3. the area a of a healing wound changes at a rate given approximately by\n\n$$\frac{da}{dt}=-4t^{-3}, quad…

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

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

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\neq - 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.