Answer
a. $\frac{dy}{dt} = -ky; \quad y(0) = 10$
b. $y(t) = 10e^{-0.0050t}$
c. Approximately 7 milligrams
d. $t \approx 240.79$ days
Work Step by Step
Radioactive Decay Analysis
Part a
Since we are dealing with radioactive decay, the formula is: \[ \frac{dy}{dt} = -ky; \quad y(0) = 10 \]
Part b
Recall the formula for half-life is $\frac{1}{2} = e^{-kt}$, so we have: \[ \ln(2) = kt \quad \Rightarrow \quad k = \frac{\ln(2)}{140} \approx 0.0050 \] Hence, the formula for $y(t)$ is: \[ y(t) = 10e^{-0.0050t} \]
Part c The milligrams after 70 weeks (which is 70 days) is: \[ y(70) = 10e^{-0.0050 \cdot 70} \approx 7 \text{ milligrams} \] Part d If 70% decay from a 10 milligram container, it will leave 3 milligrams in the container, and we have: \[ 3 = 10e^{-0.0050t} \quad \Rightarrow \quad \ln(0.3) = -0.0050t \quad \Rightarrow \quad t = -\frac{\ln(0.3)}{0.0050} \approx 240.79 \text{ days} \] Result a. $\frac{dy}{dt} = -ky; \quad y(0) = 10$ b. $y(t) = 10e^{-0.0050t}$ c. Approximately 7 milligrams d. $t \approx 240.79$ days
