An object of mass m is dropped into the sea at the surface. It is subject to a constant gravitational acceleration g and water resistance. (a) (4 points) If the water resistance is proportional to the object's velocity v(t) with coefficient b ≫ 0, so that Newton's second law gives b dv/dt =g- b/m v show that the object's velocity as a function of time t is given by v(t) = mg/b (1-e^(-bt/m)). We take downward to be the positive direction and the object is at rest when it is dropped at t=0. (b) (4 points) If instead the water resistance is quadratic in the velocity, so that Newton's second law gives dv/dt =g- b/m v^2 dt m dv dt find the object's velocity as a function of time t, again assuming that the object is dropped from rest. [Hint: Write Newtons' law as = g(1 − k^2v^2) where k = √b/mg. Then the algebra is a little less messy. After obtaining the final answer in terms of k, remember to replace k by √b/mg.] (c) (2 points) If the object could fall forever under these conditions without hitting the bottom, show that as t→ ∞ the velocity in part (a) approaches terminal velocity v → mg/b. What terminal velocity does the velocity in part (b) approach? #CalculusHelp #Integrals #differentialequations #PartialFractions