Math  /  Calculus

QuestionYou go to the doctor and he gives you 17 milligrams of radioactive dye which decays exponentially. After 24 minutes, 5.75 milligrams of dye remain in your system. To leave the doctor's office, you must pass through a radiation detector without sounding the alarm. If the detector will sound the alarm if more than 2 milligrams of the dye are in your system, how long will you have to remain at the doctor's office, assuming you were given the dye as soon as you arrived? Give your answer to the nearest minute.
You will spend \square minutes at the doctor's office.

Studdy Solution

STEP 1

What is this asking? How long until only 2 milligrams of radioactive dye are left in my system, if I start with 17 milligrams and it takes 24 minutes to decay to 5.75 milligrams? Watch out! Don't mix up the initial and final amounts!
Also, remember to round to the nearest minute at the end!

STEP 2

1. Find the decay constant
2. Find the time to reach 2 milligrams

STEP 3

We know the general formula for exponential decay is A(t)=A0ektA(t) = A_0 \cdot e^{kt}, where A(t)A(t) is the amount at time tt, A0A_0 is the **initial amount**, kk is the **decay constant**, and tt is the time.
We're given that the **initial amount** A0A_0 is **17** milligrams.

STEP 4

We know that after **24** minutes, **5.75** milligrams remain.
So, we can plug in A(24)=5.75A(24) = 5.75 and t=24t = 24 into our formula: 5.75=17e24k5.75 = 17 \cdot e^{24k}

STEP 5

To solve for kk, we first **divide both sides by 17**: 5.7517=e24k\frac{5.75}{17} = e^{24k} 2368=e24k\frac{23}{68} = e^{24k}Now, we take the **natural logarithm** of both sides: ln(2368)=ln(e24k)\ln\left(\frac{23}{68}\right) = \ln(e^{24k}) ln(2368)=24k\ln\left(\frac{23}{68}\right) = 24kFinally, we **divide by 24** to isolate kk: k=ln(2368)24k = \frac{\ln\left(\frac{23}{68}\right)}{24} k0.0433k \approx -0.0433

STEP 6

Now we want to find the time tt when A(t)=2A(t) = 2.
We'll use the same formula, A(t)=A0ektA(t) = A_0 \cdot e^{kt}, but this time we know A(t)=2A(t) = 2, A0=17A_0 = 17, and k0.0433k \approx -0.0433.

STEP 7

Let's plug in those values: 2=17e0.0433t2 = 17 \cdot e^{-0.0433t} Now, we **divide both sides by 17**: 217=e0.0433t\frac{2}{17} = e^{-0.0433t} Take the **natural logarithm** of both sides: ln(217)=ln(e0.0433t)\ln\left(\frac{2}{17}\right) = \ln(e^{-0.0433t}) ln(217)=0.0433t\ln\left(\frac{2}{17}\right) = -0.0433tFinally, **divide by -0.0433** to find tt: t=ln(217)0.0433t = \frac{\ln\left(\frac{2}{17}\right)}{-0.0433} t49.76t \approx 49.76

STEP 8

Since we need to round to the nearest minute, our final answer is approximately **50** minutes.

STEP 9

You will spend **50** minutes at the doctor's office.

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