a. fnInt(2+5sin(4(pie)t)/25),t,0,6=31.815 yards^3/hr
b.Y(t)=fnInt(2+5sin(4(pie)t/25)-fnInt(15t/1-3t)+2500
c.Y(4)=fnInt(2+5sin(4(pie)(4)/25-fnInt(15(4)/1-3(4))+2500=2426.1824
d.The minimum is at t=0 because at that point the output is 2 and then the points before 0 were negative makign 0 a minimum. Also all the other times for t thre output are all positive.
quick and to the point, =]
ReplyDeletei like it. you made it easy to see. the d part too. thats the one i had trouble with
how sure are you about part D?
ReplyDeletehaha
seems reasonable, good job
but which equation are you using?