Trapezium rule applied to real-world data

6. \begin{figure}[h] \end{figure} A river is being studied.
At one particular place, the river is 15 m wide.
The depth, \(y\) metres, of the river is measured at a point \(x\) metres from one side of the river. Figure 1 shows a plot of the cross-section of the river and the coordinate values \(( x , y )\)

1. Use the trapezium rule with all the \(y\) values given in Figure 1 to estimate the cross-sectional area of the river. The water in the river is modelled as flowing at a constant speed of \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) across the whole of the cross-section.
2. Use the model and the answer to part (a) to estimate the volume of water flowing through this section of the river each minute, giving your answer in \(\mathrm { m } ^ { 3 }\) to 2 significant figures. Assuming the model,
3. state, giving a reason for your answer, whether your answer for part (b) is an overestimate or an underestimate of the true volume of water flowing through this section of the river each minute.

6. A river, running between parallel banks, is 20 m wide. The depth, \(y\) metres, of the river measured at a point \(x\) metres from one bank is given by the formula $$y = \frac { 1 } { 10 } x \sqrt { } ( 20 - x ) , \quad 0 \leqslant x \leqslant 20$$

1. Complete the table below, giving values of \(y\) to 3 decimal places.

   \(x\)	0	4	8	12	16	20
   \(y\)	0		2.771			0

2. Use the trapezium rule with all the values in the table to estimate the cross-sectional area of the river. Given that the cross-sectional area is constant and that the river is flowing uniformly at \(2 \mathrm {~ms} ^ { - 1 }\),
3. estimate, in \(\mathrm { m } ^ { 3 }\), the volume of water flowing per minute, giving your answer to 3 significant figures.

8. \begin{figure}[h] \end{figure} Figure 4 is a graph showing the velocity of a sprinter during a 100 m race.
The sprinter's velocity during the race, \(v \mathrm {~ms} ^ { - 1 }\), is modelled by the equation $$v = 12 - \mathrm { e } ^ { t - 10 } - 12 \mathrm { e } ^ { - 0.75 t } \quad t \geqslant 0$$ where \(t\) seconds is the time after the sprinter begins to run. According to the model,

1. find, using calculus, the sprinter's maximum velocity during the race. Given that the sprinter runs 100 m in \(T\) seconds, such that $$\int _ { 0 } ^ { T } v \mathrm {~d} t = 100$$
2. show that \(T\) is a solution of the equation $$T = \frac { 1 } { 12 } \left( 116 - 16 \mathrm { e } ^ { - 0.75 T } + \mathrm { e } ^ { T - 10 } - \mathrm { e } ^ { - 10 } \right)$$ The iteration formula $$T _ { n + 1 } = \frac { 1 } { 12 } \left( 116 - 16 \mathrm { e } ^ { - 0.75 T _ { n } } + \mathrm { e } ^ { T _ { n } - 10 } - \mathrm { e } ^ { - 10 } \right)$$ is used to find an approximate value for \(T\) Using this iteration formula with \(T _ { 1 } = 10\)
3. find, to 4 decimal places,
   1. the value of \(T _ { 2 }\)
   2. the time taken by the sprinter to run the race, according to the model.

8. \begin{figure}[h] \end{figure} Figure 4 is a graph showing the path of a golf ball after the ball has been hit until it first hits the ground. The vertical height, \(h\) metres, of the ball above the ground has been plotted against the horizontal distance travelled, \(x\) metres, measured from where the ball was hit. The ball travels a horizontal distance of \(d\) metres before it first hits the ground.
The ball is modelled as a particle travelling in a vertical plane above horizontal ground.
The path of the ball is modelled by the equation $$h = 1.5 x - 0.5 x \mathrm { e } ^ { 0.02 x } \quad 0 \leqslant x \leqslant d$$ \section*{Use the model to answer parts (a), (b) and (c).}

1. Find the value of \(d\), giving your answer to 2 decimal places.
   (Solutions relying entirely on calculator technology are not acceptable.)
2. Show that the maximum value of \(h\) occurs when $$x = 50 \ln \left( \frac { 150 } { x + 50 } \right)$$ Using the iteration formula $$x _ { n + 1 } = 50 \ln \left( \frac { 150 } { x _ { n } + 50 } \right) \quad \text { with } x _ { 1 } = 30$$
3. 1. find the value of \(x _ { 2 }\) to 2 decimal places,
   2. find, by repeated iteration, the horizontal distance travelled by the golf ball before it reaches its maximum height. Give your answer to 2 decimal places. \includegraphics[max width=\textwidth, alt={}, center]{5a695b86-1660-4c06-ac96-4cdb07af9a2e-26_2270_56_309_1981}

10 Fig. 10 shows the speed of a car, in metres per second, during one minute, measured at 10-second intervals. \begin{figure}[h] \end{figure} The measured speeds are shown below.

1. Use the trapezium rule with 6 strips to find an estimate of the area of the region bounded by the curve, the line \(t = 60\) and the axes. [This area represents the distance travelled by the car.]
2. Explain why your calculation in part (i) gives an overestimate for this area. Use appropriate rectangles to calculate an underestimate for this area. The speed of the car may be modelled by \(v = 28 - t + 0.015 t ^ { 2 }\).
3. Show that the difference between the value given by the model when \(t = 10\) and the measured value is less than \(3 \%\) of the measured value.
4. According to this model, the distance travelled by the car is $$\int _ { 0 } ^ { 60 } \left( 28 - t + 0.015 t ^ { 2 } \right) \mathrm { d } t$$ Find this distance.

1 Oskar is designing a building. Fig. 12 shows his design for the end wall and the curve of the roof. The units for \(x\) and \(y\) are metres. \begin{figure}[h] \end{figure}

1. Use the trapezium rule with 5 strips to estimate the area of the end wall of the building.
2. Oskar now uses the equation \(y = - 0.001 x ^ { 3 } - 0.025 x ^ { 2 } + 0.6 x + 9\), for \(0 \leqslant x \leqslant 15\), to model the curve of the roof.
   (A) Calculate the difference between the height of the roof when \(x = 12\) given by this model and the data shown in Fig. 12.
   (B) Use integration to find the area of the end wall given by this model.

2 Fig. 7 shows a curve and the coordinates of some points on it. \begin{figure}[h] \end{figure} Use the trapezium rule with 6 strips to estimate the area of the region bounded by the curve and the positive \(x\) - and \(y\)-axes.

3 A farmer digs ditches for flood relief. He experiments with different cross-sections. Assume that the surface of the ground is horizontal.

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_432_640_410_745} \captionsetup{labelformat=empty} \caption{Fig. 9.1}
   \end{figure} Fig. 9.1 shows the cross-section of one ditch, with measurements in metres. The width of the ditch is 1.2 m and Fig. 9.1 shows the depth every 0.2 m across the ditch. Use the trapezium rule with six intervals to estimate the area of cross-section. Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.
2. Another ditch is 0.9 m wide, with cross-section as shown in Fig. 9.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-3_579_813_1336_656} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} With \(x\) - and \(y\)-axes as shown in Fig. 9.2, the curve of the ditch may be modelled closely by \(y = 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x\).
   (A) The actual ditch is 0.6 m deep when \(x = 0.2\). Calculate the difference between the depth given by the model and the true depth for this value of \(x\).
   (B) Find \(\int \left( 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x \right) \mathrm { d } x\). Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.

1 Fig. 10 shows the speed of a car, in metres per second, during one minute, measured at 10 -second intervals. \begin{figure}[h] \end{figure} The measured speeds are shown below.

1. Use the trapezium rule with 6 strips to find an estimate of the area of the region bounded by the curve, the line \(t = 60\) and the axes. [This area represents the distance travelled by the car.]
2. Explain why your calculation in part (i) gives an overestimate for this area. Use appropriate rectangles to calculate an underestimate for this area. The speed of the car may be modelled by \(v = 28 - t + 0.015 t ^ { 2 }\).
3. Show that the difference between the value given by the model when \(t = 10\) and the measured value is less than \(3 \%\) of the measured value.
4. According to this model, the distance travelled by the car is $$\int _ { 0 } ^ { 60 } \left( 28 \quad t + 0.015 t ^ { 2 } \right) \mathrm { d } t$$ Find this distance.

2 At a place where a river is 7.5 m wide, its depth is measured every 1.5 m across the river. The table shows the results. Use the trapezium rule with 5 strips to estimate the area of cross-section of the river.

3 Fig. 11 shows the cross-section of a school hall, with measurements of the height in metres taken at 1.5 m intervals from O . \begin{figure}[h] \end{figure}

1. Use the trapezium rule with 8 strips to calculate an estimate of the area of the cross-section.
2. Use 8 rectangles to calculate a lower bound for the area of the cross-section. The curve of the roof may be modelled by \(y = - 0.013 x ^ { 3 } + 0.16 x ^ { 2 } - 0.082 x + 2.4\), where \(x\) metres is the horizontal distance from O across the hall, and \(y\) metres is the height.
3. Use integration to find the area of the cross-section according to this model.
4. Comment on the accuracy of this model for the height of the hall when \(x = 7.5\).

4 Fig. 2 shows the coordinates at certain points on a curve. \begin{figure}[h] \end{figure} Use the trapezium rule with 6 strips to calculate an estimate of the area of the region bounded by this curve and the axes.

3

1. A tunnel is 100 m long. Its cross-section, shown in Fig. 9.1, is modelled by the curve $$y = \frac { 1 } { 4 } \left( 10 x - x ^ { 2 } \right) ,$$ where \(x\) and \(y\) are horizontal and vertical distances in metres. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_512_819_493_700} \captionsetup{labelformat=empty} \caption{Figure 9.1}
   \end{figure} Using this model,
   (A) find the greatest height of the tunnel,
   (B) explain why \(100 \int _ { 0 } ^ { 10 } y \mathrm {~d} x\) gives the volume, in cubic metres, of earth removed to make the tunnel. Calculate this volume.
   [0pt] [5]
2. The roof of the tunnel is re-shaped to allow for larger vehicles. Fig. 9.2 shows the new crosssection. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{1a6d059d-8ab8-41e0-8bf3-54e248f820e4-3_506_942_1703_629} \captionsetup{labelformat=empty} \caption{Not to scale}
   \end{figure} Fig. 9.2 Use the trapezium rule with 5 strips to estimate the new cross-sectional area.
   Hence estimate the volume of earth removed when the tunnel is re-shaped.

2 \begin{figure}[h] \end{figure}

1. Jean is designing a model aeroplane. Fig. 9.1 shows her first sketch of the wing's cross-section. Calculate angle A and the area of the cross-section.
2. Jean then modifies her design for the wing. Fig. 9.2 shows the new cross-section, with 1 unit for each of \(x\) and \(y\) representing one centimetre. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4dcf71fc-2585-4247-a21d-8b14f11ce0d0-1_415_1662_1081_240} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} Here are some of the coordinates that Jean used to draw the new cross-section.

   Upper surface		Lower surface
   \(x\)	\(y\)	\(x\)	\(y\)
   0	0	0	0
   4	1.45	4	- 0.85
   8	1.56	8	- 0.76
   12	1.27	12	- 0.55
   16	1.04	16	- 0.30
   20	0	20	0

   Use the trapezium rule with trapezia of width 4 cm to calculate an estimate of the area of this cross-section.

2 Fig. 2 shows the coordinates at certain points on a curve. \begin{figure}[h] \end{figure} Use the trapezium rule with 6 strips to calculate an estimate of the area of the region bounded by this curve and the axes.

3 At a place where a river is 7.5 m wide, its depth is measured every 1.5 m across the river. The table shows the results. Use the trapezium rule with 5 strips to estimate the area of cross-section of the river.

9 A farmer digs ditches for flood relief. He experiments with different cross-sections. Assume that the surface of the ground is horizontal.

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-4_437_640_470_715} \captionsetup{labelformat=empty} \caption{Fig. 9.1}
   \end{figure} Fig. 9.1 shows the cross-section of one ditch, with measurements in metres. The width of the ditch is 1.2 m and Fig. 9.1 shows the depth every 0.2 m across the ditch. Use the trapezium rule with six intervals to estimate the area of cross-section. Hence estimate the volume of water that can be contained in a 50-metre length of this ditch.
2. Another ditch is 0.9 m wide, with cross-section as shown in Fig. 9.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-4_574_808_1402_632} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} With \(x\) - and \(y\)-axes as shown in Fig. 9.2, the curve of the ditch may be modelled closely by \(y = 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x\).
   (A) The actual ditch is 0.6 m deep when \(x = 0.2\). Calculate the difference between the depth given by the model and the true depth for this value of \(x\).
   (B) Find \(\int \left( 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x \right) \mathrm { d } x\). Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.

9 \begin{figure}[h] \end{figure}

1. Jean is designing a model aeroplane. Fig. 9.1 shows her first sketch of the wing's cross-section. Calculate angle A and the area of the cross-section.
2. Jean then modifies her design for the wing. Fig. 9.2 shows the new cross-section, with 1 unit for each of \(x\) and \(y\) representing one centimetre. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{5c7ac296-a911-451b-ad18-5ade3ac23e74-3_431_1682_970_194} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} Here are some of the coordinates that Jean used to draw the new cross-section.

   Upper surface		Lower surface
   \(x\)	\(y\)	\(x\)	\(y\)
   0	0	0	0
   4	1.45	4	-0.85
   8	1.56	8	-0.76
   12	1.27	12	-0.55
   16	1.04	16	-0.30
   20	0	20	0

   Use the trapezium rule with trapezia of width 4 cm to calculate an estimate of the area of this cross-section.

1. The speed of a small jet aircraft was measured every 5 seconds, starting from the time it turned onto a runway, until the time when it left the ground.

The results are given in the table below with the time in seconds and the speed in \(\mathrm { ms } ^ { - 1 }\). Using all of this information,

1. estimate the length of runway used by the jet to take off. Given that the jet accelerated smoothly in these 25 seconds,
2. explain whether your answer to part (a) is an underestimate or an overestimate of the length of runway used by the jet to take off.

1. A measure of the effective voltage, \(M\) volts, in an electrical circuit is given by

$$M ^ { 2 } = \int _ { 0 } ^ { 1 } V ^ { 2 } \mathrm {~d} t$$ where \(V\) volts is the voltage at time \(t\) seconds. Pairs of values of \(V\) and \(t\) are given in the following table. Use the trapezium rule with five values of \(V ^ { 2 }\) to estimate the value of \(M\).
(6)

7 Sam decided to go on a high-protein diet. Sam's mass in \(\mathrm { kg } , M\), after \(t\) days of following the diet is recorded in Fig. 7.1. \begin{table}[h] \end{table} A difference table for the data is shown in Fig. 7.2. \begin{table}[h] \end{table}

1. Complete the copy of the difference table in the Printed Answer Booklet. Sam's doctor uses these data to construct a cubic interpolating polynomial to model Sam's mass at time \(t\) days after starting the diet.
2. Find the model in the form \(\mathrm { M } = \mathrm { at } ^ { 3 } + \mathrm { bt } ^ { 2 } + \mathrm { ct } + \mathrm { d }\), where \(a , b , c\) and \(d\) are constants to be determined. Subsequently it is found that when \(\mathrm { t } = 40 , \mathrm { M } = 78.7\) and when \(\mathrm { t } = 50 , \mathrm { M } = 80.05\).
3. Determine whether the model is a good fit for these data.
4. By completing the extended copy of Fig. 7.2 in the Printed Answer Booklet, explain why a quartic model may be more appropriate for the data.
5. Refine the doctor's model to include a quartic term.
6. Explain whether the new model for Sam's mass is likely to be appropriate over a longer period of time.

1 You are given that \(\left( x _ { 1 } , y _ { 1 } \right) = ( 0.9,2.3 )\) and \(\left( x _ { 2 } , y _ { 2 } \right) = ( 1.1,2.7 )\).
The values of \(x _ { 1 }\) and \(x _ { 2 }\) have been rounded to \(\mathbf { 1 }\) decimal place.

1. Determine the range of possible values of \(x _ { 2 } - x _ { 1 }\). The values of \(y _ { 1 }\) and \(y _ { 2 }\) have been chopped to \(\mathbf { 1 }\) decimal place.
2. Determine the range of possible values of \(y _ { 2 } - y _ { 1 }\). You are given that \(m = \frac { y _ { 2 } - y _ { 1 } } { x _ { 2 } - x _ { 1 } }\).
3. Determine the range of possible values of \(m\).
4. Explain why your answer to part (c) is much larger than your answer to part (a) and your answer to part (b).

2 A car tyre has a slow puncture. Initially the tyre is inflated to a pressure of 34.5 psi . The pressure is checked after 3 days and then again after 5 days. The time \(t\) in days and the pressure, \(P\) psi, are shown in the table below. You are given that the pressure in a car tyre is measured in pounds per square inch (psi). The owner of the car believes the relationship between \(P\) and \(t\) may be modelled by a polynomial.

1. Explain why it is not possible to use Newton's forward difference interpolation method for these data.
2. Use Lagrange's form of the interpolating polynomial to find an interpolating polynomial of degree 2 for these data. The car owner uses the polynomial found in part (b) to model the relationship between \(P\) and \(t\).
   Subsequently it is found that when \(t = 6 , P = 26.0\) and when \(t = 10 , P = 24.4\).
3. Determine whether the owner's model is a good fit for these data.
4. Explain why the model would not be suitable in the long term.

3 The diagram shows the graph of \(y = f ( x )\) for values of \(x\) from 1 to 3.5. \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-03_945_1248_312_244} The table shows some values of \(x\) and the associated values of \(y\).

1. Use the forward difference method to calculate an approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\).
2. Use the central difference method to calculate an approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\).
3. On the copy of the diagram in the Printed Answer Booklet, show how the central difference method gives the approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\) which was found in part (b).
4. Explain whether your answer to part (a) or your answer to part (b) is likely to give a better approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\).

6

1. 1. Calculate the relative error when \(\pi\) is chopped to \(\mathbf { 2 }\) decimal places in approximating $$\pi ^ { 2 } + 2 .$$
   2. Without doing any calculation, explain whether the relative error would be the same when \(\pi\) is chopped to 2 decimal places when approximating \(( \pi + 2 ) ^ { 2 }\). The table shows some spreadsheet output. The values of \(x\) in column A are exact.

      	A	B	C
      1	\(x\)	\(10 ^ { x }\)	\(\log _ { 10 } 10 ^ { x }\)
      2	\(1 \mathrm { E } - 12\)	1	\(1.00001 \mathrm { E } - 12\)
      3	\(1 \mathrm { E } - 11\)	1	\(9.99998 \mathrm { E } - 12\)

      The formula in cell B2 is \(= 10 ^ { \wedge } \mathrm { A } 2\).
      This has been copied down to cell B3.
      The formula in cell C2 is \(\quad =\) LOG(B2) .
      This formula has been copied down to cell C3.
2. 1. Write the value displayed in cell C 2 in standard mathematical notation.
   2. Explain why the values in cells C 2 and C 3 are neither zero nor the same as the values in cells A2 and A3 respectively.