Question

find the particular antiderivative of the following derivative that satisfies the given condition.
\\( \frac { d r } { d t } = \frac { 27 } { t ^ { 4 } }, r ( 1 ) = 75 \\)
\\( r = \square \\)

Explanation:

Step1: Find the general antiderivative

We know that if \(\frac{dR}{dt}=\frac{27}{t^{4}} = 27t^{- 4}\).
Using the power - rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have:
\(R(t)=\int\frac{27}{t^{4}}dt=27\int t^{-4}dt\)
\(R(t)=27\times\frac{t^{-4 + 1}}{-4 + 1}+C\)
\(R(t)=27\times\frac{t^{-3}}{-3}+C\)
\(R(t)=- 9t^{-3}+C=\frac{-9}{t^{3}}+C\)

Step2: Use the initial condition to find \(C\)

Given \(R(1) = 75\). Substitute \(t = 1\) and \(R(1)=75\) into \(R(t)=\frac{-9}{t^{3}}+C\)
\(75=\frac{-9}{1^{3}}+C\)
\(75=-9 + C\)
Add \(9\) to both sides: \(C=75 + 9=84\)

Answer:

\(R=\frac{-9}{t^{3}}+84\)