Chapter 1.3, Problem 23E

(a)

To determine

To express: The amount of phosphorus-32 remaining with a function of $t$ .

Answer to Problem 23E

The amount of phosphorus- $32$ remaining with a function of $t$ is $6.6{\left(\frac{1}{2}\right)}^{\frac{t}{14}}$ .

Explanation of Solution

Given information: The half-life of phosphorus-32 is about $14$ days. There are $6.6\text{g}$ present initially.

Calculation:

The initial amount of phosphorus- $32$ was $6.6\text{g}$ . The amount of phosphorus- $32$ becomes half after every $14$ days.

The exponential function for the amount of phosphorus- $32$ left is:

  $\begin{array}{c}y=y_0{a}^t\\ y=6.6{\left(\frac{1}{2}\right)}^{\frac{t}{14}}\end{array}$

Therefore, the amount of phosphorus- $32$ remaining with a function of $t$ is $6.6{\left(\frac{1}{2}\right)}^{\frac{t}{14}}$ .

(b)

To determine

To find: The time taken to reach the amount of phosphorus- $32$ to $1\text{g}$ .

Answer to Problem 23E

The time taken to reach the amount of phosphorus- $32$ to $1\text{g}$ is $38.1145$ days.

Explanation of Solution

Given information: The half-life of phosphorus- $32$ is about $14$ days. There are $6\text{g}$ present initially.

Calculation:

As calculated in part (a), the exponential function for the remaining amount of Phosphorus is $y=6.6{\left(\frac{1}{2}\right)}^{\frac{t}{14}}$ .

Substitute 1 for $y$ in the exponential function.

  $\begin{array}{l}y=6.6{\left(\frac{1}{2}\right)}^{\frac{t}{14}}\\ 1=6.6{\left(\frac{1}{2}\right)}^{\frac{t}{14}}\end{array}$

Follow the steps to solve the equation by calculator:

• First press “ON” button on graphical calculator and press $\boxed{\text{Y}=}$ key.
• Enter right hand side of the equation $6.6{\left(\frac{1}{2}\right)}^{\frac{t}{14}}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{6.6}\boxed{\times }\boxed{(}\boxed{1}\boxed{÷}\boxed{2}\boxed{)}\boxed{^}\boxed{(}\boxed{\text{X}}\boxed{÷}\boxed{14}\boxed{)}$ . Now, press the key $\boxed{\text{ENTER}}$ .
• Enter left hand side of the equation $1$ after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{1}$ .
• Press $\boxed{\text{WINDOW}}$ key and adjust the window at $\left[-20,60\right]$ by $\left[-100,100\right]$ .
• Now, press keystrokes given below to find the value of $t$ as:

  $\boxed{2\text{nd}}\to \boxed{\text{TRACE}}\to \boxed{5}\to \boxed{\text{ENTER}}\to \boxed{\text{ENTER}}\to \boxed{\text{ENTER}}$

  

So, the value of t is $38.1145$ years.

Therefore, the time taken to reach the amount of phosphorus- $32$ to $1\text{g}$ is $38.1145$ days.