Only part (d) and (e) please:)
A tap on a large tank is gradually turned off so as not to create any hydraulic shock. As a consequence, the flow rate while the tap is being turned off is given by dV/dt = −2 + 1/10t m3/s.
A) What is the initial flow rate when the tap is fully on?
B) How long does it take to turn the tap off?
C) Given that when the tap has been turned off there are still 500 m3 of water left in the tank, find V as a function of t.
D) Hence find how much water is released during the time it takes to turn the tap off.
E) Suppose that it is necessary to let out a total of 300 m3 from the tank. How long should the tap be left fully on before gradually turning it off?
dv/dt = -2 + 1/10 t
v = -2t + 1/20 t^2 + C
since dv/dt = 0 when t=20, that means that
v(20) = 500, so
-2*20 + 1/20 * 20^2 + C = 500
C = 27
Thus, v(t) = 1/20 t^2 - 2t + 27
(D) ∫[0,20] v(t) dt = 620/3
(E) with the tap fully on, at 2 m^3/s, it would take (300 - 620/3)/2 = 140/3 seconds to drain the extra 280/3 m^3
The answer for D) is apparently 20m^3 released
and for E) is 140 seconds= 2 mins and 20 seconds
Thank you:)
To solve parts (d) and (e), we'll need to use the information and equations provided. Let's break it down step by step.
D) To find how much water is released during the time it takes to turn the tap off, we need to integrate the flow rate function over the given time interval. The flow rate function is given as dV/dt = -2 + 1/10t.
Integrating both sides with respect to t, we get:
∫ dV/dt dt = ∫ (-2 + 1/10t) dt
Using the integral properties, we simplify the equation:
∫ dV = ∫ (-2 + 1/10t) dt
V = -2t + 1/10 ln|t| + C
Since we don't have any initial conditions given, we'll just consider the constant of integration, C, as a constant for now.
Next, let's move on to part (e).
E) Suppose it is necessary to let out a total of 300 m3 from the tank. We need to determine how long the tap should be left fully on before gradually turning it off.
To find the time it takes to let out a total of 300 m3, we'll set V = 300 and solve for t:
300 = -2t + 1/10 ln|t| + C
Since we only need to find t, we'll ignore the constant C for now. Rearranging the equation, we have:
-2t + 1/10 ln|t| = 300
This is a transcendental equation, so it's a bit more complicated to solve analytically. We'll need to use numerical methods such as Newton's method or a graphing calculator to find an approximate solution.