Question

The family of functions \(f\left( t \right) = C\left( {{e^{ - at}} - {e^{ - bt}}} \right)\), where a, b, and C are positive numbers and \(b > a\), has been used to model the concentration of a drug injected into a blood stream at time \(t = 0\). Graph several numbers of this family. What do they have in common? For fixed values of C and a, discover graphically what happens b increases. Then use calculus to prove what you have discovered.

Short answer

With \(b\) increases, the local maximum value increases, and the slope of the tangent increases at the origin.

Step-by-step solution

Graph several numbers of this family

The function \(f\left( t \right) = C\left( {{e^{ - at}} - {e^{ - bt}}} \right)\), the vertical stretching will be affected by \(C\).

Consider that\(C = 1\)then the function is\(f\left( t \right) = \left( {{e^{ - at}} - {e^{ - bt}}} \right)\)

The procedure to draw the graph of the function by using the graphing calculator is as follows:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(f\left( t \right) = \left( {{e^{ - at}} - {e^{ - bt}}} \right)\)in the tab.
2. Set the values as\(a = 1,b = 8\),\(a = 1,b = 2\)and\(a = 3,b = 8\). Observe the graph of the function.
3. Enter the “GRAPH” button in the graphing calculator to obtain the first graph of the function.
4. Set the values as\(a = 2\),\(b = 3,\)7, 12. Enter the “GRAPH” button in the graphing calculator to obtain the second graph of the function.

Visualization of the two graphs of the functions as shown below:

It is observed from the graph that all the graphs cross the origin, approach the \(t\)-axis with \(t\) increases, and approach \( - \infty \) as \(t \to - \infty \).

With \(b\) increases, the local maximum value increases, and the slope of the tangent increases at the origin.

Explain what happen b increases

Consider \(a = 2\) then the function is \(f\left( t \right) = \left( {{e^{ - 2t}} - {e^{ - bt}}} \right)\).

Obtain the first derivative of the function as shown below:

\(\begin{aligned}{c}f'\left( t \right) = \frac{d}{{dx}}\left( {{e^{ - 2t}} - {e^{ - bt}}} \right)\\ = b{e^{ - bt}} - 2{e^{ - 2t}}\end{aligned}\)

Take \(t = 0\) to obtain as shown below:

\(\begin{aligned}{c}f'\left( 0 \right) = b{e^{ - 0}} - 2{e^{ - 0}}\\ = b - 2\end{aligned}\)

The above values increase when \(b\) increases.

Take \(f'\left( t \right) = 0\) to obtain as shown below:

\(\begin{aligned}{c}b{e^{ - bt}} - 2{e^{ - 2t}} = 0\\b{e^{ - bt}} = 2{e^{ - 2t}}\\\frac{b}{2} = {e^{\left( {b - 2} \right)t}}\\\ln \frac{b}{2} = \left( {b - 2} \right)t\\t = {t_1} = \frac{{\ln b - \ln 2}}{{b - 2}}\end{aligned}\)

This decreases when \(b\) increases (the maximum becomes closer to the \(y\)-axis). \(f\left( {{t_1}} \right) = \frac{{\left( {b - 2} \right){2^{\frac{2}{{\left( {b - 2} \right)}}}}}}{{{b^{\frac{{1 + 2}}{{b - 2}}}}}}\). Consider it to be a function of \(b\) and graph the derivative with respect to \(b\) that is always positive. We can demonstrate that it increases when \(b\) increases.