m = 5 t^.8 - 3 t + 20
dm/dt = (5*.8) t^-.2 - 3
so at max or min
0 = 4/T^.2 - 3
T^.2 = 4/3 = 1.333
.2 log T = log 1.333
Log T = .6246
T = 10^.6246 = 4.21 seconds
find m at t = 4.21
and find dm/dt at 3 and 5 (signs better be different)
Water is poured into a container that has a leak.The
mass 'm' of the water is given as a function of time 't'
by, m=5t^0.8 -3t + 20 , with t>=0, mass in grams and t in seconds
A) at what time is the water mass greatest,and B) what is the greatest mass? In kilogram per minute what is the rate of mass change at t=3 s, and t=5 s.
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