Modelling with rate proportional to amount

Questions where a quantity changes at a rate proportional to the quantity itself or a simple function of it, leading to exponential growth/decay models.

5 questions · Standard +0.4

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CAIE P3 2005 November Q8
8 marks Standard +0.3
8 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 $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - k x t$$ where \(k\) is a positive constant. At the start of the reaction, when \(t = 0 , x = 100\).
  1. Solve this differential equation, obtaining a relation between \(x , k\) and \(t\).
  2. 20 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.
CAIE P3 2022 November Q8
8 marks Standard +0.3
8 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 $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k x \mathrm { e } ^ { - 0.1 t }$$ where \(k\) is a positive constant. It is given that \(x = 20\) at the start of the reaction.
  1. Solve the differential equation, obtaining a relation between \(x , t\) and \(k\).
  2. Given that \(x = 40\) when \(t = 10\), find the value of \(k\) and find the value approached by \(x\) as \(t\) becomes large.
CAIE Further Paper 3 2020 June Q2
6 marks Standard +0.3
2 A particle \(Q\) of mass \(m \mathrm {~kg}\) falls from rest under gravity. The motion of \(Q\) is resisted by a force of magnitude \(m k v \mathrm {~N}\), where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of \(Q\) at time \(t \mathrm {~s}\) and \(k\) is a positive constant. Find an expression for \(v\) in terms of \(g , k\) and \(t\).
Edexcel PMT Mocks Q10
9 marks Standard +0.3
10. a. Find \(\int \frac { 1 } { 30 } \cos \frac { \pi } { 6 } t \mathrm {~d} t\). The height above ground, \(X\) metres, of the passenger on a wooden roller coaster can be modelled by the differential equation $$\frac { d \mathrm { X } } { \mathrm {~d} t } = \frac { 1 } { 30 } X \cos \left( \frac { \pi } { 6 } t \right)$$ where \(t\) is the time, in seconds, from the start of the ride.
At time \(t = 0\), the passenger is 6 m above the ground.
b. Show that \(X = k e ^ { \frac { 1 } { 5 \pi } \sin \left( \frac { \pi } { 6 } t \right) }\) where the value of the constant \(k\) should be found.
c. Show that the maximum height of the passenger above the ground is 6.39 m . The passenger reaches the maximum height, for the second time, \(T\) seconds after the start of the ride.
d. Find the value of \(T\).
OCR Further Pure Core 1 2018 March Q9
8 marks Standard +0.8
9 In an experiment, at time \(t\) minutes there is \(Q\) grams of substance present.
It is known that the substance decays at a rate that is proportional to \(1 + Q ^ { 2 }\). Initially there are 100 grams of the substance present and after 100 minutes there are 50 grams present. Find the amount of the substance present after 400 minutes.