Question

1. the area a of a healing wound changes at a rate given approximately by (\frac{da}{dt} = -4t^{-3}), (1 leq t leq 10), where t is time in days and (a(1) = 2) square centimeters. what will the area of the wound be in 10 days?

Explanation:

Step1: Integrate the rate function

To find the area function \( A(t) \), we integrate \( \frac{dA}{dt} = -4t^{-3} \) with respect to \( t \). The integral of \( t^n \) is \( \frac{t^{n + 1}}{n+1} + C \) (for \( n
eq - 1\)). So,
\[

$$\begin{align*} A(t)&=\int - 4t^{-3}dt\\ &=-4\times\frac{t^{-3 + 1}}{-3+1}+C\\ &=-4\times\frac{t^{-2}}{-2}+C\\ & = 2t^{-2}+C\\ &=\frac{2}{t^{2}}+C \end{align*}$$

\]

Step2: Find the constant \( C \)

We know that \( A(1) = 2 \). Substitute \( t = 1 \) and \( A(1)=2 \) into the equation \( A(t)=\frac{2}{t^{2}}+C \):
\[
2=\frac{2}{1^{2}}+C
\]
\[
2 = 2 + C
\]
Subtract 2 from both sides: \( C=2 - 2=0 \). So the area function is \( A(t)=\frac{2}{t^{2}} \).

Step3: Find \( A(10) \)

Substitute \( t = 10 \) into \( A(t)=\frac{2}{t^{2}} \):
\[
A(10)=\frac{2}{10^{2}}=\frac{2}{100}=0.02
\]

Answer:

The area of the wound in 10 days will be \( 0.02 \) square centimeters.