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

7 A particle \(P\) of mass 0.1 kg is projected vertically upwards from a point \(O\) with speed \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Air resistance of magnitude \(0.1 v \mathrm {~N}\) opposes the motion, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of \(P\) at time \(t \mathrm {~s}\) after projection.

1. Show that, while \(P\) is moving upwards, \(\frac { 1 } { v + 10 } \frac { \mathrm {~d} v } { \mathrm {~d} t } = - 1\).
2. Hence find an expression for \(v\) in terms of \(t\), and explain why it is valid only for \(0 \leqslant t \leqslant \ln 3\).
3. Find the initial acceleration of \(P\).

7 A particle \(P\) of mass 0.3 kg is projected vertically upwards from the ground with an initial speed of \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). When \(P\) is at height \(x \mathrm {~m}\) above the ground, its upward speed is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It is given that $$3 v - 90 \ln ( v + 30 ) + x = A ,$$ where \(A\) is a constant.

1. Differentiate this equation with respect to \(x\) and hence show that the acceleration of the particle is \(- \frac { 1 } { 3 } ( v + 30 ) \mathrm { m } \mathrm { s } ^ { - 2 }\).
2. Find, in terms of \(v\), the resisting force acting on the particle.
3. Find the time taken for \(P\) to reach its maximum height.

7 A particle \(P\) of mass \(m \mathrm {~kg}\) moves in a horizontal straight line against a resistive force of magnitude \(\mathrm { mkv } ^ { 2 } \mathrm {~N}\), where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of \(P\) after it has moved a distance \(x \mathrm {~m}\) and \(k\) is a positive constant. The initial speed of \(P\) is \(u \mathrm {~ms} ^ { - 1 }\).

1. Show that \(\mathrm { x } = \frac { 1 } { \mathrm { k } } \ln 2\) when \(\mathrm { v } = \frac { 1 } { 2 } \mathrm { u }\).
   Beginning at the instant when the speed of \(P\) is \(\frac { 1 } { 2 } u\), an additional force acts on \(P\). This force has magnitude \(\frac { 5 \mathrm {~m} } { \mathrm { v } } \mathrm { N }\) and acts in the direction of increasing \(x\).
2. Show that when the speed of \(P\) has increased again to \(u \mathrm {~ms} ^ { - 1 }\), the total distance travelled by \(P\) is given by an expression of the form $$\frac { 1 } { 3 k } \ln \left( \frac { A - k u ^ { 3 } } { B - k u ^ { 3 } } \right) ,$$ stating the values of the constants \(A\) and \(B\).
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

3 A particle \(P\) is moving in a horizontal straight line. Initially \(P\) is at the point \(O\) on the line and is moving with velocity \(25 \mathrm {~ms} ^ { - 1 }\). At time \(t \mathrm {~s}\) after passing through \(O\), the acceleration of \(P\) is \(\frac { 4000 } { ( 5 t + 4 ) ^ { 3 } } \mathrm {~ms} ^ { - 2 }\) in the direction \(P O\). The displacement of \(P\) from \(O\) at time \(t\) is \(x \mathrm {~m}\). Find an expression for \(x\) in terms of \(t\). \includegraphics[max width=\textwidth, alt={}, center]{c486c59a-2493-4dd3-bf1e-dde57fe744d9-06_894_809_260_628} An object is composed of a hemispherical shell of radius \(2 a\) attached to a closed hollow circular cylinder of height \(h\) and base radius \(a\). The hemispherical shell and the hollow cylinder are made of the same uniform material. The axes of symmetry of the shell and the cylinder coincide. \(A B\) is a diameter of the lower end of the cylinder (see diagram).

1. Find, in terms of \(a\) and \(h\), an expression for the distance of the centre of mass of the object from \(A B\). [4]
   The object is placed on a rough plane which is inclined to the horizontal at an angle \(\theta\), where \(\tan \theta = \frac { 2 } { 3 }\). The object is in equilibrium with \(A B\) in contact with the plane and lying along a line of greatest slope of the plane.
2. Find the set of possible values of \(h\), in terms of \(a\). \includegraphics[max width=\textwidth, alt={}, center]{c486c59a-2493-4dd3-bf1e-dde57fe744d9-08_629_1358_269_367} A light inextensible string \(A B\) passes through two small holes \(C\) and \(D\) in a smooth horizontal table where \(A C = 3 a\) and \(D B = a\). A particle of mass \(m\) is attached at the end \(A\) and moves in a horizontal circle with angular velocity \(\omega\). A particle of mass \(\frac { 3 } { 4 } m\) is attached to the end \(B\) and moves in a horizontal circle with angular velocity \(k \omega\). \(A C\) makes an angle \(\theta\) with the downward vertical and \(D B\) makes an angle \(\theta\) with the horizontal (see diagram). Find the value of \(k\).

6 A cyclist and his bicycle have a total mass of 81 kg . The cyclist starts from rest and rides in a straight line. The cyclist exerts a constant force of 135 N and the motion is opposed by a resistance of magnitude \(9 v \mathrm {~N}\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the cyclist's speed at time \(t \mathrm {~s}\) after starting.

1. Show that \(\frac { 9 } { 15 - v } \frac { \mathrm {~d} v } { \mathrm {~d} t } = 1\).
2. Solve this differential equation to show that \(v = 15 \left( 1 - \mathrm { e } ^ { - \frac { 1 } { 9 } t } \right)\).
3. Find the distance travelled by the cyclist in the first 9 s of the motion.

3 A particle \(P\) of mass 0.4 kg is released from rest at a point \(O\) on a smooth plane inclined at \(30 ^ { \circ }\) to the horizontal. \(P\) moves down the line of greatest slope through \(O\). The velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when its displacement from \(O\) is \(x \mathrm {~m}\). A retarding force of magnitude \(0.2 v ^ { 2 } \mathrm {~N}\) acts on \(P\) in the direction \(P O\).

1. Show that \(v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 5 - 0.5 v ^ { 2 }\).
2. Express \(v\) in terms of \(x\).

3 A particle \(P\) of mass 0.4 kg is projected horizontally along a smooth horizontal plane from a point \(O\). After projection the velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and its displacement from \(O\) is \(x \mathrm {~m}\). A force of magnitude \(8 x \mathrm {~N}\) directed away from \(O\) acts on \(P\) and a force of magnitude ( \(2 \mathrm { e } ^ { - x } + 4\) ) N opposes the motion of \(P\). One end of a light elastic string of natural length 0.5 m is attached to \(O\) and the other end of the string is attached to \(P\).

1. Show that \(v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 20 x - 10 - 5 \mathrm { e } ^ { - x }\) before the elastic string becomes taut.
2. Given that the initial velocity of \(P\) is \(6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), find \(v\) when the string first becomes taut.
   When the string is taut, the acceleration of \(P\) is proportional to \(\mathrm { e } ^ { - x }\).
3. Find the modulus of elasticity of the string.

3 A smooth horizontal surface has two fixed points \(O\) and \(A\) which are 0.8 m apart. A particle \(P\) of mass 0.25 kg is projected with velocity \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) horizontally from \(A\) in the direction away from \(O\). The velocity of \(P\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when the displacement of \(P\) from \(O\) is \(x \mathrm {~m}\). A force of magnitude \(k v ^ { 2 } x ^ { - 2 } \mathrm {~N}\) opposes the motion of \(P\).

1. Show that \(v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 4 k v ^ { 2 } x ^ { - 2 }\).
2. Express \(v\) in terms of \(k\) and \(x\).

9. (a) Use the substitution \(u = 4 - \sqrt { } x\) to find $$\int \frac { \mathrm { d } x } { 4 - \sqrt { } x }$$ A team of scientists is studying a species of slow growing tree.
The rate of change in height of a tree in this species is modelled by the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 4 - \sqrt { } h } { 20 }$$ where \(h\) is the height in metres and \(t\) is the time measured in years after the tree is planted.
(b) Find the range in values of \(h\) for which the height of a tree in this species is increasing.
(c) Given that one of these trees is 1 metre high when it is planted, calculate the time it would take to reach a height of 10 metres. Write your answer to 3 significant figures. \includegraphics[max width=\textwidth, alt={}, center]{5b698944-41ac-4072-b5e1-c580b7752c39-31_154_145_2599_1804}

1. (a) Express \(\frac { 3 x ^ { 2 } - 4 } { x ^ { 2 } ( 3 x - 2 ) }\) in partial fractions.
   (b) Given that \(x > \frac { 2 } { 3 }\), find the general solution of the differential equation

$$x ^ { 2 } ( 3 x - 2 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = y \left( 3 x ^ { 2 } - 4 \right)$$ Give your answer in the form \(y = \mathrm { f } ( x )\).

1. In freezing temperatures, ice forms on the surface of the water in a barrel. At time \(t\) hours after the start of freezing, the thickness of the ice formed is \(x \mathrm {~mm}\). You may assume that the thickness of the ice is uniform across the surface of the water.

At 4 pm there is no ice on the surface, and freezing begins.
At 6pm, after two hours of freezing, the ice is 1.5 mm thick.
In a simple model, the rate of increase of \(x\), in mm per hour, is assumed to be constant for a period of 20 hours. Using this simple model,

1. express \(t\) in terms of \(x\),
2. find the value of \(t\) when \(x = 3\) In a second model, the rate of increase of \(x\), in mm per hour, is given by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { \lambda } { ( 2 x + 1 ) } \text { where } \lambda \text { is a constant and } 0 \leqslant t \leqslant 20$$ Using this second model,
3. solve the differential equation and express \(t\) in terms of \(x\) and \(\lambda\),
4. find the exact value for \(\lambda\),
5. find at what time the ice is predicted to be 3 mm thick.

10. \begin{figure}[h] \end{figure} Diagram not drawn to scale Figure 3 shows a container in the shape of an inverted right circular cone which contains some water. The cone has an internal radius of 3 m and a vertical height of 5 m as shown in Figure 3. At time \(t\) seconds,the height of the water is \(h\) metres,the volume of the water is \(V \mathrm {~m} ^ { 3 }\) and water is leaking from a hole in the bottom of the container at a constant rate of \(0.02 \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }\) [The volume of a cone of radius \(r\) and height \(h\) is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\) .]

1. Show that,while the water is leaking, $$h ^ { 2 } \frac { \mathrm {~d} h } { \mathrm {~d} t } = - \frac { 1 } { \mathrm { k } \pi }$$ where \(k\) is a constant to be found. Given that the container is initially full of water,
2. express \(h\) in terms of \(t\) .
3. Find the time taken for the container to empty,giving your answer to the nearest minute.

6. (a) Express \(\frac { 5 - 4 x } { ( 2 x - 1 ) ( x + 1 ) }\) in partial fractions.
(b) (i) Find a general solution of the differential equation $$( 2 x - 1 ) ( x + 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = ( 5 - 4 x ) y , \quad x > \frac { 1 } { 2 }$$ Given that \(y = 4\) when \(x = 2\),
(ii) find the particular solution of this differential equation. Give your answer in the form \(y = \mathrm { f } ( x )\).

1. (a) Prove by differentiation that

$$\frac { \mathrm { d } } { \mathrm {~d} y } ( \ln \tan 2 y ) = \frac { 4 } { \sin 4 y } , \quad 0 < y < \frac { \pi } { 4 }$$ (b) Given that \(y = \frac { \pi } { 6 }\) when \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \cos x \sin 4 y , \quad 0 < y < \frac { \pi } { 4 }$$ Give your answer in the form \(\tan 2 y = A \mathrm { e } ^ { B \sin x }\), where \(A\) and \(B\) are constants to be determined.

12. \begin{figure}[h] \end{figure} Figure 4 shows a right cylindrical water tank. The diameter of the circular cross section of the tank is 4 m and the height is 2.25 m . Water is flowing into the tank at a constant rate of \(0.4 \pi \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\). There is a tap at a point \(T\) at the base of the tank. When the tap is open, water leaves the tank at a rate of \(0.2 \pi \sqrt { h } \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\), where \(h\) is the height of the water in metres.

1. Show that at time \(t\) minutes after the tap has been opened, the height \(h \mathrm {~m}\) of the water in the tank satisfies the differential equation $$20 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 2 - \sqrt { h }$$ At the instant when the tap is opened, \(t = 0\) and \(h = 0.16\)
2. Use the differential equation to show that the time taken to fill the tank to a height of 2.25 m is given by $$\int _ { 0.16 } ^ { 2.25 } \frac { 20 } { 2 - \sqrt { h } } \mathrm {~d} h$$ Using the substitution \(h = ( 2 - x ) ^ { 2 }\), or otherwise,
3. find the time taken to fill the tank to a height of 2.25 m . Give your answer in minutes to the nearest minute.
   (Solutions based entirely on graphical or numerical methods are not acceptable.)

13. (a) Express \(\frac { 1 } { ( 4 - x ) ( 2 - x ) }\) in partial fractions. The mass, \(x\) grams, of a substance at time \(t\) seconds after a chemical reaction starts is modelled by the differential equation
where \(k\) is a constant.
(b) solve the differential equation and show that the solution can be written as $$x = \frac { 4 - 4 \mathrm { e } ^ { 2 k t } } { 1 - 2 \mathrm { e } ^ { 2 k t } }$$ Given that \(k = 0.1\) (c) find the value of \(t\) when \(x = 1\), giving your answer, in seconds, to 3 significant figures. The mass, \(x\) grams, of a substance at time \(t\) seconds after a chemical reaction starts is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 4 - x ) ( 2 - x ) , \quad t \geqslant 0,0 \leqslant x < 2$$ where \(k\) is a constant. $$\text { Given that when } t = 0 , x = 0$$ (b) solve the differential equation and show that the solution can be written as

2. (a) Find \(\int \frac { 4 x + 3 } { x } \mathrm {~d} x , \quad x > 0\) (b) Given that \(y = 25\) at \(x = 1\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 4 x + 3 ) y ^ { \frac { 1 } { 2 } } } { x } \quad x > 0 , y > 0$$ giving your answer in the form \(y = [ \mathrm { g } ( x ) ] ^ { 2 }\).
VJYV SIHI NITIIYIM ION OC
VI4V SIHI NI JAHMA ION OC
VEYV SIHI NI JIIIM ION OO \includegraphics[max width=\textwidth, alt={}, center]{a9870c94-0910-46ec-a54a-44a431cb324e-05_52_49_2777_1886}

1. (a) Given \(0 \leqslant h < 25\), use the substitution \(u = 5 - \sqrt { h }\) to show that

$$\int \frac { \mathrm { d } h } { 5 - \sqrt { h } } = - 10 \ln ( 5 - \sqrt { h } ) - 2 \sqrt { h } + k$$ where \(k\) is a constant.
(6) A team of scientists is studying a species of tree.
The rate of change in height of a tree of this species is modelled by the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { t ^ { 0.2 } ( 5 - \sqrt { h } ) } { 5 }$$ where \(h\) is the height of the tree in metres and \(t\) is the time in years after the tree is planted.
One of these trees is 2 metres high when it is planted.
(b) Use integration to calculate the time it would take for this tree to reach a height of 15 metres, giving your answer to one decimal place.
(c) Hence calculate the rate of change in height of this tree when its height is 15 metres. Write your answer in centimetres per year to the nearest centimetre.

9. A rare species of mammal is being studied. The population \(P\), \(t\) years after the study started, is modelled by the formula $$P = \frac { 900 \mathrm { e } ^ { \frac { 1 } { 4 } t } } { 3 \mathrm { e } ^ { \frac { 1 } { 4 } t } - 1 } , \quad t \in \mathbb { R } , \quad t \geqslant 0$$ Using the model,

1. calculate the number of mammals at the start of the study,
2. calculate the exact value of \(t\) when \(P = 315\) Give your answer in the form \(a \ln k\), where \(a\) and \(k\) are integers to be determined.
3. 1. Find \(\frac { \mathrm { d } P } { \mathrm {~d} t }\)
   2. Hence find the value of \(\frac { \mathrm { d } P } { \mathrm {~d} t }\) when \(t = 8\), giving your answer to 2 decimal places.

13. The volume of a spherical balloon of radius \(r \mathrm {~m}\) is \(V \mathrm {~m} ^ { 3 }\), where \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\)

1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\) Given that the volume of the balloon increases with time \(t\) seconds according to the formula $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 20 } { V ( 0.05 t + 1 ) ^ { 3 } } , \quad t \geqslant 0$$
2. find an expression in terms of \(r\) and \(t\) for \(\frac { \mathrm { d } r } { \mathrm {~d} t }\) Given that \(V = 1\) when \(t = 0\)
3. solve the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 20 } { V ( 0.05 t + 1 ) ^ { 3 } }$$ giving your answer in the form \(V ^ { 2 } = \mathrm { f } ( t )\).
4. Hence find the radius of the balloon at time \(t = 20\), giving your answer to 3 significant figures.
   \includegraphics[max width=\textwidth, alt={}]{c6bde466-61ec-437d-a3b4-84511a98d788-48_2632_1828_121_121}

11. A team of conservationists is studying the population of meerkats on a nature reserve. The population is modelled by the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 15 } P ( 5 - P ) , \quad t \geqslant 0$$ where \(P\), in thousands, is the population of meerkats and \(t\) is the time measured in years since the study began. Given that when \(t = 0 , P = 1\),

1. solve the differential equation, giving your answer in the form $$P = \frac { a } { b + c \mathrm { e } ^ { - \frac { 1 } { 3 } t } }$$ where \(a\), \(b\) and \(c\) are integers.
2. Hence show that the population cannot exceed 5000

10. (a) Write \(\frac { 1 } { ( H - 5 ) ( H + 3 ) }\) in partial fraction form. The depth of water in a storage tank is being monitored.
The depth of water in the tank, \(H\) metres, is modelled by the differential equation $$\frac { \mathrm { d } H } { \mathrm {~d} t } = - \frac { ( H - 5 ) ( H + 3 ) } { 40 }$$ where \(t\) is the time, in days, from when monitoring began.
Given that the initial depth of water in the tank was 13 m ,
(b) solve the differential equation to show that $$H = \frac { 10 + 3 \mathrm { e } ^ { - 0.2 t } } { 2 - \mathrm { e } ^ { - 0.2 t } }$$ (c) Hence find the time taken for the depth of water in the tank to fall to 8 m .
(Solutions relying entirely on calculator technology are not acceptable.) According to the model, the depth of water in the tank will eventually fall to \(k\) metres.
(d) State the value of the constant \(k\).

7. Water is flowing into a large container and is leaking from a hole at the base of the container. At time \(t\) seconds after the water starts to flow, the volume, \(V \mathrm {~cm} ^ { 3 }\), of water in the container is modelled by the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = 300 - k V$$ where \(k\) is a constant.

1. Solve the differential equation to show that, according to the model, $$V = \frac { 300 } { k } + A \mathrm { e } ^ { - k t }$$ where \(A\) is a constant.
   (5) Given that the container is initially empty and that when \(t = 10\), the volume of water is increasing at a rate of \(200 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\)
2. find the exact value of \(k\).
3. Hence find, according to the model, the time taken for the volume of water in the container to reach 6 litres. Give your answer to the nearest second.

13
- 1
4 \end{array} \right) + \mu \left( \begin{array} { r } 5
1
- 3 \end{array} \right)$$ where \(\lambda\) and \(\mu\) are scalar parameters.

1. Show that \(l _ { 1 }\) and \(l _ { 2 }\) meet and find the position vector of their point of intersection \(A\).
2. Find the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\), giving your answer in degrees to one decimal place. A circle with centre \(A\) and radius 35 cuts the line \(l _ { 1 }\) at the points \(P\) and \(Q\). Given that the \(x\) coordinate of \(P\) is greater than the \(x\) coordinate of \(Q\),
3. find the coordinates of \(P\) and the coordinates of \(Q\). 6. Use integration by parts to show that $$\int \mathrm { e } ^ { 2 x } \cos 3 x \mathrm {~d} x = p \mathrm { e } ^ { 2 x } \sin 3 x + q \mathrm { e } ^ { 2 x } \cos 3 x + k$$ where \(p\) and \(q\) are rational numbers to be found and \(k\) is an arbitrary constant.\\ (6)\\ 7. Water is flowing into a large container and is leaking from a hole at the base of the container. At time \(t\) seconds after the water starts to flow, the volume, \(V \mathrm {~cm} ^ { 3 }\), of water in the container is modelled by the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = 300 - k V$$ where \(k\) is a constant.
4. Solve the differential equation to show that, according to the model, $$V = \frac { 300 } { k } + A \mathrm { e } ^ { - k t }$$ where \(A\) is a constant.\\ (5) Given that the container is initially empty and that when \(t = 10\), the volume of water is increasing at a rate of \(200 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\)
5. find the exact value of \(k\).
6. Hence find, according to the model, the time taken for the volume of water in the container to reach 6 litres. Give your answer to the nearest second.\\ 8. Use proof by contradiction to prove that, for all positive real numbers \(x\) and \(y\), $$\frac { 9 x } { y } + \frac { y } { x } \geqslant 6$$ 9. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{594542dd-ee2d-49b6-9fab-77b2d1a44f8c-24_632_734_214_607} \captionsetup{labelformat=empty} \caption{Figure 1}
   \end{figure} Figure 1 shows a sketch of a closed curve with parametric equations $$x = 5 \cos \theta \quad y = 3 \sin \theta - \sin 2 \theta \quad 0 \leqslant \theta < 2 \pi$$ The region enclosed by the curve is rotated through \(\pi\) radians about the \(x\)-axis to form a solid of revolution.
7. Show that the volume, \(V\), of the solid of revolution is given by $$V = 5 \pi \int _ { \alpha } ^ { \beta } \sin ^ { 3 } \theta ( 3 - 2 \cos \theta ) ^ { 2 } \mathrm {~d} \theta$$ where \(\alpha\) and \(\beta\) are constants to be found.
8. Use the substitution \(u = \cos \theta\) and algebraic integration to show that \(V = k \pi\) where \(k\) is a rational number to be found. \includegraphics[max width=\textwidth, alt={}, center]{594542dd-ee2d-49b6-9fab-77b2d1a44f8c-28_2649_1889_109_178}

1. The volume \(V \mathrm {~cm} ^ { 3 }\) of a spherical balloon with radius \(r \mathrm {~cm}\) is given by the formula

$$V = \frac { 4 } { 3 } \pi r ^ { 3 }$$

1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\) giving your answer in simplest form. At time \(t\) seconds, the volume of the balloon is increasing according to the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 900 } { ( 2 t + 3 ) ^ { 2 } } \quad t \geqslant 0$$ Given that \(V = 0\) when \(t = 0\)
2. 1. solve this differential equation to show that $$V = \frac { 300 t } { 2 t + 3 }$$
   2. Hence find the upper limit to the volume of the balloon.
3. Find the radius of the balloon at \(t = 3\), giving your answer in cm to 3 significant figures.
4. Find the rate of increase of the radius of the balloon at \(t = 3\), giving your answer to 2 significant figures. Show your working and state the units of your answer.