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The Natural Base e

The Natural Base e 1.16 - Solution

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a
If an element decays continuously, we can use the following equation to model the situation.

y=Iert\begin{gathered} \textcolor{darkorange}{y}={\color{#0000FF}{I}}e^{\textcolor{magenta}{r}{\color{#009600}{t}}} \end{gathered} Its components can be defined as shown below. y: remaining amountI: initial amounte: Euler’s numberr: decay ratet: number of years\begin{aligned} \textcolor{darkorange}{y}\text{:}& \text{ remaining amount}\\ {\color{#0000FF}{I}}\text{:}& \text{ initial amount}\\ e\text{:}& \text{ Euler's number}\\ \textcolor{magenta}{r}\text{:}& \text{ decay rate}\\ {\color{#009600}{t}}\text{:}& \text{ number of years} \end{aligned} With this, we can analyze the equation that models the tritium decay. y=10e-0.0562t\begin{gathered} \textcolor{darkorange}{y}={\color{#0000FF}{10}}e^{\textcolor{magenta}{\text{-}0.0562}{\color{#009600}{t}}} \end{gathered} Therefore, the initial amount of uranium is 1010 milligrams. Looking at the graph, we can read the initial value as 15.15.

Radium has the highest amount in the beginning.

b
We can also find the remaining amount after 10{\color{#009600}{10}} years by substituting t=10{\color{#009600}{t}}={\color{#009600}{10}} in the given equation.
y=10e-0.0562ty=10e^{\text{-}0.0562t}
y=10e-0.0562(10)y=10e^{\text{-}0.0562({\color{#009600}{10}})}
Simplify right-hand side
y=10e-0.562y=10e^{\text{-}0.562}
y=10(0.57006)y=10(0.57006\ldots)
y=5.7006y=5.7006\ldots
y5.70y\approx5.70
The initial amount of uranium is about 5.70\textcolor{darkorange}{5.70} milligrams. Next, let's examine the graph that models the radium decay.

The graph also shows that the remaining amount is about 1\textcolor{darkorange}{1} milligram when t=10.t=10. Therefore, uranium has a greater remaining amount.