Question

The normal healing of wounds can be modeled by an exponential function. If \(A_{0}\) represents the original area of the wound, and if \(A\) equals the area of the wound, then the function $$A(n)=A_{0} e^{-0.35 n}$$ describes the area of a wound after \(n\) days following an injury when no infection is present to retard the healing. Suppose that a wound initially had an area of 100 square millimeters. (a) If healing is taking place, after how many days will the wound be one-half its original size? (b) How long before the wound is \(10 \%\) of its original size?

Short answer

The wound will be half its original size after about 1.98 days and 10% of its original size after about 6.58 days.

Step-by-step solution

Write down the given function and parameters

The exponential function describing the wound's healing is given by \[A(n) = A_{0} e^{-0.35 n}\].The initial area of the wound is \( A_{0} = 100 \) square millimeters.

Solve for half the original size of the wound

We want to find when the area is half of its original size, so set \( A(n) = \frac{100}{2} = 50 \) square millimeters. Set up the equation: \[ 50 = 100 e^{-0.35 n} \].

Isolate the exponential term

Divide both sides by 100 to isolate the exponential term: \[ \frac{50}{100} = e^{-0.35 n} \].This simplifies to: \[ 0.5 = e^{-0.35 n} \].

Solve for n using natural logarithms

Take the natural logarithm of both sides to solve for \( n \):\[ \ln(0.5) = \,\ln(e^{-0.35 n}) \].Since \( \ln(e^x) = x \), this becomes: \[ \ln(0.5) = -0.35 n \].

Calculate the value of n for half the original size

Solve for \( n \) by dividing both sides by \( -0.35 \):\[ n = \frac{\ln(0.5)}{-0.35} \].Using a calculator, \[ n \approx \frac{-0.6931}{-0.35} \approx 1.98 \, \text{days} \].

Solve for 10% of the original size of the wound

We want to find when the area is \( 10\text{\textpercent} \) of its original size, set \( A(n) = 0.1 \times 100 = 10 \) square millimeters.Set up the equation: \[ 10 = 100 e^{-0.35 n} \].

Isolate the exponential term

Divide both sides by 100 to isolate the exponential term: \[ \frac{10}{100} = e^{-0.35 n} \].This simplifies to: \[ 0.1 = e^{-0.35 n} \].

Solve for n using natural logarithms

Take the natural logarithm of both sides to solve for \( n \): \[ \ln(0.1) = \,\ln(e^{-0.35 n}) \].Since \( \ln(e^x) = x \), this becomes: \[ \ln(0.1) = -0.35 n \].

Calculate the value of n for 10% of the original size

Solve for \( n \) by dividing both sides by \( -0.35 \): \[ n = \frac{\ln(0.1)}{-0.35} \].Using a calculator, \[ n \approx \frac{-2.3026}{-0.35} \approx 6.58 \, \text{days} \].

Key concepts

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. They often appear in real-life situations like population growth, radioactive decay, and, as in our case, wound healing. The general form of an exponential function is \[f(x) = A \times b^{x}\]. Here, \(A\) is the initial value, \(b\) is the base (which determines the rate of growth or decay), and \(x\) is the exponent. In the wound healing model, the function is \[ A(n) = A_{0} e^{-0.35 n}\]. The base \( e \), approximately equal to 2.718, ensures that the rate of decay is continuous and natural, reflecting real-world processes closely.

Natural Logarithms

Natural logarithms (ln) are the inverse operations of exponentials with base \( e \). If you have an equation in the form \( e^x = y \), taking the natural logarithm of both sides \( \text{ln}(e^x) = \text{ln}(y) \) gives \( x = \text{ln}(y) \). In the wound healing example, solving \( 0.5 = e^{-0.35 n} \) involves taking the natural logarithm: \[ \text{ln}(0.5) = \text{ln}(e^{-0.35 n})\]. This equation simplifies to \( \text{ln}(0.5) = -0.35 n \), making it possible to solve for \( n \) by isolating the variable, \( n \). Natural logarithms are essential for solving exponential equations because they allow us to linearize the problem.

Wound Healing Model

The wound healing model described by the exponential function \(A(n) = A_{0} e^{-0.35 n}\) helps predict how quickly a wound decreases in size over time. In this context, \(A_0\) is the initial area of the wound, and \(A(n)\) represents the area after \( n \) days. Exponential decay occurs when the quantity decreases at a rate proportional to its current value. Here, the decay constant \( -0.35 \) indicates how rapidly the wound heals. The negative sign shows that the area is decreasing instead of increasing.

Half-Life Calculation

Calculating the half-life involves finding the time it takes for a quantity to reduce to half its initial value. For the wound healing model, we set the wound's area to 50% of its initial value: \[ 50 = 100 e^{-0.35 n} \]. By isolating the exponential term and solving with natural logarithms, you find: \( \text{ln}(0.5) = -0.35 n \) which simplifies to \[ n = \frac{\text{ln}(0.5)}{-0.35} \]. Using a calculator, you determine that the half-life is approximately 1.98 days. This means that in about two days, the wound's area will be half of its original size. Understanding half-life helps predict how quickly changes occur in exponential decay processes.

Decay Constant

The decay constant, in this case, is \( 0.35 \). It plays a crucial role in the rate at which the exponential function decreases. The larger the decay constant, the faster the decay process happens. In the equation \(A(n) = A_{0} e^{-0.35 n}\), \( 0.35 \) directly affects how steeply the exponential function declines. A negative exponent shows that the function is decaying over time rather than growing. By understanding the decay constant, we can better grasp the speed at which the value, like the area of a wound, diminishes. In summary, this constant enables us to model how physical processes, like healing, unfold over time.