Chapter 3.2, Problem 37E

To determine

a.

The number of tumor cells present at diagnosis.

To determine

b.

To graph:

The graph of the function both before and after chemotherapy if patients receives the chemotherapy after the tumor has grown for $40$ months and $99.9\%$ of the cells are instantaneously killed.

Solution:

The graph of the function both and before chemotherapy is:

Explanation:

Given information:

A very aggressive tumor is growing according to the function $N\left(t\right)=2^t$ where $N\left(t\right)$ represent the number of cells at time $t$ (in months).

Calculation:

The number of tumor cells at $40$ months are $1\text{099511627776}$.

After chemotherapy $99.9\%$ of the cells are instantaneously killed.

So the $\text{0}\text{.1\%}$ tumor cells are remaining.

Hence the remaining tumor cells are $1\text{099511627776}\times \text{0}\text{.1\%=}1\text{099511627}\text{.776}$.

Here use ${\log }_{2}N\left(t\right)$ on vertical axis that is, ${\log }_2{2}^t=t$.

So take time $t$ on vertical axis and also $t$ on horizontal axis.

The tumor is continues grow for first $40$ months, hence graph is like below:

After chemotherapy, the cells are $1\text{099511627}\text{.776}$.

Thus, ${\log }_2\left(1\text{099511627}\text{.776}\right)=30.03$.

Hence after chemotherapy the graph look like as below:

To determine

c.

The discontinuous point in graph of part (b).