A tank contains 100 grams of a substance dissolved in a large amount of water. The tank is filtered in such a way that water drains from the tank, leaving the substance behind in the tank. Consider the volume of the dissolved substance to be negligible. At what rate is the concentration (grams/liter) of the substance changing with respect to time in each scenario?
(b) the rate at the instant when 20 liters remain, if the water is draining at 2.4 L/hr at that instant.....gl/hr=?
(c) the rate in scenario (b), if the unknown substance is also being added at a rate of 20 g/hr (and there are 100 grams in the tank at that instant)..... gl/hr=?
Try changing the equations in your earlier post to reflect the new parameters. How far do you get?
To find the rate at which the concentration (grams/liter) of the substance is changing with respect to time, we need to use the concept of differentiation. The rate of change of concentration with respect to time is given by the derivative of the concentration function.
Let's denote the concentration of the substance in the tank at any given time as C (in grams/liter) and the time as t (in hours).
In scenario (b), the volume of water is draining at a constant rate of 2.4 L/hr when 20 liters remain. This means that the volume of water in the tank can be expressed as V(t) = 20 + 2.4t liters.
Since the concentration C is defined as grams of substance per liter of water, we need to find the rate at which the concentration changes with respect to time. To do this, we differentiate the concentration function with respect to time, which gives us dC/dt.
Now, let's find the rate of change of concentration in scenario (b).
Given:
Volume of water draining, dV/dt = 2.4 L/hr
Volume of water remaining, V(t) = 20 + 2.4t liters
To find dC/dt, we can use the chain rule of differentiation. The chain rule states that if we have a function C(t) that is composed with another function V(t), then the derivative of C with respect to t is equal to the derivative of C with respect to V multiplied by the derivative of V with respect to t.
dC/dt = (dC/dV) * (dV/dt)
To find dC/dV, we need to express C in terms of V. We know that the total mass of the substance in the tank is constant at 100 grams. Since the volume of the substance is negligible, the concentration can be approximated as C = 100/V.
Taking the derivative of C with respect to V, we get:
dC/dV = -100/V^2
Now, we substitute the values into the chain rule equation:
dC/dt = (dC/dV) * (dV/dt)
dC/dt = (-100/V^2) * (dV/dt)
At the instant when 20 liters remain, we have V(t) = 20 + 2.4t. Substituting this into the equation, we have:
dC/dt = (-100/(20 + 2.4t)^2) * 2.4
When evaluating at this instant, we also need to substitute t = 0 in the equation to find the rate at the instant:
dC/dt = (-100/(20 + 2.4(0))^2) * 2.4
dC/dt = (-100/20^2) * 2.4
dC/dt = -0.6 g/L/hr
So, in scenario (b), at the instant when 20 liters remain and the water is draining at a rate of 2.4 L/hr, the concentration of the substance is changing at a rate of -0.6 grams per liter per hour.
In scenario (c), if the unknown substance is also being added at a rate of 20 g/hr, we need to consider this additional contribution to the concentration change.
The rate of change of concentration in scenario (b) with the substance being added can be found by adding the rate of addition to the rate of change in scenario (b).
dC'/dt = dC/dt + (rate of addition)
The rate of addition is given as 20 g/hr. Therefore,
dC'/dt = -0.6 + 20
dC'/dt = 19.4 g/L/hr
So, in scenario (c), at the instant when 20 liters remain, with the water draining at 2.4 L/hr and the substance being added at a rate of 20 g/hr, the concentration of the substance is changing at a rate of 19.4 grams per liter per hour.