find the particular antiderivative of the following derivative that satisfies the given condition.
$ \frac { d r } { d t } = \frac { 135 } { t ^ { 4 } }, r ( 1 ) = 9 $
$ r = $

Question

Accepted Answer

# Explanation:
## Step1: Integrate the derivative
We know that if $\frac{dR}{dt}=\frac{135}{t^{4}} = 135t^{- 4}$, then by the power - rule of integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $R(t)=\int135t^{-4}dt$.
$$
\begin{align*}
R(t)&=135\int t^{-4}dt\
&=135\times\frac{t^{-4 + 1}}{-4+1}+C\
&=135\times\frac{t^{-3}}{-3}+C\
&=- 45t^{-3}+C\
&=-\frac{45}{t^{3}}+C
\end{align*}
$$
## Step2: Use the initial condition to find $C$
Given $R(1) = 9$, substitute $t = 1$ and $R(1)$ into $R(t)=-\frac{45}{t^{3}}+C$.
$$
\begin{align*}
9&=-\frac{45}{1^{3}}+C\
9&=-45 + C\
C&=9 + 45\
C&=54
\end{align*}
$$

# Answer:
$R=-\frac{45}{t^{3}}+54$

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QUESTION IMAGE

Question

Explanation:

Step1: Integrate the derivative

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

Answer: