Questions — OCR MEI C4 (332 questions)

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OCR MEI C4 Q7
7 A skydiver drops from a helicopter. Before she opens her parachute, her speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after time \(t\) seconds is modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$ When \(t = 0 , v = 0\).
  1. Find \(v\) in terms of \(t\).
  2. According to this model, what is the speed of the skydiver in the long term? She opens her parachute when her speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Her speed \(t\) seconds after this is \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and is modelled by the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
  3. Express \(\frac { 1 } { ( w - 4 ) ( w + 5 ) }\) in partial fractions.
  4. Using this result, show that \(\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }\).
  5. According to this model, what is the speed of the skydiver in the long term?
OCR MEI C4 Q1
1 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 $$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$ 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.
OCR MEI C4 Q2
2 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Its terminal (long-term) velocity is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A model of the particle's motion is proposed. In this model, \(v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)\).
  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.
  2. Verify that \(v\) satisfies the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v\). In a second model, \(v\) satisfies the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$ As before, when \(t = 0 , v = 0\).
  3. Show that this differential equation may be written as $$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$ Using partial fractions, solve this differential equation to show that $$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$ This can be re-arranged to give \(v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }\). [You are not required to show this result.]
  4. Verify that this model also gives a terminal velocity of \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Calculate the velocity after 0.5 seconds as given by this model. The velocity of the particle after 0.5 seconds is measured as \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  5. Which of the two models fits the data better?
OCR MEI C4 Q3
3 Some years ago an island was populated by red squirrels and there were no grey squirrels. Then grey squirrels were introduced. The population \(x\), in thousands, of red squirrels is modelled by the equation $$x = \frac { a } { 1 + k t }$$ where \(t\) is the time in years, and \(a\) and \(k\) are constants. When \(t = 0 , x = 2.5\).
  1. Show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { k x ^ { 2 } } { a }\).
  2. Given that the initial population of 2.5 thousand red squirrels reduces to 1.6 thousand after one year, calculate \(a\) and \(k\).
  3. What is the long-term population of red squirrels predicted by this model? The population \(y\), in thousands, of grey squirrels is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} t } = 2 y - y ^ { 2 }$$ When \(t = 0 , y = 1\).
  4. Express \(\frac { 1 } { 2 y - y ^ { 2 } }\) in partial fractions.
  5. Hence show by integration that \(\ln \left( \frac { y } { 2 y } \right) = 2 t\). Show that \(y = \frac { 2 } { 1 + \mathrm { e } ^ { - 2 t } }\).
  6. What is the long-term population of grey squirrels predicted by this model?
OCR MEI C4 Q1
3 marks
1 A drug is administered by an intravenous drip. The concentration, \(x\), of the drug in the blood is measured as a fraction of its maximum level. The drug concentration after \(t\) hours is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k \left( 1 + x - 2 x ^ { 2 } \right)$$ where \(0 \leqslant x < 1\), and \(k\) is a positive constant. Initially, \(x = 0\).
  1. Express \(\frac { 1 } { ( 1 + 2 x ) ( 1 - x ) }\) in partial fractions.
    [0pt] [3]
  2. Hence solve the differential equation to show that \(\frac { 1 + 2 x } { 1 - x } = \mathrm { e } ^ { 3 k t }\).
  3. After 1 hour the drug concentration reaches \(75 \%\) of its maximum value and so \(x = 0.75\). Find the value of \(k\), and the time taken for the drug concentration to reach \(90 \%\) of its maximum value.
  4. Rearrange the equation in part (ii) to show that \(x = \frac { 1 - \mathrm { e } ^ { - 3 k t } } { 1 + 2 \mathrm { e } ^ { - 3 k t } }\). Verify that in the long term the drug concentration approaches its maximum value.
OCR MEI C4 Q2
2 A curve has parametric equations \(x = \mathrm { e } ^ { 3 t } , y = t \mathrm { e } ^ { 2 t }\).
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). Hence find the exact gradient of the curve at the point with parameter \(t = 1\).
  2. Find the cartesian equation of the curve in the form \(y = a x ^ { b } \ln x\), where \(a\) and \(b\) are constants to be determined.
OCR MEI C4 Q3
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]
\includegraphics[alt={},max width=\textwidth]{11d26af4-19d0-4310-a64e-9888285c9980-2_260_447_281_824} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
\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 .
OCR MEI C4 Q4
4 The motion of a particle is modelled by the differential equation $$v \frac { \mathrm {~d} v } { \mathrm {~d} x } + 4 x = 0$$ where \(x\) is its displacement from a fixed point, and \(v\) is its velocity. Initially \(x = 1\) and \(v = 4\).
  1. Solve the differential equation to show that \(v ^ { 2 } = 20 - 4 x ^ { 2 }\). Now consider motion for which \(x = \cos 2 t + 2 \sin 2 t\), where \(x\) is the displacement from a fixed point at time \(t\).
  2. Verify that, when \(t = 0 , x = 1\). Use the fact that \(v = \frac { \mathrm { d } x } { \mathrm {~d} t }\) to verify that when \(t = 0 , v = 4\).
  3. Express \(x\) in the form \(R \cos ( 2 t - \alpha )\), where \(R\) and \(\alpha\) are constants to be determined, and obtain the corresponding expression for \(v\). Hence or otherwise verify that, for this motion too, \(v ^ { 2 } = 20 - 4 x ^ { 2 }\).
  4. Use your answers to part (iii) to find the maximum value of \(x\), and the earliest time at which \(x\) reaches this maximum value.
OCR MEI C4 Q5
5 The total value of the sales made by a new company in the first \(t\) years of its existence is denoted by \(\pounds V\). A model is proposed in which the rate of increase of \(V\) is proportional to the square root of \(V\). The constant of proportionality is \(k\).
  1. Express the model as a differential equation. Verify by differentiation that \(V = \left( \frac { 1 } { 2 } k t + c \right) ^ { 2 }\), where \(c\) is an arbitrary constant, satisfies this differential equation.
  2. The value of the company’s sales in its first year is \(\pounds 10000\), and the total value of the sales in the first two years is \(\pounds 40000\). Find \(V\) in terms of \(t\).
OCR MEI C4 Q1
1 Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { x ( x + 1 ) }\), given that when \(x = 1 , y = 1\). Your answer should express \(y\) explicitly in terms of \(x\).
OCR MEI C4 Q2
2 Water is leaking from a container. After \(t\) seconds, the depth of water in the container is \(x \mathrm {~cm}\), and the volume of water is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 1 } { 3 } x ^ { 3 }\). The rate at which water is lost is proportional to \(x\), so that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = - k x\), where \(k\) is a constant.
  1. Show that \(x \frac { \mathrm {~d} x } { \mathrm {~d} t } = - k\). Initially, the depth of water in the container is 10 cm .
  2. Show by integration that \(x = \sqrt { 100 - 2 k t }\).
  3. Given that the container empties after 50 seconds, find \(k\). Once the container is empty, water is poured into it at a constant rate of \(1 \mathrm {~cm} ^ { 3 }\) per second. The container continues to lose water as before.
  4. Show that, \(t\) seconds after starting to pour the water in, \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 - x } { x ^ { 2 } }\).
  5. Show that \(\frac { 1 } { 1 - x } - x - 1 = \frac { x ^ { 2 } } { 1 - x }\). Hence solve the differential equation in part (iv) to show that $$t = \ln \left( \frac { 1 } { 1 - x } \right) - \frac { 1 } { 2 } x ^ { 2 } - x$$
  6. Show that the depth cannot reach 1 cm .
OCR MEI C4 Q3
3 A curve satisfies the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } y\), and passes through the point \(( 1,1 )\). Find \(y\) in terms of \(x\).
OCR MEI C4 Q4
4 A skydiver drops from a helicopter. Before she opens her parachute, her speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) after time \(t\) seconds is modelled by the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$ When \(t = 0 , v = 0\).
  1. Find \(v\) in terms of \(t\).
  2. According to this model, what is the speed of the skydiver in the long term? She opens her parachute when her speed is \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Her speed \(t\) seconds after this is \(w \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and is modelled by the differential equation $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
  3. Express \(\frac { 1 } { ( w - 4 ) ( w + 5 ) }\) in partial fractions.
  4. Using this result, show that \(\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }\).
  5. According to this model, what is the speed of the skydiver in the long term?
OCR MEI C4 Q5
5 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 $$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$ 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.
OCR MEI C4 Q6
6
  1. The number of bacteria in a colony is increasing at a rate that is proportional to the square root of the number of bacteria present. Form a differential equation relating \(x\), the number of bacteria, to the time \(t\).
  2. In another colony, the number of bacteria, \(y\), after time \(t\) minutes is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} t } = \frac { 10000 } { \sqrt { y } }$$ Find \(y\) in terms of \(t\), given that \(y = 900\) when \(t = 0\). Hence find the number of bacteria after 10 minutes.
OCR MEI C4 Q1
1 In a chemical process, the mass \(M\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac { \mathrm { d } M } { \mathrm {~d} t } = \frac { M } { t \left( 1 + t ^ { 2 } \right) } .$$
  1. Find \(\int \frac { t } { 1 + t ^ { 2 } } \mathrm {~d} t\).
  2. Find constants \(A , B\) and \(C\) such that $$\frac { 1 } { t \left( 1 + t ^ { 2 } \right) } = \frac { A } { t } + \frac { B t + C } { 1 + t ^ { 2 } }$$
  3. Use integration, together with your results in parts (i) and (ii), to show that $$M = \frac { K t } { \sqrt { 1 + t ^ { 2 } } } ,$$ where \(K\) is a constant.
  4. When \(t = 1 , M = 25\). Calculate \(K\). What is the mass of the chemical in the long term?
OCR MEI C4 Q2
2 The growth of a tree is modelled by the differential equation $$10 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 20 - h$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).
  1. Write down the value of \(h\) for which \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 0\), and interpret this in terms of the growth of the tree.
  2. Verify that \(h = 20 \left( 1 - \mathrm { e } ^ { - 0.1 t } \right)\) satisfies this differential equation and its initial condition. The alternative differential equation $$200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 400 - h ^ { 2 }$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).
  3. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac { 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) } { 1 + \mathrm { e } ^ { - 0.2 t } }$$
  4. What does this solution indicate about the long-term height of the tree?
  5. After a year, the tree has grown to a height of 2 m . Which model fits this information better? \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{5699ea45-6a4d-4681-a044-4a337e30588c-3_273_461_190_830} \captionsetup{labelformat=empty} \caption{Fig. 9}
    \end{figure} Fig. 9 shows a hemispherical bowl, of radius 10 cm , filled with water to a depth of \(x \mathrm {~cm}\). It can be shown that the volume of water, \(V \mathrm {~cm} ^ { 3 }\), is given by $$V = \pi \left( 10 x ^ { 2 } - \frac { 1 } { 3 } x ^ { 3 } \right) .$$ Water is poured into a leaking hemispherical bowl of radius 10 cm . Initially, the bowl is empty. After \(t\) seconds, the volume of water is changing at a rate, in \(\mathrm { cm } ^ { 3 } \mathrm {~s} ^ { - 1 }\), given by the equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = k ( 20 - x )$$ where \(k\) is a constant.
  6. Find \(\frac { \mathrm { d } V } { \mathrm {~d} x }\), and hence show that \(\pi x \frac { \mathrm {~d} x } { \mathrm {~d} t } = k\).
  7. Solve this differential equation, and hence show that the bowl fills completely after \(T\) seconds, where $$T = \frac { 50 \pi } { k }$$ Once the bowl is full, the supply of water to the bowl is switched off, and water then leaks out at a rate of \(k x \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
  8. Show that, \(t\) seconds later, \(\pi ( 20 - x ) \frac { \mathrm { d } x } { \mathrm {~d} t } = - k\).
  9. Solve this differential equation. Hence show that the bowl empties in \(3 T\) seconds.
OCR MEI C4 Q4
4 A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Its terminal (long-term) velocity is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A model of the particle's motion is proposed. In this model, \(v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)\).
  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.
  2. Verify that \(v\) satisfies the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 2 v\). In a second model, \(v\) satisfies the differential equation $$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 - 0.4 v ^ { 2 }$$ As before, when \(t = 0 , v = 0\).
  3. Show that this differential equation may be written as $$\frac { 10 } { ( 5 - v ) ( 5 + v ) } \frac { \mathrm { d } v } { \mathrm {~d} t } = 4$$ Using partial fractions, solve this differential equation to show that $$t = \frac { 1 } { 4 } \ln \left( \frac { 5 + v } { 5 - v } \right)$$ This can be re-arranged to give \(v = \frac { 5 \left( 1 - \mathrm { e } ^ { - 4 t } \right) } { 1 + \mathrm { e } ^ { - 4 t } }\). [You are not required to show this result.]
  4. Verify that this model also gives a terminal velocity of \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Calculate the velocity after 0.5 seconds as given by this model. The velocity of the particle after 0.5 seconds is measured as \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  5. Which of the two models fits the data better?
OCR MEI C4 Q1
1 Fig. 7 illustrates the growth of a population with time. The proportion of the ultimate (long term) population is denoted by \(x\), and the time in years by \(t\). When \(t = 0 , x = 0.5\), and as \(t\) increases, \(x\) approaches 1 . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7278ce82-f710-44a7-945e-c194a4fb1744-1_598_940_366_633} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure} One model for this situation is given by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = x ( 1 - x )$$
  1. Verify that \(x = \frac { 1 } { 1 + \mathrm { e } ^ { - t } }\) satisfies this differential equation, including the initial condition.
  2. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value. An alternative model for this situation is given by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = x ^ { 2 } ( 1 - x )$$ with \(x = 0.5\) when \(t = 0\) as before.
  3. Find constants \(A , B\) and \(C\) such that \(\frac { 1 } { x ^ { 2 } ( 1 - x ) } = \frac { A } { x ^ { 2 } } + \frac { B } { x } + \frac { C } { 1 - x }\).
  4. Hence show that \(t = 2 + \ln \left( \frac { x } { 1 - x } \right) - \frac { 1 } { x }\).
  5. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value.
OCR MEI C4 Q2
2 Scientists can estimate the time elapsed since an animal died by measuring its body temperature.
  1. Assuming the temperature goes down at a constant rate of 1.5 degrees Fahrenheit per hour, estimate how long it will take for the temperature to drop
    (A) from \(98 ^ { \circ } \mathrm { F }\) to \(89 ^ { \circ } \mathrm { F }\),
    (B) from \(98 ^ { \circ } \mathrm { F }\) to \(80 ^ { \circ } \mathrm { F }\). In practice, rate of temperature loss is not likely to be constant. A better model is provided by Newton's law of cooling, which states that the temperature \(\theta\) in degrees Fahrenheit \(t\) hours after death is given by the differential equation $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k \left( \theta - \theta _ { 0 } \right)$$ where \(\theta _ { 0 } { } ^ { \circ } \mathrm { F }\) is the air temperature and \(k\) is a constant.
  2. Show by integration that the solution of this equation is \(\theta = \theta _ { 0 } + A \mathrm { e } ^ { - k t }\), where \(A\) is a constant. The value of \(\theta _ { 0 }\) is 50 , and the initial value of \(\theta\) is 98 . The initial rate of temperature loss is \(1.5 ^ { \circ } \mathrm { F }\) per hour.
  3. Find \(A\), and show that \(k = 0.03125\).
  4. Use this model to calculate how long it will take for the temperature to drop
    (A) from \(98 ^ { \circ } \mathrm { F }\) to \(89 ^ { \circ } \mathrm { F }\),
    (B) from \(98 ^ { \circ } \mathrm { F }\) to \(80 ^ { \circ } \mathrm { F }\).
  5. Comment on the results obtained in parts (i) and (iv).
OCR MEI C4 Q4
4 A curve has equation $$x ^ { 2 } + 4 y ^ { 2 } = k ^ { 2 }$$ where \(k\) is a positive constant.
  1. Verify that $$x = k \cos \theta , \quad y = \frac { 1 } { 2 } k \sin \theta ,$$ are parametric equations for the curve.
  2. Hence or otherwise show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { x } { 4 y }\).
  3. Fig. 8 illustrates the curve for a particular value of \(k\). Write down this value of \(k\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{7278ce82-f710-44a7-945e-c194a4fb1744-4_666_1080_886_522} \captionsetup{labelformat=empty} \caption{Fig. 8}
    \end{figure}
  4. Copy Fig. 8 and on the same axes sketch the curves for \(k = 1 , k = 3\) and \(k = 4\). On a map, the curves represent the contours of a mountain. A stream flows down the mountain. Its path on the map is always at right angles to the contour it is crossing.
  5. Explain why the path of the stream is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 y } { x } .$$
  6. Solve this differential equation. Given that the path of the stream passes through the point \(( 2,1 )\), show that its equation is \(y = \frac { x ^ { 4 } } { 16 }\).
OCR MEI C4 Q1
1 Using partial fractions, find \(\int \frac { x } { ( x + 1 ) ( 2 x + 1 ) } \mathrm { d } x\).
  1. Express \(\cos \theta + \sqrt { 3 } \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(\alpha\) is acute, expressing \(\alpha\) in terms of \(\pi\).
  2. Write down the derivative of \(\tan \theta\). Hence show that \(\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \frac { 1 } { ( \cos \theta + \sqrt { 3 } \sin \theta ) ^ { 2 } } \mathrm {~d} \theta = \frac { \sqrt { 3 } } { 4 }\).
OCR MEI C4 Q3
3 In a chemical process, the mass \(\boldsymbol { M }\) grams of a chemical at time \(\boldsymbol { t }\) minutes is modelled by the differential equation $$\underset { d t } { d \underline { \underline { M } } } - \underset { \mathrm { z } \left( \left. \right| ^ { \prime } + \mathrm { z } ^ { 2 } \right) ^ { \prime } } { M _ { - } }$$
  1. Find \({ } _ { \mathrm { f } } \overline { 1 } ; \overline { 12 } d t\)
    (li) Find constants \(\boldsymbol { A } , \boldsymbol { B }\) and \(\boldsymbol { C }\) such lhat $$\begin{array} { c c } 1 & B t + C
    \hdashline t \left( I + t ^ { 2 } \right) & I \end{array} . + \begin{array} { c c } I & I + 1 ^ { 2 } . \end{array}$$
  2. Use integration, together with your results in parts (i) and (ii), to show that $$M = \frac { K t } { J \sqrt { + , 2 } } ,$$ where \(\boldsymbol { K }\) is a constant.
  3. When \(\boldsymbol { t } = \mathrm { I } , \boldsymbol { M } = 25\). Calculate \(\boldsymbol { K }\) What is the mass of the chemical in the long term?
OCR MEI C4 Q4
4 The growth of a tree is modelled by the differential equation $$10 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 20 - h$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).
  1. Write down the value of \(h\) for which \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 0\), and interpret this in terms of the growth of the tree.
  2. Verify that \(h = 20 \left( 1 - \mathrm { e } ^ { - 0.1 t } \right)\) satisfies this differential equation and its initial condition. The alternative differential equation $$200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 400 - h ^ { 2 }$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).
  3. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac { 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) } { 1 + \mathrm { e } ^ { - 0.2 t } }$$
  4. What does this solution indicate about the long-term height of the tree?
  5. After a year, the tree has grown to a height of 2 m . Which model fits this information better?
OCR MEI C4 Q5
4 marks
5
  1. Find the first three non-zero terms of the binomial expansion of \(\frac { 1 } { \sqrt { 4 } x ^ { 2 } }\) for \(| x | < 2\). [4]
  2. Use this result to find an approximation for \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 x ^ { 2 } } } \mathrm {~d} x\), rounding your answer to
    4 significant figures.
  3. Given that \(\int \frac { 1 } { \sqrt { 4 x ^ { 2 } } } \mathrm {~d} x = \arcsin \left( \frac { 1 } { 2 } x \right) + c\), evaluate \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 4 x ^ { 2 } } } \mathrm {~d} x\), rounding your answer to 4 significant figures.