Chapter 5.5, Problem 44PS

Solution

a.

To graph: The equation $A=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}$ .

Given information:

The equation: $A=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}$

Graph:

  

Interpretation:

Obtained a graph of equation: $A=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}$

b.

To calculate: A second dose of the drug is taken $1$ hour after the first dose. Write an equation to model the amount of the second dose in the bloodstream.

Equation to model the amount of the second dose in the bloodstream is: $B(t)=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}+\frac{391{(t+1)}^2+0.112}{0.218{(t+1)}^4+0.991{(t+1)}^2+1}$

Given information:

The equation: $A=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}$

Concept used:

Algebraically simplifying the equation $A=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}$ .

Calculation:

Since $A=A\left(t\right)$ is the amount of aspirin in the bloodstream $t$ hours after one dose was taken,

It means that the dose was taken at the moment $t=0$ .

The amount $B=B\left(t\right)$ of the aspirin in the bloodstream $t$ hours after the second dose was taken.

The second dose is taken one hour after the first one, it follows that if $t$ hours have passed since the second dose, $\left(t+1\right)$ hours have passed since the first dose.

The amount of the aspirin from the second dose in the bloodstream after $t$ hours is $A=A\left(t\right)$ and from the first dose $A\left(t+1\right)$ .

Since the second dose is just added to the amount of aspirin already circulating from the first dose,

Add these two amounts together. Hence, the formula is;

  $\begin{array}{l}B(t)\text{}=A(t)+A(t+1)\\ \\ B(t)=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}+\frac{391{(t+1)}^2+0.112}{0.218{(t+1)}^4+0.991{(t+1)}^2+1}\end{array}$

Conclusion:

  $B(t)=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}+\frac{391{(t+1)}^2+0.112}{0.218{(t+1)}^4+0.991{(t+1)}^2+1}$

c.

To graph: A model for the total amount of aspirin in the bloodstream after the second dose is taken.

Given information:

The equation: $A=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}$

Graph:

Let, $A\left(t\right)=A_1$ and $A(t+1)=A_2$

  $A_1=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}$

  $A_2=\frac{391{(t+1)}^2+0.112}{0.218{(t+1)}^4+0.991{(t+1)}^2+1}$

Total: $A_T=A_1+A_2$

  $A_T=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}+\frac{391{(t+1)}^2+0.112}{0.218{(t+1)}^4+0.991{(t+1)}^2+1}$

Graph of the total $A_T$:

  

Interpretation:

  $A_T=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}+\frac{391{(t+1)}^2+0.112}{0.218{(t+1)}^4+0.991{(t+1)}^2+1}$

d.

To graph: Model About how long after the second dose has been taken is the greatest amount of aspirin in the bloodstream.

Given information:

The equation: $A=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}$

Graph:

  $A_T=\frac{391t^2+0.112}{0.218t^4+0.991t^2+1}+\frac{391{(t+1)}^2+0.112}{0.218{(t+1)}^4+0.991{(t+1)}^2+1}$

  

Interpretation:

The maximum of $A_T$ occurs at about $t=\text{2}\text{.2}\text{hr}\text{.}$ with about $\text{369}\text{miligram}$ .