Mathematical question

Question

A tank initially contains 100 liters of water. Water flows into the tank at a rate of , and simultaneously leaks out at a rate proportional to the amount of water in the tank. If the proportionality constant is , find the volume of water in the tank at any time (in minutes). What is the long-term behavior of the water volume?

@Muruca · Answer

a tank initially contains 100 liters of water. water flows into the tank at a constant rate r liters per minute, and at the same time, water leaks out at a rate proportional to the current volume of water in the tank. let k be the proportionality constant for the leakage, with units of per minute. the change in the volume of water in the tank over time can be modeled by the following first-order linear differential equation: dv/dt = r - kv(t) where v(t) represents the volume of water in the tank at time t, measured in minutes. to solve this equation, we first rewrite it in standard linear form: dv/dt + kv = r the integrating factor for this equation is: μ(t) = e^(kt) multiplying both sides of the equation by the integrating factor gives: e^(kt) dv/dt + k e^(kt) v = r e^(kt) this simplifies to: d/dt [e^(kt) v] = r e^(kt) integrating both sides with respect to t results in: e^(kt) v = (r/k) e^(kt) + c where c is the constant of integration. solving for v(t): v(t) = (r/k) + c e^(-kt) using the initial condition v(0) = 100, we find the value of c: 100 = (r/k) + c therefore: c = 100 - (r/k) the general solution for the volume of water in the tank at any time t is: v(t) = (r/k) + [100 - (r/k)] e^(-kt) as time progresses towards infinity, the exponential term e^(-kt) tends towards zero. this means that the long-term behavior of the volume is to stabilize at: v(t) → (r/k)

ae6178881 · Answer

JangoTheOfficial · Answer

solution to the water tank problemproblem statement
a tank initially contains 100 liters of water. water flows into the tank at a constant rate of r liters per minute, while simultaneously leaking out at a rate proportional to the amount of water present. the leakage rate is given by kv(t), where k is the proportionality constant and v(t) is the volume of water at time t.
objective
to determine:1. the volume v(t) of water in the tank at any time t2. the long term behavior of the water volume
mathematical formulation
the problem can be modeled by a first order ordinary differential equation. the rate of change of water volume in the tank is given by the difference between the inflow rate and the outflow rate:
dv/dt = r - kv(t)
where:- dv/dt represents the rate of change of volume with respect to time- r is the constant inflow rate (liters/minute)- kv(t) is the outflow rate, proportional to the current volume
solution method this is a linear first order differential equation of the form:
dv/dt + kv = r
we solve this using an integrating factor approach.
step 1: find the integrating factor
the integrating factor μ(t) is given by:
μ(t) = e^(∫k dt) = e^(kt)
step 2: multiply through by the integrating factor
e^(kt)(dv/dt) + ke^(kt)v = re^(kt)
the left side is now the derivative of v(t)e^(kt):
d/dt(v(t)e^(kt)) = re^(kt)
step 3: integrate both sides
∫d(v(t)e^(kt)) = ∫re^(kt) dt
v(t)e^(kt) = (r/k)e^(kt) + c
step 4: solve for v(t)
v(t) = r/k + ce^(-kt)
step 5: apply initial condition
at t = 0, v(0) = 100:
100 = r/k + c
c = 100 - r/k
final solution
the volume of water in the tank at any time t is:
v(t) = r/k + (100 - r/k)e^(-kt)
long term behavior analysis
to determine the long term behavior, we examine the limit as t approaches infinity:
lim(t→∞) v(t) = lim(t→∞) [r/k + (100 - r/k)e^(-kt)] = r/k
the exponential term e^(-kt) approaches zero as t becomes large, leaving only the constant term r/k.
interpretation of results1. the solution shows that the volume approaches a steady state value of r/k over time.2. the rate at which it approaches this steady state depends on the value of k.3. the initial condition affects the transient behavior but not the final steady state value.
verification
we can verify the solution by:1. checking that it satisfies the original differential equation2. confirming it meets the initial condition3. examining that the long term behavior makes physical senseconclusion the volume of water in the tank follows an exponential approach from its initial value of 100 liters to a final equilibrium value of r/k liters. the time constant of this approach is determined by the leakage proportionality constant k.
final answer
the volume of water in the tank at time t is:
v(t) = (r/k) + (100 - r/k)e^(-kt)
the long term volume approaches:
lim(t→∞) v(t) = r/k

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