Solve the problem.
-Suppose that the amount in grams of a radioactive substance present at time $t$ (in years) is given by $A(t)=120 e^{-.48 t}$. Find the rate of change of the quantity present at the time when $t=6$.
A) -3.2 grams per year
B) -17.7 grams per year
C) 17.7 grams per year
D) 3.2 grams per year

Question

Quizplus · Accepted Answer

The rate of change is given by the derivative of $A(t)$, which is $A'(t)=-57.6 e^{-.48 t}$. At $t=6$, the rate of change is $A'(6)=-57.6 e^{-.48 \cdot 6} = -3.2$ grams per year.

Solve the Problem ttt (In Years) Is Given By A(t)=120e−.48tA(t)=120 e^{-.48 t}A(t)=120e−.48t

Solve the Problem $t$ (In Years) Is Given By $A(t)=120 e^{-.48 t}$