Chapter 3.1, Problem 78E

a.

To determine

Find the intial quantity when $t=0$ .

Answer to Problem 78E

The quantity needed is $25$ grams to make $t=0$ .

Explanation of Solution

Given:

It is given in the question that the quantity of radium present after t years is $Q=25{(\frac{1}{2})}^{t/1600}$ , where Q represents a mass, in grams, of radioactive radium ${}^{226}Ra$ ,whose half − life is $1600$ years.

Concept Used:

In this, use the concept of putting the value in the given function $Q=25{(\frac{1}{2})}^{t/1600}$ and solve it very accurately.

Calculation:

The quantity of radium present after t years is given by the function: $Q(t)=25{(\frac{1}{2})}^{t/1600}$ .

At , $t=0$ ,

  $\begin{array}{l}$ Q(t)=25{(\frac{1}{2})}^{t/1600}$$ Q(0)=25{(\frac{1}{2})}^{0/1600}$$ Q(0)=25{(\frac{1}{2})}^0$Q(0)=25\cdot 1\\ Q(0)=25\end{array}$

Conclusion:

The Quantity is $25$ grams.

b.

To determine

Calculate the quantity present after $1000$ years.

Answer to Problem 78E

The quantity needed is $0.0975$ grams to make $t=1000$ .

Explanation of Solution

Given:

It is given in the question that the quantity of radium present after t years is $Q=25{(\frac{1}{2})}^{t/1600}$ , where Q represents a mass, in grams, of radioactive radium ${}^{226}Ra$ ,whose half − life is $1600$ years.

The time is $1000$ years.

Concept Used:

In this, use the concept of putting the value in the given function $Q=25{(\frac{1}{2})}^{t/1600}$ and solve it very accurately.

Calculation: The quantity of radium present after t years is given by the function: $Q(t)=25{(\frac{1}{2})}^{t/1600}$ .

At , $t=1000$

  $\begin{array}{l}$ Q(t)=25{(\frac{1}{2})}^{t/1600}$$ Q(1000)=25{(\frac{1}{2})}^{1000/1600}$$ Q(1000)=25{(\frac{1}{2})}^{\frac{5}{8}}$Q(1000)=25\cdot 0.0039\\ Q(1000)=0.0975\end{array}$

Conclusion:

The Quantity is $0.0975$ grams.

c.

To determine

Draw the graph of the function in the interval $t=0$ to $t=5000$ .

Explanation of Solution

Given:

It is given in the question that the function is $Q(t)=25{(\frac{1}{2})}^{t/1600}$ .

Graph:

  

Interpretation:

The graph is shown above for the given function.

Simply,put the function in the graphic calculator,it gives the correct graph.And in the graph the curves linearly touch at the x axis at some certain points.And also the interval for $Q(t)$ be $t\in [0,5000]$ .

d.

To determine

Find in which time the quantity of radium be $0$ gram,explain.

Answer to Problem 78E

The time at which $Q(t)\to 0$ will be $t\to \infty$ .

Explanation of Solution

Given:

It is given in the question that the function is $Q(t)=25{(\frac{1}{2})}^{t/1600}$ .

Concept Used:

In this, use the concept of putting the limit on the both side and tending both the side to $0$ and then find the time.

Calculation:

Since , the function is $Q(t)=25{(\frac{1}{2})}^{t/1600}$ .

Now,

  $\begin{array}{l}$ Q(t)=25{(\frac{1}{2})}^{t/1600}$Q(t)=\frac{25}{2^{t/1600}}\end{array}$

Put limit to both sides,

  $\underset{t\to \infty }{\lim }Q(t)=\underset{t\to \infty }{\lim }\frac{25}{2^{t/1600}}$

The function is decreasing on $[0,\infty )$ .The quantity of radium will get closer to $0$ grams when t increases a lot: $\underset{t\to \infty }{\lim }\frac{25}{2^{t/1600}}=0$

Conclusion:

The time at which $Q(t)\to 0$ will be $t\to \infty$ .