3. the area a of a healing wound changes at a rate given approximately by

$$\frac{da}{dt}=-4t^{-3}, quad 1 \leq t \leq 10,$$

where t is time in days and a(1) = 2 square centimeters. what will the area of the wound be in 10 days?

Question

3. the area a of a healing wound changes at a rate given approximately by

$$\frac{da}{dt}=-4t^{-3}, quad 1 \leq t \leq 10,$$

where t is time in days and a(1) = 2 square centimeters. what will the area of the wound be in 10 days?

Accepted Answer

# Explanation:
## Step1: Integrate the rate function
We know that $A(t)=\int\frac{dA}{dt}dt$. Given $\frac{dA}{dt}=- 4t^{-3}$, then $A(t)=\int-4t^{-3}dt$.
Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have:
$$
\begin{align*}
A(t)&=-4\int t^{-3}dt\
&=-4\times\frac{t^{-3 + 1}}{-3+1}+C\
&=-4\times\frac{t^{-2}}{-2}+C\
&=2t^{-2}+C=\frac{2}{t^{2}}+C
\end{align*}
$$
## Step2: Find the constant of integration
We are given that $A(1) = 2$. Substitute $t = 1$ and $A(1)=2$ into $A(t)=\frac{2}{t^{2}}+C$:
$$
\begin{align*}
2&=\frac{2}{1^{2}}+C\
2&=2 + C\
C&=0
\end{align*}
$$
So the area function is $A(t)=\frac{2}{t^{2}}$
## Step3: Calculate the area at $t = 10$
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 after $10$ days is $0.02$ square centimeters.

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QUESTION IMAGE

Question

Explanation:

Step1: Integrate the rate function

Step2: Find the constant of integration

Step3: Calculate the area at \(t = 10\)

Answer: