Verification of solutions

10. The amount of an antibiotic in the bloodstream, from a given dose, is modelled by the formula $$x = D \mathrm { e } ^ { - 0.2 t }$$ where \(x\) is the amount of the antibiotic in the bloodstream in milligrams, \(D\) is the dose given in milligrams and \(t\) is the time in hours after the antibiotic has been given. A first dose of 15 mg of the antibiotic is given.

1. Use the model to find the amount of the antibiotic in the bloodstream 4 hours after the dose is given. Give your answer in mg to 3 decimal places. A second dose of 15 mg is given 5 hours after the first dose has been given. Using the same model for the second dose,
2. show that the total amount of the antibiotic in the bloodstream 2 hours after the second dose is given is 13.754 mg to 3 decimal places. No more doses of the antibiotic are given. At time \(T\) hours after the second dose is given, the total amount of the antibiotic in the bloodstream is 7.5 mg .
3. Show that \(T = a \ln \left( b + \frac { b } { \mathrm { e } } \right)\), where \(a\) and \(b\) are integers to be determined.
   

7 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.

1. Suppose that the number of cases, \(P\) thousand, after time \(t\) months is modelled by the equation \(P = \frac { 2 } { 2 - \sin t }\). Thus, when \(t = 0 , P = 1\).
   1. By considering the greatest and least values of \(\sin t\), write down the greatest and least values of \(P\) predicted by this model.
   2. Verify that \(P\) satisfies the differential equation \(\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t\).
2. An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, \(P = 1\) when \(t = 0\).
   1. Express \(\frac { 1 } { P ( 2 P - 1 ) }\) in partial fractions.
   2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P - 1 } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give \(P = \frac { 1 } { 2 - \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }\).
   3. Find the greatest and least values of \(P\) predicted by this model. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{9296c786-a42a-4aa5-b326-39adbb544cbc-05_609_622_301_719} \captionsetup{labelformat=empty} \caption{Fig. 8}
      \end{figure} In a theme park ride, a capsule C moves in a vertical plane (see Fig. 8). With respect to the axes shown, the path of C is modelled by the parametric equations $$x = 10 \cos \theta + 5 \cos 2 \theta , \quad y = 10 \sin \theta + 5 \sin 2 \theta , \quad ( 0 \leqslant \theta < 2 \pi )$$ where \(x\) and \(y\) are in metres.
      1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { \cos \theta + \cos 2 \theta } { \sin \theta + \sin 2 \theta }\). Verify that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(\theta = \frac { 1 } { 3 } \pi\). Hence find the exact coordinates of the highest point A on the path of C .
      2. Express \(x ^ { 2 } + y ^ { 2 }\) in terms of \(\theta\). Hence show that $$x ^ { 2 } + y ^ { 2 } = 125 + 100 \cos \theta$$
      3. Using this result, or otherwise, find the greatest and least distances of C from O . You are given that, at the point B on the path vertically above O , $$2 \cos ^ { 2 } \theta + 2 \cos \theta - 1 = 0$$
      4. Using this result, and the result in part (ii), find the distance OB. Give your answer to 3 significant figures. \section*{ADVANCED GCE UNIT MATHEMATICS (MEI)} Applications of Advanced Mathematics (C4) \section*{Paper B: Comprehension} \section*{THURSDAY 14 JUNE 2007} Afternoon
         Time: Up to 1 hour
         Additional materials:
         Rough paper
         MEI Examination Formulae and Tables (MF2) Candidate
         Name □
         Centre
         Number sufficient detail of the working to indicate that a correct method is being used. 1 This basic cycloid has parametric equations $$x = a \theta - a \sin \theta , \quad y = a - a \cos \theta$$
         \includegraphics[max width=\textwidth, alt={}]{9296c786-a42a-4aa5-b326-39adbb544cbc-10_307_1138_445_411}
         Find the coordinates of the points M and N , stating the value of \(\theta\) at each of them. Point M Point N 2 A sea wave has parametric equations (in suitable units) $$x = 7 \theta - 0.25 \sin \theta , \quad y = 0.25 \cos \theta$$ Find the wavelength and height of the wave.
         3 The graph below shows the profile of a wave.
         1. Assuming that it has parametric equations of the form given on line 68 , find the values of \(a\) and \(b\).
         2. Investigate whether the ratio of the trough length to the crest length is consistent with this shape. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-11_312_1141_623_415}
         3. \(\_\_\_\_\)
         4. \(\_\_\_\_\) 4 This diagram illustrates two wave shapes \(U\) and \(V\). They have the same wavelength and the same height. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-12_423_1552_356_205} One of the curves is a sine wave, the other is a curtate cycloid.
         5. State which is which, justifying your answer.
         6. \(\_\_\_\_\) The parametric equations for the curves are: $$x = a \theta , \quad y = b \cos \theta ,$$ and $$x = a \theta - b \sin \theta , \quad y = b \cos \theta .$$
         7. Show that the distance marked \(d\) on the diagram is equal to \(b\).
         8. Hence justify the statement in lines 109 to 111: "In such cases, the curtate cycloid and the sine curve with the same wavelength and height are very similar and so the sine curve is also a good model."
         9. \(\_\_\_\_\)
         10. \(\_\_\_\_\) 5 The diagram shows a curtate cycloid with scales given. Show that this curve could not be a scale drawing of the shape of a stable sea wave. \includegraphics[max width=\textwidth, alt={}, center]{9296c786-a42a-4aa5-b326-39adbb544cbc-13_289_1310_397_331}

3 Fig. 8.1 shows an upright cylindrical barrel containing water. The water is leaking out of a hole in the side of the barrel. \begin{figure}[h] \end{figure} The height of the water surface above the hole \(t\) seconds after opening the hole is \(h\) metres, where $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - A \sqrt { h }$$ and where \(A\) is a positive constant. Initially the water surface is 1 metre above the hole.

1. Verify that the solution to this differential equation is $$h = \left( 1 - \frac { 1 } { 2 } A t \right) ^ { 2 } .$$ The water stops leaking when \(h = 0\). This occurs after 20 seconds.
2. Find the value of \(A\), and the time when the height of the water surface above the hole is 0.5 m . Fig. 8.2 shows a similar situation with a different barrel; \(h\) is in metres. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{11d26af4-19d0-4310-a64e-9888285c9980-2_235_455_1425_820} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
   \end{figure} For this barrel, $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - B \frac { \sqrt { h } } { ( 1 + h ) ^ { 2 } } ,$$ where \(B\) is a positive constant. When \(t = 0 , h = 1\).
3. Solve this differential equation, and hence show that $$h ^ { \frac { 1 } { 2 } } \left( 30 + 20 h + 6 h ^ { 2 } \right) = 56 - 15 B t .$$
4. Given that \(h = 0\) when \(t = 20\), find \(B\). Find also the time when the height of the water surface above the hole is 0.5 m .

Fig. 8.1 shows an upright cylindrical barrel containing water. The water is leaking out of a hole in the side of the barrel. \includegraphics{figure_8.1} The height of the water surface above the hole \(t\) seconds after opening the hole is \(h\) metres, where $$\frac{dh}{dt} = -A\sqrt{h}$$ and where \(A\) is a positive constant. Initially the water surface is 1 metre above the hole.

1. Verify that the solution to this differential equation is $$h = \left(1 - \frac{1}{2}At\right)^2.$$ [3]

The water stops leaking when \(h = 0\). This occurs after 20 seconds.

1. Find the value of \(A\), and the time when the height of the water surface above the hole is 0.5 m. [4]

Fig. 8.2 shows a similar situation with a different barrel; \(h\) is in metres. \includegraphics{figure_8.2} For this barrel, $$\frac{dh}{dt} = -B\frac{\sqrt{h}}{(1+h)^2},$$ where \(B\) is a positive constant. When \(t = 0\), \(h = 1\).

1. Solve this differential equation, and hence show that $$h^{\frac{1}{2}}(30 + 20h + 6h^2) = 56 - 15Bt.$$ [7]
2. Given that \(h = 0\) when \(t = 20\), find \(B\). Find also the time when the height of the water surface above the hole is 0.5 m. [4]

A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v\) m s\(^{-1}\). Its terminal (long-term) velocity is 5 m s\(^{-1}\). A model of the particle's motion is proposed. In this model, \(v = 5(1 - e^{-2t})\).

1. Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model. [3]
2. Verify that \(v\) satisfies the differential equation \(\frac{dv}{dt} = 10 - 2v\). [3]

In a second model, \(v\) satisfies the differential equation $$\frac{dv}{dt} = 10 - 0.4v^2.$$ As before, when \(t = 0\), \(v = 0\).

1. Show that this differential equation may be written as $$\frac{10}{(5-v)(5+v)} \frac{dv}{dt} = 4.$$ Using partial fractions, solve this differential equation to show that $$t = \frac{1}{4} \ln\left(\frac{5+v}{5-v}\right).$$ [8] This can be re-arranged to give \(v = \frac{5(1-e^{-4t})}{1+e^{-4t}}\). [You are not required to show this result.]
2. Verify that this model also gives a terminal velocity of 5 m s\(^{-1}\). Calculate the velocity after 0.5 seconds as given by this model. [3]

The velocity of the particle after 0.5 seconds is measured as 3 m s\(^{-1}\).

1. Which of the two models fits the data better? [1]