#a)# The total flow into the pipe over a given time period is effectively the sum of the water flowing into the pipe at every point in time throughout that period. A "continuous sum" like this is just another way of thinking about an integral, and so the total flow can be represented by the integral of #R# over that time period. Doing so from #0 <= t <= 8#, we have
#int_0^8 R(t)dt = int_0^8 20sin(t^2/35)dt#
This integral cannot easily be solved analytically, but can be approximated with the graphing calculator and found to be approximately #76.5704 " ft"^3#
#b)# The change in the amount of water in the pipe at a given time can be expressed as the difference between the water flowing into the pipe and the water flowing out. In terms of the given function, that means that the change in the amount of water is given by #R(t)-D(t)#.
At #t = 3#, we have the change as
#20sin(3^2/35)-(-0.04(3)^3+0.4(3)^2+0.96(3))#
#~~ -0.314" ft"^3#
Thus the amount of water is decreasing.
#c)# The amount of water in the pipe at any given time can be expressed as the sum of the initial amount and the total change in amount over time. That is, the amount of water #a(t)# is given by
#a(t) = 30 + int_0^t(R(x)-D(x))dx#
Then, to find the minimum, we must look for the extrema of #a(t)# with #0 <= t <= 8#, that is, the points in that range at which #a'(t) = 0# (as #a'(t)# is defined on all of #RR#, we do not need to look for points at which it is undefined).
#a'(t) = d/dt(30 + int_0^t(R(x)-D(x))dx)#
#= R(t)-D(t)#
#= 20sin(t^2/35)+0.04t^3-0.4t^2-0.96t = 0#
While this is not easily solved algebraically, a graphing calculator can be used to find the zeroes of #a'(t)#. Doing so finds two roots at #t = 0# and #t ~~ 3.271#.
If we pick two values #t_1 in (0,3.271...)# and #t_2 in (3.271..., 8)#, we will find that #a'(t_1) < 0# and #a'(t_2) > 0#. As there are no roots of #a'# in those intervals, this implies that #a# is decreasing on #(0,3.271...)# and increasing on #(3.271..., 8)# (we could also verify this by looking at the graph of #a'#). Thus we have the minimum for #t in [0,8]# at #t ~~ 3.271 " hours"#
#d)# A look at the graph of #a(t)# shows that there is not a maximum at #a(t)=50#, meaning the pipe will begin to overflow at that point. Then, the desired equation is simply #a(w) = 50#, or
#30+int_0^w(R(t)-D(t))dt = 50#
