Author Topic: 4U Maths Question Thread (Read 665216 times) Tweet Share

RuiAce · « **Reply #1815 on:** April 08, 2018, 09:14:30 pm »

Quote from: roygbivmagic on April 08, 2018, 03:25:58 pm

Hi, I'm stuck on this rates of change question:
A ferris wheel has radius 50m and the carriages move at a constant speed of 0.3m/s without stopping. Suppose a carriage is C, the base of the ferris wheel is B and the centre is O, and z = angle <BOC. The position of C can be represented by: [50sin(z),50cos(z)] i.e. coordinates x=50sin(z) and y=50cos(z). Also, the speed of C is given by the formula = sqrt((dx/dt)^2 + (dy/dt)^2).
a) Find dz/dt in radians per second.
b) How long does it take for carriage C to make one complete revolution? (in minutes)

$\text{A few steps were skipped. The important observation to make was}\\ 0.09 = \left( \frac{dx}{dz}\, \frac{dz}{dt} \right)^2 + \left( \frac{dy}{dz}\, \frac{dz}{dt} \right)^2$
Part b should now be easy. If it takes 1 second to travel $0.0006$ radians, then how long does it take to travel $2\pi$ radians?

Dragomistress · « **Reply #1816 on:** April 11, 2018, 08:49:07 am »

I am having trouble visualising this question.

RuiAce · « **Reply #1817 on:** April 11, 2018, 08:56:59 am »

Quote from: Dragomistress on April 11, 2018, 08:49:07 am

I am having trouble visualising this question.

This is just proving the reflection property.

What you need to prove is that the highlighted angles are equal

arii · « **Reply #1818 on:** April 11, 2018, 06:20:21 pm »

Could I get some help with this question please?

RuiAce · « **Reply #1819 on:** April 11, 2018, 07:01:06 pm »

Quote from: arii on April 11, 2018, 06:20:21 pm

Could I get some help with this question please?

$\text{Note from very classic binomial theorem proofs in 3U,}\\ \binom{n-1}{0}+\binom{n-1}{1} + \binom{n-1}{2}+\dots + \binom{n-1}{n-2}+\binom{n-1}{n-1} = 2^{n-1}.$
$\text{Using the fact that }\binom{n-1}{0} = \binom{n-1}{n-1} = 1 \text{ we then have}\\ \binom{n-1}{1} + \binom{n-1}{2} + \dots + \binom{n-1}{n-2} = 2^{n-1} - 2$
$\text{But of course, the RHS factorises to }2(2^{n-2} - 1)\\ \text{so you're done upon multiplying }n\text{ to both sides}$

arii · « **Reply #1820 on:** April 12, 2018, 09:05:16 pm »

Found another question I wasn't too sure about... Am I mean't to use double sigma? But then I wouldn't know what values to set it equal to since that's normally used when it says something like find the coefficient of x...

RuiAce · « **Reply #1821 on:** April 12, 2018, 09:15:25 pm »

Quote from: arii on April 12, 2018, 09:05:16 pm

Found another question I wasn't too sure about... Am I mean't to use double sigma? But then I wouldn't know what values to set it equal to since that's normally used when it says something like find the coefficient of x...

This was in one of the older 3U papers. It has already been addressed in the corresponding compilation.

arii · « **Reply #1822 on:** April 12, 2018, 09:55:23 pm »

The same exact question was in some 3U paper? Hmm, not sure if I see it.. Some links keep redirecting me to solutions I wrote up for others when its under the binomial heading..

RuiAce · « **Reply #1823 on:** April 12, 2018, 09:56:15 pm »

Quote from: arii on April 12, 2018, 09:55:23 pm

The same exact question was in some 3U paper? Hmm, not sure if I see it.. Some links keep redirecting me to solutions I wrote up for others when its under the binomial heading..

From memory it was in 2001

Opengangs · « **Reply #1824 on:** April 12, 2018, 09:57:57 pm »

Quote from: arii on April 12, 2018, 09:55:23 pm

The same exact question was in some 3U paper? Hmm, not sure if I see it.. Some links keep redirecting me to solutions I wrote up for others when its under the binomial heading..

The explanation can be found here.

arii · « **Reply #1825 on:** April 12, 2018, 10:10:39 pm »

Quote from: Opengangs on April 12, 2018, 09:57:57 pm

The explanation can be found here.

Life saver. Thank you

arii · « **Reply #1826 on:** April 13, 2018, 06:59:16 pm »

...and I'm back.
Another binomial question.. (idk if it's past hsc or not)

I started finding the coefficient, there's n cases for the double sigma thing (as shown in the table, after I did the binomial expansion) for the given LHS which is (x+1)^n *(x+1)^n. But I wasn't able to condense it into a closed form since I don't know how big n is.. There must be some way around it.

EDIT:
For the first and last row of the last column: nCn * nC1 = n, NOT 1.

Opengangs · « **Reply #1827 on:** April 13, 2018, 07:25:10 pm »

$\left(1 + x\right)^n \cdot \left(1 + x\right)^n = \left(1 + x\right)^{2n} \\ \sum_{k = 0}^n \binom{n}{k}x^{k} \cdot \sum_{j = 0}^n \binom{n}{j}x^{j} = \sum_{i = 0}^{2n} \binom{2n}{i}x^{i}$

You didn't have to use double sigma notation*

$\text{You could just as easily have observed the pairing that happens between } k \text{ and } j \text{ like below: } \\ \text{When } k = 1 \text{, it follows that } j \text{ must be } n. \\ \text{When } k = 2 \text{, it follows that } j \text{ must be } n - 1. \\ \text{When } k = n \text{, it follows that } j \text{ must be } 1.$
$\text{To see why we don't include } k, j = 0 \\ \text{we first need to observe the maximum value } k, j \text{ must take, which is the power of } (1 + x) \\ \text{By pairing, there are no values that when } k, j = 0 \\ \text{ the other will satisfy } n + 1.$

$\text{Note: For double sigma notation, we can combine the two together like we do below:} \\ \sum_{k = 0}^n \binom{n}{k}x^{k} \cdot \sum_{j = 0}^n \binom{n}{j}x^{j} = \sum_{k = 0}^n \sum_{j = 0}^n \binom{n}{k} \binom{n}{j}x^{k+j}$
$\text{So, we see that: } k + j = i \\ \text{Letting } i = n + 1 \text{ from the hint we're given, we see that: } \\ k + j = n + 1$
$\text{We can also see now that we're finding the following cases: } \\ k = 1 \Rightarrow j = n \\ k = 2 \Rightarrow j = n - 1 \\ k = 3 \Rightarrow j = n - 2 \\ \dots \\ k = n \Rightarrow j = 1$
$\text{This gives us the following: } \\ \binom{n}{1}\binom{n}{n} + \binom{n}{2}\binom{n}{n-1} + \binom{n}{3}\binom{n}{n-4} + \dots + \binom{n}{1}\binom{n}{n}$
$\text{We can use the identity: } \binom{n}{k} = \binom{n}{n - k} \text{ to simplify it further: } \\ \binom{n}{1}\binom{n}{n-n} + \binom{n}{2}\binom{n}{n - (n - 1)} + \dots + \binom{n}{n}\binom{n}{n-1}$
$\text{This gives us the LHS: } \\ \binom{n}{1}\binom{n}{0} + \binom{n}{2}\binom{n}{1} + \dots + \binom{n}{n}\binom{n}{n-1}$
$\text{Also, recall that } k + j \text{ is defined as the same as } i \text{ from the first line.} \\ \text{Using this identity, we see that: } \\ \begin{align*}\binom{n}{1}\binom{n}{0} + \binom{n}{2}\binom{n}{1} + \dots + \binom{n}{n}\binom{n}{n-1} &= \binom{2n}{n - 1} \\ &= \frac{(2n)!}{(n - 1)!(2n - (n - 1))!} \\ &= \frac{(2n)!}{(n - 1)!(n + 1)!}\end{align*}$
$\text{And so, we've just proven: } \\ \binom{n}{1}\binom{n}{0} + \binom{n}{2}\binom{n}{1} + \dots + \binom{n}{n}\binom{n}{n-1} = \frac{(2n)!}{(n - 1)!(n + 1)!}$

radnan11 · « **Reply #1828 on:** April 15, 2018, 10:15:56 pm »

Are we required to know the proofs for the intersection of tangents and normals of a rectangular hyperbola. My teacher said yes but noted that it had never come in an exam before but my tutor said no since it was too tedious to ask as a question. Whose right? $:-\$

RuiAce · « **Reply #1829 on:** April 15, 2018, 10:42:18 pm »

Quote from: radnan11 on April 15, 2018, 10:15:56 pm

Are we required to know the proofs for the intersection of tangents and normals of a rectangular hyperbola. My teacher said yes but noted that it had never come in an exam before but my tutor said no since it was too tedious to ask as a question. Whose right? $:-\$

That should be a proof that you can wing on the spot, not one that you should memorise.

Although I do agree with your teacher in that I doubt it has been examined before. The reason for this is ironically what your tutor says - tediousness.

Starting the intersection of tangents

\begin{align*}x+p^2y&=2cp\\ x+q^2y&=2cq\end{align*}
$\text{So upon subtracting,}\\ \begin{align*}(p^2-q^2)y &= 2c(p-q)\\ y &= \frac{2c}{p+q}\end{align*}$

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