Questions - OCR MEI C4 - A-Level Maths

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OCR MEI C4 Q7

18 marks Standard +0.3

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 $t = 0 , v = 0$.

Find $v$ in terms of $t$.
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 $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
Express $\frac { 1 } { ( w - 4 ) ( w + 5 ) }$ in partial fractions.
Using this result, show that $\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }$.
According to this model, what is the speed of the skydiver in the long term?

OCR MEI C4 Q1

20 marks Standard +0.3

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.

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 $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t$.
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

18 marks Standard +0.3

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, $v = 5 \left( 1 - \mathrm { e } ^ { - 2 t } \right)$.

Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model.
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 $t = 0 , v = 0$.
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.]
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 $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Which of the two models fits the data better?

OCR MEI C4 Q3

19 marks Standard +0.3

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 $t = 0 , x = 2.5$.

Show that $\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { k x ^ { 2 } } { a }$.
Given that the initial population of 2.5 thousand red squirrels reduces to 1.6 thousand after one year, calculate $a$ and $k$.
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 $t = 0 , y = 1$.
Express $\frac { 1 } { 2 y - y ^ { 2 } }$ in partial fractions.
Hence show by integration that $\ln \left( \frac { y } { 2 y } \right) = 2 t$. Show that $y = \frac { 2 } { 1 + \mathrm { e } ^ { - 2 t } }$.
What is the long-term population of grey squirrels predicted by this model?

OCR MEI C4 Q1

18 marks Standard +0.3

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, $x = 0$.

Express $\frac { 1 } { ( 1 + 2 x ) ( 1 - x ) }$ in partial fractions.
[0pt] [3]
Hence solve the differential equation to show that $\frac { 1 + 2 x } { 1 - x } = \mathrm { e } ^ { 3 k t }$.
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.
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

7 marks Standard +0.3

2 A curve has parametric equations $x = \mathrm { e } ^ { 3 t } , y = t \mathrm { e } ^ { 2 t }$.

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 $t = 1$.
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

18 marks Standard +0.8

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.

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.
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 $t = 0 , h = 1$.
Solve this differential equation, and hence show that $$h ^ { \frac { 1 } { 2 } } \left( 30 + 20 h + 6 h ^ { 2 } \right) = 56 - 15 B t .$$
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

18 marks Standard +0.3

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 $v = 4$.

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 $t$.
Verify that, when $t = 0 , x = 1$. Use the fact that $v = \frac { \mathrm { d } x } { \mathrm {~d} t }$ to verify that when $t = 0 , v = 4$.
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, $v ^ { 2 } = 20 - 4 x ^ { 2 }$.
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

8 marks Moderate -0.3

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 $k$.

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.
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

8 marks Standard +0.3

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

18 marks Standard +0.3

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.

Show that $x \frac { \mathrm {~d} x } { \mathrm {~d} t } = - k$. Initially, the depth of water in the container is 10 cm .
Show by integration that $x = \sqrt { 100 - 2 k t }$.
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.
Show that, $t$ seconds after starting to pour the water in, $\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 - x } { x ^ { 2 } }$.
Show that $\frac { 1 } { 1 - x } - x - 1 = \frac { x ^ { 2 } } { 1 - x }$. Hence solve the differential equation in part (iv) to show that $$t = \ln \left( \frac { 1 } { 1 - x } \right) - \frac { 1 } { 2 } x ^ { 2 } - x$$
Show that the depth cannot reach 1 cm .

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 $t = 0 , v = 0$.

Find $v$ in terms of $t$.
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 $$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
Express $\frac { 1 } { ( w - 4 ) ( w + 5 ) }$ in partial fractions.
Using this result, show that $\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }$.
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.

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 $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t$.
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

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 $t$.
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

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}

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.
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}

The following table gives some values of $x$ and $y$.
$x$ 0 0.25 0.5 0.75 1
$y$ 1 1.0308 1.25 1.4142
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.
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.
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}

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.
  $x$ 0 0.5 1 1.5 2
  $y$ 1.9283 2.8964 4.5919
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

Complete the table of values for the curve $y = \sqrt { \cos x }$.
$X$ 0 $\frac { \pi } { 8 }$ $\frac { \pi } { 4 }$ $\frac { 3 \pi } { 8 }$ $\frac { \pi } { 2 }$
$y$ 0.9612 0.8409
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}
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

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}
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.

$x$	1	1.5	2
$\sqrt { 1 + \mathrm { e } ^ { - x } }$	1.1696	1.1060	1.0655

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.
Show that, provided $\mathrm { e } ^ { - x }$ is suitably small, $\left( 1 + \mathrm { e } ^ { - x } \right) ^ { \frac { 1 } { 2 } } \approx 1 + \frac { 1 } { 2 } \mathrm { e } ^ { - x } \quad \frac { 1 } { 8 } \mathrm { e } ^ { - 2 x }$.
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}

State, with their coordinates, the points on the curve for which $\theta = - \frac { \pi } { 2 } , \theta = 0$ and $\theta = \frac { \pi } { 2 }$.
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.
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}
Express $\sin ^ { 2 } \theta$ in terms of $x$. Hence show that the cartesian equation of the curve is $y ^ { 2 } = x ^ { 2 } \left( 1 - \frac { 1 } { 4 } x ^ { 2 } \right)$.
Find the volume of the paperweight shape.