Chapter 4.3, Problem 49AYU

(a)

To determine

To Calculate:

The horizontal asymptote of $C\left(t\right)$ .what happens to the concentration of the drug as $t$ increases?

Answer to Problem 49AYU

The horizontal asymptote of $C\left(t\right)$ is $0$ and the concentration of the drug tends to zero as $t$ increases.

Explanation of Solution

Given Information:

  $C\left(t\right)=\frac{t}{2t^2+1}$

Calculation:

The horizontal asymptote of a function $h\left(x\right)$ is $\underset{x\to \infty }{\lim }h\left(x\right)$ .

The point at which the maximum value of a function occur is found out using seconf derivative test.

If $f\left(x\right)$ is the function, then find the critical points by setting $f\text{'}\left(x\right)=0$ .

Then find the values for $f\text{'}\text{'}\left(x\right)$ at the critical points.

If $f\text{'}\text{'}<0$ for a critical point, then it is the point at which function is the highest.

The horizontal asymptote is,

  $\begin{array}{l}\underset{x\to \infty }{\lim }C\left(t\right)=\underset{x\to \infty }{\lim }\left(\frac{t}{2t^2+1}\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{\frac{t}{t^2}}{\frac{2t^2}{t^2}+\frac{1}{t^2}}\right)\\ =\underset{x\to \infty }{\lim }\left(\frac{\frac{t}{t}}{2+\frac{1}{t^2}}\right)\\ =\frac{0}{2+0}\\ =0\end{array}$

The concentration of the drug tends to zero as $t$ increases.

(b)

To determine

To Graph:

  $C=C\left(t\right)$ by using graphing utility.

Answer to Problem 49AYU

The graph of $C=C\left(t\right)$ is,

  

Explanation of Solution

Given Information:

  $C\left(t\right)=\frac{t}{2t^2+1}$

By using desmos graphing utility, the graph of $C=C\left(t\right)$ as follows:

  $\begin{array}{l}C=C\left(t\right)\\ C\left(t\right)=\frac{t}{2t^2+1}\end{array}$

  

(c)

To determine

To Calculate:

The time at which the concentration is highest.

Answer to Problem 49AYU

The time at which the concentration is highest is $0.71$ hours after the injection.

Explanation of Solution

Given Information:

  $C\left(t\right)=\frac{t}{2t^2+1}$

Calculation:

The time at which the concentration is highest is $0.71$ hours after the injection