Calculus Exponential growth/ decay?

Question

The half-life of Radium-226 is 1590 years. If a sample contains 100 mg, how many mg will remain after 4000 years?

Answer 1

Equation 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 calcuator returns the result 17.5 grams.

Answer 2

Exponential Growth Calculus

Answer 3

Exponential Growth And Decay Calculus

Answer 4

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RE:

Calculus Exponential growth/ decay?

The half-life of Radium-226 is 1590 years. If a sample contains 100 mg, how many mg will remain after 4000 years?

Answer 5

We know that



e^(1590 k) = 0.5



Take logs



1590 k = -0.69315



k = -0.00043594



So the remaining fraction after 4000 years is given by



e ^(-0.00043594 * 4000) = e^-1.74376



The fraction is 0.174.



This means that there will be 17.4 grams left out of 100 grams.

Answer 6

Let 'th' be the half life, m(0) be the initial mass, and m(t) be the remaining mass after t years.

m(t) = m(0)(1/2)^(t/th)

m(4000) = 100(1/2)^(4000/1590) = 17.486 mg