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The derivative of a function of a real variable in mathematics describes the sensitivity of the function value (output value) to changes in its argument (input value).

Calculus uses derivatives as a fundamental tool. When a derivative of a single-variable function exists at a given input value, it is the slope of the tangent line to the function's graph at that point.

(a)

Let the mass of the raindrop at time\(t\)be,\(m(t) = \rho \frac{4}{3}\pi r{(t)^3}\). Here,\(\rho \)is the density of the rain drop. And, its surface area is\(A(t) = 4\pi r{(t)^2}\). Assume that there is some positive constant\(k\)such that\({m^\prime }(t) = - kA(t)\). Plugging in the formulas above and using the chain rule to get:

\(\rho 4\pi r{(t)^2}{r^\prime }(t) = - k4\pi r{(t)^2}\)

\({r^\prime }(t) = - \frac{k}{\rho }\)

This implies that, \(r(t) = {r_0} - t\frac{k}{\rho }\).

(b)

From (a), let鈥檚 have:\(\frac{{{m^\prime }(t)}}{{m(t)}} = \frac{{ - kA(t)}}{{m(t)}} = \frac{{ - 3k}}{{\rho r(t)}}\)

Also let鈥檚 have:\(\frac{d}{{dt}}m(t)v(t) = m(t)g\)

\(\begin{aligned}{l} \Rightarrow {m^\prime }(t)v(t) + m(t){v^\prime }(t) = m(t)g\\ \Rightarrow \frac{{ - 3k}}{{\rho r(t)}}v(t) + {v^\prime }(t) = g\end{aligned}\)

So, the differential equation for the velocity is \(\frac{{ - 3k}}{{\rho {r_0} - tk}}v(t) + {v^\prime }(t) = g\) with initial condition \(v(0) = 0\).