R= 2pi sin pi.tdR/dt = 2cos pi. t , find st. pts:2cos pi.t = 0 , pi.t = pi/2, 3pi/2 , t= 1/2, 3/2 .... d^2R/dt^2 = -2/pi sin pi.t , test t= 1/2 (they asked for the earliest time)turns out to be a maximum. therefore it will be filling up at its greatest rate. t is in hours so t= 1/2 is half an...