Question

Radioactive Decay Suppose that $\mathit{d}\mathit{A}\mathbf{/}\mathit{d}\mathit{t}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{0004332}\mathit{A}\left(t\right)$represents a mathematical model for the decay of radium-$\mathbf{226}$, where $\mathit{A}\left(t\right)$is the amount of radium (measured in grams) remaining at time $\mathit{t}\mathbf{>}\mathbf{0}$(measured in years). How much of the radium sample remains at the time when the sample is decaying at a rate of $\mathbf{0}\mathbf{.}\mathbf{002}$grams per year?

Short answer

Answer:

$4.6gm$of the radium sample radium remains at the time when the sample is decaying at a rate of 0.002 grams per year.

Step-by-step solution

Define the a derivative of the a function

The derivative of a function of a real variable in mathematics describes the sensitivity of the function value (output value) to changes in its argument (input value).

Calculus uses derivatives as a fundamental tool. When a derivative of a single-variable function exists at a given input value, it is the slope of the tangent line to the function's graph at that point.

Determine the mathematical model

It is given that $\frac{dA\left(t\right)}{dt}=-0.002$. When sample decays at -0.002 grams per year, the value of $\frac{dA\left(t\right)}{dt}$becomes -0.002.

Substitute -0.002 for $\frac{dA\left(t\right)}{dt}$into $\frac{dA\left(t\right)}{dt}=-0.002$.

$$\begin{gathered} 0.002=-0.0004332A\left(t\right) \\ A\left(t\right)=\frac{-0.002}{-0.0004332} \\ A\left(t\right)\approx 4.6gm \end{gathered}$$

Hence, $4.6gm$of the radium sample remains at the time when the sample is decaying at a rate of 0.002 grams per year.