Question

1. a certain radioactive isotope decays according to the function:

\\a(t) = 50(0.8)^t\\

where \\(a(t)\\) represents the remaining mass (in grams) after \\(t\\) years.

a) create a table of values for \\(t = 0, 2, 4, 6, 8, 10\\)

b) graph the function and identify the half-life.

c) write a short paragraph explaining what the asymptote represents in this situation.

Explanation:

Evaluate the exponential function

We calculate the values of \(A(t) = 50(0.8)^t\) for the given values of \(t\).

• For \(t = 0\): \(A(0) = 50(0.8)^0 = 50\)
• For \(t = 2\): \(A(2) = 50(0.8)^2 = 32\)
• For \(t = 4\): \(A(4) = 50(0.8)^4 = 20.48\)
• For \(t = 6\): \(A(6) = 50(0.8)^6 = 13.1072\)
• For \(t = 8\): \(A(8) = 50(0.8)^8 \approx 8.3886\)
• For \(t = 10\): \(A(10) = 50(0.8)^{10} \approx 5.3687\)

Determine the half-life

The initial mass is \(50\text{ g}\). Half of this mass is \(25\text{ g}\).
We solve for \(t\) when \(A(t) = 25\):

$$ 25 = 50(0.8)^t $$

$$ 0.5 = (0.8)^t $$

$$ t = \frac{\ln(0.5)}{\ln(0.8)} \approx 3.106 $$

The half-life is approximately \(3.11\text{ years}\).

Analyze the horizontal asymptote

As \(t \to \infty\), the term \((0.8)^t \to 0\).
Thus, \(A(t) \to 0\).
The horizontal asymptote is the line \(A(t) = 0\) (the \(t\)-axis).
In this physical context, it represents that the mass of the radioactive isotope continuously decreases and approaches \(0\text{ grams}\) over a very long period, but theoretically never reaches absolute zero.

Answer:

Part a)

The completed table of values is:

\(t\)	0	2	4	6	8	10

Part b)

The half-life of the isotope is approximately \(3.11\text{ years}\).

Part c)

The horizontal asymptote is \(A(t) = 0\). In this situation, it represents that as time goes on indefinitely, the remaining mass of the radioactive isotope approaches \(0\text{ grams}\). It signifies that while the substance decays continuously and becomes negligibly small, mathematically a tiny fraction of the isotope theoretically remains forever.