Chapter 5.4, Problem 120AYU

(a)

To determine

The number of days after which the wound will be one-half of its original size, where the normal healing of wounds can be modelled by an exponential function, $A\left(n\right)=A_0{e}^{-0.35n}$ , $A_0$ represents the original area of the wound and $A$ equals the area of the wound, then the function denotes the area of wound after $n$ days following an injury when no injection is present to retard the healing.

Answer to Problem 120AYU

Solution:

After approximately $2$days, the wound will be one half of its original size

Explanation of Solution

Given information:

The normal healing of wounds can be modelled by an exponential function, $A\left(n\right)=A_0{e}^{-0.35n}$ , where $A_0$ represents the original area of the wound and $A$ equals the area of the wound, then the function denotes the area of wound after $n$ days following an injury when no injection is present to retard the healing.

The wound initially had an area of $100$ square millimeters

From the given information,

The normal healing of wounds can be modelled by an exponential function $A\left(n\right)=A_0{e}^{-0.35n}$

Suppose that after $n$ days, the wound will be one half of its original size.

  $\Rightarrow \frac{A_0}{2}=A_0{e}^{-0.35n}$

Also, the wound initially had an area of $100$ square millimeters. That is, $A_0=100$

  $\Rightarrow \frac{100}{2}=100e^{-0.35n}$

By simplifying,

  $\Rightarrow 50=100e^{-0.35n}$

  $\Rightarrow \frac{1}{2}=e^{-0.35n}$

Taking natural log on both sides,

  $\Rightarrow \ln \left(\frac{1}{2}\right)=\ln \left(e^{-0.35n}\right)$

  $\Rightarrow \ln \left(\frac{1}{2}\right)=-0.35n$

  $\Rightarrow n=1.98042$

Thus, after $1.98042\approx 2$ days, the wound will be one half of its original size.

(b)

To determine

The number of days before which the wound is $10\%$ of its original size, where the normal healing of wounds can be modelled by an exponential function, $A\left(n\right)=A_0{e}^{-0.35n}$ , $A_0$ represents the original area of the wound and $A$ equals the area of the wound, then the function denotes the area of wound after $n$ days following an injury, when no injection is present to retard the healing.

Answer to Problem 120AYU

Solution:

Before approximately $6.57881\approx 7$ days, the wound is $10\%$ of its original size.

Explanation of Solution

Given information:

The normal healing of wounds can be modelled by an exponential function $A\left(n\right)=A_0{e}^{-0.35n}$ , where $A_0$ represents the original area of the wound and $A$ equals the area of the wound, then the function denotes the area of wound after $n$ days following an injury when no injection is present to retard the healing.

The wound initially had an area of $100$ square millimeters

From the given information,

The normal healing of wounds can be modelled by an exponential function $A\left(n\right)=A_0{e}^{-0.35n}$

Suppose that before $n$ days the wound will be $10\%$ of its original size.

  $\Rightarrow 10\%A_0=A_0{e}^{-0.35n}$

  $\Rightarrow \frac{1}{10}{A}_0=A_0{e}^{-0.35n}$

  $\Rightarrow \frac{1}{10}=e^{-0.35n}$

Taking natural log on both sides,

  $\Rightarrow \ln \left(\frac{1}{10}\right)=\ln \left(e^{-0.35n}\right)$

  $\Rightarrow \ln \left(\frac{1}{10}\right)=-0.35n$

  $\Rightarrow n=6.57881$

Thus, before $6.57881\approx 7$ days the wound will be $10\%$ of its original size