Problem

Painful bone metastases are common in advanced prostate cancer. Physicians often order treatment with strontium- 89 ( ${ }^{89} \mathrm{Sr}$ ), a radionuclide with a strong affinity for bone tissue. A patient is given a sample containing $4 \mathrm{mCi}$ of ${ }^{89} \mathrm{Sr}$.
Part: $0 / 2$
Part 1 of 2
(a) If $25 \%$ of the ${ }^{89} \mathrm{Sr}$ remains in the body after 100 days, write a function of the form $Q(t)=Q_{0} e^{-k t}$ to model the amount $Q(t)$ of radioactivity in the body $t$ days after the initial dose. Round the value of $k$ to five decimal places. Do not round intermediate calculations.
\[
Q(t)=
\]

Answer

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Answer

\(\boxed{Q(t) = 4e^{-0.01386t}}\) is the function that models the amount of radioactivity in the body \(t\) days after the initial dose.

Steps

Step 1 :We are given that the initial dose of radioactivity, \(Q_{0}\), is 4 mCi.

Step 2 :We are also given that 25% of the \(^{89}Sr\) remains in the body after 100 days. This means that \(Q(100) = 0.25 * Q_{0} = 1\) mCi.

Step 3 :We can use these two points to solve for the decay constant, \(k\), in the equation \(1 = 4e^{-100k}\).

Step 4 :Solving for \(k\) gives us approximately 0.01386.

Step 5 :Using this value of \(k\), we can write the function \(Q(t)\) as \(Q(t) = 4e^{-0.01386t}\).

Step 6 :\(\boxed{Q(t) = 4e^{-0.01386t}}\) is the function that models the amount of radioactivity in the body \(t\) days after the initial dose.

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