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

8

1. A pond is initially empty and is then filled gradually with water. After \(t\) minutes, the depth of the water, \(x\) metres, satisfies the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { \sqrt { 4 + 5 x } } { 5 ( 1 + t ) ^ { 2 } }$$ Solve this differential equation to find \(x\) in terms of \(t\).
2. Another pond is gradually filling with water. After \(t\) minutes, the surface of the water forms a circle of radius \(r\) metres. The rate of change of the radius is inversely proportional to the area of the surface of the water.
   1. Write down a differential equation, in the variables \(r\) and \(t\) and a constant of proportionality, which represents how the radius of the surface of the water is changing with time.
      (You are not required to solve your differential equation.)
   2. When the radius of the pond is 1 metre, the radius is increasing at a rate of 4.5 metres per second. Find the radius of the pond when the radius is increasing at a rate of 0.5 metres per second.
      [0pt] [2 marks]
      \includegraphics[max width=\textwidth, alt={}]{fdd3905e-11f7-4b20-adfe-4c686018a221-18_1277_1709_1430_153}
      \includegraphics[max width=\textwidth, alt={}]{fdd3905e-11f7-4b20-adfe-4c686018a221-20_2288_1707_221_153}

6. Liquid is poured into a container at a constant rate of \(30 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). At time \(t\) seconds liquid is leaking from the container at a rate of \(\frac { 2 } { 15 } V \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\), where \(V \mathrm {~cm} ^ { 3 }\) is the volume of liquid in the container at that time.

1. Show that $$- 15 \frac { \mathrm {~d} V } { \mathrm {~d} t } = 2 V - 450$$ Given that \(V = 1000\) when \(t = 0\),
2. find the solution of the differential equation, in the form \(V = \mathrm { f } ( t )\).
3. Find the limiting value of \(V\) as \(t \rightarrow \infty\).

6. A radioactive isotope decays in such a way that the rate of change of the number \(N\) of radioactive atoms present after \(t\) days, is proportional to \(N\).

1. Write down a differential equation relating \(N\) and \(t\).
2. Show that the general solution may be written as \(N = A \mathrm { e } ^ { - k t }\), where \(A\) and \(k\) are positive constants. Initially the number of radioactive atoms present is \(7 \times 10 ^ { 18 }\) and 8 days later the number present is \(3 \times 10 ^ { 17 }\).
3. Find the value of \(k\).
4. Find the number of radioactive atoms present after a further 8 days.

4. During a chemical reaction, a compound is being made from two other substances. At time \(t\) hours after the start of the reaction, \(x \mathrm {~g}\) of the compound has been produced. Assuming that \(x = 0\) initially, and that $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 2 ( x - 6 ) ( x - 3 )$$

1. show that it takes approximately 7 minutes to produce 2 g of the compound.
2. Explain why it is not possible to produce 3 g of the compound.
   4. continued

1. Given that \(y = - 2\) when \(x = 1\), solve the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } \sqrt { x }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).

5. A bath is filled with hot water which is allowed to cool. The temperature of the water is \(\theta ^ { \circ } \mathrm { C }\) after cooling for \(t\) minutes and the temperature of the room is assumed to remain constant at \(20 ^ { \circ } \mathrm { C }\). Given that the rate at which the temperature of the water decreases is proportional to the difference in temperature between the water and the room,

1. write down a differential equation connecting \(\theta\) and \(t\). Given also that the temperature of the water is initially \(37 ^ { \circ } \mathrm { C }\) and that it is \(36 ^ { \circ } \mathrm { C }\) after cooling for four minutes,
2. find, to 3 significant figures, the temperature of the water after ten minutes. Advice suggests that the temperature of the water should be allowed to cool to \(33 ^ { \circ } \mathrm { C }\) before a child gets in.
3. Find, to the nearest second, how long a child should wait before getting into the bath.
   5. continued

5. At time \(t = 0\), a tank of height 2 metres is completely filled with water. Water then leaks from a hole in the side of the tank such that the depth of water in the tank, \(y\) metres, after \(t\) hours satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} t } = - k \mathrm { e } ^ { - 0.2 t }$$ where \(k\) is a positive constant,

1. Find an expression for \(y\) in terms of \(k\) and \(t\). Given that two hours after being filled the depth of water in the tank is 1.6 metres,
2. find the value of \(k\) to 4 significant figures. Given also that the hole in the tank is \(h \mathrm {~cm}\) above the base of the tank,
3. show that \(h = 79\) to 2 significant figures.
   5. continued

7. When a plague of locusts attacks a wheat crop, the proportion of the crop destroyed after \(t\) hours is denoted by \(x\). In a model, it is assumed that the rate at which the crop is destroyed is proportional to \(x ( 1 - x )\). A plague of locusts is discovered in a wheat crop when one quarter of the crop has been destroyed. Given that the rate of destruction at this instant is such that if it remained constant, the crop would be completely destroyed in a further six hours,

1. show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 2 } { 3 } x ( 1 - x )\),
2. find the percentage of the crop destroyed three hours after the plague of locusts is first discovered.
   7. continued
   7. continued

14
2 \end{array} \right) , $$ and\\ where \(a\) is a constant and \(\lambda\) and \(\mu\) are scalar parameters.\\ Given that the two lines intersect,\\

1. find the position vector of their point of intersection,
2. find the value of \(a\). Given also that \(\theta\) is the acute angle between the lines,
3. find the value of \(\cos \theta\) in the form \(k \sqrt { 5 }\) where \(k\) is rational.\\ 4. continued\\ 5. A curve has the equation $$x ^ { 2 } - 4 x y + 2 y ^ { 2 } = 1$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in its simplest form in terms of \(x\) and \(y\).
2. Show that the tangent to the curve at the point \(P ( 1,2 )\) has the equation $$3 x - 2 y + 1 = 0$$ The tangent to the curve at the point \(Q\) is parallel to the tangent at \(P\).
3. Find the coordinates of \(Q\).\\ 5. continued\\ 6. The rate of increase in the number of bacteria in a culture, \(N\), at time \(t\) hours is proportional to \(N\).

\\

1. Write down a differential equation connecting \(N\) and \(t\). Given that initially there are \(N _ { 0 }\) bacteria present in a culture,
2. Show that \(N = N _ { 0 } \mathrm { e } ^ { k t }\), where \(k\) is a positive constant. Given also that the number of bacteria present doubles every six hours,
3. find the value of \(k\),
4. find how long it takes for the number of bacteria to increase by a factor of ten, giving your answer to the nearest minute. of ten, giving your answer to the nearest minute.\\ 6. continued\\ 7. A curve has parametric equations $$x = \sec \theta + \tan \theta , \quad y = \operatorname { cosec } \theta + \cot \theta , \quad 0 < \theta < \frac { \pi } { 2 } .$$

1. Show that \(x + \frac { 1 } { x } = 2 \sec \theta\). Given that \(y + \frac { 1 } { y } = 2 \operatorname { cosec } \theta\),
2. find a cartesian equation for the curve.
3. Show that \(\frac { \mathrm { d } x } { \mathrm {~d} \theta } = \frac { 1 } { 2 } \left( x ^ { 2 } + 1 \right)\).
4. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
   7. continued
   7. continued

6 Alice places a toy, of mass 0.4 kg , on a slope. The toy is set in motion with an initial velocity of \(1 \mathrm {~ms} ^ { - 1 }\) down the slope. The resultant force acting on the toy is \(( 2 - 4 v )\) newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the toy's velocity at time \(t\) seconds after it is set in motion.

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = - 10 ( v - 0.5 )\).
2. By using \(\int \frac { 1 } { v - 0.5 } \mathrm {~d} v = - \int 10 \mathrm {~d} t\), find \(v\) in terms of \(t\).
3. Find the time taken for the toy's velocity to reduce to \(0.55 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). \(7 \quad\) A small bead, of mass \(m\), is suspended from a fixed point \(O\) by a light inextensible string of length \(a\). With the string taut, the bead is at the point \(B\), vertically below \(O\), when it is set into vertical circular motion with an initial horizontal velocity \(u\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{06c3e260-8167-4616-97d4-0f360a376a0f-5_616_613_520_733} The string does not become slack in the subsequent motion. The velocity of the bead at the point \(A\), where \(A\) is vertically above \(O\), is \(v\).

5 A particle, of mass 12 kg , is moving along a straight horizontal line. At time \(t\) seconds, the particle has speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). As the particle moves, it experiences a resistance force of magnitude \(4 v ^ { \frac { 1 } { 3 } }\). No other horizontal force acts on the particle. The initial speed of the particle is \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that $$v = \left( 4 - \frac { 2 } { 9 } t \right) ^ { \frac { 3 } { 2 } }$$
2. Find the value of \(t\) when the particle comes to rest.

7 A particle of mass 20 kg moves along a straight horizontal line. At time \(t\) seconds the velocity of the particle is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A resistance force of magnitude \(10 \sqrt { v }\) newtons acts on the particle while it is moving. At time \(t = 0\) the velocity of the particle is \(25 \mathrm {~ms} ^ { - 1 }\).

1. Show that, at time \(t\) $$v = \left( \frac { 20 - t } { 4 } \right) ^ { 2 }$$
2. State the value of \(t\) when the particle comes to rest.

7 A stone of mass \(m\) is moving along the smooth horizontal floor of a tank which is filled with a viscous liquid. At time \(t\), the stone has speed \(v\). As the stone moves, it experiences a resistance force of magnitude \(\lambda m v\), where \(\lambda\) is a constant.

1. Show that $$\frac { \mathrm { d } v } { \mathrm {~d} t } = - \lambda v$$
2. The initial speed of the stone is \(U\). Show that $$v = U \mathrm { e } ^ { - \lambda t }$$

6 A car accelerates from rest along a straight horizontal road. The car's engine produces a constant horizontal force of magnitude 4000 N .
At time \(t\) seconds, the speed of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and a resistance force of magnitude \(40 v\) newtons acts upon the car. The mass of the car is 1600 kg .

1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 100 - v } { 40 }\).
2. Find the velocity of the car at time \(t\).

7 A parachutist, of mass 72 kg , is falling vertically. He opens his parachute at time \(t = 0\) when his speed is \(30 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). He then experiences an air resistance force of magnitude \(240 v\) newtons, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is his speed at time \(t\) seconds.

1. When \(t > 0\), show that \(- \frac { 3 } { 10 } \frac { \mathrm {~d} v } { \mathrm {~d} t } = v - 2.94\).
2. Find \(v\) in terms of \(t\).
3. Sketch a graph to show how, for \(t \geqslant 0\), the parachutist's speed varies with time.
   [0pt] [2 marks]

6 A stone of mass 0.125 kg falls freely under gravity, from rest, until it has travelled a distance of 10 m . The stone then continues to fall in a medium which exerts an upward resisting force of \(0.025 v \mathrm {~N}\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of the stone \(t \mathrm {~s}\) after the instant that it enters the resisting medium.

1. Show by integration that \(v = 49 - 35 \mathrm { e } ^ { - 0.2 t }\).
2. Find how far the stone travels during the first 3 seconds in the medium.

4 A particle \(P\) of mass 0.2 kg travels in a straight line on a horizontal surface. It passes through a point \(O\) on the surface with speed \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A resistive force of magnitude \(0.2 \left( v + v ^ { 2 } \right) \mathrm { N }\) acts on \(P\) in the direction opposite to its motion, where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the speed of \(P\) when it is at a distance \(x \mathrm {~m}\) from \(O\).

1. Show that \(\frac { 1 } { 1 + v } \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 1\).
2. By solving the differential equation in part (i) show that \(\frac { - \mathrm { e } ^ { x } } { 3 - \mathrm { e } ^ { x } } \frac { \mathrm {~d} x } { \mathrm {~d} t } = - 1\), where \(t\) s is the time taken for \(P\) to travel \(x \mathrm {~m}\) from \(O\).
3. Hence find the value of \(t\) when \(x = 1\).

3 A particle \(P\) of mass 0.2 kg is projected horizontally with speed \(u \mathrm {~ms} ^ { - 1 }\) from a fixed point \(O\) on a smooth horizontal surface. \(P\) moves in a straight line and, at time \(t \mathrm {~s}\) after projection, \(P\) has speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and is \(x \mathrm {~m}\) from \(O\). The only force acting on \(P\) has magnitude \(0.4 v ^ { 2 } \mathrm {~N}\) and is directed towards \(O\).

1. Show that \(\frac { 1 } { v } \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 2\).
2. Hence show that \(v = u \mathrm { e } ^ { - 2 x }\).
3. Find \(u\), given that \(x = 2\) when \(t = 4\).

4 A particle \(P\) of mass \(m \mathrm {~kg}\) is held at rest at a point \(O\) on a fixed plane inclined at an angle \(\sin ^ { - 1 } \left( \frac { 4 } { 7 } \right)\) to the horizontal. \(P\) is released and moves down the plane. The total resistance acting on \(P\) is \(0.2 m v \mathrm {~N}\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the velocity of \(P\) at time \(t \mathrm {~s}\) after leaving \(O\).

1. Show that \(5 \frac { \mathrm {~d} v } { \mathrm {~d} t } = 28 - v\) and hence find an expression for \(v\) in terms of \(t\).
2. Find the acceleration of \(P\) when \(t = 10\).

3. At time \(t = 0\), a particle of mass \(m\) is projected vertically downwards with speed \(U\) from a point above the ground. At time \(t\) the speed of the particle is \(v\) and the magnitude of the air resistance is modelled as being \(m k v\), where \(k\) is a constant. Given that \(U < \frac { \boldsymbol { g } } { \mathbf { 2 } \boldsymbol { k } }\), find, in terms of \(k , U\) and \(g\), the time taken for the particle to double its speed.
(8)

5. A light elastic spring has natural length \(l\) and modulus of elasticity \(m g\). One end of the spring is fixed to a point \(O\) on a rough horizontal table. The other end is attached to a particle \(P\) of mass \(m\) which is at rest on the table with \(O P = l\). At time \(t = 0\) the particle is projected with speed \(\sqrt { } ( g l )\) along the table in the direction \(O P\). At time \(t\) the displacement of \(P\) from its initial position is \(x\) and its speed is \(v\). The motion of \(P\) is subject to air resistance of magnitude \(2 m v \omega\), where \(\omega = \sqrt { \frac { g } { l } }\). The coefficient of friction between \(P\) and the table is 0.5 .

1. Show that, until \(P\) first comes to rest, $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \omega \frac { \mathrm {~d} x } { \mathrm {~d} t } + \omega ^ { 2 } x = - 0.5 g$$
2. Find \(x\) in terms of \(t , l\) and \(\omega\).
3. Hence find, in terms of \(\omega\), the time taken for \(P\) to first come to instantaneous rest.
   (3)

6. \begin{figure}[h] \end{figure} A river is 30 m wide and flows between two straight parallel banks. At each point of the river, the direction of flow is parallel to the banks. At time \(t = 0\), a boat leaves a point \(O\) on one bank and moves in a straight line across the river to a point \(P\) on the opposite bank. Its path \(O P\) is perpendicular to both banks and \(O P = 30 \mathrm {~m}\), as shown in Figure 2. The speed of flow of the river, \(r \mathrm {~m} \mathrm {~s} ^ { - 1 }\), at a point on \(O P\) which is at a distance \(x \mathrm {~m}\) from \(O\), is modelled as $$r = \frac { 1 } { 10 } x , \quad 0 \leq x \leq 30$$ The speed of the boat relative to the water is constant at \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). At time \(t\) seconds the boat is at a distance \(x \mathrm {~m}\) from \(O\) and is moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the direction \(O P\).

1. Show that $$100 v ^ { 2 } = 2500 - x ^ { 2 }$$
2. Hence show that $$\frac { \mathbf { d } ^ { 2 } x } { \mathbf { d } t ^ { 2 } } + \frac { x } { 100 } = 0$$
3. Find the total time taken for the boat to cross the river from \(O\) to \(P\).
   (9)

2. At time \(t = 0\), a particle \(P\) of mass \(m\) is projected vertically upwards with speed \(\sqrt { \frac { g } { k } }\), where \(k\) is a constant. At time \(t\) the speed of \(P\) is \(v\). The particle \(P\) moves against air resistance whose magnitude is modelled as being \(m k v ^ { 2 }\) when the speed of \(P\) is \(v\). Find, in terms of \(k\), the distance travelled by \(P\) until its speed first becomes half of its initial speed.
(9)

6. A light elastic spring \(A B\) has natural length \(2 a\) and modulus of elasticity \(2 m n ^ { 2 } a\), where \(n\) is a constant. A particle \(P\) of mass \(m\) is attached to the end \(A\) of the spring. At time \(t = 0\), the spring, with \(P\) attached, lies at rest and unstretched on a smooth horizontal plane. The other end \(B\) of the spring is then pulled along the plane in the direction \(A B\) with constant acceleration \(f\). At time \(t\) the extension of the spring is \(x\).

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + n ^ { 2 } x = f .$$
2. Find \(x\) in terms of \(n , f\) and \(t\). Hence find
3. the maximum extension of the spring,
4. the speed of \(P\) when the spring first reaches its maximum extension.
   \section*{June 2009}

2 A light elastic string AB has stiffness \(k\). The end A is attached to a fixed point and a particle of mass \(m\) is attached at the end B . With the string vertical, the particle is released from rest from a point at a distance \(a\) below its equilibrium position. At time \(t\), the displacement of the particle below the equilibrium position is \(x\) and the velocity of the particle is \(v\).

1. Show that $$m v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - k x$$
2. Show that, while the particle is moving upwards and the string is taut, $$v = - \sqrt { \frac { k } { m } \left( a ^ { 2 } - x ^ { 2 } \right) }$$
3. Hence use integration to find an expression for \(x\) at time \(t\) while the particle is moving upwards and the string is taut.