Leaving Cert 2023 Paper: Note that this paper will instead be composed of TEN questions, EIGHT of which must be answered.

For each of the questions below, you will find the following:
• ANSWERS, allowing you to quickly check if you did the question correctly (just like the back of a schoolbook).
• SOLUTIONS, i.e. what you should write down in the exam in order to get full marks.
• VIDEO WALKTHROUGHS in which we explain where the solutions come from in great detail. (Members only!)

Question 1

(a) A directed graph is represented by the adjacency matrix $$M=\left(\begin{array}{ccc} 0 & 0 & 2 \\ 1 & 1 & 0\\ 0 & 1 & 0 \end{array}\right)$$.

(i) Draw the graph represented by $$M$$.

(ii) Calculate $$M^2$$.

(iii) What information is provided by the elements of $$M^2$$?

(i)

(Note that any graph that is isomorphic to this graph is considered correct.)

(ii) $$\left(\begin{array}{ccc} 0 & 2 & 0 \\ 1 & 1 & 2\\ 1 & 1& 0 \end{array}\right)$$

(iii) The number of walks of length $$2$$ between any two vertices of the network shown in part (i).

(i)

(Note that any graph that is isomorphic to this graph is considered correct.)

(ii)

\begin{align}M^2&=M\times M\\&=\left(\begin{array}{ccc}
0 & 0 & 2 \\ 1 & 1 & 0\\ 0 & 1 & 0
\end{array}\right)\times\left(\begin{array}{ccc}
0 & 0 & 2 \\ 1 & 1 & 0\\ 0 & 1 & 0
\end{array}\right)\\&=\left(\begin{array}{ccc}
0 & 2 & 0 \\ 1 & 1 & 2\\ 1 & 1& 0
\end{array}\right)\end{align}

(iii) The number of walks of length $$2$$ between any two vertices of the network shown in part (i).

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(b) A gardener plants a new fruit tree which has three new branches. In a branch’s first year of growth it will not produce any additional branches. Each branch will produce one additional branch every year after that.

The gardener models this growth pattern by defining $$U_n$$ to be the number of branches on the tree $$n$$ years after planting, with $$U_0=3$$ and $$U_1=3$$.

(i) Write down the values of $$U_2$$ and $$U_3$$.

(ii) Write down a difference equation for $$U_{n+2}$$ in terms of $$U_{n+1}$$ and $$U_n$$, where $$n \geq 0$$, $$n \in \mathbb{Z}$$.

(iii) Solve this difference equation to find an expression for $$U_n$$ in terms of $$n$$.

(iv) Plants must be cut back regularly to allow them room to grow. How many of the old branches should be removed at the end of year $$4$$ to ensure that there are exactly $$14$$ branches at the end of year $$5$$?

(i) $$U_2=6$$ and $$U_3=9$$.

(ii) $$U_{n+2} = U_{n+1} + U_n$$

(iii)

\begin{align}U_n = \left(\dfrac{3\sqrt{5}+3}{2\sqrt{5}}\right)\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\left(\dfrac{3\sqrt{5}-3}{2\sqrt{5}}\right)\left(\dfrac{1-\sqrt{5}}{2}\right)^n\end{align}

(iv) $$5$$

(i) $$U_2=6$$ and $$U_3=9$$.

(ii) \begin{align}U_{n+2} = U_{n+1} + U_n\end{align}

(iii)\begin{align}U_{n+2} = U_{n+1} + U_n\end{align}

\begin{align}\downarrow\end{align}

\begin{align}U_{n+2}-U_{n+1}-U_n = 0\end{align}

Characteristic equation

\begin{align}x^2-x-1=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2(1)}\\&=\frac{1\pm\sqrt{5}}{2}\end{align}

\begin{align}U_n = p\left(\frac{1+\sqrt{5}}{2}\right)^n+q\left(\frac{1-\sqrt{5}}{2}\right)^n\end{align}

First Initial Condition ($$U_0=3$$)

\begin{align}U_0 = p+q=3\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p=3-q\end{align}

Second Initial Condition ($$U_1=3$$)

\begin{align}U_1 = p\left(\frac{1+\sqrt{5}}{2}\right)+q\left(\frac{1-\sqrt{5}}{2}\right)=3\end{align}

\begin{align}\downarrow\end{align}

\begin{align}p(1+\sqrt{5})+q(1-\sqrt{5})=6\end{align}

Inserting that $$p=3-q$$ into the above equation:

\begin{align}(3-q)(1+\sqrt{5})+q(1-\sqrt{5})=6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3+3\sqrt{5}-q-\sqrt{5}q+q-\sqrt{5}q=6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-2\sqrt{5}q=3-3\sqrt{5}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}q=\frac{3\sqrt{5}-3}{2\sqrt{5}}\end{align}

Using $$p=3-q$$ to find $$p$$:

\begin{align}p&=3-\frac{3\sqrt{5}-3}{2\sqrt{5}}\\&=\frac{6\sqrt{5}-3\sqrt{5}+3}{2\sqrt{5}}\\&=\frac{3\sqrt{5}+3}{2\sqrt{5}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}U_n = \left(\frac{3\sqrt{5}+3}{2\sqrt{5}}\right)\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{3\sqrt{5}-3}{2\sqrt{5}}\right)\left(\frac{1-\sqrt{5}}{2}\right)^n\end{align}

(iv) \begin{align}U_4&=U_3+U_2\\&=9+6\\&=15\end{align}

There are $$15$$ branches at the end of year $$4$$: $$9$$ old branches which will each produce an additional branch in year $$5$$ and $$6$$ new branches that won’t produce any new branches in year $$5$$.

Therefore, we must reduce the $$9$$ old branches to $$4$$ old branches so that we have $$6+4+4=14$$ branches at the end of year $$5$$. Therefore, we remove $$9-4=5$$ old branches.

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Question 2

The diagram below shows the scheduling network for a project to manufacture a new chemical compound. The network provides some information about the relationships between the twelve activities that have to be completed as part of the project.

The edges of the network represent these activities and are labelled with the letters $$A$$ to $$L$$.
The unlabelled edges (shown with dashed lines) do not represent real activities but they help  explain the order in which the activities must happen. The letters used to label the edges should not be taken as representing the order in which the activities happen.

The nodes of the network represent events or points in time during the project. The source node is the time when the project begins and the sink node is the time when the project ends.

(i) Complete the table below by listing, for each activity, the other activities on which it depends directly. That is, for each activity $$X\in\{C,D,E,…,L\}$$, write the smallest possible list of other activities which need to be completed before activity $$X$$ can begin.

Activities $$A$$ and $$B$$ do not depend on any prior activities, so the list is empty for these activities, as shown.

Activity Depends directly on...

$$A$$

-

$$B$$

-

$$C$$

$$D$$

$$E$$

$$F$$

$$G$$

$$H$$

$$I$$

$$J$$

For each of the statements in parts (ii) and (iii) below, state whether you agree or disagree with the statement. Use the scheduling network and/or your answer to part (i) to justify your answer in each case.

(ii) Activity $$D$$ must be completed before activity $$G$$.

(iii) Activity $$E$$ must be completed before activity $$H$$.

$\,$

The time, in days, to complete each of the activities $$A$$ to $$L$$ is given in the table below and has also been included in the network redrawn on this page.

1

Activity

$$A$$

$$B$$

$$C$$

$$D$$

$$E$$

$$F$$

$$G$$

$$H$$

$$I$$

$$J$$

$$K$$

$$L$$

Time (days)

$$5$$

$$3$$

$$6$$

$$9$$

$$4$$

$$2$$

$$8$$

$$4$$

$$11$$

$$2$$

$$7$$

$$5$$

(iv) Calculate the early time and the late time for each event.
Complete the diagram below by writing the early time (upper box) and late time (lower box) at the node representing each event.

(v) Write down the critical path for the network.

(vi) Write down the minimum time, in days, needed to complete this project.

(vii) Select any one non‐critical activity on the network and calculate its float, in days.

(viii) The project is due to begin on the morning of July 1st. The key worker needed to carry out activity $$G$$ will be away on holidays when the project begins.
What is the latest date on which this worker could return to work without necessarily causing the project to be delayed? Justify your answer.

(i)

Activity Depends directly on...

$$A$$

-

$$B$$

-

$$C$$

$$A$$

$$D$$

$$B$$

$$E$$

$$A$$

$$F$$

$$B$$

$$G$$

$$F$$

$$H$$

$$C$$, $$E$$, $$F$$

$$I$$

$$E$$, $$F$$

$$J$$

$$D$$, $$F$$

$$K$$

$$G$$, $$J$$

$$L$$

$$G$$, $$H$$, $$I$$, $$J$$

(ii) Activity $$G$$ directly depends only on activity $$F$$ which itself directly depends on activity $$B$$. Therefore, this statement is false.

(iii) Activity $$H$$ directly depends on activities $$C$$, $$E$$ and $$F$$. Therefore, this statement is true.

(iv)

(v) $$AEIL$$

(vi) $$25$$ days

(vii) Activity $$D$$: $$4$$ days.

(You are free to calculate the float of any other non-critical activity.)

(viii) 10th July

(i)

Activity Depends directly on...

$$A$$

-

$$B$$

-

$$C$$

$$A$$

$$D$$

$$B$$

$$E$$

$$A$$

$$F$$

$$B$$

$$G$$

$$F$$

$$H$$

$$C$$, $$E$$, $$F$$

$$I$$

$$E$$, $$F$$

$$J$$

$$D$$, $$F$$

$$K$$

$$G$$, $$J$$

$$L$$

$$G$$, $$H$$, $$I$$, $$J$$

(ii) Activity $$G$$ directly depends only on activity $$F$$ which itself directly depends on activity $$B$$. Therefore, this statement is false.

(iii) Activity $$H$$ directly depends on activities $$C$$, $$E$$ and $$F$$. Therefore, this statement is true.

(iv)

(v) $$AEIL$$

(vi) $$25$$ days

(vii)

Activity D

\begin{align}\mbox{float}&=16-(3+9)\\&=4\mbox{ days}\end{align}

(You are free to calculate the float of any other non-critical activity.)

(viii) The latest time at which the events following activity $$G$$ can start without delaying the completion of the project is the $$18$$th day.

As activity $$G$$ takes $$8$$ days to complete, this means that the latest that it could start without delaying the project is on the $$18-8=10$$th day, i.e. on 10th July.

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Question 3

(a) A particle has initial displacement $$s_0$$ from a fixed point $$P$$. It moves away from $$P$$ with initial velocity $$u$$ and constant acceleration $$a=\dfrac{dv}{dt}=\dfrac{d^2s}{dt^2}$$.

Use calculus to derive an expression for $$s$$, the displacement of the particle from $$P$$ at any time $$t$$.

$$s-s_0=ut+\dfrac{1}{2}at^2$$

Step 1

\begin{align}dv = a\, dt\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\int dv = \int a\, dt\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\int dv = a\int \, dt\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v+C=a(t+D)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v(t) = at + E\end{align}

Since

\begin{align}v(0)= & a(0) + E\\&=E\end{align}

we must have that $$E=u$$. Hence, our equation becomes

\begin{align}v(t) = at + u\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v(t) = u+at \end{align}

Step 2

\begin{align}v=\frac{ds}{dt}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}ds = v\, dt\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\int ds = \int v\, dt\end{align}

Inserting $$v=u+at$$, we obtain:

\begin{align}\int ds = \int (u+at)\, dt\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\int ds = \int u \, dt + \int at\, dt\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\int ds = u\int \, dt + a\int t\, dt\end{align}

\begin{align}\downarrow\end{align}

\begin{align}s+C = u(t+D) + a\left(\frac{t^2}{2}+E\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}s(t) = ut+\frac{1}{2}at^2 + F\end{align}

Therefore, since

\begin{align}s(0) &= u(0)+\frac{1}{2}a(0^2 )+ F\\&=F\end{align}

is equal to $$s_0$$, we must have that $$F=s_0$$ and therefore

\begin{align}s=ut+\frac{1}{2}at^2+s_0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}s-s_0=ut+\frac{1}{2}at^2\end{align}

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(b) Two athletes, Brian and Clara, are taking part in a relay race. Brian is preparing to hand over the baton to Clara. During the hand‐over of the baton the athletes need to be running in the same straight line and at the same velocity.

As Brian approaches Clara’s position at a constant speed of $$11\mbox{ m s}^{-1}$$, Clara starts running from rest with constant acceleration $$f$$.

A short time later Brian begins to decelerate at $$2\mbox{ m s}^{-2}$$.

Clara receives the baton $$2.5\mbox{ s}$$ after she starts running.

The baton is exchanged when Clara is $$75\mbox{ cm}$$ ahead of Brian and when both athletes have a speed of $$8\mbox{ m s}^{-1}$$.

After the baton is exchanged, Brian continues to decelerate at $$2\mbox{ m s}^{-2}$$ until he comes to rest.
Clara continues to accelerate at $$f$$ until she reaches her maximum speed of $$12\mbox{ m s}^{-1}$$, which she then maintains.

(i) Calculate the time it takes for Brian to decelerate before he exchanges the baton.

(ii) Using the axes below, draw an accurate velocity‐time graph for the motion of each runner.
Time is measured from the instant that Clara begins to run.

(iii) Calculate the distance between the two athletes when Clara begins to run.

(i) $$1.5\mbox{ s}$$

(ii)

(iii) $$16\mbox{ m}$$

(i)

\begin{align}u=11 && v=8 && a=-2 && t=?\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v=u+at\end{align}

\begin{align}\downarrow\end{align}

\begin{align}8=11+(-2)t\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{8-11}{-2}\\&=1.5\mbox{ s}\end{align}

(ii)

Brian

How long until Brian starts decelerating?

The baton exchange occurs after $$2.5$$ seconds. Brian is decelerating for $$1.5$$ seconds until this exchange. Therefore, Brian is moving at a constant speed of $$11\mbox{ m/s}$$ for $$2.5-1.5=1$$ second.

$\,$

How long until Brian comes to rest?

\begin{align}u=11 && v=0 && a=-2 && t=?\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v=u+at\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0=11+(-2)t\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{0-11}{-2}\\&=5.5\mbox{ s}\end{align}

Therefore, Brian comes to rest after $$1+5.5=6.5\mbox{ s}$$.

Clara

What is the value of $$f$$?

\begin{align}u=0 && v=8 && t=2.5 && a=f \end{align}

\begin{align}\downarrow\end{align}

\begin{align}v=u+at\end{align}

\begin{align}\downarrow\end{align}

\begin{align}8=0+(f)(2.5)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}f&=\frac{8-0}{2.5}\\&=3.2\mbox{ m/s}^2\end{align}

$\,$

When does Clara stop accelerating?

\begin{align}u=0 && v=12 && a=3.2 && t=? \end{align}

\begin{align}\downarrow\end{align}

\begin{align}v=u+at\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12=0+3.2t\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{12-0}{3.2}\\&=3.75\mbox{ s}\end{align}

(iii) In the first $$2.5$$ seconds of the above graph, Clara travels a distance of:

\begin{align}\frac{1}{2}(2.5)(8)=10\mbox{ m}\end{align}

and Brian travels a distance of:

\begin{align}2.5\times 11-\frac{1}{2}(2.5-1)(11-8)=25.25\mbox{ m}\end{align}

Therefore, as there is still $$0.75\mbox{ m}$$ separating them after this time, the distance between them when Clara begins to run is $$25.25-10+0.75=16\mbox{ m}$$.

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Question 4

(a) A ball is projected from a point on horizontal ground, with initial speed $$u$$ and at an angle $$\alpha$$ to the horizontal. The ball reaches a maximum height of $$H_0$$ above the horizontal.

Upon landing, the ball bounces with a maximum height of $$H_1$$.

The coefficient of restitution between the ball and the ground is $$e$$.

(i) Calculate $$\dfrac{H_0}{H_1}$$.

(ii) The ball continues bouncing. Find an expression (in terms of $$e$$ and $$H_0$$) for $$H_5$$, the maximum height of the ball after it lands on the ground for the fifth time.

(i) $$\dfrac{1}{e^2}$$

(ii) $$e^{10}H_0$$

(i)

First Maximum Height

\begin{align} u_y && a_y=-g && v_y=0 && s_y=H_0\end{align}

\begin{align}\downarrow\end{align}

$v_y^2 = u_y^2+2a_ys_y$

\begin{align}\downarrow\end{align}

\begin{align}0^2 = u_y^2 + 2(-g)H_0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}H_0 = \frac{u_y^2}{2g}\end{align}

Second Maximum Height

After the first bounce, $$u_y$$ reduces to $$eu_y$$.

\begin{align}H_1 &= \frac{(eu_y)^2}{2g}\\&=e^2\frac{u_y^2}{2g}\end{align}

\begin{align}\frac{H_0}{H_1}&=\frac{\frac{u_y^2}{2g}}{e^2\frac{u_y^2}{2g}}\\&=\frac{1}{e^2}\end{align}

(ii) The maximum height reduces by $$e^2$$ after each bounce.

\begin{align}H_5 &= (e^2)^5H_0 \\&=e^{10}H_0 \end{align}

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(b) Two identical smooth spheres, $$P$$ and $$Q$$, each moving with speed $$u$$, collide obliquely. The line joining their centres at the point of impact is along the $$\vec{i}$$ axis.

Before the collision, the velocity of sphere $$P$$ makes an angle $$\theta$$ with the $$\vec{i}$$ axis and the velocity of sphere $$Q$$ makes an angle $$\theta$$ with the $$\vec{j}$$ axis, as shown in the diagram.

The coefficient of restitution between the spheres is $$e$$ , where $$0\leq e \leq 1$$.

After the collision sphere $$Q$$ moves off parallel to the $$\vec{j}$$ axis.

(i) Show that $$e=\dfrac{\tan \theta-1}{\tan \theta+1}$$.

(ii) If $$25\%$$ of the spheres’ total kinetic energy is lost during the collision, calculate $$\theta$$ and $$e$$.

(ii) $$\theta=75^{\circ}$$ and $$e\approx 0.58$$

(i)

Sphere P

$$m_P = m$$

Initial Velocity

$$\vec{u}_{Px}=u\cos \theta \vec{i}$$

$$\vec{u}_{Py}=u\sin \theta \vec{j}$$

Final Velocity

$$\vec{v}_{Px}=v_{Px}\vec{i}$$

$$\vec{v}_{Py}=u\sin \theta \vec{j}$$

Sphere Q

$$m_Q = m$$

Initial Velocity

$$\vec{u}_{Qx}=-u\sin \theta \vec{i}$$

$$\vec{u}_{Qy}=u \cos \theta \vec{j}$$

Final Velocity

$$\vec{v}_{Qx}=0 \vec{i}\mbox{ m/s}$$

$$\vec{v}_{Qy}=u\cos \theta \vec{j}$$

Conservation of Momentum

\begin{align}m_Pu_{Px}+m_Qu_{Qx} = m_Pv_Px+m_Qv_{Qx}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}mu\cos\theta+m(-u\sin\theta) = mv_{Px}+m(0)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_{Px}=u\cos \theta-u\sin\theta\end{align}

Newton’s Law of Restitution

\begin{align}e=-\frac{v_{Px}-v_{Qx}}{u_{Px}-u_{Qx}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}e=-\frac{v_{Px}-0}{u\cos\theta-(-u\sin \theta)}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}e=-\frac{v_{Px}}{u\cos\theta+u\sin \theta}\end{align}

Inserting the first equation into the second:

\begin{align}e&=-\frac{u\cos \theta-u\sin\theta}{u\cos\theta+u\sin \theta}\\&=-\frac{\frac{u\cos \theta}{u\cos\theta}-\frac{u\sin\theta}{u\cos\theta}}{\frac{u\cos\theta}{u\cos\theta}+\frac{u\sin \theta}{u\cos\theta}}\\&=-\frac{1-\tan\theta}{1+\tan\theta}\\&=\frac{\tan\theta-1}{\tan\theta+1}\end{align}

as required.

(ii)

\begin{align}E_i &= K_{Pi}+K_{Qi}\\&=\frac{1}{2}mu^2+\frac{1}{2}mu^2\\&=mu^2\end{align}

and

\begin{align}E_f &= K_{Pf}+K_{Qf}\\&=\frac{1}{2}m_Pv_{Px}^2 + \frac{1}{2}m_Pv_{Py}^2 + \frac{1}{2}m_Qv_{Qx}^2 + \frac{1}{2}m_Qv_{Qy}^2\\&=\frac{1}{2}mv_{Px}^2+\frac{1}{2}m(u^2\sin^2\theta)+ \frac{1}{2}m(0^2)+\frac{1}{2}m(u^2\cos^2\theta)\\&=\frac{1}{2}mv_{Px}^2+\frac{1}{2}mu^2(\sin^2\theta+\cos^2\theta)\\&=\frac{1}{2}mv_{Px}^2+\frac{1}{2}mu^2\end{align}

As $$E_f=0.75E_i$$, we obtain

\begin{align}\frac{1}{2}mv_{Px}^2+\frac{1}{2}mu^2=0.75(mu^2)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{2}v_{Px}^2=\frac{u^2}{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_{Px}^2=\frac{u^2}{2}\end{align}

Inserting that $$v_{Px}=u\cos \theta-u\sin\theta$$:

\begin{align}(u\cos\theta-u\sin\theta)^2=\frac{u^2}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}u^2\cos^2\theta-2u^2\sin\theta\cos\theta+u^2\sin^2\theta=\frac{u^2}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}(\cos^2\theta+\sin^2\theta)-\sin 2\theta= \frac{1}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1-\sin 2\theta= \frac{1}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\sin2\theta=\frac{1}{2}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\theta &= \frac{1}{2}\sin^{-1}\left(\frac{1}{2}\right)\\&=\frac{30^{\circ}}{2}\mbox{ or } \frac{150^{\circ}}{2}\\&=15^{\circ} \mbox{ or }75^{\circ}\end{align}

If $$\theta=15^{\circ}$$:

\begin{align}e&=\dfrac{\tan \theta-1}{\tan \theta+1}\\&=\dfrac{\tan 15^{\circ}-1}{\tan15^{\circ}+1}\\&\approx -0.58\end{align}

which is not possible as $$0\leq e\leq 1$$. Therefore, $$\theta$$ must be $$75^{\circ}$$.

When $$\theta=75^{\circ}$$:

\begin{align}e&=\dfrac{\tan \theta-1}{\tan \theta+1}\\&=\dfrac{\tan 75^{\circ}-1}{\tan75^{\circ}+1}\\&\approx 0.58\end{align}

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Question 5

(a) In the network shown below, the edges represent roads and the nodes represent the junctions of two or more roads, labelled with the letters $$A$$ to $$N$$. The weight of each edge represents the distance (in $$\mbox{km}$$) between a pair of junctions.

(i) Use Dijkstra’s algorithm to find the shortest path from junction $$A$$ to junction $$N$$.
Calculate the length of the shortest path. Relevant supporting work must be shown.

(ii) A group of engineers want to close down some of the roads to carry out maintenance work.
They wish to close down as much of the road network as possible while still allowing a person to drive between any two junctions on the network.

Using an appropriate algorithm, find the minimum spanning tree for the network.
Name the algorithm you used. Relevant supporting work must be shown.

(i) $$ACGJLN$$ with a length of $$72\mbox{ km}$$.

(ii)

(i)

Shortest Paths Diagram

The shortest path from $$A$$ to $$N$$ is $$NLJGCA$$ reversed, i.e. $$ACGJLN$$, and has a length of $$72\mbox{ km}$$.

(ii)

Kruskal’s Algorithm:

Edge Weight Included?

$$KL$$

$$4$$

Yes

$$JL$$

$$6$$

Yes

$$JM$$

$$7$$

Yes

$$DF$$

$$8$$

Yes

$$LM$$

$$9$$

No (cycle)

$$BD$$

$$9$$

Yes

$$BE$$

$$10$$

Yes

$$CD$$

$$11$$

Yes

$$EF$$

$$12$$

No (cycle)

$$IJ$$

$$12$$

Yes

$$LN$$

$$12$$

Yes

$$AC$$

$$12$$

Yes

$$IL$$

$$13$$

No (cycle)

$$MN$$

$$14$$

No (cycle)

$$DE$$

$$14$$

No (cycle)

$$GI$$

$$15$$

Yes

$$KN$$

$$15$$

No (cycle)

$$HK$$

$$16$$

Yes

$$AB$$

$$17$$

No (cycle)

$$HL$$

$$17$$

No (cycle)

$$EH$$

$$18$$

Yes

$$GJ$$

$$19$$

-

$$FI$$

$$20$$

-

$$HI$$

$$21$$

-

$$CG$$

$$23$$

-

$$DG$$

$$24$$

-

$$AD$$

$$25$$

-

Minimum Spanning Tree

(You can also use Prim’s algorithm if you prefer.)

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(b) A rumour may be spread when a person who has heard the rumour interacts with a person who has not heard the rumour.

Therefore, the rate of spread of a rumour within a group can be modelled as being proportional to the product of the number of people in the group who have heard the rumour and the number of people in the group who have not heard it.

A student models the rate at which a certain rumour spreads within a school population of $$1200$$ students using the differential equation:

\begin{align}\frac{dR}{dt} = kR(1200-R)\end{align}

where $$R(t)$$ is the number of students of that school who have heard the rumour at time $$t$$, measured in days, and where $$k$$ is a positive constant.

On Monday morning ($$t=0$$), $$100$$ students had heard the rumour.

(i) Solve the differential equation to find an expression that relates $$R$$, $$k$$ and $$t$$.

Note that $$\dfrac{1}{y(x-y)}=\dfrac{1}{x}\left(\dfrac{1}{y}+\dfrac{1}{x-y}\right)$$.

(ii) By Wednesday morning $$250$$ students had heard the rumour. Calculate the value of $$k$$.

(iii) Sketch the shape of a graph of $$R$$ against $$t$$ to show how the model predicts the spread of the rumour.

(i) $$R(t) =\dfrac{1200e^{1200kt}}{11+e^{1200kt}}$$

(ii) $$k\approx 0.00044$$

(iii)

(i)

\begin{align}\frac{dR}{dt} = kR(1200-R)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{R(1200-R)}\,dR=k\,dt\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{1200}\left(\frac{1}{R}+\frac{1}{1200-R}\right)\,dR=k\,dt\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{1200}\left(\int\frac{1}{R}\,dR+\int\frac{1}{1200-R}\,dR\right)=k\int dt\end{align}

Second integral

\begin{align}\int\frac{1}{1200-R}\,dR\end{align}

\begin{align}u=1200-R && \frac{du}{dR}=-1\end{align}

With $$u=1200-R$$ and $$du=-dR$$, the original integral becomes

\begin{align}-\int\frac{1}{u}\,du&=-\ln|u|+D\\&=-\ln|1200-R|+D\end{align}

\begin{align}\frac{1}{1200}\left(\ln |R| + C-\ln|1200-R|+D\right)=k(t+E)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{1200}\ln\left|\frac{ R}{1200-R}\right| =kt+F\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\ln\left|\frac{ R}{1200-R}\right| =1200kt+G\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{ R}{1200-R} =He^{1200kt}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}R =1200He^{1200kt}-HRe^{1200kt}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}R(1+He^{1200kt}) =1200He^{1200kt}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}R =\frac{1200He^{1200kt}}{1+He^{1200kt}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}R(t) =\frac{1200e^{1200kt}}{I+e^{1200kt}}\end{align}

We are told that

\begin{align}R(0) &=\frac{1200e^{1200k(0)}}{I+e^{1200k(0)}}\\&=\frac{1200}{I+1}\end{align}

is equal to $$100$$. Therefore

\begin{align}\frac{1200}{I+1}=100\end{align}

\begin{align}\downarrow\end{align}

\begin{align}I&=\frac{1200}{100}-1\\&=11\end{align}

Hence, the solution is given by

\begin{align}R(t) =\frac{1200e^{1200kt}}{11+e^{1200kt}}\end{align}

(ii) We are told that

\begin{align}R(2) &=\frac{1200e^{1200k(2)}}{11+e^{1200k(2)}}\\&=\frac{1200e^{2400k}}{11+e^{2400k}}\end{align}

is equal to $$250$$. Therefore

\begin{align}\frac{1200e^{2400k}}{11+e^{2400k}}=250\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1200e^{2400k}=250(11+e^{2400k})\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1200e^{2400k}=2750+250e^{2400k}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}950e^{2400k}=2750\end{align}

\begin{align}\downarrow\end{align}

\begin{align}e^{2400k}=\frac{2750}{950}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2400k=\ln\left(\frac{2750}{950}\right)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k&=\frac{\ln\left(\frac{2750}{950}\right)}{2400}\\&\approx 0.00044\end{align}

(iii) $$R_0=100$$, $$R_5\approx 672$$, $$R_{10}\approx 1136$$, $$R_{15}\approx 1195$$, $$R_{20}\approx 1200$$, $$R_{25}\approx 1200$$, $$R_{30}\approx 1200$$.

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Question 6

A learner driver is practising driving around a roundabout.

The motion of the car may be modelled as horizontal circular motion around centre $$O$$, with radius $$r$$ and constant angular speed $$\omega$$, as in the diagram above.

(i) Write an expression for $$\vec{s}$$, the displacement of the car relative to $$O$$ at any time $$t$$, in terms of $$r$$, $$\omega$$ and $$t$$. Your expression should use the unit vectors $$\vec{i}$$ and $$\vec{j}$$.

Note that $$t=0$$ when $$\vec{s}$$ is along the $$\vec{i}$$ axis.

(ii) Derive an expression for $$\vec{v}$$, the velocity of the car at any time $$t$$.

(iii) Use a dot product calculation to show that the car’s velocity and displacement are always perpendicular to each other.

(iv) Show that the acceleration of the car is always directed towards $$O$$.

(v) Derive an expression for the maximum velocity the car could have as it travels around the roundabout, without slipping. Your expression should be written in terms of $$r$$, $$g$$ and $$\mu$$, the coefficient of friction between the car and the road.

(vi) Use dimensional analysis to show that the units for the expression you derived in part (v) are equivalent to the units for velocity.

(i) $$\vec{s}(t) = -r \cos (\omega t) \vec{i} +r \sin (\omega t) \vec{j}\tag{1}$$

(Your signs may be different depending on what quadrant the displacement vector is located in on your diagram at time $$t$$.)

(ii) $$\vec{v}(t)=r\omega \sin (\omega t) \vec{i} +r\omega \cos (\omega t) \vec{j}$$

(Again, you may be correct even if you have different signs.)

(v) The magnitude of the velocity is $$\sqrt{\mu rg}$$ and its direction is tangent to the circle.

(vii) Yes. Most roundabouts are close to being circular in shape and most drivers do travel at a constant linear speed when in motion at such a roundabout. Any tangential acceleration will be negligible compared to the centripetal acceleration and it is therefore acceptable that it has been ignored in this model.

(i)

Initial Location

$$\mathbf{t}$$ seconds later

$\vec{s}(\theta) = -r \cos \theta \vec{i} +r \sin \theta \vec{j}$

Inserting $$\theta=\omega t$$ gives

$\vec{s}(t) = -r \cos (\omega t) \vec{i} +r \sin (\omega t) \vec{j}\tag{1}$

(Your signs may be different depending on what quadrant the displacement vector is located in on your diagram at time $$t$$.)

(ii)

\begin{align}\vec{v}(t) &= \frac{d\vec{s}}{dt}\\&= -r[-\omega \sin (\omega t) \vec{i}] +r [\omega\cos (\omega t) \vec{j}]\\&=r\omega \sin (\omega t) \vec{i} +r\omega \cos (\omega t) \vec{j}\tag{2}\end{align}

(Again, you may be correct even if you have different signs.)

(iii) According to equations $$(1)$$ and $$(2)$$, the dot product of these two vectors is given by

\begin{align}\vec{v}\cdot\vec{s} &= [r \omega \sin (\omega t) \vec{i} +r\omega\cos (\omega t) \vec{j}] \cdot [-r \cos (\omega t) \vec{i} +r \sin (\omega t) \vec{j}]\\&=-r^2\omega \sin (\omega t) \cos (\omega t) (\overbrace{\vec{i} \cdot \vec{i}}^{=1}) +r^2\omega \sin^2 (\omega t) (\overbrace{\vec{i} \cdot \vec{j}}^{=0})-r^2\omega \cos^2 (\omega t) (\overbrace{\vec{j} \cdot \vec{i}}^{=0}) +r^2\omega \cos (\omega t) \sin (\omega t) (\overbrace{\vec{j} \cdot \vec{j}}^{=1}) \\&=-r^2\omega \sin (\omega t) \cos (\omega t) +r^2\omega \cos (\omega t) \sin (\omega t) \\&=0 \end{align}

as required.

(iv)

\begin{align}\vec{a}_c &= \frac{d\vec{v}}{dt}\\&=r \omega[\omega \cos (\omega t) \vec{i}] +r \omega[-\omega \sin (\omega t) \vec{j}]\\&=-\omega^2 [-r\cos (\omega t) \vec{i} +r\sin (\omega t) \vec{j}]\\&=-\omega^2 \vec{s}\end{align}

where we have reinserted equation $$(1)$$.

Thus, as $$\vec{a}_c$$ is a negative multiple of $$\vec{s}$$, and as $$\vec{s}$$ points away from $$O$$ towards the car, we have proven that the centripetal acceleration vector points from the car towards $$O$$.

(v)

\begin{align} \vec{f} = \mu R\vec{i} && \vec{W}=-mg\vec{j} && \vec{R}=R\vec{j}\end{align}

and

\begin{align} \vec{a}_c = \frac{v^2}{r}\vec{i} \end{align}

Newton’s 2nd law gives:

$\mu R\vec{i}-mg\vec{j}+R\vec{j} = \frac{mv^2}{r}\vec{i}$

Dotting with $$\vec{j}$$:

\begin{align}-mg+R=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}R=mg\end{align}

Dotting with $$\vec{i}$$:

\begin{align}\mu R=\frac{mv^2}{r}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\mu (mg)=\frac{mv^2}{r}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\mu g =\frac{v^2}{r}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v=\pm\sqrt{\mu rg}\end{align}

The magnitude of the velocity is $$\sqrt{\mu rg}$$ and its direction is tangent to the circle (as was proven in part (iii)).

(vi)

$v=\sqrt{\mu rg}$

\begin{align}
\begin{gathered}[t]
v\\ \, \\ \Downarrow\\ \, \\\mbox{m/s}
\end{gathered}&& \begin{gathered}[t]
\sqrt{\mu rg}\\ \, \\ \Downarrow\\ \, \\\sqrt{(\mbox{m})(\mbox{m/s}^2)}\\=\sqrt{(\mbox{m/s})^2}\\=\mbox{m/s}\end{gathered}\end{align}

As both terms have the same units, this equation is indeed dimensionally correct.

(vii) Yes. Most roundabouts are close to being circular in shape and most drivers do travel at a constant linear speed when in motion at such a roundabout. Any tangential acceleration will be negligible compared to the centripetal acceleration and it is therefore acceptable that it has been ignored in this model.

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Question 7

(a) A bungee jumper of mass $$75\mbox{ kg}$$ jumps from a height of $$35\mbox{ m}$$ above water. The jumper is tied to an elastic rope of natural length $$12\mbox{ m}$$ and elastic constant $$100\mbox{ N m}^{-1}$$.

(i) Derive an expression for the work done when a spring of elastic constant $$k\mbox{ N m}^{-1}$$ is stretched by $$x\mbox{ m}$$.

(ii) The motion of the bungee jumper may be modelled using the principle of conservation of energy. Using this model, calculate the distance between the water and the bungee jumper when the bungee jumper is at the lowest point of their motion.

(i) $$W=\dfrac{1}{2}kx^2$$

(ii) $$0.47\mbox{ m}$$

(i)

\begin{align}W &= \int_a^b F(s)\,ds\\&=\int_0^x ks\,ds\\&=k\int_0^x s\,ds\\&=k\left(\left.\frac{s^2}{2}\right|_0^x\right)\\&=k\left(\frac{x^2}{2}-\frac{0^2}{2}\right)\\&=\frac{1}{2}kx^2\end{align}

(ii)

Location A

\begin{align}E_A &= U_A + K_A\\&=mgh_A+\frac{1}{2}mv_A^2\\&=(75)(9.8)(35)+\frac{1}{2}(75)(0^2)\\&=25{,}725\mbox{ J}\end{align}

Location C

\begin{align}E_C &= U_C + K_C + \frac{1}{2}kx^2\\&=mgh_C+\frac{1}{2}mv_C^2+\frac{1}{2}kx^2\\&=(75)(9.8)(35-12-x)+\frac{1}{2}(75)(0^2)+\frac{1}{2}(100)x^2\\&=16{,}905-735x+50x^2\end{align}

Conservation of energy

\begin{align}E_A=E_C\end{align}

\begin{align}\downarrow\end{align}

\begin{align}25{,}725=16{,}905-735x+50x^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}50x^2-735x-8{,}820=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=\frac{-(-735)\pm\sqrt{(-735)^2-4(50)(-8{,}820)}}{2(50)}\\&=\frac{735\pm\sqrt{2{,}304{,}225}}{100}\\&\approx -7.83\mbox{ m or } 22.53\mbox{ m}\end{align}

Negative root ignored. Therefore, the distance between the water and the bungee jumper when the bungee jumper is at the lowest point of their motion is approximately $$35-12-22.53=0.47\mbox{ m}$$.

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(b) A small smooth moveable disk $$D$$, of mass $$0.2\mbox{ kg}$$, rests on a light inextensible string. One end of the string is connected to block $$B$$, of mass $$4\mbox{ kg}$$, which rests on a rough plane inclined at $$30^{\circ}$$ to the horizontal.
The other end of the string is connected vertically to a fixed point.

The coefficient of friction between block $$B$$ and the inclined plane is $$\dfrac{1}{10}$$.

When the system is released from rest, $$D$$ moves upwards with acceleration $$a$$.

The tension in the string is $$T$$.

(i) Show, on separate diagrams, the forces acting on block $$B$$ and disk $$D$$ while they are moving.

(ii) Explain why the acceleration of $$B$$ is $$2a$$.

(iii) Calculate $$a$$ and $$T$$.

(i)

Block

Disk

(iii) $$a=1.88\mbox{ m/s}^2$$ and $$T=1.17\mbox{ N}$$

(i)

Block

Disk

(ii) Twice as much string is needed to move the pulley the same distance as the string is on both sides of the pulley.

Therefore, the block moves twice as far as the pulley in the same time.

As $$s\propto a$$ (according to $$s=\frac{1}{2}at^2$$), it is therefore also the case that the block’s acceleration is twice as large as the pulley.

(iii)

Block

$W_{1x}\vec{i} + W_{1y}\vec{j}+ T\vec{i}+ f\vec{i}+ R_1\vec{j} = 4(-2a\vec{i})$

We know that $$W_{1x}=-4g\sin30^{\circ}= -19.6$$, $$W_{1y}=-4g\cos30^{\circ}\approx-33.95$$ and $$f=\mu R_1$$. Therefore

$-19.6\vec{i} -33.95\vec{j}+ T\vec{i}+ \mu R_1\vec{i}+ R_1\vec{j} = -8a\vec{i}$

Dotting with $$\vec{j}$$:

$-33.95+ R_1 = 0$

\begin{align}\downarrow\end{align}

\begin{align}R_1=33.95\end{align}

Dotting with $$\vec{i}$$:

\begin{align}-19.6 + T+ \mu R_1  = -8a\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-19.6 + T+ \frac{1}{10}(33.95)  = -8a\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-19.6 + T+3.395 = -8a\end{align}

\begin{align}\downarrow\end{align}

\begin{align}T-16.205 = -8a\end{align}

Disk

$2T\vec{j} + W_{2}\vec{j} = 0.2(a\vec{j})$

We know that $$W_{2}=-0.2g= -1.96$$. Therefore

$2T\vec{j} -1.96\vec{j} = 0.2a\vec{j}$

Dotting with $$\vec{j}$$:

$2T -1.96 = 0.2a$

We therefore have the following two equations with two unknowns:

\begin{align}T-16.205 = -8a\end{align}

\begin{align}2T -1.96 = 0.2a\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2T-32.41 = -16a\end{align}

\begin{align}2T -1.96 = 0.2a\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-30.45=-16.2a\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a&=\frac{30.45}{16.2}\\&\approx 1.88\mbox{ m/s}^2\end{align}

Inserting this into $$T-16.205 = -8a$$:

\begin{align}T&=16.205-8a\\&\approx 16.205-8(1.88)\\&\approx 1.17\mbox{ N}\end{align}

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Question 8

A group of scientists are investigating the population, $$P$$ , of rabbits on a certain island. They estimate that there are $$8000$$ rabbits on the island and that the population is growing at a constant rate of $$3\%$$ per year.

The scientists plan to remove a number of rabbits from the island every year, to help populate another habitat. They develop mathematical models to predict how $$P$$ will change if $$B$$ rabbits are removed from the island every year.

The first model which the scientists develop uses a difference equation to express the population of rabbits in year $$n+1$$ in terms of the population in year $$n$$.

The difference equation is:

\begin{align}P_{n+1}=1.03P_n-B\end{align}

where $$n\geq0$$, $$n\in\mathbb{Z}$$ and $$P_0=8000$$.

(i) Solve this difference equation to find an expression for $$P_n$$ in terms of $$n$$ and $$B$$.

$\,$

The second model which the scientists develop uses a differential equation to express the rate of change of $$P$$ with respect to $$n$$, time measured in years.

The differential equation is:

\begin{align}\frac{dP}{dn}=0.03P-B\end{align}

where $$n\geq0$$, $$n\in\mathbb{R}$$ and $$P(0)=8000$$.

(ii) Solve this differential equation to find an expression for $$P$$ in terms of $$n$$ and $$B$$.

The scientists want to know what each model predicts the rabbit population on the island will be after $$50$$ years, if $$200$$ rabbits are removed each year.

(iii) Calculate $$P_{50}$$ using the first model and $$P(50)$$ using the second model, when $$B=200$$.

(iv) Each of these models makes a different assumption about the removal of the rabbits from the island. What are the two different assumptions?

(v) The scientists want to know what value of $$B$$ should be chosen so as to keep the rabbit population on the island constant. Calculate this value of $$B$$ using either model.

(i) $$P_n=8000(1.03^n)+\dfrac{B}{0.03}(1-1.03^n)$$

(ii) $$P(n)=\dfrac{1}{0.03}[(240-B)e^{0.03n}+B]$$

(iii) $$P_{50}\approx 12{,}512$$ and $$P(50)\approx 12{,}642$$

(iv) The difference equation assumes that a total of $$B$$ rabbits are removed in one instant each year whereas the differential equation assumes that a total of $$B$$ rabbits are removed gradually throughout each year.

(v) $$B=240$$

(i)

\begin{align}P_{n+1}=1.03P_n-B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P_n&=P_0(1.03^n)+\frac{-B(1-1.03^n)}{1-1.03}\\&=8000(1.03^n)+\frac{B}{0.03}(1-1.03^n)\end{align}

(ii)

\begin{align}\frac{dP}{dn}=0.03P-B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{dP}{0.03P-B}=dn\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\int\frac{dP}{0.03P-B}=\int dn\end{align}

\begin{align}u=0.03P-B && \frac{du}{dP}=0.03\end{align}

With $$u=0.03P-B$$ and $$du=0.03\,dP$$, the original integral becomes

\begin{align}\frac{1}{0.03}\int\frac{1}{u}\,du=\int dn\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{0.03}(\ln|u|+C)=n+D\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{0.03}\ln|0.03P-B|=n+E\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.03P-B=Fe^{0.03n}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(n)=\frac{1}{0.03}(Fe^{0.03n}+B)\end{align}

As we are told that:

\begin{align}P(0)&=\frac{1}{0.03}(Fe^{0.03(0)}+B)\\&=\frac{1}{0.03}(F+B)\end{align}

is equal to $$8000$$, we obtain

\begin{align}\frac{1}{0.03}(F+B)=8000\end{align}

\begin{align}\downarrow\end{align}

\begin{align}F=240-B\end{align}

and therefore

\begin{align}P(n)=\frac{1}{0.03}[(240-B)e^{0.03n}+B]\end{align}

(iii)

\begin{align}P_n=8000(1.03^n)+\frac{200}{0.03}(1-1.03^n)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P_{50}&=8000(1.03^{50})+\frac{200}{0.03}(1-1.03^{50})\\&\approx 12{,}512\end{align}

and

\begin{align}P(n)=\frac{1}{0.03}[40e^{0.03n}+200]\end{align}

\begin{align}\downarrow\end{align}

\begin{align}P(50)&=\frac{1}{0.03}[40e^{0.03(50)}+200]\\&\approx 12{,}642\end{align}

(iv) The difference equation assumes that a total of $$B$$ rabbits are removed in one instant each year whereas the differential equation assumes that a total of $$B$$ rabbits are removed gradually throughout each year.

(v)

\begin{align}P(n)=\frac{1}{0.03}[(240-B)e^{0.03n}+B]\end{align}

If $$B=240$$, then

\begin{align}P(n)&=\frac{1}{0.03}[(240-240)e^{0.03n}+240]\\&=\frac{240}{0.03}\\&=8000\end{align}

which is a constant.

(You can obtain a similar result by instead using the difference equation if you prefer.)

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