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For parts a & b below, derive only the initial value problem set up.
A tank of capacity 200 gallons holds 100 gallons of water with 25 pounds of sugar in solution. Pure water is poured into it at the rate of 2 gallons per minute, and the well-stirred mixture flows out of this tank at a rate of gallon per minute and into the second tank that initially contained 50 gallons of fresh water. The well-stirred mixture in the second tank is allowed to leave the tank at the same rate (1 gal/min).
1. Determine an Initial value problem that satisfied by the amount of sugar in the first tank (Tank A).
2. Determine an Initial value problem that satisfied by the amount of sugar in the second tank (Tank B).

Sagot :

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Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

[tex]\dfrac{dy}{dt}= (rate \ in ) - ( rate \ out )[/tex]

The rate of the sugar coming into the tank is 0

[tex]\text{rate of sugar going out }= \dfrac{y(t) \ pound}{100}\times \dfrac{1 }{min} \\ \\ = \dfrac{y}{100} \ pound/min[/tex]

[tex]So; \\ \\ \\ \dfrac{dy}{dt} = 0 - \dfrac{y}{100} \\ \\ \implies \dfrac{dy}{y }= -\dfrac{dt}{100 } \\ \\ \implies dx|y| = -\dfrac{t}{100}+C \\ \\ \implies e*{In|y|}= e^{-\dfrac{t}{100}+C} \\ \\ \implies y = Ce^{-\dfrac{t}{100}}[/tex]

Initial amount of sugar = 25 Pounds

Now; y(0) = 25

25 = Ce⁰

C = 25

So;  y(t) = 25 [tex]e^{-\dfrac{t}{100}[/tex]

Thus, the amount of sugar at any time t = [tex]\mathbf{25 e^{^{-\dfrac{t}{100}}}}[/tex]

B) For tank B :

[tex]\dfrac{dy}{dt}= 1 - \dfrac{y}{50 } \\ \\ \dfrac{dy}{dt}= \dfrac{50-y}{50} \\ \\ \dfrac{dy}{50-y }= \dfrac{dt}{50}[/tex]

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