Friday

December 9, 2016
Posted by **Anonymous ** on Friday, November 16, 2012 at 6:43pm.

- Calculus -
**PayDay**, Friday, November 16, 2012 at 6:48pmEquation for exponential growth/decay:

y = Ce^(kt)

Where y is the amount leftover and t is the time elapsed.

Notice if you plug in time (4000 years) right away, you get

y = Ce^(4000k), and you can't solve for y (the amount)--there's too many unknowns (C and k).

So first you have to find C and k in order to find y.

Use what you know.

At time t=0 years, you still have the whole sample of 100 mg. Therefore (0,100) is a solution to the equation. Plug these values in.

100 = Ce^(0*k) so 100 = Ce^(0)

Anything to the zero power is just one,

so 100 = C*1 or C = 100

Now you have to find k. Think of what else we know about half life. At time 1590 years, we'll have half the sample, or 50 mg. Therefore (1590, 50) is a solution to the equation. Plug it in.

y = Ce^(kt)

we know C=100, t=1590, and y=50, so we can solve for k.

50 = 100*e^(1590k)

1/2 = e^(1590k)

take the natural log of both sides to bring down the exponent

ln (1/2) = ln [ e^(1590k)]

ln (1/2) = 1590k

therefore k = ln(1/2) / 1590

You now have your final equation

y = Ce^(kt) or y = 100e^[(ln(1/2) / 1590)t]

Now you can plug in any value for time and find the amount left over:

y = 100e^[(ln(1/2) / 1590)*4000]

Plugging this into a calculator returns the result 17.5 grams. - Calculus -
**Steve**, Friday, November 16, 2012 at 8:16pmHmmm. If you start with 100mg, I don't see how it will grow to 17.5g after 4000 years.

Since we are dealing with half-life, it's really not necessary to go through the gyrations of a general exponential function.

The amount remaining after t years is

a(t) = 100 (1/2)^(t/1590)

a(4000) = 17.5 mg

I guess you are correct numerically. Just the units are off.