Questions — OCR MEI (4333 questions)

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OCR MEI C4 Q3
4 marks Moderate -0.5
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
18 marks Standard +0.3
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
20 marks Standard +0.3
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
8 marks Moderate -0.3
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
18 marks Standard +0.3
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
19 marks Moderate -0.3
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
18 marks Standard +0.3
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
20 marks Standard +0.8
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
17 marks Standard +0.3
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
18 marks Moderate -0.3
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
7 marks Standard +0.3
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
18 marks Standard +0.3
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
19 marks Standard +0.3
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
7 marks Standard +0.3
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.
OCR MEI C4 Q1
5 marks Moderate -0.3
1 Fig. 3 shows the curve \(y = x ^ { 3 } + \sqrt { ( \sin x ) }\) for \(0 \leqslant x \leqslant \frac { \pi } { 4 }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-1_587_540_393_768} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure}
  1. Use the trapezium rule with 4 strips to estimate the area of the region bounded by the curve, the \(x\)-axis and the line \(x = \frac { \pi } { 4 }\), giving your answer to 3 decimal places.
  2. Suppose the number of strips in the trapezium rule is increased. Without doing further calculations, state, with a reason, whether the area estimate increases, decreases, or it is not possible to say.
OCR MEI C4 Q2
8 marks Moderate -0.3
2 Fig. 2 shows the curve \(y = \overline { 1 + x ^ { 2 } }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-2_577_941_549_636} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure}
  1. The following table gives some values of \(x\) and \(y\).
    \(x\)00.250.50.751
    \(y\)11.03081.251.4142
    Find the missing value of \(y\), giving your answer correct to 4 decimal places. Hence show that, using the trapezium rule with four strips, the shaded area is approximately 1.151 square units.
  2. Jenny uses a trapezium rule with 8 strips, and obtains a value of 1.158 square units. Explain why she must have made a mistake.
  3. The shaded area is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Find the exact volume of the solid of revolution formed.
OCR MEI C4 Q3
9 marks Standard +0.3
3 Fig. 4 shows the curve \(y = \sqrt { 1 + \mathrm { e } ^ { 2 x } }\), and the region between the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-3_656_736_482_665} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
  1. Find the exact volume of revolution when the shaded region is rotated through \(360 ^ { \circ }\) about the \(x\)-axis.
    1. Complete the table of values, and use the trapezium rule with 4 strips to estimate the area of the shaded region.
      \(x\)00.511.52
      \(y\)1.92832.89644.5919
    2. The trapezium rule for \(\int _ { 0 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { 2 x } } \mathrm {~d} x\) with 8 and 16 strips gives 6.797 and 6.823, although not necessarily in that order. Without doing the calculations, say which result is which, explaining your reasoning.
OCR MEI C4 Q4
4 marks Moderate -0.3
4
  1. Complete the table of values for the curve \(y = \sqrt { \cos x }\).
    \(X\)0\(\frac { \pi } { 8 }\)\(\frac { \pi } { 4 }\)\(\frac { 3 \pi } { 8 }\)\(\frac { \pi } { 2 }\)
    \(y\)0.96120.8409
    Hence use the trapezium rule with strip width \(h = \frac { \pi } { 8 }\) to estimate the value of the integral \(\int _ { 0 } ^ { \frac { \pi } { 2 } } \sqrt { \cos x } \mathrm {~d} x\), giving your answer to 3 decimal places. Fig. 4 shows the curve \(y = \sqrt { \cos x }\) for \(0 \leqslant x \leqslant \frac { \pi } { 2 }\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-4_459_751_799_638} \captionsetup{labelformat=empty} \caption{Fig. 4}
    \end{figure}
  2. State, with a reason, whether the trapezium rule with a strip width of \(\frac { \pi } { 16 }\) would give a larger or smaller estimate of the integral.
OCR MEI C4 Q5
6 marks Moderate -0.8
5
  1. Use the trapezium rule with four strips to estimate \(\int _ { - 2 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x\), showing your working. Fig. 1 shows a sketch of \(y = \sqrt { 1 + \mathrm { e } ^ { x } }\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{ce44db53-2ec8-497b-a1d5-a8adf85e3929-5_533_1074_441_565} \captionsetup{labelformat=empty} \caption{Fig. 1}
    \end{figure}
  2. Suppose that the trapezium rule is used with more strips than in part (i) to estimate \(\int _ { - 2 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x\). State, with a reason but no further calculation, whether this would give a larger or smaller estimate.
    [0pt] [2]
OCR MEI C4 Q6
8 marks Standard +0.3
6 Two students are trying to evaluate the integral \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\).
Sarah uses the trapezium rule with 2 strips, and starts by constructing the following table.
\(x\)11.52
\(\sqrt { 1 + \mathrm { e } ^ { - x } }\)1.16961.10601.0655
  1. Complete the calculation, giving your answer to 3 significant figures. Anish uses a binomial approximation for \(\sqrt { 1 + \mathrm { e } ^ { - x } }\) and then integrates this.
  2. Show that, provided \(\mathrm { e } ^ { - x }\) is suitably small, \(\left( 1 + \mathrm { e } ^ { - x } \right) ^ { \frac { 1 } { 2 } } \approx 1 + \frac { 1 } { 2 } \mathrm { e } ^ { - x } \quad \frac { 1 } { 8 } \mathrm { e } ^ { - 2 x }\).
  3. Use this result to evaluate \(\int _ { 1 } ^ { 2 } \sqrt { 1 + \mathrm { e } ^ { - x } } \mathrm {~d} x\) approximately, giving your answer to 3 significant figures.
OCR MEI C4 Q1
6 marks Standard +0.3
1 Fig. 6 shows the region enclosed by the curve \(y = \left( 1 + 2 x ^ { 2 } \right) ^ { \frac { 1 } { 3 } }\) and the line \(y = 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-1_426_664_277_714} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure} This region is rotated about the \(y\)-axis. Find the volume of revolution formed, giving your answer as a multiple of \(\pi\).
OCR MEI C4 Q2
19 marks Standard +0.3
2 Fig. 7a shows the curve with the parametric equations $$x = 2 \cos \theta , \quad y = \sin 2 \theta , \quad - \frac { \pi } { 2 } \leqslant \theta \leqslant \frac { \pi } { 2 } .$$ The curve meets the \(x\)-axis at O and P . Q and R are turning points on the curve. The scales on the axes are the same. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-2_509_660_571_714} \captionsetup{labelformat=empty} \caption{Fig. 7a}
\end{figure}
  1. State, with their coordinates, the points on the curve for which \(\theta = - \frac { \pi } { 2 } , \theta = 0\) and \(\theta = \frac { \pi } { 2 }\).
  2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(\theta\). Hence find the gradient of the curve when \(\theta = \frac { \pi } { 2 }\), and verify that the two tangents to the curve at the origin meet at right angles.
  3. Find the exact coordinates of the turning point Q . When the curve is rotated about the \(x\)-axis, it forms a paperweight shape, as shown in Fig. 7b. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-2_324_389_1692_857} \captionsetup{labelformat=empty} \caption{Fig. 7b}
    \end{figure}
  4. Express \(\sin ^ { 2 } \theta\) in terms of \(x\). Hence show that the cartesian equation of the curve is \(y ^ { 2 } = x ^ { 2 } \left( 1 - \frac { 1 } { 4 } x ^ { 2 } \right)\).
  5. Find the volume of the paperweight shape.
OCR MEI C4 Q3
7 marks Challenging +1.2
3 Fig. 6 shows the region enclosed by part of the curve \(y = 2 x ^ { 2 }\), the straight line \(x + y = 3\), and the \(y\)-axis. The curve and the straight line meet at \(\mathrm { P } ( 1,2 )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-3_640_923_399_613} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure} The shaded region is rotated through \(360 ^ { \circ }\) about the \(y\)-axis. Find, in terms of \(\pi\), the volume of the solid of revolution formed.
[0pt] [You may use the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) for the volume of a cone.]
OCR MEI C4 Q4
5 marks Standard +0.3
4 The part of the curve \(y = 4 - x ^ { 2 }\) that is above the \(x\)-axis is rotated about the \(y\)-axis. This is shown in Fig. 4. Find the volume of revolution produced, giving your answer in terms of \(\pi\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-3_534_595_1831_785} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure}
OCR MEI C4 Q5
4 marks Standard +0.3
5 Fig. 2 shows the curve \(y = \sqrt { } 1 + \mathrm { e } ^ { 2 x }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8d786d33-c5c2-44a6-8273-7a3e43e552ef-4_434_873_306_665} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} The region bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 1\) is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. Show that the volume of the solid of revolution produced is \(\frac { 1 } { 2 } \pi \left( 1 + \mathrm { e } ^ { 2 } \right)\).