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Differential Equations · Paper 1 — Free Maths AA SL Practice Questions

1FoundationSAQ-SSolving first-order differential equationss7 marksPaper 1~11 min
The number of bacteria, NN, in a culture grows at a rate proportional to NN. This modelled by the differential equation dNdt=kN\dfrac{dN}{dt} = kN, where tt is the time in hours and kk is a positive constant.
(a)
Show that the general solution to dNdt=kN\dfrac{dN}{dt} = kN is N=AektN = Ae^{kt}, where AA is a positive constant. [2 marks]
(b)
Initially there are 500500 bacteria. After 33 hours there are 40004000 bacteria. Find the exact value of kk. [3 marks]
(c)
A scientist claims the culture will exceed 50 00050\,000 bacteria within 1010 hours of the start. Determine whether this claim is correct. [2 marks]
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Solutions

2FoundationSAQ-SSolving first-order differential equationss7 marksPaper 1~11 min
Consider the differential equation dydx=3x2y\dfrac{dy}{dx} = \dfrac{3x^2}{y}, where y>0y > 0.
(a)
Show that the general solution can be written as y2=2x3+Cy^2 = 2x^3 + C, where CC is a constant. [3 marks]
(b)
Given that y=3y = 3 when x=2x = 2, find the value of CC. [1 mark]
(c)
Hence find the value of xx when y=5y = 5, giving your answer correct to 3 significant figures. [3 marks]
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3MasterySAQ-SSolving first-order differential equationss5 marksPaper 1~8 min
The population of a colony of ants is modelled by the differential equation dPdt=150P(500−P)\frac{dP}{dt} = \frac{1}{50}P(500 - P) where PP is the population at time tt years, t≥0t \geq 0. Initially, the population is 100100 ants.
(a)
Show that 1P(500−P)=1500(1P+1500−P)\frac{1}{P(500-P)} = \frac{1}{500}\left(\frac{1}{P} + \frac{1}{500-P}\right) [2 marks]
(b)
Hence solve the differential equation, giving your answer in the form P=a1+be−ctP = \frac{a}{1 + be^{-ct}} where aa, bb, cc are integers. [3 marks]
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4ChallengeSAQ-LSolving first-order differential equationss8 marksPaper 1~12 min
A cylindrical water tank has a constant cross-sectional area of 2 m22\,\text{m}^2 and a small hole at its base. Water leaks such that the rate of change of volume VV (in m3\text{m}^3) at time tt (in hours) is proportional to the square root of the depth hh (in metres). The tank is initially full with depth 4 m4\,\text{m}. After 11 hour, the depth has decreased to 3 m3\,\text{m}. Using V=2hV = 2h, the differential equation for the depth is dhdt=−kh,k>0.\frac{dh}{dt} = -k\sqrt{h}, \quad k > 0.
(a)
Show that k=2(2−3)k = 2(2 - \sqrt{3}). [4 marks]
(b)
Determine the time TT at which the depth reaches 2 m2\,\text{m}. Give your answer in exact form. [2 marks]
(c)
Determine whether the tank empties completely according to this model. Justify your answer mathematically. [2 marks]
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Solutions

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