Tank 1 initial contains 70 L (liters) of water
and 450 g of salt, while tank 2 initially contains 20 L of water
and 200 g of salt. Water containing 10 g/L of salt
is poured into tank1 at a rate of 1 L/min while the mixture
flowing into tank 2 contains a salt concentration of 50 g/L
of salt and is flowing at the rate of 4 L/min.
The two connecting tubes have a flow rate of 4 L/min from
tank 1 to tank 2; and of 3 L/min from tank 2 back to tank 1.
Tank 2 is drained at the rate of 5 L/min.
how do you find q'(t)?
how do you find the initial values of p and q?
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R{in}-R{out}=dA/dt, R=rate
Initial values are initial amounts of salt.
To find q'(t), we need to differentiate the equation that represents the volume of water in tank 2 with respect to time (t).
But before we do that, let's break down the problem and define some variables:
- Let q(t) be the volume of water in tank 2 at time t (in liters).
- Let p(t) be the volume of water in tank 1 at time t (in liters).
- Let s(t) be the amount of salt in tank 2 at time t (in grams).
Based on the given information, we can derive the following equations:
1. The rate of change in the volume of water in tank 2 is given by the difference in flows:
q'(t) = 4 L/min * (p(t)/q(t)) - 5 L/min
2. The rate of change in the volume of water in tank 1 is given by the net flow in and out:
p'(t) = 1 L/min - 4 L/min + 3 L/min * (s(t)/p(t))
Now, to find q'(t), we differentiate equation 1 with respect to t:
d(q'(t))/dt = d(4 L/min * (p(t)/q(t)))/dt - d(5 L/min)/dt
To differentiate the first term on the right side, we can apply the quotient rule:
d(4 L/min * (p(t)/q(t)))/dt = 4 L/min * (q(t) * p'(t) - p(t) * q'(t))/(q(t))^2
Since d(5 L/min)/dt is 0 (as it's a constant rate of drain), we can simplify the equation:
d(q'(t))/dt = 4 L/min * (q(t) * p'(t) - p(t) * q'(t))/(q(t))^2
To find p and q initial values, we need to evaluate p(0) and q(0), which represent the initial volumes of water in tanks 1 and 2, respectively. These values are given in the problem statement.
- p(0) = 70 L (liters) (Volume in tank 1 initially)
- q(0) = 20 L (liters) (Volume in tank 2 initially)
These are the initial values of p and q in the problem.