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Answer:
The differential equation for the amount of salt A(t) in the tank at a time t > 0 is .
Step-by-step explanation:
We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.
The concentration of the solution entering is 4 lb/gal.
Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;
where, = concentration of salt in the inflow
input rate of brine solution
and = concentration of salt in the outflow
outflow rate of brine solution
So, = 4 lb/gal
3 gal/min = 12 lb/gal
Now, the rate of accumulation = Rate of input of solution - Rate of output of solution
= 3 gal/min - 2 gal/min
= 1 gal/min.
It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.
So, = concentration of salt in the outflow
outflow rate of brine solution
= =
Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;
=
or .
Answer:
The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.
Step-by-step explanation:
The area of a sphere is given by the following formula:
In which A is the area, measured in cm², and r is the radius, measured in cm.
Assume that the radius r of a sphere is expanding at a rate of 40 cm/min.
This means that
Determine the rate of change in surface area when r = 20 cm.
This is when
. So
Applying implicit differentiation.
We have two variables, A and r, so:
The rate of change in surface area when r = 20 cm is 20,106.19 cm²/min.