P8.8Differential equations in context: interpreting solutions
Edexcel A level Maths (9MA0) · Pure mathematics › Integration
Practise Differential equations in context: interpreting solutions. 4 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy9 marks
The number of bacteria, \(N\), in a culture \(t\) hours after the start of an experiment is modelled by the differential equation \[\frac{\mathrm{d}N}{\mathrm{d}t} = kN\] where \(k\) is a positive constant. When \(t = 0\), \(N = 200\).
(a) Solve the differential equation to show that \(N = 200\mathrm{e}^{kt}\).[3]
(b) Given that \(N = 400\) when \(t = 2\), show that \(k = \frac{1}{2}\ln 2\).[2]
(c) Find, according to the model, the number of bacteria when \(t = 6\), giving your answer to the nearest hundred.[2]
(d) Find the time taken for the number of bacteria to reach 1000, giving your answer in hours to 3 significant figures.[2]
Show the answer and mark scheme
(a)Answer: Shown
M1 for separating the variables and integrating: \(\int \frac{1}{N}\,\mathrm{d}N = \int k\,\mathrm{d}t\)
A1 for \(\ln N = kt + c\)
A1* for using \(N = 200\) at \(t = 0\) to obtain \(N = 200\mathrm{e}^{kt}\) (cso)
Worked solution: \(\int \frac{1}{N}\,\mathrm{d}N = \int k\,\mathrm{d}t\), so \(\ln N = kt + c\) and \(N = B\mathrm{e}^{kt}\). At \(t = 0\), \(N = 200\), so \(B = 200\) and \(N = 200\mathrm{e}^{kt}\).
(b)Answer: Shown
M1 for substituting: \(400 = 200\mathrm{e}^{2k}\) and taking logarithms
A1* for \(k = \frac{1}{2}\ln 2\) (cso)
Worked solution: \(400 = 200\mathrm{e}^{2k}\), so \(\mathrm{e}^{2k} = 2\) and \(2k = \ln 2\), giving \(k = \frac{1}{2}\ln 2\).
M1 for \(1000 = 200\mathrm{e}^{kt}\) leading to \(t = \frac{\ln 5}{k}\)
A1 for 4.64 hours
Worked solution: \(\mathrm{e}^{kt} = 5\), so \(t = \frac{\ln 5}{k} = \frac{2\ln 5}{\ln 2} \approx 4.64\) hours.
Question 2Medium10 marks
A gardener models the height, \(h\) metres, of a sunflower \(t\) days after it is planted out by \[\frac{\mathrm{d}h}{\mathrm{d}t} = 0.02(3 - h)\] When \(t = 0\), \(h = 0.1\).
(a) Explain the significance of the number 3 in this model.[1]
(b) Solve the differential equation to find \(h\) in terms of \(t\).[4]
(c) Find the height of the sunflower after 60 days, according to the model, giving your answer to 2 decimal places.[2]
(d) Give two criticisms of this model.[2]
(e) Suggest a refinement to the model that would address one of your criticisms.[1]
Show the answer and mark scheme
(a)Answer: 3 m is the height the sunflower approaches: the growth rate is zero when \(h = 3\).
B1 for: 3 m is the limiting (maximum) height of the sunflower according to the model, since the growth rate is zero when \(h = 3\)
Worked solution: When \(h = 3\), \(\frac{\mathrm{d}h}{\mathrm{d}t} = 0\); for \(h \lt 3\) the sunflower grows, so its height approaches 3 m.
(b)Answer: \(h = 3 - 2.9\mathrm{e}^{-0.02t}\)
M1 for separating: \(\int \frac{1}{3 - h}\,\mathrm{d}h = \int 0.02\,\mathrm{d}t\)
A1 for \(-\ln(3 - h) = 0.02t + c\)
M1 for using \(h = 0.1\) at \(t = 0\)
A1 for \(h = 3 - 2.9\mathrm{e}^{-0.02t}\)
Worked solution: \(-\ln(3 - h) = 0.02t + c\), so \(3 - h = A\mathrm{e}^{-0.02t}\). At \(t = 0\), \(h = 0.1\): \(A = 2.9\). So \(h = 3 - 2.9\mathrm{e}^{-0.02t}\).
(c)Answer: \(2.13\) m
M1 for substituting \(t = 60\)
A1 for 2.13 m
Worked solution: \(h = 3 - 2.9\mathrm{e}^{-1.2} = 2.1265\ldots \approx 2.13\) m
(d)Answer: e.g. fastest growth at the start; the plant is predicted to grow for ever.
B1 for one valid criticism, e.g. the model gives the fastest growth when the plant is smallest, whereas a young plant grows slowly at first
B1 for a second valid criticism, e.g. the model predicts the sunflower keeps growing (towards 3 m) indefinitely, but it stops growing and dies at the end of the season; the growth rate would depend on weather
Worked solution: The growth rate \(0.02(3 - h)\) is greatest when \(h\) is smallest, but young plants grow slowly at first; the model also predicts growth for ever, ignoring the end of the season and changes in weather.
(e)Answer: e.g. \(\frac{\mathrm{d}h}{\mathrm{d}t} = kh(3 - h)\).
B1 for a sensible refinement, e.g. use \(\frac{\mathrm{d}h}{\mathrm{d}t} = kh(3 - h)\) (logistic growth), so growth is slow when \(h\) is small; or let the constant depend on \(t\)
Worked solution: A logistic model \(\frac{\mathrm{d}h}{\mathrm{d}t} = kh(3 - h)\) gives slow growth when the plant is small, faster growth in the middle and slow growth as \(h\) approaches 3.
Question 3Hard10 marks
Water drains from a tank through a small hole in its base. The depth of water in the tank, \(h\) metres, \(t\) minutes after the hole is opened is modelled by \[\frac{\mathrm{d}h}{\mathrm{d}t} = -k\sqrt{h}\]where \(k\) is a positive constant. Initially the depth is 4 m, and after 10 minutes the depth is 1 m.
(a) Solve the differential equation to show that \(h = \left(2 - \dfrac{t}{10}\right)^2\).[5]
(b) Find the time taken for the tank to empty, according to the model.[1]
(c) Explain why the formula in part (a) should not be used for \(t \gt 20\).[1]
(d) The tank is refilled to a depth of 4 m. Water is now also poured into the tank at a constant rate, so that the depth is modelled by \[\frac{\mathrm{d}h}{\mathrm{d}t} = q - 0.2\sqrt{h}\]where \(q\) is a positive constant. Find the value of \(q\) for which the depth stays at 4 m.[2]
(e) State one assumption about the shape of the tank that is needed for a model in which \(\frac{\mathrm{d}h}{\mathrm{d}t}\) depends only on \(h\) in this way to be reasonable.[1]
M1 for separating the variables: \(\int h^{-\frac{1}{2}}\,\mathrm{d}h = -\int k\,\mathrm{d}t\)
A1 for \(2h^{\frac{1}{2}} = -kt + c\)
M1 for using both conditions: \(t = 0, h = 4\) and \(t = 10, h = 1\)
A1 for \(c = 4\) and \(k = 0.2\)
A1* for \(h = \left(2 - \frac{t}{10}\right)^2\)
Worked solution: \(\displaystyle\int h^{-\frac{1}{2}}\,\mathrm{d}h = -\int k\,\mathrm{d}t \Rightarrow 2\sqrt{h} = -kt + c\). \(t = 0, h = 4\): \(c = 4\). \(t = 10, h = 1\): \(2 = -10k + 4\), \(k = 0.2\). So \(\sqrt{h} = 2 - 0.1t\) and \(h = \left(2 - \frac{t}{10}\right)^2\).
(b)Answer: 20 minutes
B1 for 20 (minutes)
Worked solution: \(h = 0 \Rightarrow 2 - \frac{t}{10} = 0 \Rightarrow t = 20\) minutes.
(c)Answer: For \(t \gt 20\) the formula gives \(h \gt 0\) and increasing, but the tank is empty and stays empty.
B1 for the formula predicts the depth increasing again after \(t = 20\), which is impossible (the tank remains empty)
Worked solution: For \(t \gt 20\), \(\left(2 - \frac{t}{10}\right)^2\) increases again, which would mean the tank refilling. In reality the tank stays empty, so the formula only applies for \(0 \le t \le 20\).
(d)Answer: 0.4
M1 for \(\frac{\mathrm{d}h}{\mathrm{d}t} = 0\) when \(h = 4\): \(q = 0.2\sqrt{4}\)
A1 for 0.4
Worked solution: The depth stays constant when \(\frac{\mathrm{d}h}{\mathrm{d}t} = 0\): \(q = 0.2\sqrt{4} = 0.4\) (metres of depth per minute).
(e)Answer: The tank has a constant horizontal cross-section (e.g. a vertical cylinder or cuboid), so a volume flow rate corresponds to a proportional rate of change of depth.
B1 for the tank has a uniform (constant) horizontal cross-sectional area, e.g. it is a vertical cylinder or cuboid
Worked solution: The rate of outflow of volume depends on the depth. For the rate of change of depth to be a fixed multiple of that outflow, the tank must have the same horizontal cross-sectional area at every depth (for example a vertical cylinder).