[Maximum number: 2]

This question uses differential equations to model the maximum velocity of a skydiver in free fall.
In 2012, Felix Baumgartner jumped from a height of 40000 m . He was attempting to travel at the speed of sound, $330 \mathrm{~m} \mathrm{~s}^{-1}$ , whilst free-falling to the Earth.
Before making his attempt, Felix used mathematical models to check how realistic his attempt would be. The simplest model he used suggests that

\frac{\mathrm{d} v}{\mathrm{~d} t}=g

where $v \mathrm{~m} \mathrm{~s}^{-1}$ is Felix's velocity and $g \mathrm{~ms}^{-2}$ is the acceleration due to gravity. The time since he began to free-fall is t seconds and the displacement from his initial position is s metres.

Throughout this question, the direction towards the centre of the Earth is taken to be positive and v is a positive quantity.

When s=0, it is given that Felix jumps with an initial velocity v=10.

(a)

To test the model

\frac{\mathrm{d} v}{\mathrm{~d} t}=g

Felix conducted a trial jump from a lower height, and data for v against t was found.

[ 2 ]

(i)

If the model is correct, describe the shape of the graph of v against t.

Felix's data are plotted on the following graph.

[ 2 ]

[Maximum number: 12]

In this question you will use a historic method of calculating the cost of a barrel of wine to determine which shape of barrel gives the best value for money.
In Austria in the 17th century, one method for measuring the volume of a barrel of wine, and hence determining its cost, was by inserting a straight stick into a hole in the side, as shown in the following diagram, and measuring the length SD . The longer the length, the greater the cost to the customer.

Let SD be d metres and the cost be C gulden (the local currency at the time). When the length of SD was 0.5 metres, the cost was 0.80 gulden.

(a)

Show that $V=\frac{\pi}{2}\left(d^{2} h-h^{3}\right)$ .

The remainder of this question considers the shape of barrel that gives the best value when d=0.6.

[ 2 ]

(b)

(i)

Use the formula from part (d) to find the volume of this barrel when h=0.4.

[ 2 ]

(ii)

Use differentiation to show that $h=\sqrt{0.12}$ when $\frac{\mathrm{d} V}{\mathrm{~d} h}=0$ .

[ 3 ]

(iii)

Given that this value of h maximizes the volume, find the largest possible volume of this barrel.

Kepler then considered a non-cylindrical barrel whose base and lid are circles with radius 0.2 m and whose length is 0.8 m .

He modelled the curved surface of this barrel by rotating a quadratic curve, ASB , with equation $y=a x^{2}+b x+c, 0 \leq x \leq 0.8$ , about the x-axis. The origin of the coordinate system is at the centre of one of the circular faces as shown in the following diagram. S is at the vertex of the quadratic curve and SD=0.6.

Kepler wished to find out if his barrel would give him more wine than any cylindrical barrel with d=0.6.

The coordinates of A and B are ( 0,0.2 ) and ( 0.8,0.2 ) respectively.

[ 2 ]

(c)

Show that the volume of this barrel is greater than the maximum volume of any cylindrical barrel with d=0.6.

[ 3 ]

[Maximum number: 6]

This question considers the optimal route between two points, separated by several regions where different speeds are possible.
Huw lives in a house, H , and he attends a school, S , where H and S are marked on the following diagram. The school is situated 1.2 km south and 4 km east of Huw's house. There is a boundary [MN], going from west to east, 0.4 km south of his house. The land north of [MN] is a field over which Huw runs at 15 kilometres per hour $\left(\mathrm{km} \mathrm{h}^{-1}\right)$ . The land south of [MN] is rough ground over which Huw walks at $5 \mathrm{~km} \mathrm{~h}^{-1}$ . The two regions are shown in the following diagram.

diagram not to scale

(a)

Huw realizes that his journey time could be reduced by taking a less direct route. He therefore defines a point P on [MN] that is $x \mathrm{~km}$ east of M. Huw decides to run from H to P and then walk from P to S . Let T(x) represent the time, in hours, taken by Huw to complete the journey along this route.

[ 3 ]

(i)

Hence determine the value of x that minimizes T(x).

[ 1 ]

(ii)

Find by how much Huw's journey time is reduced when he takes this optimal route, compared to travelling in a straight line from H to S . Give your answer correct to the nearest minute.

[ 2 ]

(b)

(i)

Determine an expression for the derivative $T^{\prime}(x)$ .

[ 3 ]

(ii)

Hence show that T(x) is minimized when

\frac{x}{\sqrt{0.16+x^{2}}}=\frac{3(4-x)}{\sqrt{0.64+(4-x)^{2}}}

[Maximum number: 6]

Kailash manufactures drink containers in the shape of a cuboid. The container has a square top and a square base of length, $l \mathrm{~cm}$ . Its height, $d \mathrm{~cm}$ , is three times the length of the base.

(a)

Show that $A=2 \pi r^{2}+\frac{750}{r}$ .

(b)

Find $\frac{\mathrm{d} A}{\mathrm{~d} r}$ .
(g) Hence or otherwise

[ 3 ]

(c)

(i)

find the value of r that will minimize A.

(ii)

find the minimum value of A needed for the cylinder.

[ 3 ]

[Maximum number: 3]

Inspectors are investigating the carbon dioxide emissions of a power plant. Let R be the rate, in tonnes per hour, at which carbon dioxide is being emitted and t be the time in hours since the inspection began.
When R is plotted against t, the total amount of carbon dioxide produced is represented by the area between the graph and the horizontal t-axis.
The rate, R, is measured over the course of two hours. The results are shown in the following table.

(a)

Use the trapezoidal rule with an interval width of 0.4 to estimate the total amount of carbon dioxide emitted during these two hours.

The real amount of carbon dioxide emitted during these two hours was 72 tonnes.

[ 3 ]

[Maximum number: 12]

The following diagram shows a model of the side view of a water slide. All lengths are measured in metres.

The curved edge of the slide is modelled by

f(x)=-\frac{1}{4} x^{2}+2 x \text { for } 0 \leq x \leq 4

The remainder of the slide is modelled by

g(x)=\left\{\begin{array}{c} 4, \text { for } 4 \leq x \leq 5 \\ \frac{48}{7}-\frac{4 x}{7}, \text { for } 5 \leq x \leq 12 \end{array}\right.

(a)

Use the trapezoidal rule with an interval width of 1 to calculate the approximate area under the model of the slide in the interval $0 \leq x \leq 4$ .

[ 5 ]

(b)

Find $\int\left(-\frac{1}{4} x^{2}+2 x\right) \mathrm{d} x$ .

[ 3 ]

(c)

Calculate the exact area under the entire model of the slide, for $0 \leq x \leq 12$ .

[ 4 ]

[Maximum number: 26]

The cross-sectional view of a tunnel is shown on the axes below. The line $[\mathrm{AB}]$ represents a vertical wall located at the left side of the tunnel. The height, in metres, of the tunnel above the horizontal ground is modelled by $y=-0.1 x^{3}+0.8 x^{2}, 2 \leq x \leq 8$ , relative to an origin O .

Point A has coordinates ( 2,0 ), point B has coordinates ( 2,2.4 ), and point C has coordinates (8,0).

(a)

(i)

Find $\frac{\mathrm{d} y}{\mathrm{~d} x}$ .

[ 6 ]

(ii)

Hence find the maximum height of the tunnel.

[ 6 ]

(b)

Find the height of the tunnel when

[ 3 ]

(i)

$\quad x=4$ .

(ii)

$\quad x=6$ .

[ 3 ]

(c)

Use the trapezoidal rule, with three intervals, to estimate the cross-sectional area of the tunnel.

[ 3 ]

(d)

(i)

Write down the integral which can be used to find the cross-sectional area of the tunnel.

[ 4 ]

(ii)

Hence find the cross-sectional area of the tunnel.

[ 4 ]

[Maximum number: 12]

A cafe makes x litres of coffee each morning. The cafe's profit each morning, C, measured in dollars, is modelled by the following equation

C=\frac{x}{10}\left(k^{2}-\frac{3}{100} x^{2}\right)

where k is a positive constant.

(a)

Find an expression for $\frac{\mathrm{d} C}{\mathrm{~d} x}$ in terms of k and x.

[ 3 ]

(b)

Hence find the maximum value of C in terms of k. Give your answer in the form $p k^{3}$ , where p is a constant.

The cafe's manager knows that the cafe makes a profit of $ 426 when 20 litres of coffee are made in a morning.

[ 4 ]

(c)

(i)

Find the value of k.

(ii)

Use the model to find how much coffee the cafe should make each morning to maximize its profit.

[ 3 ]

(d)

Determine the maximum amount of coffee the cafe can make that will not result in a loss of money for the morning.

[ 2 ]

[Maximum number: 7]

The graphs of y=6-x and $y=1.5 x^{2}-2.5 x+3$ intersect at (2,4) and (-1,7), as shown in the following diagrams.
In diagram 1, the region enclosed by the lines y=6-x, x=-1, x=2 and the x-axis has been shaded.

Diagram 1

(a)

Calculate the area of the shaded region in diagram 1.

[ 2 ]

(b)

In diagram 2, the region enclosed by the curve $y=1.5 x^{2}-2.5 x+3$ , and the lines x=-1, x=2 and the x-axis has been shaded.
Diagram 2

[ 3 ]

(i)

Write down an integral for the area of the shaded region in diagram 2.

(ii)

Calculate the area of this region.

[ 3 ]

(c)

Hence, determine the area enclosed between y=6-x and $y=1.5 x^{2}-2.5 x+3$ .

[ 2 ]

[Maximum number: 11]

The height of a plant, $h \mathrm{~cm}$ , depends on the time, t days, after it is planted.
For the first 15 days, the rate of change of the height of the plant is modelled as

\frac{\mathrm{d} h}{\mathrm{~d} t}=1.2-0.1 t, \text { where } 0 \leq t \leq 15 .

(a)

(i)

Find the value of $\frac{\mathrm{d} h}{\mathrm{~d} t}$ at t=3.

(ii)

Interpret the meaning of your answer to part (a) (i) in context.

[ 2 ]

(b)

Use $\frac{\mathrm{d} h}{\mathrm{~d} t}$ to find the value of t when the plant reaches its maximum height.

Two days after it is planted, the height of the plant is 6.5 cm .

[ 2 ]

(c)

Find an expression for h in terms of t, for $0 \leq t \leq 15$ .

[ 5 ]

(d)

Hence, find the maximum height of the plant.

A second plant was planted at the same time as the first plant. Its height, $H \mathrm{~cm}$ , is modelled by $H=2(1.2)^{t}$ , where $0 \leq t \leq 15$ .

[ 2 ]