As a raindrop falls, it evaporates while retaining its spherical shape. If we make the further assumptions that the rate at which the raindrop evaporates is proportional to its surface area and that air resistance is negligible, then a model for the velocity v(t) of the raindrop is 3(k/p) (k/p)t + ro Here p is the density of water, ro is the radius of the raindrop at t= 0, k < 0 is the constant of proportionality, and the downward direction is taken to be the positive direction. (a) Solve for v(t) if the raindrop falls from rest. v(t) = dv dt -v=g. (b) This model assumes that the rate at which the raindrop evaporates-that is, the rate at which it loses mass-is proportional to its surface area, with constant of proportionality k<0. This assumption implies that the rate at which the radius r of the raindrop decreases is a constant. Show that the radius of the raindrop at time t is r(t) = (k/p)t + 50₁ Letting A denote the surface area of the raindrop and m the mass, the model assumes that and r. dmk A. dt Since mass equals density times Select and a sphere of radius r has volume m = A- and surface area we obtain the following formulas form and A in terms of p