends angles of equal size at \( P\) and \(S\), which are both on the same side of \(QR\). That'll help you prove that the points are concyclic.

To do so, you may need \(b^2=a^2(1-e^2)\), but it's likely you'll also need the angle between two lines formula to find \( \tan \angle QPR \) and \(\tan \angle QSR\). I haven't tried the problem yet, but I wouldn't be surprised by some nasty algebra.

Damn, im still struggling through the algebra of this question. Maybe i haven't gotten the right y coordinate for R, -sinΘ(a^2-b^2)/b?

RuiAce

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Re: 4U Maths Question Thread
« Reply #2324 on: June 21, 2019, 03:57:38 pm »
+2
Damn, im still struggling through the algebra of this question. Maybe i haven't gotten the right y coordinate for R, -sinΘ(a^2-b^2)/b?
\[ \angle QPR = 90^\circ\\ \text{because the normal is perpendicular to the tangent by definition.} \\ \text{So we just have to show that }\angle QSR = 90^\circ. \]
\[ \begin{align*} m_{QS} &= \frac{0 - \frac{b}{\sin\theta}}{ae - 0}\\ &= -\frac{b}{ae\sin\theta}\\ m_{RS} &= \frac{0 + \frac{(a^2-b^2)\sin\theta}{b}}{ae-0} \\ &= \frac{(a^2-b^2)\sin\theta}{abe}\end{align*} \]
\begin{align*}
\therefore m_{QS} m_{RS} &= -\frac{b}{ae\sin\theta} \cdot \frac{(a^2-b^2)\sin\theta}{abe}\\
&= -\frac{a^2-b^2}{a^2e^2}\\
&= -\frac{a^2-a^2(1-e^2)}{a^2e^2}\\
&= -\frac{a^2e^2}{a^2e^2}\\
&= -1
\end{align*}
Hence \( QS \perp RS\) as required. You should now be able to puzzle it all together.