Exponential Growth and Decay

Example: If a population of rabbits doubles every month, we would have 2, then 4, then 8, 16, 32, 64, 128, 256, etc!

Example: 2 months ago you had 3 mice, you now have 18.

 

Start with the formula:

y(t) = a × e^kt

We know a=3 mice, t=2 months, and right now y(2)=18 mice:

18 = 3 × e^2k

Now some algebra to solve for k:

Notes:

• The step where we used ln(e^x)=x is explained at Exponents and Logarithms.
• we could calculate k ≈ 0.896, but it is best to keep it as k = ln(6)/2 until we do our final calculations.

 

We can now put k = ln(6)/2 into our formula from before:

y(t) = 3 e^(ln(6)/2)t

Now let's calculate the population in 2 more months (at t=4 months):

y(4) = 3 e^(ln(6)/2)×4 = 108

And in 1 year from now (t=14 months):

y(14) = 3 e^{(ln(6)/2)×14} = 839,808

That's a lot of mice! I hope you will be feeding them properly.

Example: Atmospheric pressure (the pressure of air around you) decreases as you go higher.

It decreases about 12% for every 1000 m: an exponential decay.

The pressure at sea level is about 1013 hPa (depending on weather).

 

• Write the formula (with its "k" value),
• Find the pressure on the roof of the Empire State Building (381 m),
• and at the top of Mount Everest (8848 m)

 

Start with the formula:

y(t) = a × e^kt

We know

• a (the pressure at sea level) = 1013 hPa
• t is in meters (distance, not time, but the formula still works)
• y(1000) is a 12% reduction on 1013 hPa = 891.44 hPa

So:

891.44 = 1013 e^k×1000

Now some algebra to solve for k:

 

Now we know "k" we can write:

y(t) = 1013 e^{(ln(0.88)/1000)×t}

 

And finally we can calculate the pressure at 381 m, and at 8848 m:

 

y(381) = 1013 e^{(ln(0.88)/1000)×381} = 965 hPa

y(8848) = 1013 e^{(ln(0.88)/1000)×8848} = 327 hPa

 

In fact pressures at the top of Mount Everest are around 337 hPa ... good calculations!

Example: The half-life of caffeine in your body is about 6 hours. If you had 1 cup of coffee 9 hours ago how much is left in your system?

Start with the formula:

y(t) = a × e^kt

We know:

• a (the starting dose) = 1 cup of coffee!
• t is in hours
• at y(6) we have a 50% reduction (because 6 is the half life)

So:

0.5 = 1 cup × e^6k

Now some algebra to solve for k:

Now we can write:

y(t) = 1 e^{(ln(0.5)/6)×t}

In 6 hours:

y(6) = 1 e^{(ln(0.5)/6)×6} = 0.5

Which is correct as 6 hours is the half life

And in 9 hours:

y(9) = 1 e^{(ln(0.5)/6)×9} = 0.35

After 9 hours the amount left in your system is about 0.35 of the original amount. Have a nice sleep :)