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A tank contains 1000 L of brine with 15 kg of dissolved salt. Pure water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank (a) after $ t $ minutes and (b) after 20 minutes?

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Calculus 2 / BC

Chapter 9

Differential Equations

Section 3

Separable Equations

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Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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A tank contains 1000 L of …

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A tank contains 1000 $\mat…

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A tank initially contains …

In the problem we have part 10 contains the 100 liter brawn with 15 dissolved side. Therefore, it is like this. We are then, the later here at 1000 later over here and fiftenow. It is coming out from here, and that is therefore we have over the x equal to rate in minus rate, that is at what rate is coming inside, the tenawa. It is going on side the time so we have in that is rate in is 10 into 0 equal to 0 and out that i rate is equal to minus y over 1000 into 10 equal to minus y. Over 100 point now, this equal to d y over 18 is given as minus y over 100 point now further, we have what is equal minus y over 100 or dy over y is equal to integral minus d. Upon 100 point, therefore, it is la y, equal to minus t of 100 plus c, or we have y equals to the power minus t upon 108 point the power. This is equal to d power minton to equal. Now we have further as after dina. That is after minutes, we have y 0, equal 15 or l 15, equal minus 0 over 108 or c, equal l, n 15 point. Therefore, we have e to the power n y equal to e to the power minus t upon 100 plus l 15. Therefore, y equal, you power, minus 3 upon 100 and out power, l 15 y equal to 15 point. You bow minus t upon 100 point. So this is the answer for part of the problem. Now we have. This is part a now. We have part b of the problem where, after 20 minutes 20 is even as 15 e power minus 20 upon 100, it responds to 12.28. That is, equality with 12.3 k. This is the answer to the given problem.

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