Exponential growth/decay model setup

A question is this type if and only if it requires finding constants in an exponential model from given initial conditions and one other data point.

33 questions · Moderate -0.6

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OCR MEI Paper 3 2021 November Q5
8 marks Moderate -0.8
5
  1. The diagram shows the curve \(\mathrm { y } = \mathrm { e } ^ { \mathrm { x } }\). \includegraphics[max width=\textwidth, alt={}, center]{a0d9573f-8273-4562-a2d3-07f15d9da1af-5_574_682_315_328} On the axes in the Printed Answer Booklet, sketch graphs of
    1. \(\frac { \mathrm { dy } } { \mathrm { dx } }\) against \(x\),
    2. \(\frac { \mathrm { dy } } { \mathrm { dx } }\) against \(y\).
  2. Wolves were introduced to Yellowstone National Park in 1995. The population of wolves, \(y\), is modelled by the equation \(y = A e ^ { k t }\),
    where \(A\) and \(k\) are constants and \(t\) is the number of years after 1995.
    1. Give a reason why this model might be suitable for the population of wolves.
    2. When \(t = 0 , y = 21\) and when \(t = 1 , y = 51\). Find values of \(A\) and \(k\) consistent with the data.
    3. Give a reason why the model will not be a good predictor of wolf populations many years after 1995.
AQA C4 2011 June Q2
6 marks Moderate -0.8
2 The average weekly pay of a footballer at a certain club was \(\pounds 80\) on 1 August 1960. By 1 August 1985, this had risen to \(\pounds 2000\). The average weekly pay of a footballer at this club can be modelled by the equation $$P = A k ^ { t }$$ where \(\pounds P\) is the average weekly pay \(t\) years after 1 August 1960, and \(A\) and \(k\) are constants.
    1. Write down the value of \(A\).
    2. Show that the value of \(k\) is 1.137411 , correct to six decimal places.
  1. Use this model to predict the year in which, on 1 August, the average weekly pay of a footballer at this club will first exceed \(\pounds 100000\).
AQA C4 2006 January Q4
9 marks Moderate -0.8
4 On 1 January 1900, a sculpture was valued at \(\pounds 80\).
When the sculpture was sold on 1 January 1956, its value was \(\pounds 5000\).
The value, \(\pounds V\), of the sculpture is modelled by the formula \(V = A k ^ { t }\), where \(t\) is the time in years since 1 January 1900 and \(A\) and \(k\) are constants.
  1. Write down the value of \(A\).
  2. Show that \(k \approx 1.07664\).
  3. Use this model to:
    1. show that the value of the sculpture on 1 January 2006 will be greater than £200 000;
    2. find the year in which the value of the sculpture will first exceed \(\pounds 800000\).
AQA C4 2008 January Q4
9 marks Moderate -0.3
4 David is researching changes in the selling price of houses. One particular house was sold on 1 January 1885 for \(\pounds 20\). Sixty years later, on 1 January 1945, it was sold for \(\pounds 2000\). David proposes a model $$P = A k ^ { t }$$ for the selling price, \(\pounds P\), of this house, where \(t\) is the time in years after 1 January 1885 and \(A\) and \(k\) are constants.
    1. Write down the value of \(A\).
    2. Show that, to six decimal places, \(k = 1.079775\).
    3. Use the model, with this value of \(k\), to estimate the selling price of this house on 1 January 2008. Give your answer to the nearest \(\pounds 1000\).
  1. For another house, which was sold for \(\pounds 15\) on 1 January 1885, David proposes the model $$Q = 15 \times 1.082709 ^ { t }$$ for the selling price, \(\pounds Q\), of this house \(t\) years after 1 January 1885. Calculate the year in which, according to these models, these two houses would have had the same selling price.
AQA C4 2009 June Q4
6 marks Moderate -0.3
4 A car depreciates in value according to the model $$V = A k ^ { t }$$ where \(\pounds V\) is the value of the car \(t\) months from when it was new, and \(A\) and \(k\) are constants. Its value when new was \(\pounds 12499\) and 36 months later its value was \(\pounds 7000\).
    1. Write down the value of \(A\).
    2. Show that the value of \(k\) is 0.984025 , correct to six decimal places.
  1. The value of this car first dropped below \(\pounds 5000\) during the \(n\)th month from new. Find the value of \(n\).
AQA AS Paper 1 2023 June Q10
8 marks Moderate -0.3
10
  1. Charlie says:
    • because the car has lost \(\pounds 6000\) after two years, after another two years it will be worth £6000.
    Charlie's friend Kaya says:
    • because the car has lost one third of its value after two years, after another two years it will be worth \(\pounds 8000\).
    Explain whose statement is correct, justifying the value they have stated.
    10
  2. The value of Charlie's car, \(\pounds V , t\) years after 1 January 2016 may be modelled by the equation \(V = A \mathrm { e } ^ { - k t }\)
    where \(A\) and \(k\) are positive constants.
    Find the value of \(t\) when the car has a value of \(\pounds 10000\), giving your answer to two significant figures.
    [5 marks] \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)
    [1 mark] \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)
AQA AS Paper 2 2018 June Q12
8 marks Moderate -0.3
12 Trees in a forest may be affected by one of two types of fungal disease, but not by both. The number of trees affected by disease \(\mathrm { A } , n _ { \mathrm { A } }\), can be modelled by the formula $$n _ { \mathrm { A } } = a \mathrm { e } ^ { 0.1 t }$$ where \(t\) is the time in years after 1 January 2017.
The number of trees affected by disease \(\mathrm { B } , n _ { \mathrm { B } }\), can be modelled by the formula $$n _ { \mathrm { B } } = b \mathrm { e } ^ { 0.2 t }$$ On 1 January 2017 a total of 290 trees were affected by a fungal disease.
On 1 January 2018 a total of 331 trees were affected by a fungal disease.
12
  1. Show that \(b = 90\), to the nearest integer, and find the value of \(a\).
    12
  2. Estimate the total number of trees that will be affected by a fungal disease on 1 January 2020.
    [1 mark]
    12
  3. Find the year in which the number of trees affected by disease B will first exceed the number affected by disease A.
    12
  4. Comment on the long-term accuracy of the model.
AQA AS Paper 2 2020 June Q7
2 marks Moderate -0.8
7 The population of a country was 3.6 million in 1989. It grew exponentially to reach 6 million in 2019.
Estimate the population of the country in 2049 if the exponential growth continues unchanged.