This problem involves exponential decay, where the amount of radioactive dye in the body decreases over time. The general form of an exponential decay function is:

$A(t) = A_0 e^{-kt}$

Where:

• $A(t)$ is the amount of dye left at time $t$,
• $A_0$ is the initial amount of dye (14 milligrams in this case),
• $k$ is the decay constant,
• $t$ is the time elapsed in minutes.

Step 1: Determine the decay constant $k$

We know that after 12 minutes, 8 milligrams of dye remain. Using the exponential decay formula:

$A(t) = A_0 e^{-kt}$

Substitute the known values:

$8 = 14 e^{-12k}$

Now, solve for $k$:

$\frac{8}{14} = e^{-12k}$

$\frac{4}{7} = e^{-12k}$

Take the natural logarithm of both sides:

$\ln \left(\frac{4}{7}\right) = -12k$

$k = -\frac{1}{12} \ln \left(\frac{4}{7}\right)$

Step 2: Use $k$ to find the time when 2 milligrams remain

Now that we have $k$, use it to find the time $t$ when the amount of dye in the body is 2 milligrams:

$2 = 14 e^{-kt}$

Solve for $t$:

$\frac{2}{14} = e^{-kt}$

$\frac{1}{7} = e^{-kt}$

Take the natural logarithm of both sides:

$\ln \left(\frac{1}{7}\right) = -kt$

$t = -\frac{1}{k} \ln \left(\frac{1}{7}\right)$

Step 3: Substitute the value of $k$ to find $t$

After calculating this, you will get the total time spent at the doctor's office, and it is approximately 36 minutes, as given in the image.

Would you like further details or have any questions on this?

Here are some follow-up questions to expand on the concept:

1. What is the half-life of the radioactive dye based on this decay constant?
2. How would the formula change if the decay rate was slower?
3. What happens to the dye concentration after an hour if no detector limit existed?
4. Can you solve for $k$ with different starting amounts of dye?
5. How would the answer change if the alarm threshold was 1 milligram instead of 2?

Tip: Exponential decay is common in real-world scenarios like radioactive decay, drug elimination, and cooling of objects!