Math Ext 1 - Got Some Questions (1 Viewer)

Life'sHard · Jul 16, 2021

Hey guys, was feeling bored and decided to create a math ext 1 thread for everyone to contribute in posting questions here. Hopefully, some brilliant minds out there will be able to solve your questions.

This thread is for anyone who needs help. Or anyone bored of studies to do some questions

Life'sHard · Jul 16, 2021

Heres a differential question:

notme123 · Jul 16, 2021

B1andB2 · Jul 16, 2021

@notme123

B1andB2 · Jul 16, 2021

Life'sHard said:
Heres a differential question:

View attachment 31120

I'd like to second this question. Never quite understood these weird leaking ones. Could someone help? @CM_Tutor mb?

jimmysmith560 · Jul 16, 2021

This question is from the James Ruse Agricultural High School Year 12 Mathematics Extension 1 2020 Trial HSC Examination:

Life'sHard · Jul 17, 2021

jimmysmith560 said:
This question is from the James Ruse Agricultural High School Year 12 Mathematics Extension 1 2020 Trial HSC Examination:

View attachment 31144
View attachment 31145

I still believe there should be an easier way to do this. So hoping someone else can decipher this mess of working. @CM_Tutor hahahaha

CM_Tutor · Jul 17, 2021

We know that water is entering at 3 L per min and being removed at 2 L per min. It follows that

$\begin{align*} \cfrac{dV}{dt} &= \text{rate of inflow of fluid}\ -\ \text{rate of outflow of fluid} \\ &= +3 - 2 \\ &= +1\ \text{L min}^{-1} \end{align*}$

And hence, the volume, as a function of time, is given by:

$\begin{align*} V &= \int 1\ dt \\ V &= t + C,\ \text{for some constant $C$} \end{align*}$

Taking the origin of time, $t=0$ , as the "initial" point when there was $Q = 20\ \text{kg}$ of salt in $V = 10\ \text{L}$ of water:

$\begin{align*} \text{At $t = 0$:} \quad 10 &= 0 + C \\ C &= 10 \\ \text{Thus,} \quad V &= t + 10 \end{align*}$

Now, the quantity of salt at time $t$ is changing as salt is removed from the tank and water is added, and thus:

$\begin{align*} \cfrac{dQ}{dt} &= \text{rate of inflow of salt}\ -\ \text{rate of outflow of salt} \\ &= +3\ \text{L min}^{-1}\ \times 0\ \text{kg L}^{-1} - 2\ \text{L min}^{-1}\ \times \cfrac{Q}{V}\ \text{kg L}^{-1} \\ &= -\cfrac{2Q}{V}\ \text{kg min}^{-1} \\ \cfrac{dQ}{dt} &= -\cfrac{2Q}{t + 10} \end{align*}$

Trebla · Jul 17, 2021

By conservation of mass we must have that

$Q_{\text{tank}}=Q_{\text{initial}}+Q_{\text{in}}-Q_{\text{out}}$

However, since there is no salt coming into the tank (i.e. $Q_{\text{in}}=0$ ) the problem now reduces to figuring out how much salt is coming out the tank at time t, so

$\dfrac{dQ_{\text{tank}}}{dt}=-\dfrac{dQ_{\text{out}}}{dt}=-\dfrac{dV_{\text{out}}}{dt}\times\dfrac{dQ_{\text{out}}}{dV_{\text{out}}}=-2\dfrac{dQ_{\text{out}}}{dV_{\text{out}}}$
where V is the volume of water.

The concentration C in the tank is the same as the concentration of the water coming out at an instantaneous time t (note this is slightly "hand-wavey" in terms of rigour, the Ruse solution has a bit more rigour by using differentials though that is a tad outside of the syllabus), so

$C_{\text{tank}}=C_{\text{out}}=\dfrac{Q_{\text{out}}}{V_{\text{out}}}\Rightarrow Q_{\text{out}}=C_{\text{tank}}\times V_{\text{out}}\Rightarrow \dfrac{dQ_{\text{out}}}{dV_{\text{out}}}=C_{\text{tank}}$

But by definition

$C_{\text{tank}}=\dfrac{Q_{\text{tank}}}{V_{\text{tank}}}$

We know that the tank's net change of volume per minute is given by

$\dfrac{dV_{\text{tank}}}{dt}=3-2=1\Rightarrow V_{\text{tank}}=t+c = t+10$
since the volume is initially at 10L so c = 10.

Substitute it all in to get

$\dfrac{dQ_{\text{tank}}}{dt}=-2C_{\text{tank}}=-\dfrac{2Q_{\text{tank}}}{t+10}$

CM_Tutor · Jul 17, 2021

These tank problems can be expanded into quite challenging longer questions. Here's an example to ponder (no exam would ask all of this at once, but to show what's possible):

A tank that can hold 100 L of liquid is designed with a built-in stirrer to thoroughly mix its contents and with a tap to allow fluid to be removed at rates up to six litres per minute. The tank is initially filled with a solution in which $S$ kg of sugar is dissolved. At time $t=0$ minutes, the tap is fully opened and simultaneously pure water is added to the tank at a rate of four litres per minute.

Let $m$ kg be the mass of sugar in the tank at time $t$ min. Also at time $t$ min, let $C$ kg/L be the concentration of sugar in the tank, and let $V$ L be the volume of solution in the tank. It follows that $C$ , $m$ , and $V$ are related by the equation $m = CV$ .

$\text{(i)} \qquad \text{Show that} \quad \cfrac{dm}{dt} = \cfrac{-3m}{50-t}.$

$\text{(ii)} \qquad \text{Explain why $m = 0$ kg when $t = 50$ min.}$

$\text{(iii)} \qquad \text{Find an expression for $V$ at time $t$ and state its domain, $D$, in set notation.}$

$\text{(iv)} \qquad \text{Show that} \quad m = S\left(\cfrac{50-t}{50}\right)^3 \quad \text{for $t \in D$.}$

$\text{(v)} \qquad \text{Show that} \quad C = \cfrac{S}{100}\left(\cfrac{50-t}{50}\right)^2 \quad \text{for $t \in D$.}$

$\text{(vi)} \qquad \text{\emph{Without} performing any calculations, state the value of $\left|\int_0^{50} \cfrac{3m}{50-t}\ dt\right|$ and explain your answer.}$

$\text{(vii)} \qquad \text{Confirm your answer to part (vi) by evaluating the stated integral.}$

$\text{(viii)} \qquad \text{Prove that} \quad \cfrac{dm}{dV} = 3C \quad \text{for $t \in D$.}$

$\text{(ix)} \qquad \text{By simplifying the RHS, prove that} \quad V\cfrac{dm}{dV} = V^2\cfrac{dC}{dV} + CV.$

$\begin{align*} \text{(x)} &\qquad \text{Use the equation $C = \cfrac{m}{V}$ to confirm that} \quad V\cfrac{dm}{dV} = V^2\cfrac{dC}{dV} + CV \\ &\qquad \text{\emph{without} using the formulae you have derived for $C$, $m$, or $V$ in terms of $t$.} \end{align*}$

CM_Tutor · Jul 17, 2021

Life'sHard said:
I still believe there should be an easier way to do this. So hoping someone else can decipher this mess of working. @CM_Tutor hahahaha

For an MCQ, you can shortcut the process. If there is only water coming in then the time derivative will be proportional to the rate that solution is leaving, proportional to the amount at time t, and inversely proportional to the volume. This can be seen from the units:

Solution leaving is in L per min (say), quantity is in kg, volume is in L, so you get

$\cfrac{dq}{dt} \propto \cfrac{\text{outflow} \times \text{quantity}}{\text{volume}} \equiv \cfrac{\text{L min}^{-1} \times \text{kg}}{\text{L}} = \text{kg min}^{-1}$

Rate of outflow is negative and the volume may be increasing or decreasing.

You should be able to eliminate the incorrect options in an MCQ fairly easily, and to check if you have to do the calculation whether your result is reasonable in form if the question is not in "Show that ..." form.

If the solution entering has solute in it, you'd need to add a term for rate of inflow (in L / min) time concentration in inflow (in kg / L) to get a change term in kg / min.

@B1andB2, you now have the working from the approach used in the solutions of the trial, the approach I use, and the one Trebla has provided... does this make sense now?

B1andB2 · Jul 17, 2021

CM_Tutor said:
For an MCQ, you can shortcut the process. If there is only water coming in then the time derivative will be proportional to the rate that solution is leaving, proportional to the amount at time t, and inversely proportional to the volume. This can be seen from the units:

Solution leaving is in L per min (say), quantity is in kg, volume is in L, so you get

$\cfrac{dq}{dt} \propto \cfrac{\text{outflow} \times \text{quantity}}{\text{volume}} \equiv \cfrac{\text{L min}^{-1} \times \text{kg}}{\text{L}} = \text{kg min}^{-1}$

Rate of outflow is negative and the volume may be increasing or decreasing.

You should be able to eliminate the incorrect options in an MCQ fairly easily, and to check if you have to do the calculation whether your result is reasonable in form if the question is not in "Show that ..." form.

If the solution entering has solute in it, you'd need to add a term for rate of inflow (in L / min) time concentration in inflow (in kg / L) to get a change term in kg / min.

@B1andB2, you now have the working from the approach used in the solutions of the trial, the approach I use, and the one Trebla has provided... does this make sense now?

yeah, thanks a lot! Are these type of concentration questions something we could expect to be in the HSC?

CM_Tutor · Jul 17, 2021

B1andB2 said:
yeah, thanks a lot! Are these type of concentration questions something we could expect to be in the HSC?

Tank problems with constant volume have been asked for ages. The ones with variable volume have been appearing in trials under the new syllabus, so I think you need to have tried them. The expanded question that I posted is then using them in differential equations questions that are certainly within the new syllabus.

Will the HSC ask one? I think they could... if I knew for sure what was on this year's HSC, I couldn't tell anyone!

... and for the record, I don't have any inside information on the content of the HSC exams beyond what is publicly known.

Math Ext 1 - Got Some Questions (1 Viewer)

Life'sHard

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Life'sHard

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notme123

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B1andB2

oui oui baguette

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jimmysmith560

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Life'sHard

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CM_Tutor

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Trebla

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B1andB2

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