Author Topic: 3U Maths Question Thread (Read 1235748 times) Tweet Share

bsdfjnlkasn · « **Reply #2040 on:** May 20, 2017, 04:34:05 pm »

Quote from: RuiAce on May 20, 2017, 03:57:23 pm

Ironically this question appears from just Google searching the first sentence.
$\text{Let }x\text{ be the total number of fish in the dam}\\ \textit{just prior to the sample of 30 being removed}\\ \text{i.e. after 50 tagged, before 30 removed}\\ \text{This question is then just a ratio analysis question}$
$\text{At any time, the ratio }\frac{\text{No. of tagged fish}}{\text{No. of untagged fish}}\text{ is equal (for this estimate)}\\ \text{Hence, }\frac{8}{22} = \frac{50}{x-50}\\ \text{The solution is }x=187.5$

Thanks Rui, again a probably stupid question but how do you find to the probably of P(A and B) generally?
I was just working through a question and multiplied the compliments of each event and got the answer... and if this is the right method, could you possibly explain why because I can't seem to make sense of it?

Thank you!

RuiAce · « **Reply #2041 on:** May 20, 2017, 04:57:29 pm »

Quote from: bsdfjnlkasn on May 20, 2017, 04:34:05 pm

Thanks Rui, again a probably stupid question but how do you find to the probably of P(A and B) generally?
I was just working through a question and multiplied the compliments of each event and got the answer... and if this is the right method, could you possibly explain why because I can't seem to make sense of it?

Thank you!

$\text{If you know that the events are independent then a shortcut is}\\ \mathbb{P}(A\text{ and }B) = \mathbb{P}(A) \times \mathbb{P}(B)$
Otherwise it really depends on the question.

armtistic · « **Reply #2042 on:** May 20, 2017, 10:59:06 pm »

Hey guys, how do you do part b) of this question without doing c) first?

RuiAce · « **Reply #2043 on:** May 20, 2017, 11:24:45 pm »

Quote from: armtistic on May 20, 2017, 10:59:06 pm

Hey guys, how do you do part b) of this question without doing c) first?

Personally, I feel this question lacks information. It's probably implied that the volume of the vessel is A but it isn't stated.

jamonwindeyer · « **Reply #2044 on:** May 21, 2017, 11:29:40 am »

Quote from: armtistic on May 20, 2017, 10:59:06 pm

Hey guys, how do you do part b) of this question without doing c) first?

Assuming it is as Rui says, we would do this!

First, we know one quarter of the vessel is filled in 5 minutes:

$V=\frac{A}{4},t=5\implies\frac{A}{4}=A\left(1-e^{-5k}\right)\\e^{-5k}=\frac{3}{4}\\k=\frac{-1}{5}\ln\left(\frac{3}{4}\right)\\k\approx0.0575$

So that gives us the value of $k$, then we want the volume after 10 minutes:

$V=A\left(1-e^{-10*0.0575}\right)$

Now we just want a percentage/fraction of $A$, so we just find the value of the bracketed term:

$\left(1-e^{-10*0.0575}\right)\approx0.4373$

So after 10 minutes, about 44% of the container is filled (about 19% on top of the 25% we had at five minutes)

does that make sense?

chelseam · « **Reply #2045 on:** May 21, 2017, 04:53:21 pm »

Hi! Can someone please help me with this question? I don't know what I'm supposed to be doing with the percentages they've given. Thank you so much

jakesilove · « **Reply #2046 on:** May 21, 2017, 07:16:40 pm »

Quote from: chelseam on May 21, 2017, 04:53:21 pm

Hi! Can someone please help me with this question? I don't know what I'm supposed to be doing with the percentages they've given. Thank you so much

Hey! So we have an equation
$P=500+6500e^{-kt}$

Before we can solve for anything, we need to find k! All we know is that, after 5 years, the mine's production decreases by 18%. We know that, initially, there is
$P=500+6500e^0=7000$
And, after 5 years,
$0.82*P=0.82*7000=500+6500e^{-kt}$

We can use the above equation to solve for k! Then, use that same k value and sub in t=10 years to answer part a). Finally, sub in 0.1*P on the LHS and solve for t to solve part b)

Let us know if you need a more extensive walkthrough!

chelseam · « **Reply #2047 on:** May 21, 2017, 07:54:42 pm »

Quote from: jakesilove on May 21, 2017, 07:16:40 pm

Let us know if you need a more extensive walkthrough!

Got it! Thanks heaps Jake!

marioedd · « **Reply #2048 on:** May 21, 2017, 08:39:53 pm »

Can someone please do this question. No answers

pikachu975 · « **Reply #2049 on:** May 21, 2017, 08:59:19 pm »

Quote from: marioedd on May 21, 2017, 08:39:53 pm

Can someone please do this question. No answers

First we need to find the rate the radius expands:

dr/dt = dr/dV * dV/dt
= 60/4pi*r^2
= 15/pi*r^2

Now,
dA/dt = dA/dr * dr/dt
= 8pi*r * 15/pi*r^2
= 120/r
= 120/15
= 8 mm^2 / s

RuiAce · « **Reply #2050 on:** May 21, 2017, 09:16:40 pm »

$\text{Some steps were skipped in the above solution}\\ \text{Note that some formulae were used. For the sake of clarity they are mentioned here.}$
$V=\frac{4}{3}\pi r^3 \implies \frac{dV}{dr}=4\pi r^2\implies \frac{dr}{dV}=\frac{1}{4\pi r^2}$
$A=4\pi r^2 \implies \frac{dA}{dr} = 8\pi r$
And yes, the crux of this question was the need to apply the chain rule twice. Alternatively, use the trivariate chain rule
$\frac{dA}{dt} = \frac{dA}{dr}\, \frac{dr}{dV}\, \frac{dV}{dt}$

chelseam · « **Reply #2051 on:** May 21, 2017, 10:32:26 pm »

Hi again! Could someone please hepl me with this question? Not sure if I'm approaching this right because right now I'm stuck with a fraction with a bunch of logs that I can't figure out how to simplify... Thank you

RuiAce · « **Reply #2052 on:** May 21, 2017, 10:38:17 pm »

Quote from: chelseam on May 21, 2017, 10:32:26 pm

Hi again! Could someone please hepl me with this question? Not sure if I'm approaching this right because right now I'm stuck with a fraction with a bunch of logs that I can't figure out how to simplify... Thank you

$\text{Note that the initial population is 1800}$
$P=900\text{ when }t=56\text{ so}\\ \begin{align*}900&=200+1600e^{-56k}\\ \frac{7}{16}&=e^{-56k}\\ -k&=\frac{1}{56}\ln \frac{7}{16}\end{align*}$
$\text{And we want when }P=450\text{, i.e.}\\ \begin{align*}450&=200+1600e^{-kt}\\ \frac{25}{160}&=e^{-kt}\\ t&= \frac{\ln \frac{25}{160}}{-k}\\ &= \frac{\ln \frac{25}{160}}{\frac{1}{56}\ln \frac{7}{16}}\end{align*}$

chelseam · « **Reply #2053 on:** May 21, 2017, 10:48:02 pm »

Quote from: RuiAce on May 21, 2017, 10:38:17 pm

$\text{Note that the initial population is 1800}$

Thank you so much Rui!

J.B · « **Reply #2054 on:** May 23, 2017, 07:49:56 am »

Hi I was wondering if someone would be able to help me with 20a? This is my solution but the answer is -1cm/s?

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