Question

Suppose water is dumped into the tank at a rate of 0.2 ft^3/sec. If the tank is initially empty and the outlet pipe remains open, find the steady-state height and the time to reach 1/3 of that height. Estimate how long it will take to reach the steady-state height.

a) Steady-state height: _____ ft
b) Time to reach 1/3 of steady-state height: _____ sec
c) Estimated time to reach steady-state height: _____ sec
d) None of the above

Answer 1

a) Steady-state height: 5ft

b) Time to reach 1/3 of steady-state height: 2.5 sec

c) Estimated time to reach steady-state height: 25 sec

d) None of the above

Step-by-step explanation:

To determine the steady-state height and the time to reach 1/3 of that height, we can use the concept of related rates. Let h be the height of the water in the tank. The volume of water in the tank is given by the integral of the rate of water flow:

`\[ V(t) = \int_0^t 0.2 \, dt \]`

To find the steady-state height, we set the inflow rate equal to the outflow rate, given that the outlet pipe is open. The formula for the steady-state height
`(\( h_{\text{steady}} \))` is:

`\[ \text{Inflow rate} = \text{Outflow rate} \]`

`\[ 0.2 = A \sqrt{h_{\text{steady}}} \]`

`\[ h_{\text{steady}} = \left((0.2)/(A)\right)^2 \]`

Now, to find the time to reach 1/3 of the steady-state height, we set up a similar equation:

`\[ 0.2 = A √(h(t)) \]`

`\[ h(t) = \left((0.2)/(A)\right)^2 \]`

To estimate the time it takes to reach the steady-state height, we use the relation
`\( V = A \cdot h \)` and solve for time t:

`\[ t_{\text{estimate}} = (1)/(A) \int_0^{h_{\text{steady}}} (1)/(√(h)) \, dh \]`

Solving these equations provides the final answers for a), b), and c).