# Exponential Decay

Exponential decay models apply to any situation where the decay (decrease) is proportional to the current size of the quantity of interest.  Such situations are encountered in biology, business, chemistry and the social sciences.

Exponential decay models are also used very commonly, especially for radioactive decay, drug concentration in the bloodstream, of depreciation of value.

Radioactive decay problems are often given in terms of half-life of a radioactive element. This is modeled by the equation:

where N0 is the initial amount of the element, N is the amount remaining after t years, and τ is the half-life.

Example:

If you start with a quantity of the unstable element Potassium-40, it takes 1.26 billion years for half of it to decay into Argon-40. So the half-life of Potassium-40 is 1.26 billion years.

Write an exponential decay model to find the number of Potassium-40 atoms remaining after t years, if you start with 2000 Potassium-40 atoms.

Here, N0 = 2000 and τ = 1,260,000,000. So the model is:

## Drug Concentration

For drug concentration problems, you may be given the fraction p of the original amount of the drug left in the bloodstream after a unit of time. In this case, the situation is modeled by the equation

y = Apt,

where y is the concentration remaining after time t, and A is the initial amount.

Example:

If a person takes A milligrams of a drug at time 0, then y = A(0.8)t gives the concentration left in the bloodstream after t hours. If the initial dose is 200 mg, what is the concentration of the drug in the bloodstream after 4 hours?

Substitute.

y = 200(0.8)4

You might want a calculator!

y = 200(0.4096)

y = 81.92

So there are about 82 milligrams of the drug left in the bloodstream after four hours.

## Depreciation

If the value of some article (for example, a car), originally \$C, depreciates x% per year, then the value after t years is given by the formula:

y = C(1 – x/100)t

Example:

The original value of a car is \$28,000. If it depreciates by 15% each year, find its value in 4 years.

Substitute.

y = 28000(1 – 0.15)4

y = 28000(0.85)4

y = 28000(0.52200625)

y = 14616.175

So after four years, the car is worth about \$14,616.