= a solution which (a) (2 pt) Take limit ba and show that we recover the b we wrote down in a previous class where Qin Qout = Q and tank volume stays at its initial value V = Vo. Use the same technique as you used in the previous problem in (e) without L'Hopital's rule. Qin (b) (8 pt) Sketch the full solution where ab for a < b which corresponds to Qout so that the tank empties out from its initial volume Vo > 0. i. (1 pt) Show that the tank empties in a time t = Vo/(b-a). ii. (1 pt) Factor -1 from a b and rewrite the solution in terms of b - a, e.g. the exponent -a/(a - b) is a/(ba) and now we can see that it is positive. Do the same thing and replace (a - b)t with -(b-a)t. Show that - M(t) = VoCin [1 (b a)t/Vo] [1 (1 (b a)t/Vo)/(b-a)] - - - (3) iii. (2 pt) Early Times Approximate the term (1-(b-a)t/Vo)a/(b-a) at small time using the Taylor Expansion (1+X) 1+aX for aX < < 1 and setting X = -(ba)t/Vo. Insert the approximation into the provided solution and once the dust settles on the algebra and show that at small times MatCin and remember that a = Qin is the input flow rate. This little result means that the M is increasing at small times as we expect and we know how rapidly that is happening too! it and M = iv. (2 pt) Nearly Empty As the tank empties of fluid all the mass goes with = 0 when the tank is empty. When the tank is nearly empty 1-(b-a)t/Vo is small and given that the exponent a/(b-a) >0 show that M V(t)Cin where V(t) = Vo - (b-a)t is the tank volume. v. (2 pt) Now you can sketch M(t). Be sure to explicitly label your sketch with the results from above for full points. The only thing we have not found out is an approximation to the solution near its peak. But that is okay!