1.08k Separable differential equations: dy/dx = f(x)g(y)

9 A tank contains water which is heated by an electric water heater working under the action of a thermostat. The temperature of the water, \(\theta ^ { \circ } \mathrm { C }\), may be modelled as follows. When the water heater is first switched on, \(\theta = 40\). The heater causes the temperature to increase at a rate \(k _ { 1 } { } ^ { \circ } \mathrm { C }\) per second, where \(k _ { 1 }\) is a constant, until \(\theta = 60\). The heater then switches off.

1. Write down, in terms of \(k _ { 1 }\), how long it takes for the temperature to increase from \(40 ^ { \circ } \mathrm { C }\) to \(60 ^ { \circ } \mathrm { C }\). The temperature of the water then immediately starts to decrease at a variable rate \(k _ { 2 } ( \theta - 20 ) ^ { \circ } \mathrm { C }\) per second, where \(k _ { 2 }\) is a constant, until \(\theta = 40\).
2. Write down a differential equation to represent the situation as the temperature is decreasing.
3. Find the total length of time for the temperature to increase from \(40 ^ { \circ } \mathrm { C }\) to \(60 ^ { \circ } \mathrm { C }\) and then decrease to \(40 ^ { \circ } \mathrm { C }\). Give your answer in terms of \(k _ { 1 }\) and \(k _ { 2 }\). 4

8

1. Find the quotient and the remainder when \(x ^ { 2 } - 5 x + 6\) is divided by \(x - 1\).
2. (a) Find the general solution of the differential equation $$\left( \frac { x - 1 } { x ^ { 2 } - 5 x + 6 } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } = y - 5 .$$ (b) Given that \(y = 7\) when \(x = 8\), find \(y\) when \(x = 6\).

7 The gradient of a curve at the point \(( x , y )\), where \(x > - 2\), is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 3 y ^ { 2 } ( x + 2 ) }$$ The points \(( 1,2 )\) and \(( q , 1.5 )\) lie on the curve. Find the value of \(q\), giving your answer correct to 3 significant figures.

4 Solve the differential equation $$\mathrm { e } ^ { 2 y } \frac { \mathrm {~d} y } { \mathrm {~d} x } + \tan x = 0 ,$$ given that \(x = 0\) when \(y = 0\). Give your answer in the form \(y = \mathrm { f } ( x )\).

8 At time \(t\) seconds, the radius of a spherical balloon is \(r \mathrm {~cm}\). The balloon is being inflated so that the rate of increase of its radius is inversely proportional to the square root of its radius. When \(t = 5 , r = 9\) and, at this instant, the radius is increasing at \(1.08 \mathrm {~cm} \mathrm {~s} ^ { - 1 }\).

1. Write down a differential equation to model this situation, and solve it to express \(r\) in terms of \(t\).
2. How much air is in the balloon initially?
   [0pt] [The volume of a sphere is \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\).]

10 A container in the shape of an inverted cone of radius 3 metres and vertical height 4.5 metres is initially filled with liquid fertiliser. This fertiliser is released through a hole in the bottom of the container at a rate of \(0.01 \mathrm {~m} ^ { 3 }\) per second. At time \(t\) seconds the fertiliser remaining in the container forms an inverted cone of height \(h\) metres.
[0pt] [The volume of a cone is \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\).]

1. Show that \(h ^ { 2 } \frac { \mathrm {~d} h } { \mathrm {~d} t } = - \frac { 9 } { 400 \pi }\).
2. Express \(h\) in terms of \(t\).
3. Find the time it takes to empty the container, giving your answer to the nearest minute.

8 In the year 2000 the population density, \(P\), of a village was 100 people per \(\mathrm { km } ^ { 2 }\), and was increasing at the rate of 1 person per \(\mathrm { km } ^ { 2 }\) per year. The rate of increase of the population density is thought to be inversely proportional to the size of the population density. The time in years after the year 2000 is denoted by \(t\).

1. Write down a differential equation to model this situation, and solve it to express \(P\) in terms of \(t\).
2. In 2008 the population density of the village was 108 people per \(\mathrm { km } ^ { 2 }\) and in 2013 it was 128 people per \(\mathrm { km } ^ { 2 }\). Determine how well the model fits these figures.

10

1. Express \(\frac { 16 + 5 x - 2 x ^ { 2 } } { ( x + 1 ) ^ { 2 } ( x + 4 ) }\) in partial fractions.
2. It is given that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \left( 16 + 5 x - 2 x ^ { 2 } \right) y } { ( x + 1 ) ^ { 2 } ( x + 4 ) }$$ and that \(y = \frac { 1 } { 256 }\) when \(x = 0\). Find the exact value of \(y\) when \(x = 2\). Give your answer in the form \(A \mathrm { e } ^ { n }\).

7 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).

7 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] \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.

7 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?

8 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?

7 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.
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. \section*{END OF QUESTION PAPER} \section*{Tuesday 16 J une 2015 - Afternoon} \section*{A2 GCE MATHEMATICS (MEI)} 4754/01B Applications of Advanced Mathematics (C4) Paper B: Comprehension \section*{QUESTION PAPER} \section*{Candidates answer on the Question Paper.} \section*{OCR supplied materials:}
   \section*{Other materials required:}
   Duration: Up to 1 hour \includegraphics[max width=\textwidth, alt={}, center]{132ae754-bd4c-4819-80ef-4823ac2ead4f-05_117_495_1014_1308} PLEASE DO NOT WRITE IN THIS SPACE 2 In line 79 it says "For most journeys, more than half the journey time is composed of load time and transfer time". For what percentage of the journey time for the round trip made by car A in Table 4 is the car stationary?
   \includegraphics[max width=\textwidth, alt={}]{132ae754-bd4c-4819-80ef-4823ac2ead4f-07_645_1746_388_164}
   3 Using the expression on line 51, work out the answer to the question on lines 39 and 40 for the case where there are 10 upper floors and 7 people. Give your answer to 2 decimal places.
   \includegraphics[max width=\textwidth, alt={}]{132ae754-bd4c-4819-80ef-4823ac2ead4f-07_488_1746_1233_164}
   4 In lines 89 and 90 it says "... on average there will be approximately 8 stops per trip. A round trip with 8 stops would take between 188 and 200 seconds". Explain how the figure of 188 seconds has been derived.

5

1. Referring to Strategy 3 and lines 99 to 101, complete the table below for car C .
2. Calculate the time car C will take to transport all the people who work on floors 7 and 8 , and return to the ground floor.

   5	\includegraphics[max width=\textwidth, alt={}]{132ae754-bd4c-4819-80ef-4823ac2ead4f-08_1095_816_484_700}

   68 people make independent visits to any one of the upper floors of a building with 10 upper floors. What is the probability that at least one of the visitors goes to the top floor?

   6

   7 On lines 94 and 95 it says "Table 4 gives the timings for round trips in which the cars are required to stop at every floor they serve; Table 2 suggests this is a common occurrence in this case". Explain how Table 2 is used to make this claim. \includegraphics[max width=\textwidth, alt={}, center]{132ae754-bd4c-4819-80ef-4823ac2ead4f-09_1093_1740_1238_166} END OF QUESTION PAPER

7 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?

13 A scientist is attempting to model the number of insects, \(N\), present in a colony at time \(t\) weeks. When \(t = 0\) there are 400 insects and when \(t = 1\) there are 440 insects.

1. A scientist assumes that the rate of increase of the number of insects is inversely proportional to the number of insects present at time \(t\).
   1. Write down a differential equation to model this situation.
   2. Solve this differential equation to find \(N\) in terms of \(t\).
   3. In a revised model it is assumed that \(\frac { \mathrm { d } N } { \mathrm {~d} t } = \frac { N ^ { 2 } } { 3988 \mathrm { e } ^ { 0.2 t } }\). Solve this differential equation to find \(N\) in terms of \(t\).
   4. Compare the long-term behaviour of the two models.

8 A cylindrical tank is initially full of water. There is a small hole at the base of the tank out of which the water leaks. The height of water in the tank is \(x \mathrm {~m}\) at time \(t\) seconds. The rate of change of the height of water may be modelled by the assumption that it is proportional to the square root of the height of water. When \(t = 100 , x = 0.64\) and, at this instant, the height is decreasing at a rate of \(0.0032 \mathrm {~ms} ^ { - 1 }\).

1. Show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - 0.004 \sqrt { x }\).
2. Find an expression for \(x\) in terms of \(t\).
3. Hence determine at what time, according to this model, the tank will be empty.

12 Find the general solution of the differential equation \(\left( 2 x ^ { 3 } - 3 x ^ { 2 } - 11 x + 6 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 20 x - 35 )\).
Give your answer in the form \(y = \mathrm { f } ( x )\). \section*{END OF QUESTION PAPER} \section*{OCR
Oxford Cambridge and RSA}

12 A cake is cooling so that, \(t\) minutes after it is removed from an oven, its temperature is \(\theta ^ { \circ } \mathrm { C }\). When the cake is removed from the oven, its temperature is \(160 ^ { \circ } \mathrm { C }\). After 10 minutes its temperature has fallen to \(125 ^ { \circ } \mathrm { C }\).

1. In a simple model, the rate of decrease of the temperature of the cake is assumed to be constant.
   1. Write down a differential equation for this model.
   2. Solve this differential equation to find \(\theta\) in terms of \(t\).
   3. State one limitation of this model.
2. In a revised model, the rate of decrease of the temperature of the cake is proportional to the difference between the temperature of the cake and the temperature of the room. The temperature of the room is a constant \(20 ^ { \circ } \mathrm { C }\).
   1. Write down a differential equation for this revised model.
   2. Solve this differential equation to find \(\theta\) in terms of \(t\).
3. The cake can be decorated when its temperature is \(25 ^ { \circ } \mathrm { C }\). Find the difference in time between when the two models would predict that the cake can be decorated, giving your answer correct to the nearest minute. \section*{END OF QUESTION PAPER}

11 The gradient function of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 x ^ { 2 } \ln x } { \mathrm { e } ^ { 3 y } }\).
The curve passes through the point (e, 1).

1. Find the equation of this curve, giving your answer in the form \(\mathrm { e } ^ { 3 y } = \mathrm { f } ( x )\).
2. Show that, when \(x = \mathrm { e } ^ { 2 }\), the \(y\)-coordinate of this curve can be written as \(y = a + \frac { 1 } { 3 } \ln \left( b \mathrm { e } ^ { 3 } + c \right)\), where \(a , b\) and \(c\) are constants to be determined.

6 Helga invests \(\pounds 4000\) in a savings account.
After \(t\) days, her investment is worth \(\pounds y\).
The rate of increase of \(y\) is \(k y\), where \(k\) is a constant.

1. Write down a differential equation in terms of \(t , y\) and \(k\).
2. Solve your differential equation to find the value of Helga's investment after \(t\) days. Give your answer in terms of \(k\) and \(t\). It is given that \(k = \frac { 1 } { 365 } \ln \left( 1 + \frac { r } { 100 } \right)\) where \(r \%\) is the rate of interest per annum. During the first year the rate of interest is \(6 \%\) per annum.
3. Find the value of Helga's investment after 90 days. After one year (365 days), the rate of interest drops to 5\% per annum.
4. Find the total time that it will take for Helga's investment to double in value.

7 The gradient of the curve \(y = \mathrm { f } ( x )\) is given by the differential equation $$( 2 x - 1 ) ^ { 3 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y ^ { 2 } = 0$$ and the curve passes through the point \(( 1,1 )\). By solving this differential equation show that $$f ( x ) = \frac { a x ^ { 2 } - a x + 1 } { b x ^ { 2 } - b x + 1 }$$ where \(a\) and \(b\) are integers to be determined.

1. A circular stain is growing.

The rate of increase of its radius is inversely proportional to the square of the radius. At time \(t\) seconds the circular stain has radius \(r \mathrm {~cm}\) and area \(A \mathrm {~cm} ^ { 2 }\).
a. Show that \(\frac { \mathrm { d } A } { \mathrm {~d} t } = \frac { k } { \sqrt { A } }\). Given that

• the initial area of the circular stain is \(0.09 \mathrm {~cm} ^ { 2 }\).
• after 10 seconds the area of the circular stain is \(0.36 \mathrm {~cm} ^ { 2 }\).
  b. Solve the differential equation to find a complete equation linking \(A\) and \(t\).

10. a. Find \(\int \frac { 1 } { 30 } \cos \frac { \pi } { 6 } t \mathrm {~d} t\). The height above ground, \(X\) metres, of the passenger on a wooden roller coaster can be modelled by the differential equation $$\frac { d \mathrm { X } } { \mathrm {~d} t } = \frac { 1 } { 30 } X \cos \left( \frac { \pi } { 6 } t \right)$$ where \(t\) is the time, in seconds, from the start of the ride.
At time \(t = 0\), the passenger is 6 m above the ground.
b. Show that \(X = k e ^ { \frac { 1 } { 5 \pi } \sin \left( \frac { \pi } { 6 } t \right) }\) where the value of the constant \(k\) should be found.
c. Show that the maximum height of the passenger above the ground is 6.39 m . The passenger reaches the maximum height, for the second time, \(T\) seconds after the start of the ride.
d. Find the value of \(T\).

10. Given that \(y = 8\) at \(x = 1\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 12 x + 9 ) y ^ { \frac { 1 } { 3 } } } { x }$$ Giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

14. a. Express \(\frac { 1 } { ( 3 - x ) ( 1 - x ) }\) in partial fractions.
(2) A scientist is studying the mass of a substance in a laboratory.
The mass, \(x\) grams, of a substance at time \(t\) seconds after a chemical reaction starts is modelled by the differential equation $$2 \frac { d x } { d t } = ( 3 - x ) ( 1 - x ) \quad t \geq 0,0 \leq x < 1$$ Given that when \(t = 0 , x = 0\) b. solve the differential equation and show that the solution can be written as $$x = \frac { 3 \left( e ^ { t } - 1 \right) } { 3 e ^ { t } - 1 }$$ c. Find the mass, \(x\) grams, which has formed 2 seconds after the start of the reaction. Give your answer correct to 3 significant figures.
d. Find the limiting value of \(x\) as \(t\) increases.