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castortr0y [4]
3 years ago
11

A large mixing tank initially contains 1000 gallons of water in which 30 pounds of salt have been dissolved. Another brine solut

ion is pumped into the tank at the rate of 4 gallons per minute, and the resulting mixture is pumped out at the same rate. The concentration of the incoming brine solution is 2 pounds of salt per gallon. If represents the amount of salt in the tank at time t, the correct differential equation for A is:__________.A.) dA/dt = 4 - .08AB.) dA/dt = 8 -.04AC.) dA/dt = 4-.04AD.) dA/dt = 2-.04AE.) dA/dt = 8-.02A
Mathematics
1 answer:
serg [7]3 years ago
5 0

Answer:

(B)\dfrac{dA}{dt}=8-0.004A

Step-by-step explanation:

Volume of fluid in the tank =1000 gallons

Initial Amount of Salt in the tank, A(0)= 30 pounds

Incoming brine solution of concentration 2 pounds of salt per gallon is pumped in at a rate of 4 gallons per minute.

Rate In=(concentration of salt in inflow)(input rate of brine)

=(2\frac{lbs}{gal})( 4\frac{gal}{min})=8\frac{lbs}{min}

The resulting mixture is pumped out at the same rate, therefore:

Rate Out =(concentration of salt in outflow)(output rate of brine)

=(\frac{A(t)}{1000})( 4\frac{gal}{min})=\frac{A}{250}

Therefore:

The rate of change of amount of salt in the tank,

\dfrac{dA}{dt}=$Rate In-Rate out\\\dfrac{dA}{dt}=8-\dfrac{A}{250}\\\dfrac{dA}{dt}=8-0.004A

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