Exponential growth/decay - direct proportionality (dN/dt = kN)

8. A population growth is modelled by the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = k P ,$$ where \(P\) is the population, \(t\) is the time measured in days and \(k\) is a positive constant.
Given that the initial population is \(P _ { 0 }\),

1. solve the differential equation, giving \(P\) in terms of \(P _ { 0 } , k\) and \(t\). Given also that \(k = 2.5\),
2. find the time taken, to the nearest minute, for the population to reach \(2 P _ { 0 }\). In an improved model the differential equation is given as $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \lambda P \cos \lambda t$$ where \(P\) is the population, \(t\) is the time measured in days and \(\lambda\) is a positive constant.
   Given, again, that the initial population is \(P _ { 0 }\) and that time is measured in days,
3. solve the second differential equation, giving \(P\) in terms of \(P _ { 0 } , \lambda\) and \(t\). Given also that \(\lambda = 2.5\),
4. find the time taken, to the nearest minute, for the population to reach \(2 P _ { 0 }\) for the first time, using the improved model.

4. The rate of decay of the mass of a particular substance is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { 5 } { 2 } x , \quad t \geqslant 0$$ where \(x\) is the mass of the substance measured in grams and \(t\) is the time measured in days.
Given that \(x = 60\) when \(t = 0\),

1. solve the differential equation, giving \(x\) in terms of \(t\). You should show all steps in your working and give your answer in its simplest form.
2. Find the time taken for the mass of the substance to decay from 60 grams to 20 grams. Give your answer to the nearest minute.

6 Helga invests \(\pounds 4000\) in a savings account.
After \(t\) days, her investment is worth \(\pounds y\).
The rate of increase of \(y\) is \(k y\), where \(k\) is a constant.

1. Write down a differential equation in terms of \(t , y\) and \(k\).
2. Solve your differential equation to find the value of Helga's investment after \(t\) days. Give your answer in terms of \(k\) and \(t\). It is given that \(k = \frac { 1 } { 365 } \ln \left( 1 + \frac { r } { 100 } \right)\) where \(r \%\) is the rate of interest per annum. During the first year the rate of interest is \(6 \%\) per annum.
3. Find the value of Helga's investment after 90 days. After one year (365 days), the rate of interest drops to 5\% per annum.
4. Find the total time that it will take for Helga's investment to double in value.

11. In a science experiment, a radio active particle, \(N\), decays over time, \(t\), measured in minutes. The rate of decay of a particle is proportional to the number of particles remaining. Write down a suitable equation for the rate of change of the number of particles, \(N\) in terms of \(t\).

1. A new smartphone was released by a company.

The company monitored the total number of phones sold, \(n\), at time \(t\) days after the phone was released. The company observed that, during this time,
the rate of increase of \(n\) was proportional to \(n\) Use this information to write down a suitable equation for \(n\) in terms of \(t\).
(You do not need to evaluate any unknown constants in your equation.)

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.

7 As a spherical snowball melts its volume decreases. The rate of decrease of the volume of the snowball at any given time is modelled as being proportional to its volume at that time. Initially the volume of the snowball is \(500 \mathrm {~cm} ^ { 3 }\) and the rate of decrease of its volume is \(20 \mathrm {~cm} ^ { 3 }\) per hour.

1. Find the time that this model would predict for the snowball's volume to decrease to \(250 \mathrm {~cm} ^ { 3 }\).
2. Write down one assumption made when using this model.
3. Comment on how realistic this model would be in the long term.

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]
2. Show that the general solution may be written as \(N = Ae^{-kt}\), where \(A\) and \(k\) are positive constants. [5]

Initially the number of radioactive atoms present is \(7 \times 10^{18}\) and 8 days later the number present is \(3 \times 10^{17}\).

1. Find the value of \(k\). [3]
2. Find the number of radioactive atoms present after a further 8 days. [2]

A Pancho car has value \(£V\) at time \(t\) years. A model for \(V\) assumes that the rate of decrease of \(V\) at time \(t\) is proportional to \(V\).

1. By forming and solving an appropriate differential equation, show that \(V = Ae^{-kt}\), where \(A\) and \(k\) are positive constants. [3]

The value of a new Pancho car is \(£20\,000\), and when it is 3 years old its value is \(£11\,000\).

1. Find, to the nearest \(£100\), an estimate for the value of the Pancho when it is 10 years old. [5]

A Pancho car is regarded as 'scrap' when its value falls below \(£500\).

1. Find the approximate age of the Pancho when it becomes 'scrap'. [3]

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 \(t\). [1]

Given that initially there are \(N_0\) bacteria present in a culture,

1. Show that \(N = N_0 e^{kt}\), where \(k\) is a positive constant. [6]

Given also that the number of bacteria present doubles every six hours,

1. find the value of \(k\), [3]
2. Find how long it takes for the number of bacteria to increase by a factor of ten, giving your answer to the nearest minute. [2]

A circular ornamental garden pond, of radius 2 metres, has weed starting to grow and cover its surface. As the weed grows, it covers an area of \(A\) square metres. A simple model assumes that the weed grows so that the rate of increase of its area is proportional to \(A\).

1. Show that the area covered by the weed can be modelled by $$A = Be^{kt}$$ where \(B\) and \(k\) are constants and \(t\) is time in days since the weed was first noticed. [4 marks]
2. When it was first noticed, the weed covered an area of 0.25 m². Twenty days later the weed covered an area of 0.5 m²
   1. State the value of \(B\). [1 mark]
   2. Show that the model for the area covered by the weed can be written as $$A = 2^{\frac{t}{20} - 2}$$ [4 marks]
   3. How many days does it take for the weed to cover half of the surface of the pond? [2 marks]
3. State one limitation of the model. [1 mark]
4. Suggest one refinement that could be made to improve the model. [1 mark]

The rate of change of a variable \(y\) with respect to \(x\) is directly proportional to \(y\).

1. Write down a differential equation satisfied by \(y\). [1]
2. When \(x = 1\) and \(y = 0.5\), the rate of change of \(y\) with respect to \(x\) is 2. Find \(y\) when \(x = 3\). [6]

Helga invests £4000 in a savings account. After \(t\) days, her investment is worth \(£y\). The rate of increase of \(y\) is \(ky\), where \(k\) is a constant.

1. Write down a differential equation in terms of \(t\), \(y\) and \(k\). [1]
2. Solve your differential equation to find the value of Helga's investment after \(t\) days. Give your answer in terms of \(k\) and \(t\). [4]

It is given that \(k = \frac{r}{365}\ln\left(1 + \frac{r}{100}\right)\) where \(r\%\) is the rate of interest per annum. During the first year the rate of interest is 6% per annum.

1. Find the value of Helga's investment after 90 days. [2]

After one year (365 days), the rate of interest drops to 5% per annum.

1. Find the total time that it will take for Helga's investment to double in value. [5]