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September 23, 2019, 06:53:45 am

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#### Kombmail

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« Reply #4200 on: May 12, 2019, 07:25:29 pm »
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So I was doing 6.2 ( exponential growth and decay) for mathing and came across this:
'The rate of percentage of leakage of water out of a container is proportional to the amount of water in the container at any one time. If the container is 60% empty after 5 minutes, find how long it will take for the container to be 90% empty'

I got an answer of 6.82 minutes but the actual answer is 12.6 minutes. Please assist in telling me how the answer 12.6 is found.

thanks!
-KgkG-

#### RuiAce

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« Reply #4201 on: May 12, 2019, 07:46:19 pm »
+1
So I was doing 6.2 ( exponential growth and decay) for mathing and came across this:
'The rate of percentage of leakage of water out of a container is proportional to the amount of water in the container at any one time. If the container is 60% empty after 5 minutes, find how long it will take for the container to be 90% empty'

I got an answer of 6.82 minutes but the actual answer is 12.6 minutes. Please assist in telling me how the answer 12.6 is found.

thanks!

$\text{If }V_0\text{ is the initial quantity}\\ \text{then }V = V_0 e^{-kt}.$
$\text{When }t=5, \quad V = 0.4V_0\\ \text{Therefore }0.4V_0 = V_0 e^{-5k} \implies \boxed{k = -\frac15 \ln 0.4}$
$\text{Therefore }V = V_0 e^{\frac{t}5\ln 0.4}\\ \text{We want when }V = 0.1 V_0.\\ \text{Solving }0.1V_0 = V_0 e^{\frac{t}{5}\ln 0.4}\text{ gives }\boxed{t = \frac{\ln 0.1}{\frac15 \ln 0.4} \approx 12.6}.$
There is a slight trap in this question worth mentioning. It is in that the analysis must somewhat be done in reverse. The question gives what percentage the tank is empty at, but actually measuring how much of the initial quantity we have means that our calculations must be based on what percentage the tank is full at.

Aside from that, if you have any further questions you should post relevant working out, or ask for key areas to clarify.

#### Kombmail

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« Reply #4202 on: May 17, 2019, 09:57:13 am »
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$\text{If }V_0\text{ is the initial quantity}\\ \text{then }V = V_0 e^{-kt}.$
$\text{When }t=5, \quad V = 0.4V_0\\ \text{Therefore }0.4V_0 = V_0 e^{-5k} \implies \boxed{k = -\frac15 \ln 0.4}$
$\text{Therefore }V = V_0 e^{\frac{t}5\ln 0.4}\\ \text{We want when }V = 0.1 V_0.\\ \text{Solving }0.1V_0 = V_0 e^{\frac{t}{5}\ln 0.4}\text{ gives }\boxed{t = \frac{\ln 0.1}{\frac15 \ln 0.4} \approx 12.6}.$
There is a slight trap in this question worth mentioning. It is in that the analysis must somewhat be done in reverse. The question gives what percentage the tank is empty at, but actually measuring how much of the initial quantity we have means that our calculations must be based on what percentage the tank is full at.

Aside from that, if you have any further questions you should post relevant working out, or ask for key areas to clarify.

Thanks!
-KgkG-

#### spnmox

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« Reply #4203 on: May 19, 2019, 07:10:57 pm »
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Hey guys, I need a bit of help with this question. I understand part i) but not part ii). I don't know how to figure it out and don't understand the solutions either Any help appreciated!

PS. please ignore the scribbles down the side of the page!

#### RuiAce

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« Reply #4204 on: May 19, 2019, 08:42:35 pm »
+1
Hey guys, I need a bit of help with this question. I understand part i) but not part ii). I don't know how to figure it out and don't understand the solutions either Any help appreciated!

PS. please ignore the scribbles down the side of the page!
This will be added to the compilation. This is still NESA's solution but with more added depth (no pun intended)
$\textbf{2010 HSC Mathematics - Q10 b) ii)}$
$\text{In part (1), a subtle thing to note is how they mention that the}\\ \text{hemispherical container is initially laid out }\textbf{horizontal.}\\ \text{That, and it is }\textbf{full.}$
$\text{So initially it must've looked something like this.}$
For reference I intentionally plot the centre of the circle.

$\text{Hence the initial depth is literally just the distance between the}\\ \text{centre of a circle, and any point on its circumference.}\\ \text{Which is of course, the radius, and hence equal to }r.$
Note therefore that half of the initial depth must be $\boxed{\frac{r}{2}}$.
$\text{Now after doing that rotation, the situation is as follows.}\\ \text{We wish to understand why.}$

$\text{The }r\text{ towards the right should be reasonably clear.}\\ \text{However the red line segment }\textbf{also}\text{ has length }r.$
$\text{This is because }\textbf{any}\text{ line from the centre of the circle to its circumference}\\ \text{has length }r\text{, where }r\text{ is the radius.}\\ \text{Note that the red line segment }\textbf{also}\text{ satisfies this.}$
It just so happens that the red line is special, in that it's perpendicular to the ground (or whatever surface the container is lying on. And consequently the red line is vertical.
$\text{Again, the blue region shaded reflects the water in the bowl.}\\ \text{Note that the depth of the water is then just from the point where the bowl is at the ground}\\ \text{up to the black line, i.e. where the water goes up to.}$
$\text{Since the question instructs us to make this depth equal to }\frac{r}{2}\\ \text{we do so. But then because the red line segment}\\ \text{has length }r\text{, everything above the water to the centre of the circle}\\ \text{must }\textbf{also}\text{ have length }\frac{r}{2}.$
$\text{The last thing to justify is where the }\theta\text{ is placed.}\\ \text{This relies on the fact that }\theta\text{ measures simply what angle the bowl is }\textbf{tilted at}\\ \text{and therefore must be the angle drawn from the horizontal, as required.}$
Or alternatively, if the diagram they gave was clear enough, you should be able to automatically see why $\theta$ is there. I literally just copied $\theta$ across - it was the $r$ that required more deduction.
$\text{With the set-up done, the trigonometry is now easy.}\\ \text{Because we have opposite/hypotenuse here, we just consider}\\ \sin \theta = \frac{\frac{r}{2}}{r} \implies \sin \theta = \frac12 \implies \boxed{\theta = \frac\pi6}.$
__________________________________________________________
$\text{Put simply, part (2) now wishes us to simplify the following fraction:}\\ \frac{\text{Remaining volume left}}{\text{Initial volume}}$
Because we're given that the water initially fills the entire hemisphere, we just need to compute the volume of said hemisphere. Which is of course half that of the volume of the entire sphere, i.e. $\frac12 \times \frac{4\pi r^3}{3}$, which simplifies to $\boxed{\frac{2\pi r^3}{3}}$.
$\text{The remaining bit is to make use of part i).}\\ \text{It was more or less up to you to recognise that the 'solid' that the water manifests in}\\ \text{is basically the same as what you do in that volumes question above.}$
By simply rotating NESA's own diagram by $90^\circ$ anticlockwise this may be clearer. Focus on the grey shaded area, because that reflects the volume of the water.

Note that in the diagram in part i), only the top half is shaded. But recall that when you rotate to form your solid, you go through a full $360^\circ$ rotation about the $x$-axis. By doing so, the bottom half of the minor segment does get included as well.
$\text{So in our case, the volume of the water is just the answer in part i)}\\ \text{except thanks to (1), we know now that }\theta = \frac\pi6.$
\text{Thus we get}\\ \begin{align*}V&= \frac{\pi r^3}{3} \left(2-3\sin \frac\pi6 +\sin^3\frac\pi6 \right)\\ &= \frac{\pi r^3}{3} \times \frac{5}{8}\\ &= \frac{5\pi r^3}{24}. \end{align*}
$\text{Finally subbing back, the required fraction is}\\ \frac{ \frac{5\pi r^3}{24}}{\frac{2\pi r^3}{3}} = \frac{5}{16}.$

#### spnmox

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« Reply #4205 on: May 19, 2019, 09:31:42 pm »
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This will be added to the compilation. This is still NESA's solution but with more added depth (no pun intended)
$\textbf{2010 HSC Mathematics - Q10 b) ii)}$
$\text{In part (1), a subtle thing to note is how they mention that the}\\ \text{hemispherical container is initially laid out }\textbf{horizontal.}\\ \text{That, and it is }\textbf{full.}$
$\text{So initially it must've looked something like this.}$
For reference I intentionally plot the centre of the circle.

$\text{Hence the initial depth is literally just the distance between the}\\ \text{centre of a circle, and any point on its circumference.}\\ \text{Which is of course, the radius, and hence equal to }r.$
Note therefore that half of the initial depth must be $\boxed{\frac{r}{2}}$.
$\text{Now after doing that rotation, the situation is as follows.}\\ \text{We wish to understand why.}$

$\text{The }r\text{ towards the right should be reasonably clear.}\\ \text{However the red line segment }\textbf{also}\text{ has length }r.$
$\text{This is because }\textbf{any}\text{ line from the centre of the circle to its circumference}\\ \text{has length }r\text{, where }r\text{ is the radius.}\\ \text{Note that the red line segment }\textbf{also}\text{ satisfies this.}$
It just so happens that the red line is special, in that it's perpendicular to the ground (or whatever surface the container is lying on. And consequently the red line is vertical.
$\text{Again, the blue region shaded reflects the water in the bowl.}\\ \text{Note that the depth of the water is then just from the point where the bowl is at the ground}\\ \text{up to the black line, i.e. where the water goes up to.}$
$\text{Since the question instructs us to make this depth equal to }\frac{r}{2}\\ \text{we do so. But then because the red line segment}\\ \text{has length }r\text{, everything above the water to the centre of the circle}\\ \text{must }\textbf{also}\text{ have length }\frac{r}{2}.$
$\text{The last thing to justify is where the }\theta\text{ is placed.}\\ \text{This relies on the fact that }\theta\text{ measures simply what angle the bowl is }\textbf{tilted at}\\ \text{and therefore must be the angle drawn from the horizontal, as required.}$
Or alternatively, if the diagram they gave was clear enough, you should be able to automatically see why $\theta$ is there. I literally just copied $\theta$ across - it was the $r$ that required more deduction.
$\text{With the set-up done, the trigonometry is now easy.}\\ \text{Because we have opposite/hypotenuse here, we just consider}\\ \sin \theta = \frac{\frac{r}{2}}{r} \implies \sin \theta = \frac12 \implies \boxed{\theta = \frac\pi6}.$
__________________________________________________________
$\text{Put simply, part (2) now wishes us to simplify the following fraction:}\\ \frac{\text{Remaining volume left}}{\text{Initial volume}}$
Because we're given that the water initially fills the entire hemisphere, we just need to compute the volume of said hemisphere. Which is of course half that of the volume of the entire sphere, i.e. $\frac12 \times \frac{4\pi r^3}{3}$, which simplifies to $\boxed{\frac{2\pi r^3}{3}}$.
$\text{The remaining bit is to make use of part i).}\\ \text{It was more or less up to you to recognise that the 'solid' that the water manifests in}\\ \text{is basically the same as what you do in that volumes question above.}$
By simply rotating NESA's own diagram by $90^\circ$ anticlockwise this may be clearer. Focus on the grey shaded area, because that reflects the volume of the water.

Note that in the diagram in part i), only the top half is shaded. But recall that when you rotate to form your solid, you go through a full $360^\circ$ rotation about the $x$-axis. By doing so, the bottom half of the minor segment does get included as well.
$\text{So in our case, the volume of the water is just the answer in part i)}\\ \text{except thanks to (1), we know now that }\theta = \frac\pi6.$
\text{Thus we get}\\ \begin{align*}V&= \frac{\pi r^3}{3} \left(2-3\sin \frac\pi6 +\sin^3\frac\pi6 \right)\\ &= \frac{\pi r^3}{3} \times \frac{5}{8}\\ &= \frac{5\pi r^3}{24}. \end{align*}
$\text{Finally subbing back, the required fraction is}\\ \frac{ \frac{5\pi r^3}{24}}{\frac{2\pi r^3}{3}} = \frac{5}{16}.$

Thank you so much for this lengthy explanation, and especially for rotating the diagram! It made things a lot clearer. I just don't understand the solutions provided:

if depth of water remaining = half original
r-rsintheta = 1/2 r

where did the r-rsintheta come from?

#### RuiAce

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« Reply #4206 on: May 19, 2019, 10:21:11 pm »
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Thank you so much for this lengthy explanation, and especially for rotating the diagram! It made things a lot clearer. I just don't understand the solutions provided:

if depth of water remaining = half original
r-rsintheta = 1/2 r

where did the r-rsintheta come from?
That's what you get if you don't do what I did and sub $\frac{r}{2}$ straight in. The length of that red line segment will still be $r$, but if you choose to work backwards, then you first conclude using $\sin = \frac{opp}{hyp}$ that the opposite side of that right angle triangle has length $r\sin \theta$. But this is also just the top part of the red line, so the bottom part of the red line (which corresponds to the water's depth) has length $r - r\sin\theta$.

#### Thankunext

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« Reply #4207 on: May 21, 2019, 12:38:45 pm »
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Use the trapezoidal rule with 4 subintervals to find an approximation to the volume of the solid formed by rotating the curve y=square root cosx about the x axis from x=0 to x= pi/2.

#### Thankunext

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« Reply #4208 on: May 21, 2019, 12:49:20 pm »
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And also this:
A curve has double derivative 18sin3x and a stationary point at (pi/6, -2). Find the equation of the curve.

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« Reply #4209 on: May 21, 2019, 03:11:52 pm »
-1
Heyy!
My friend's and I are really stuck on this rates question.

A tap releases liquid A into a tank at the rate of (2+t^2/t+1) litres per minute, where t is time in minutes.
A second tap releases liquid B into the same tank at the rate of (1+1/t+1) litres per minute.
The tapes are opened at the same time and release the liquids into an empty tank.

ii) The taps are closed after 4 minutes. By how many litres is the volume of liquid A greater than the volume of liquid B in the tank when the taps are closed?

#### fun_jirachi

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« Reply #4210 on: May 21, 2019, 05:12:09 pm »
+1
Hey there!

Just a friendly reminder to please post any working you have so we can target the areas you're struggling with (unless you have no idea, then at least tell us what you've learned!) It really helps us help you

Use the trapezoidal rule with 4 subintervals to find an approximation to the volume of the solid formed by rotating the curve y=square root cosx about the x axis from x=0 to x= pi/2.

With this question, remember that the trapezoidal rule for one application is (f(a)+f((a+b)/2)+f(b)) * (b-a)/2, and with multiple applications it's (b-a)/2 * first + last + 2(everything in between). Here, since your function f(x)=cos1/2 x, your function that you're using for the trapezoidal rule will just be cos x (remembering to square the function prior to doing any work with volumes).
From here, you're given four subintervals ie. the boundaries should be 0, pi/8, pi/4, 3pi/8, pi/2.
Using that, you have the area being roughly equal to:
$\pi \times \left(\cos 0 + \cos \frac{\pi}{2} + 2\left(\cos \frac{\pi}{8} + \cos \frac{\pi}{4} +\cos \frac{3\pi}{8} \right) \right) \times \frac{\pi}{16} = 3.1 (2SF)$

And also this:
A curve has double derivative 18sin3x and a stationary point at (pi/6, -2). Find the equation of the curve.

Integrating once, then twice like so:
$\int 18\sin{3x} dx = -6\cos{3x} + C_1 \\ \int -6\cos{3x} dx = -2\sin{3x} + C_1x + C_2$
Substitute the x value given into the equation ie. pi/6 to find that C1 and C2 are zero, since f(x) at that point is equal to -2.

Heyy!
My friend's and I are really stuck on this rates question.

A tap releases liquid A into a tank at the rate of (2+t^2/t+1) litres per minute, where t is time in minutes.
A second tap releases liquid B into the same tank at the rate of (1+1/t+1) litres per minute.
The tapes are opened at the same time and release the liquids into an empty tank.

ii) The taps are closed after 4 minutes. By how many litres is the volume of liquid A greater than the volume of liquid B in the tank when the taps are closed?

Welcome to atarnotes
Firstly, I'm not actually sure I'm solving this correctly because I might not have the question down right
Right now, they're reading like (given no intuition and worst case scenario)
$2+t^{\frac{2}{t}+1} \ \text{and} \ 1+\frac{1}{t}+1$
Please make sure you use parentheses properly so it's clear what the fractions are If I've got it down wrong, point it out and I'll happily do it again, but I'm going to be making a few assumptions as I solve this due to that (the process should still be the same though)

Essentially they've made this question a bit too wordy. All they want you to do is integrate the functions given for A and B with respect to t, with bounds 4 minutes and 0 minutes, then find the difference. So for a), then b)
$\int_0^4 \frac{2+t^2}{t+1} dt \\ = \int_0^4 \frac{t^2-1+3}{t+1} dt \\ = \int_0^4 t-1+\frac{3}{t+1} dt \\ = \left[\frac{t^2}{2}-t+3\ln{|t+1|}\right]_0^4 \\ =4+3\ln 5$
$\int_0^4 1+\frac{1}{t+1} dt \\ = \left[t+\ln{|t+1|}\right]_0^4 \\ = 4+\ln 5$
So basically they differ by 2ln5, which you can verify is roughly 3.2L.

Hope this helps
« Last Edit: May 21, 2019, 05:26:34 pm by fun_jirachi »
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#### Thankunext

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« Reply #4211 on: May 21, 2019, 06:50:45 pm »
0
Hey there!

Just a friendly reminder to please post any working you have so we can target the areas you're struggling with (unless you have no idea, then at least tell us what you've learned!) It really helps us help you

Integrating once, then twice like so:
$\int 18\sin{3x} dx = -6\cos{3x} + C_1 \\ \int -6\cos{3x} dx = -2\sin{3x} + C_1x + C_2$
Substitute the x value given into the equation ie. pi/6 to find that C1 and C2 are zero, since f(x) at that point is equal to -2.

I wish I could post my working but I dont know how to use the italic text with the integration signs and the like.

I dont really understand the +c part and substituting into where. I got the answer just by integrating twice but dont know what to do with the points given.

#### fun_jirachi

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« Reply #4212 on: May 21, 2019, 08:28:37 pm »
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LaTeX isn't required as working If you're worried about formatting, take a picture of what you've done!

With the +C, you have to have a constant when evaluating an indefinite integral. This is because say 4x-4 has the same derivative as 4x, and thus the primitive of 4 'differs by a constant', given as the +C. Since the C is just a constant, we can then integrate that term to get a linear term as seen in the second line (it's like integrating 1, pi, e, 3/2, sqrt2 or any other constant!) That's how you get to the integral in the second line You shouldn't in any case be 'just getting the answer' by integrating twice, because you're omitting constants you don't know exist or not without new information. It's so important that you remember +C (not only because of the memes about it) but because of the marks you get docked for forgetting it in the HSC.

As for the substitution, the fact that the point (pi/6, 2) lies on the curve tells you that f(pi/6)=-2. In this case, your f(x)becomes -2sin3x+C1x+C2, and you sub in x=pi/6. By already being given that f(pi/6)=-2, and that result becoming -2+pi/6 * C1+C2, none of the constants should exist to maintain equality with the already known fact. Therefore, C1 and C2 are both zero and the answer is just -2sin3x.

Hope thois helps
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