Chapter 6.4, Problem 64E

(a)

To determine

The function that represents the amount y of plutonium 238 after t years.

Answer to Problem 64E

The value is $y=3{\left(0.5\right)}^{t/88}$

Explanation of Solution

Given information:

The given function as shown below,

  $y=a{\left(0.5\right)}^{1/x}$

Formula used:

Substitution method is used.

Calculation:

Substitute a = 3 grams & x = 88 to the given model:

  $y=3{\left(0.5\right)}^{t/88}$

Conclusion:

The value is $y=3{\left(0.5\right)}^{t/88}$

(b)

To determine

the yearly percent decreases of plutonium 238.

Answer to Problem 64E

The yearly percent decrease is about 0.8%

Explanation of Solution

Given information:

The given function as shown below,

  $y=a{\left(0.5\right)}^{1/x}$

Formula used:

Substitution method is used.

Calculation:

Use the fact that $t=\frac{1}{88}\left(88t\right)$ & the properties of exponents to rewrite the function in a form that reveals the yearly rate of decay.

Rewrite the model as shown below,

  $y=3{\left(0.5\right)}^{\left(1/88\right)\left(88t\right)}$

Using Power of a Power Property,

  $\begin{array}{l}y=3{\left({0.5}^{1/88}\right)}^{\left(88t\right)}\\ y=3{\left(0.992\right)}^{\left(88t\right)}\end{array}$

Use the growth factor (1 - r) to find the growth rate of $\approx$ 0.008.

Therefore, the yearly percent decrease is

About 0.8%

Conclusion:

The yearly percent decrease is about 0.8%

(c)

To determine

The amount remaining after 12 years.

Answer to Problem 64E

The amount at t = 12 is approximately 2.7 grams

Explanation of Solution

Given information:

The given function as shown below,

  $y=a{\left(0.5\right)}^{1/x}$

Formula used:

The graph is plotted against x axis and y axis.

Calculation:

Solve for a few points, plot & connect with a smooth line as shown below,

  

From the graph, the amount at t = 12 is approximately 2.7 grams

Conclusion:

The amount at t = 12 is approximately 2.7 grams