Calculus
posted by Grant .
Fuel is flowing into a storage tank which can be filled to a depth of 6 metres.
When the fuel started flowing the tank was already filled to a depth of 2.5m. If the rate at which the depth of fuel in the tank is increasing, in metres per hour is given by d'(t) = 4t +5, find:
(i) the rate at which the height is increasing after 20 minutes
(ii) the height of water in the tank after 20 minutes
(iii) the time it takes to fill the tank

Let the depth of the water be d(t)
if d ' (t) = 4t+5 , where t is in hours
d(t) = 2t^2 + 5t + c, where c is a constant
given: when t=0, d(0) = 2.5
2.5 = 0 + 0 + c , > c = 2.5
d(t) = 2t^2 + 5t + 2.5
a) when t = 20 min, t = 1/3 hrs
d ' (1/3) = 4(1/3) + 5 = 19/3 m/hr
b) after 20 min or after 1/3 hr
d(1/3) = 2(1/9) + 5(1/3) + 2.5 = 79/18 or appr 4.39 minutes
c) to fill the tank ...
2t^2 + 5t + 2.5 = 6
2t^2 + 5t  3.5 = 0
by the formula ...
t = (5 ± √53)/4
= .57 hrs or a negative time, which we will reject
.57 hrs = 34.2 minutes