Fluorodeoxyglucose is a derivative of glucose that contains the radionuclide fluorine-18 ( $\left.{ }^{18} \mathrm{~F}\right)$. A patient is given a sample of this material containing $300 \mathrm{MBq}$ of ${ }^{18} \mathrm{~F}$ (a megabecquerel is a unit of radioactivity). The patient then undergoes a PET scan (positron emission tomography). to detect areas of metabolic activity indicative of cancer. After $174 \mathrm{~min}$, one-third of the original dose remains in the body.
Part: $0 / 2$
Part 1 of 2
(a) Write a function of the form $Q(t)=Q_{0} e^{-k t}$ to model the radioactivity level $Q(t)$ of fluorine-18 at a time $t$ minutes after the intial dose. Round the value of $k$ to five decimal places. Do not round intermediate calculations.
\[
Q(t)=\square \quad \square^{\square}
\]
Answer
So, the function that models the radioactivity level of fluorine-18 at a time \(t\) minutes after the initial dose is: \(Q(t) = Q0 * e^{-0.00699t}\)
Step 1 ：Substitute \(Q(174) = \frac{Q0}{3}\) into the exponential decay function: \(\frac{Q0}{3} = Q0 * e^{-k*174}\)
Step 2 ：Divide both sides by \(Q0\) to isolate the exponential term: \(\frac{1}{3} = e^{-k*174}\)
Step 3 ：Take the natural logarithm of both sides to solve for \(-k*174\): \(ln(\frac{1}{3}) = -k*174\)
Step 4 ：Divide both sides by \(-174\) to solve for \(k\): \(k = -\frac{ln(\frac{1}{3})}{174}\)
Step 5 ：Using a calculator, we find that \(k \approx 0.00699\) (rounded to five decimal places)
Step 6 ：\(\boxed{k \approx 0.00699}\)
Step 7 ：So, the function that models the radioactivity level of fluorine-18 at a time \(t\) minutes after the initial dose is: \(Q(t) = Q0 * e^{-0.00699t}\)
