Author Topic: 3U Maths Question Thread (Read 1239929 times) Tweet Share

winstondarmawan · « **Reply #2205 on:** June 15, 2017, 07:53:18 pm »

Would appreciate help for the Q below, TIA.
A fever is responsible for a rise in temperature of a body of 1 degree per hour, if no cooling took place. However an adult currently measuring 40 degrees is placed in a 12 degrees cool room which cools the temperature of the body according to Newton's rate of cooling. If cooling stops when the body reaches 37 degrees, calculate when the body will reach 38 degrees.

RuiAce · « **Reply #2206 on:** June 15, 2017, 08:17:36 pm »

Quote from: winstondarmawan on June 15, 2017, 07:53:18 pm

Would appreciate help for the Q below, TIA.
A fever is responsible for a rise in temperature of a body of 1 degree per hour, if no cooling took place. However an adult currently measuring 40 degrees is placed in a 12 degrees cool room which cools the temperature of the body according to Newton's rate of cooling. If cooling stops when the body reaches 37 degrees, calculate when the body will reach 38 degrees.

$\text{This is a 'weird' question on Newton's law of cooling}\\ \text{When nothing is given, immediately assume the equation}\\ T = P + Ae^{-kt}$
$\text{The ambient temperature is 12, so }P=12\\ \text{When }t=0, T=40\text{ so upon substitution}\\ 40 = 12 + A \implies A = 28$
$\text{So }T = 12 + 28e^{-kt}\\ \text{We would like to follow the usual process to find }k\text{ by substituting some other condition}\\ \textbf{The question is unlikely to appear in the HSC due to its ambiguity}\\ \text{But we will need to assume 'useful' conditions}\text{ (they're honestly shit)}$
$\text{The question hints that a fever causes the temperature to rise by 1 degree per hour}\\ \text{It would make sense for us to reverse this in this model}\\ \text{Assume that in the cool room, }\boxed{\text{when }t=3\text{ we have }T=37}$
$\therefore 25 = 28e^{-3k} \implies k=-\frac13 \ln \frac{25}{28}\\ \text{The last bit is left as an exercise}$
Of course, this assumption is absolutely ridiculous. But again, this sort of question causes nothing but ambiguity and will not appear in the HSC exam.

Sukakadonkadonk · « **Reply #2207 on:** June 15, 2017, 10:00:31 pm »

Hello,
How do you do a general solution question like this?

Thanks.

pikachu975 · « **Reply #2208 on:** June 15, 2017, 10:10:29 pm »

Quote from: Sukakadonkadonk on June 15, 2017, 10:00:31 pm

Hello,
How do you do a general solution question like this?

Thanks.

sin3x = sinx
3x = npi + (-1)^n * sin^-1 (sinx)
3x = npi + (-1)^n * x

if n is even, 3x = npi + x, x = npi/2
if n is odd, 3x = npi - x, 4x = npi, x = npi/4

winstondarmawan · « **Reply #2209 on:** June 15, 2017, 10:25:24 pm »

Hi! I need help with Q10:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19206465_1596278383748088_94351669_n.jpg?oh=0c0567b639176b464f5f12ee734c4054&oe=594481EB

RuiAce · « **Reply #2210 on:** June 16, 2017, 09:48:24 am »

Quote from: winstondarmawan on June 15, 2017, 10:25:24 pm

Hi! I need help with Q10:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19206465_1596278383748088_94351669_n.jpg?oh=0c0567b639176b464f5f12ee734c4054&oe=594481EB

$\text{Denote the masses as }M_X\text{ and }M_Y\\ \text{Let the initial mass of }X\text{ be }M\\ \text{Then, when }t=0, M_Y =2M$
$\text{Then, we have the system}\\ \begin{align*}M_X&=Me^{-kt}\\ M_Y&=2Me^{-\ell t}\end{align*}$
$\text{When }t = T_1, M_X = \frac{M}{2}\text{ by definition of half life}\\ \therefore \frac{1}{2}=e^{-kT_1} \implies k = \frac{1}{T_1}\ln 2$
$\text{When }t=T_2, M_Y = M\\ \text{So similarly, }\ell = \frac{1}{T_2}\ln 2$
$\text{The system is now}\\ \begin{align*}M_X&=Me^{-\frac{t}{T_1}\ln 2}\\ M_Y&=2Me^{-\frac{t}{T_2}\ln 2}\end{align*}$
$\text{We wish to find the point where }M_X=M_y\\ \text{Hence upon equating}\\ \begin{align*}Me^{-\frac{t}{T_1}\ln 2}&=2Me^{-\frac{t}{T_2}\ln 2}\\ e^{-\frac{t}{T_1}\ln 2}&=2e^{-\frac{t}{T_2}\ln 2} \end{align*}$
$\text{Taking ln of both sides and using relevant log laws}\\ \begin{align*}-\frac{t}{T_1}\ln 2&= \ln 2 -\frac{t}{T_2}\ln 2\\ t \left(\frac{1}{T_2}-\frac{1}{T_1}\right)&=1\\ t\left(\frac{T_1-T_2}{T_1T_2}\right)&=1\\ t&=\frac{T_1T_2}{T_1-T_2}\end{align*}$

jakesilove · « **Reply #2211 on:** June 16, 2017, 09:49:38 am »

Quote from: winstondarmawan on June 15, 2017, 10:25:24 pm

Hi! I need help with Q10:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/19206465_1596278383748088_94351669_n.jpg?oh=0c0567b639176b464f5f12ee734c4054&oe=594481EB

Rui I was literally about to post my answer. So rude.

Nah bro you know I love you

RuiAce · « **Reply #2212 on:** June 16, 2017, 09:50:09 am »

Quote from: jakesilove on June 16, 2017, 09:49:38 am

Rui I was literally about to post my answer. So rude.

Maybe we need the system back...

<3

jakesilove · « **Reply #2213 on:** June 16, 2017, 09:52:07 am »

Quote from: RuiAce on June 16, 2017, 09:50:09 am

Maybe we need the system back...

<3

Or maybe I just need to get better at LaTex

chelseam · « **Reply #2214 on:** June 18, 2017, 04:36:06 pm »

Hi! Could I please get some help with this question? The answer is k=2 and n=6! Thank you

RuiAce · « **Reply #2215 on:** June 18, 2017, 04:42:48 pm »

Quote from: chelseam on June 18, 2017, 04:36:06 pm

Hi! Could I please get some help with this question? The answer is k=2 and n=6! Thank you

$\text{The coefficient of }x^2\text{ in the expansion of }(1+kx)^n\text{ is just }\binom{n}{2}k^2\\ \text{It is the expansion of }\left(1+3x+\frac{3x^2}{2}\right)^4\text{ that needs more work}$
$\text{The power on 4 isn't the worst to handle}\\ \text{Hence, grouping }1+3x\text{ as }(1+3x),\text{ direct use of the binomial theorem produces}\\ (1+3x)^4 + 4(1+3x)^3 \frac{3x^2}{2} + \dots$
$\textbf{The trick is that the rest of the terms don't even matter.}\\ \text{We're only interested in the coefficient of }x\\ \text{But if you look closely enough, the remainder terms}\\ \textbf{will only have }x^4\textbf{ or higher}$
$\text{The problem reduces to finding the coefficient on }x^2\text{ in }(1+3x)^4 + 6x^2(1+3x)^3\\ \text{This is much easier. It will be }\binom{4}{2}3^2 + 6\binom{3}{0} = 60$
$\text{So now we would like to consider}\\ \binom{n}{2}k^2 = 60$
____________________________
$\textbf{Here's where a bit of ambiguity comes in}\\ \text{Presumably, to fulfil the criteria }\textbf{uniquely}\text{,, we want }\boxed{k \in \mathbb{N}}$
$\text{If }k\text{ is a natural number, then we can just list out values for }k^2:\\ 1,4,9,16,25,36,49\\ \text{Note that }64 > 60\text{ so we no longer care for beyond }k=7$
$\text{Clearly, }4\text{ and }1\text{ are the only factors of }60\\ \text{So either }k^2 = 4\implies k = 2\text{ or }k^2 = 1 \implies k = 1$
$\text{If }k=2\text{, we have }\binom{n}{2} = 15 \implies n(n-1) = 30 \implies n=6\\ \text{If }k=1\text{, we have }\binom{n}{2}=60 \implies n(n-1) = 120\\ \text{which only has irrational solutions}\\ \text{Note that it is }\textbf{implied}\text{ that }n\text{ is an integer}$
$\text{Hence, }k=2\text{ and }n=6$

chelseam · « **Reply #2216 on:** June 18, 2017, 04:54:46 pm »

Quote from: RuiAce on June 18, 2017, 04:42:48 pm

$\text{The coefficient of }x^2\text{ in the expansion of }(1+kx)^n\text{ is just }\binom{n}{2}k^2\\ \text{It is the expansion of }\left(1+3x+\frac{3x^2}{2}\right)^4\text{ that needs more work}$

Thank you so much Rui!

teapancakes08 · « **Reply #2217 on:** June 19, 2017, 06:16:07 pm »

Not sure where to start on this one: "An object falling in a vertical line passes a window 3m high in 1/6 second. Find the distance above the top of the window from which the object was let fall."

jakesilove · « **Reply #2218 on:** June 19, 2017, 07:05:36 pm »

Quote from: teapancakes08 on June 19, 2017, 06:16:07 pm

Not sure where to start on this one: "An object falling in a vertical line passes a window 3m high in 1/6 second. Find the distance above the top of the window from which the object was let fall."

Hey! You can derive the vertical displacement equation in projectile motion as
$\Delta y=ut+\frac{1}{2}at^2$
Where u is the initial velocity, t is the time, a is acceleration due to gravity and delta y is the change in y displacement. Note that, usually, there is a 'sin theta' term, which I've neglected here because its dropping straight down. Initially, the object must have started from rest. Therefore, u=0, and
$\Delta y=\frac{1}{2}(9.8)(\frac{1}{6})^2$

Dragomistress · « **Reply #2219 on:** June 21, 2017, 08:30:16 pm »

I would like to know how you simplify:
1. (sin3x-cos3x)^2
2. 1-2sin^2(10(Theta))

Thanks

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Author Topic: 3U Maths Question Thread (Read 1239929 times) Tweet Share

winstondarmawan

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