Question

find the particular antiderivative of the following derivative that satisfies the given condition.

\\( \frac { d r } { d t } = \frac { 75 } { 4 t } \cdot r ( 1 ) = 25 \\)

\\( r = \square \\)

Explanation:

Step1: Integrate the derivative

We know that if \(\frac{dR}{dt}=\frac{75}{4t}\), then by the rule of integration \(\int\frac{1}{t}dt=\ln|t| + C\). So, \(R=\int\frac{75}{4t}dt\). Using the constant - multiple rule of integration \(\int kf(t)dt = k\int f(t)dt\) (\(k = \frac{75}{4}\) and \(f(t)=\frac{1}{t}\)), we get \(R=\frac{75}{4}\ln|t|+C\).

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

Given \(R(1) = 25\). Substitute \(t = 1\) into \(R(t)=\frac{75}{4}\ln|t|+C\). Since \(\ln(1)=0\), we have \(25=\frac{75}{4}\ln(1)+C\). So, \(C = 25\).

Answer:

\(R=\frac{75}{4}\ln|t| + 25\)