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

To answer these questions, we need to integrate the given derivative function and use it for calculations. Let's break down each question and explain how to find the answers step by step.

(i) To find the rate at which the height is increasing after 20 minutes, we need to evaluate the derivative function at that time. The given derivative function is d'(t) = 4t + 5.

Step 1: Convert 20 minutes to hours. Since there are 60 minutes in an hour, 20 minutes is equal to 20/60 = 1/3 hours.

Step 2: Substitute the value of t = 1/3 into the derivative function:
d'(1/3) = 4(1/3) + 5 = 4/3 + 5 = 19/3 meters per hour.

Therefore, the rate at which the height is increasing after 20 minutes is 19/3 meters per hour.

(ii) To find the height of water in the tank after 20 minutes, we need to integrate the derivative function and evaluate it at t = 1/3.

Step 1: Integrate the derivative function.
To integrate 4t + 5, we get:
∫(4t + 5) dt = 2t^2 + 5t + C.

Step 2: Evaluate the indefinite integral at t = 1/3.
2(1/3)^2 + 5(1/3) + C = 2/9 + 5/3 + C = 19/9 + C.

Since the constant of integration (C) is not given, we can't find the exact height. We can only obtain a relative value.

(iii) To find the time it takes to fill the tank, we need to determine when the height of the tank reaches 6 meters. For this, we need to set up and solve an equation.

Step 1: Set height (h) = 6 meters in the indefinite integral we obtained in part (ii):
2t^2 + 5t + C = 6.

Step 2: Solve the quadratic equation for t. Subtract 6 from both sides and set the equation to zero:
2t^2 + 5t + C - 6 = 0.

Since we don't have the specific value of the constant of integration (C), we won't be able to solve for t and determine the exact time it takes to fill the tank.

In summary:
(i) The rate at which the height is increasing after 20 minutes is 19/3 meters per hour.
(ii) The height of water in the tank after 20 minutes is 19/9 + C meters (with C being the constant of integration).
(iii) The time it takes to fill the tank cannot be determined without the specific value of the constant of integration (C).