Precalc Question

Question

A radioactive substance decays exponentially. A scientist begins with 130 milligrams of a radioactive substance. After 32 hours, 65 mg of the substance remains. How many milligrams will remain after 41 hours?

Give your answer accurate to at least one decimal place

______mg?

Answer 1

Hello, Andrew,

We can see from the data that the half life of the mystery radioactive substance is 32 hours. In that time the sample decayed from 130 mg to 65 mg, or half it's initial value.

What we need to calculate the sample size after 41 hours is the decay constant, the Greek letter lambda. Since I can't type that easily from my keyboard, I'll use "L," instead.

The relationship between half-life and Lambda is:

L=ln(2)/(t^1/2) ≈ 0.693/(t^1/2)

t^1/2 is 32 hours for our mystery .substance.

The equation that tells us the amount remaining as a function of time is:

N(t) = N₀e^-(L*t)

Where N_o is the initial amount, L is he decay constant and t is the time in hours ( we calculated the decay constant using hours).

We arrive at the following:

N(41) = (130)*e^-(0.022*41)

A calculator or spreadsheet will help find the solution I find a value of 53.5 mg, but I did it quickly. It is less than what we had at 32 hours, so it at least is trending in the right direction.

Bob

Answer 2

65 = 130e^32k

0.5 = e^32k

ln 0.5 = 32k

-0.6931 ≈ 32k

-0.0217 ≈ k

A(41) = 130e^-0.0217(41)

Answer 3

It's a bit difficult to explain this one, but here goes anyway. A function that decays is just like one that grows, but has been transformed. Begin with a basic exponential function P(t) = e^kt. This function goes through the point (0,1) and grows exponentially. However, if the we add a negative sign to the k, P(t) = e^(-kt), this transforms the function by flipping it around the y-axis. It still goes through the point (0,1) but now is a decay function. Note that the function is asymptotic to x-axis - that is, the bigger t gets, the closer the function moves towards 0. Now our function does not go through (0,1) but rather through (0,130). Another transformation can stretch the function verticallyor P(t) = 130e^(-kt). Notice that at t=0, e^(-kt) = 1, so the point on the function is the starting point (0,130). Now for the next point. We know both t and P (32,65). So we plug this into the function or:

65 = 130e^(-k32).

Now solve for k

65/130 =e^(-k32).

ln(65/130) = ln(e^(-k32) = -k*32

so k = ln(65/130)/32

Incidentally, this is how all exponential growth decay functions are solved. You have two points, which allow you to solve for the bending factor K.

Now plug in the final time (41) and you get your answer. I came up with 53.487.