1 / 9

Exponential growth and decay

Delve into the world of exponential growth and decay by exploring how to predict mould growth on a lovely block of cheese left in the fridge. Learn how to extract information to form a model for mould population with time, finding constants A and k, and making predictions using exponential growth equations.

mnagle
Download Presentation

Exponential growth and decay

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. www.mathssandpit.co.uk/blog Exponential growth and decay

  2. www.mathssandpit.co.uk/blog Let us begin • You go on holiday and leave a particularly lovely block of cheese in the fridge. • Describe what will happen to it over time?

  3. www.mathssandpit.co.uk/blog • Is the cheese decaying or is the mould growing? • Mould growth can be predicted using an exponential growth model. • The trickiest bit is extracting the information to form the model.

  4. www.mathssandpit.co.uk/blog The Growth Model How can you find A if you have P0? t represents time P0 represents the population when t = 0 P = A × e kt P represents population How can you find k if you have P1? What is happening if k is negative? A and k are constants

  5. www.mathssandpit.co.uk/blog The Growth Model: Cheese Find A The initial population of mould is 500 spores P = A × e kt 500 = A × e k × 0 500 = A × e 0 500 = A × 1 A = 500

  6. www.mathssandpit.co.uk/blog The Growth Model: Cheese The initial population of mould is 500 spores After 2 days the population is 1300 P = 500 × e kt 1300 = 500 × e k ×2 13 = 5 × e 2k 13 ÷ 5 = e 2k Find k ln (2.6) = ln (e 2k) ln (2.6) = 2k k = 0.5 ln (2.6) = 0.478 (3dp)

  7. www.mathssandpit.co.uk/blog The Growth Model: Cheese The initial population of mould is 500 spores After 2 days the population is 1300 P = 500 × e 0.478t P = 500 × e 0.478 × 14 How many spores are there by the end of the second week? P = 500 × e 6.692 P = 402966 (nearest whole number)

  8. www.mathssandpit.co.uk/blog Would you eat the cheese?

  9. www.mathssandpit.co.uk/blog Image credits https://crowarrowinc.wordpress.com www.thegrocer.co.uk Microsoft clipart www.basenow.net

More Related