Question

The mass m milligrams of a radioactive substance at time t days, is given by:m(t) = Ae^-tkwhere A and k are constants.The initial mass of the substance is 120 milligrams.The mass of the substance after 10 days is 90 milligrams.a)Sketch the graph to show the relationship between tand mfor t≥ 0 days.

Answer 1

Since the initial mass of the substance is 120 milligrams, then:

$m(0)=120.$

Therefore, the y-intercept of the graph of the given function is (0,120), also, evaluating the given function at t=0 we get:

$m(0)=Ae^{-0k}=Ae^0=A\text{.}$

Therefore A=120.

Now, we know that after 10 days, the mass of the substance is 90 milligrams, then the graph of the function passes through the point (10,90), also, evaluating the given function at t=10 we get:

$m(10)=120e^{-10k}.$

Therefore:

$90=120e^{-10k}\text{.}$

Dividing the above equation by 120 we get:

$\begin{gathered} \frac{90}{120}=\frac{120e^{-10k}}{120}, \\ \frac{3}{4}=e^{-10k}\text{.} \end{gathered}$

Applying the natural logarithm to the above equation we get:

$\begin{gathered} \ln (\frac{3}{4})=\ln (e^{-10k}), \\ \ln (\frac{3}{4})=-10k\text{.} \end{gathered}$

Dividing the above equation by -10 we get:

$k=-\frac{\ln (\frac{3}{4})}{10}\text{.}$

Therefore, the function that models this behavior is:

$\begin{gathered} m(e)=120e^{-(-\ln (\frac{3}{4}))\cdot\frac{t}{10}}=120e^{\ln (\frac{3}{4})\cdot\frac{t}{10}}\text{.} \\ \text{for t}\ge0. \end{gathered}$

And its graph is:

Answer: