Question

Suppose that the amount in grams of a radioactive substance present at time t (in years) is given by A(t) = 800e -0.86t. Find the rate of change of the quantity present at the time when t = 5. O 9.3 grams per year 0 -72.7 grams per year 0-9.3 grams per year O 72.7 grams per year

Answer 1

Let's begin by listing out the information given to us:

`\begin{gathered} A\mleft(t\mright)=800e^(-0.86t) \\ \Delta A(t)=-0.86\cdot800e^(-0.86t) \\ \Delta A(5)=-0.86\cdot800e^(-0.86\mleft(5\mright)) \\ A(5)=-9.34 \\ A(5)=-9.3g\text{ / year} \end{gathered}`

Hence, the correct option is the third option -9.3 grams per year