A large mixing tank currently contains 100 gallons of water into which 6 pounds of sugar have been mixed. A tap will open pouring 20 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 10 minutes. Is that a greater concentration than at the beginning?
Write a function that models this:
C(t)=
What will the concentration be in 10 minutes?
C(10)=
Watch the amount of sugar and water:
c(t) = (6+t)/(100+20t)
Now you can answer the question.
Leg8?22o?no lice?
To model the concentration of sugar in the tank over time, we can use the rate of change of sugar in the tank.
Let C(t) represent the concentration of sugar in the tank at time t (in minutes).
The rate of change of sugar in the tank can be represented by the difference between the rate at which sugar is added and the rate at which water is added.
Given that 1 pound of sugar is added per minute and 20 gallons of water are added per minute, the rate of change of sugar in the tank is 1 pound per minute minus 0 pounds per minute (as no sugar is added with the water).
Therefore, we can write the differential equation for C(t) as follows:
C'(t) = 1 - 0
Simplifying,
C'(t) = 1
To solve this differential equation, we need an initial condition. The initial concentration of sugar in the tank is 6 pounds in 100 gallons of water, so we have:
C(0) = 6 / 100
Using this initial condition, we can solve the differential equation:
C(t) = t + C(0)
Plugging in the values, we have:
C(t) = t + 6 / 100
Now, we can find the concentration of sugar in the tank after 10 minutes by substituting t = 10 into the equation:
C(10) = 10 + 6 / 100
Simplifying,
C(10) = 10.06 pounds per gallon
Therefore, the concentration of sugar in the tank after 10 minutes is 10.06 pounds per gallon.
Yes, the concentration after 10 minutes is greater than at the beginning. At the beginning, the concentration was 6 / 100 = 0.06 pounds per gallon, while after 10 minutes, the concentration is 10.06 pounds per gallon.
To solve this problem, we need to consider the rate at which water and sugar are being added to the tank, and how the concentration of sugar changes over time.
Let's break down the problem step by step:
1. Initially, the tank contains 100 gallons of water mixed with 6 pounds of sugar. So the initial concentration of sugar is 6 pounds / 100 gallons = 0.06 pounds per gallon.
2. The tap is opened, pouring 20 gallons per minute of water into the tank. At the same time, 1 pound per minute of sugar is poured into the tank.
3. The rate of change of sugar in the tank can be represented by the equation dS(t)/dt = 1, where S(t) represents the amount of sugar in the tank at time t. This means that the amount of sugar is increasing by 1 pound per minute.
4. The volume of water in the tank is increasing by 20 gallons per minute, so the total volume of water in the tank after t minutes is 100 + 20t gallons.
5. To find the concentration of sugar in the tank at any given time t, we divide the amount of sugar (which is increasing at a constant rate) by the total volume of water in the tank. Therefore, the concentration C(t) can be represented as: C(t) = (6 + t) / (100 + 20t) pounds per gallon.
Now, let's calculate the concentration of sugar in the tank after 10 minutes:
C(10) = (6 + 10) / (100 + 20*10)
= 16 / 300
= 0.0533 pounds per gallon.
Comparing this concentration to the initial concentration of 0.06 pounds per gallon, we see that the concentration after 10 minutes is lower than the initial concentration, which means the concentration has decreased over time.