Chapter 12.4, Problem 33E

a.

To determine

Find the number of cells that have grown in culture by the end of the seventh day if no cells die.

Answer to Problem 33E

  $1400and713$

Explanation of Solution

Given information:

Leticia Cox is testing two different types of drugs that induce cell growth. She has selected two cultures of $1000cells$ each. To culture $A$ she administers a drug that raises the number of cells by $200$ each day and every day thereafter. Culture $B$ gets a drug that increases cell growth by $8\%$ each day and everyday thereafter.

Calculation:

Here, for culture $A$ :

Culture $A$ generates an arithmetic series where $a_1=1000,d=200,n=8$

  $\begin{array}{l}{a}_n=a_1+\left(n-1\right)\\ After7days:\\ a_8=1000+\left(8-1\right)200\\ a_8=2400\\ Thenumberofnewcells=2400-100=1400\end{array}$

Now, for culture $B$ :

Culture $B$ generates an arithmetic series where $a_1=1000,r=1.08,n=8$

  $\begin{array}{l}{a}_n=a_1{r}^{n-1}\\ After7days:\\ a_8=1000\times {1.08}^{8-1}\\ a_8=1713\\ Thenumberofnewcells=1713-1000=713\end{array}$

Hence the required answer is $1400and713$ .

b.

To determine

Find the drug that will be more effective in promoting cell growth.

Answer to Problem 33E

  $6000and9062$

Explanation of Solution

Given information:

Leticia Cox is testing two different types of drugs that induce cell growth. She has selected two cultures of $1000cells$ each. To culture $A$ she administers a drug that raises the number of cells by $200$ each day and every day thereafter. Culture $B$ gets a drug that increases cell growth by $8\%$ each day and everyday thereafter.

Calculation:

Here, for culture $A$ :

  $\begin{array}{l}After31days:\\ a_{31}=1000+\left(31-1\right)200\\ a_{31}=7000\\ Thenumberofnewcells=7000-1000=6000\end{array}$

Now, for culture $B$ :

  $\begin{array}{l}{a}_n=a_1{r}^{n-1}\\ After30days:\\ a_{31}=1000\times {1.08}^{31-1}\\ a_{31}=10062\\ Thenumberofnewcells=10062-1000=9062\end{array}$

Hence, the required answer is $6000and9062$ .