Question

You go to the doctor and he injects you with 13 milligrams of radioactive dye. After 12 minutes, 4.75 milligrams of dye remain in you system. To leave the doctor's office, you must pass through a radiation detector without sounding the alarm. If the detector will sound will sound the alarm whenever more than 2 milligrams of the dye are in your system, how long will your visit to the doctor take, assuming you were given the dye as soon as you arrived and the amount of dye decayse exponentially?

Answer 1

Answer:

Step-by-step explanation:

You need to find the model for this situation using the points given. If you are given 13 mg immediately, the coordinate point for that is (0, 13), where the x coordinate is the time in minutes and the y coordinate is the amount of dye in mg.

If after 12 minutes, 4.75 mg remain, the coordinate point for that is (12, 4.75). Just so you know, the time unit always goes in for x, never for y.

Now that we have these coordinates we are going to use them as a system and find the model. Begin with the point (0, 13) in

$A=a(b)^t$ where A is the amount left in the body, a is the initial amount injected, b is the rate at which it decays, and t is the time in minutes. Plug in the first set of coordinates:

$13=a(b)^0$ and since anything raised to the power of 0 is equal to 1, we have

13 = a. Now we can use that and write another equation with the other set of coordinates to solve for b:

$4.75=13(b)^{12}$ and divide both sides by 13 (don't round anything yet) to get

.3653846154 = b¹². Now take the 12th root of both sides (you need a scientific calculator for this) to get that

b = .9195228407  Yikes.

Now for the model, using the fact that a = 13 and b = .9195228407:

$A=13(.9195228407)^t$  

We are told that the amount that we are allowed to leave with still remaining in our system cannot be more than 2 mg, so we sub in a 2 for A:

$2=13(.9195228407)^t$ and begin by dividing away 13 to get

$.1538461538=.9195228407^t$  Take the natural log of both sides, which allows us to bring the t out front:

ln(.1538461538) = t ln(.9195228407) and finally divide by ln(.9195228407) to get

t = 22.3 minutes