Exponential Growth and Decay

Exponential growth can be amazing!

The idea: something always grows in relation to its current value, such as always doubling.

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

Amazing Tree

Tall green tree against a blue sky

Let's say we have this special tree.

It grows exponentially , following this formula:

Height (in mm) = e^x

e is Euler's number, about 2.718

At 1 year old it is:e¹ = 2.7 mm high ... really tiny! At 5 years it is:e⁵ = 148 mm high ... as high as a cup At 10 years:e¹⁰ = 22 m high ... as tall as a building At 15 years:e¹⁵ = 3.3 km high ... 10 stacked Eiffel Towers At 20 years:e²⁰ = 485 km high ... up into space!

Line graph showing the exponential function e raised to the power of x

No tree could ever grow that tall.
So when people say "it grows exponentially" ... just think what that means.

Growth and Decay

But sometimes things can grow (or the opposite: decay) exponentially, at least for a while.

So we have a generally useful formula:

y(t) = a × e^kt

Where y(t) = value at time "t"
a = value at the start
k = rate of growth (when >0) or decay (when <0)
t = time

See it here (try the sliders):

images/function-graph.js?fn0=a*e^(kx)&xmin=-5.294&xmax=5.259&ymin=-1.447&ymax=4.553&vara=1|0|10&vark=1|0|2

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

Assuming the growth continues like that

What's the "k" value?
How many mice 2 months from now?
How many mice 1 year from now?

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:

Divide both sides by 3:6 = e^2k Take the natural logarithm of both sides:ln(6) = ln(e^2k) ln(e^x)=x, so:ln(6) = 2k Swap sides:2k = ln(6) Divide by 2:k = ln(6)/2

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.

Exponential Decay

Some things "decay" (get smaller) exponentially.

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).

Snow-capped peak of Mount Everest

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:

Divide both sides by 1013:0.88 = e^1000k Take the natural logarithm of both sides:ln(0.88) = ln(e^1000k) ln(e^x)=x, so:ln(0.88) = 1000k Swap sides:1000k = ln(0.88) Divide by 1000:k = ln(0.88)/1000

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!

Half Life

The "half life" is how long it takes for a value to halve with exponential decay.

Commonly used with radioactive decay, but it has many other applications!

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?

White cup of coffee with foam on top

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:

Take the natural logarithm of both sides:ln(0.5) = ln(e^6k) ln(e^x)=x, so:ln(0.5) = 6k Swap sides:6k = ln(0.5) Divide by 6:k = ln(0.5)/6

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 :)

Have a play with the Half Life of Medicine Tool to get a good understanding of this.

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