Past Exam Questions

In a certain chemical reaction, the amount, \(x\) grams, of a substance present is decreasing. The rate of decrease of \(x\) is proportional to the product of \(x\) and the time, \(t\) seconds, since the start of the reaction. Thus \(x\) and \(t\) satisfy the differential equation

\[ \frac{dx}{dt} = -kxt, \]

where \(k\) is a positive constant. At the start of the reaction, when \(t = 0\), \(x = 100\).

(i) Solve this differential equation, obtaining a relation between \(x\), \(k\), and \(t\).

(ii) Twenty seconds after the start of the reaction, the amount of substance present is 90 grams. Find the time after the start of the reaction at which the amount of substance present is 50 grams.

Compressed air is escaping from a container. The pressure of the air in the container at time \(t\) is \(P\), and the constant atmospheric pressure of the air outside the container is \(A\). The rate of decrease of \(P\) is proportional to the square root of the pressure difference \(P - A\). Thus the differential equation connecting \(P\) and \(t\) is

\[ \frac{dP}{dt} = -k\sqrt{P - A}, \]

where \(k\) is a positive constant.

1. Find, in any form, the general solution of this differential equation. [3]
2. Given that \(P = 5A\) when \(t = 0\), and that \(P = 2A\) when \(t = 2\), show that \[ k = \sqrt{A}. \] [4]
3. Find the value of \(t\) when \(P = A\). [2]
4. Obtain an expression for \(P\) in terms of \(A\) and \(t\). [2]

In a certain chemical reaction, the amount, \(x\) grams, of a substance is increasing. The differential equation satisfied by \(x\) and \(t\), the time in seconds since the reaction began, is

\[ \frac{dx}{dt} = kx e^{-0.1t}, \]

where \(k\) is a positive constant. It is given that \(x = 20\) at the start of the reaction.

(a) Solve the differential equation, obtaining a relation between \(x\), \(t\), and \(k\).

(b) Given that \(x = 40\) when \(t = 10\), find the value of \(k\) and find the value approached by \(x\) as \(t\) becomes large.

At time \(t\) days after the start of observations, the number of insects in a population is \(N\). The variation in the number of insects is modelled by the differential equation

\[ \frac{dN}{dt} = kN^{\frac{3}{2}} \cos(0.02t), \]

where \(k\) is a constant and \(N\) is treated as a continuous variable. It is given that when \(t = 0\), \(N = 100\).

(a) Solve the differential equation, obtaining a relation between \(N\), \(k\), and \(t\).

(b) Given also that \(N = 625\) when \(t = 50\), find the value of \(k\).

(c) Obtain an expression for \(N\) in terms of \(t\), and find the greatest value of \(N\) predicted by this model.

The variables x and t satisfy the differential equation \(5 \frac{dx}{dt} = (20-x)(40-x)\). It is given that \(x = 10\) when \(t = 0\).

(i) Using partial fractions, solve the differential equation, obtaining an expression for x in terms of t.

(ii) State what happens to the value of x when t becomes large.

In a certain chemical reaction, the amount, \(x\) grams, of a substance is decreasing. The differential equation relating \(x\) and \(t\), the time in seconds since the reaction started, is

\[ \frac{dx}{dt} = -\frac{kx}{\sqrt{t}}, \]

where \(k\) is a positive constant. It is given that \(x = 100\) at the start of the reaction.

1. Solve the differential equation, obtaining a relation between \(x\), \(t\), and \(k\).
2. Given that \(t = 25\) when \(x = 80\), find the value of \(t\) when \(x = 40\).

A large field of area 4 km² is becoming infected with a soil disease. At time t years the area infected is x km² and the rate of growth of the infected area is given by the differential equation \(\frac{dx}{dt} = kx(4-x)\), where k is a positive constant. It is given that when t = 0, x = 0.4 and that when t = 2, x = 2.

1. Solve the differential equation and show that \(k = \frac{1}{4} \ln 3\).
2. Find the value of t when 90% of the area of the field is infected.

The number of micro-organisms in a population at time \(t\) is denoted by \(M\). At any time, the variation in \(M\) is assumed to satisfy the differential equation

\[ \frac{dM}{dt} = k\sqrt{M}\cos(0.02t), \]

where \(k\) is a constant and \(M\) is taken to be a continuous variable. It is given that when \(t = 0\), \(M = 100\).

1. Solve the differential equation, obtaining a relation between \(M\), \(k\), and \(t\).
2. Given also that \(M = 196\) when \(t = 50\), find the value of \(k\).
3. Obtain an expression for \(M\) in terms of \(t\) and find the least possible number of micro-organisms.

The number of organisms in a population at time t is denoted by x. Treating x as a continuous variable, the differential equation satisfied by x and t is

\(\frac{dx}{dt} = \frac{xe^{-t}}{k + e^{-t}},\)

where k is a positive constant.

1. Given that x = 10 when t = 0, solve the differential equation, obtaining a relation between x, k, and t.
2. Given also that x = 20 when t = 1, show that k = 1 - \(\frac{2}{e}\).
3. Show that the number of organisms never reaches 48, however large t becomes.

In a certain country the government charges tax on each litre of petrol sold to motorists. The revenue per year is \(R\) million dollars when the rate of tax is \(x\) dollars per litre. The variation of \(R\) with \(x\) is modelled by the differential equation

\(\frac{dR}{dx} = R \left( \frac{1}{x} - 0.57 \right),\)

where \(R\) and \(x\) are taken to be continuous variables. When \(x = 0.5, R = 16.8\).

(i) Solve the differential equation and obtain an expression for \(R\) in terms of \(x\). [6]

(ii) This model predicts that \(R\) cannot exceed a certain amount. Find this maximum value of \(R\). [3]

A large plantation of area 20 km² is becoming infected with a plant disease. At time t years the area infected is x km² and the rate of increase of x is proportional to the ratio of the area infected to the area not yet infected.

When t = 0, x = 1 and \(\frac{dx}{dt} = 1\).

(a) Show that x and t satisfy the differential equation \(\frac{dx}{dt} = \frac{19x}{20-x}\).

(b) Solve the differential equation and show that when t = 1 the value of x satisfies the equation \(x = e^{0.9 + 0.05x}\).

(c) Use an iterative formula based on the equation in part (b), with an initial value of 2, to determine x correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

(d) Calculate the value of t at which the entire plantation becomes infected.

A water tank has vertical sides and a horizontal rectangular base, as shown in the diagram. The area of the base is \(2\text{ m}^2\). At time \(t = 0\), the tank is empty and water begins to flow into it at a rate of \(1\text{ m}^3\) per hour. At the same time, water begins to flow out from the base at a rate of \(0.2\sqrt{h}\text{ m}^3\) per hour, where \(h\) m is the depth of water in the tank at time \(t\) hours.

(i) Form a differential equation satisfied by \(h\) and \(t\), and show that the time \(T\) hours taken for the depth of water to reach \(4\) m is given by

\[ T = \int_0^4 \frac{10}{5 - \sqrt{h}} \, dh. \]

(ii) Using the substitution \(u = 5 - \sqrt{h}\), find the value of \(T\).

The diagram shows a variable point \(P\) with coordinates \((x, y)\) and the point \(N\) which is the foot of the perpendicular from \(P\) to the \(x\)-axis. \(P\) moves on a curve such that, for all \(x \geq 0\), the gradient of the curve is equal in value to the area of the triangle \(OPN\), where \(O\) is the origin.

(i) State a differential equation satisfied by \(x\) and \(y\).

The point with coordinates \((0, 2)\) lies on the curve.

(ii) Solve the differential equation to obtain the equation of the curve, expressing \(y\) in terms of \(x\).

(iii) Sketch the curve.

Naturalists are managing a wildlife reserve to increase the number of plants of a rare species. The number of plants at time t years is denoted by N, where N is treated as a continuous variable.

(i) It is given that the rate of increase of N with respect to t is proportional to (N - 150). Write down a differential equation relating N, t and a constant of proportionality.

(ii) Initially, when t = 0, the number of plants was 650. It was noted that, at a time when there were 900 plants, the number of plants was increasing at a rate of 60 per year. Express N in terms of t.

(iii) The naturalists had a target of increasing the number of plants from 650 to 2000 within 15 years. Will this target be met?

The population of a country at time \(t\) years is \(N\) millions. At any time, \(N\) is assumed to increase at a rate proportional to the product of \(N\) and \((1 - 0.01N)\). When \(t = 0\), \(N = 20\) and

\[ \frac{dN}{dt} = 0.32. \]

(i) Treating \(N\) and \(t\) as continuous variables, show that they satisfy the differential equation

\[ \frac{dN}{dt} = 0.02N(1 - 0.01N). \]

(ii) Solve the differential equation, obtaining an expression for \(t\) in terms of \(N\).

(iii) Find the time at which the population will be double its value at \(t = 0\).

A particular solution of the differential equation

\(3y^2 \frac{dy}{dx} = 4(y^3 + 1) \cos^2 x\)

is such that \(y = 2\) when \(x = 0\). The diagram shows a sketch of the graph of this solution for \(0 \leq x \leq 2\pi\); the graph has stationary points at \(A\) and \(B\). Find the \(y\)-coordinates of \(A\) and \(B\), giving each coordinate correct to 1 decimal place.

A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi-vertical angle is 60°, as shown in the diagram. At time t = 0, the tank is full and the depth of water is H. At this instant, a tap at C is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to \(\sqrt{h}\), where h is the depth of water at time t. The tank becomes empty when \(t = 60\).

(i) Show that h and t satisfy a differential equation of the form \(\frac{dh}{dt} = -Ah^{-\frac{3}{2}}\), where A is a positive constant.

(ii) Solve the differential equation given in part (i) and obtain an expression for t in terms of h and H.

(iii) Find the time at which the depth reaches \(\frac{1}{2}H\).

[The volume V of a cone of vertical height h and base radius r is given by \(V = \frac{1}{3} \pi r^2 h\).]

Liquid is flowing into a small tank which has a leak. Initially the tank is empty and, t minutes later, the volume of liquid in the tank is V cm³. The liquid is flowing into the tank at a constant rate of 80 cm³ per minute. Because of the leak, liquid is being lost from the tank at a rate which, at any instant, is equal to kV cm³ per minute where k is a positive constant.

(i) Write down a differential equation describing this situation and solve it to show that \(V = \frac{1}{k}(80 - 80e^{-kt})\).

(ii) It is observed that \(V = 500\) when \(t = 15\), so that \(k\) satisfies the equation \(k = \frac{4 - 4e^{-15k}}{25}\). Use an iterative formula, based on this equation, to find the value of \(k\) correct to 2 significant figures. Use an initial value of \(k = 0.1\) and show the result of each iteration to 4 significant figures.

(iii) Determine how much liquid there is in the tank 20 minutes after the liquid started flowing, and state what happens to the volume of liquid in the tank after a long time.

In a certain chemical process a substance A reacts with another substance B. The masses in grams of A and B present at time t seconds after the start of the process are x and y respectively. It is given that \(\frac{dy}{dt} = -0.6xy\) and \(x = 5e^{-3t}\). When \(t = 0\), \(y = 70\).

(i) Form a differential equation in y and t. Solve this differential equation and obtain an expression for y in terms of t.

(ii) The percentage of the initial mass of B remaining at time t is denoted by p. Find the exact value approached by p as t becomes large.

In a chemical reaction, a compound X is formed from two compounds Y and Z. The masses in grams of X, Y and Z present at time t seconds after the start of the reaction are x, 10 − x and 20 − x respectively. At any time the rate of formation of X is proportional to the product of the masses of Y and Z present at the time. When t = 0, x = 0 and \(\frac{dx}{dt} = 2\).

(i) Show that x and t satisfy the differential equation \(\frac{dx}{dt} = 0.01(10-x)(20-x)\).

(ii) Solve this differential equation and obtain an expression for x in terms of t.

(iii) State what happens to the value of x when t becomes large.