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A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15L/min. How much salt is in the tank (a) after $ t $ minutes and (b) after one hour?
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Calculus 2 / BC
Chapter 9
Differential Equations
Section 3
Separable Equations
Oregon State University
Baylor University
University of Nottingham
Idaho State University
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 $\mat…
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A tank contains 1000 L of …
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At the d y upon d t equals 113 minus 3 y over 200 point and integration dy over 113 minus 3 y equals to integration dt over 200 point. Therefore, it is made 130 minus 3 y over marstins 3, equal 1 upon 200 t plus c. Therefore, we have minus 1 upon 310 made 130 is equal to c or minus 1 upon 3 l, mad 130 minus 3 y equal to t upon 200 minus 1 upon 3 l, n 113 point. So it is minus 1. Upon 3 l n mo 130 minus 3 y plus 1 upon 3, an not 130 equal to g. Upon 200 point now it is l, n, mad 130 minus 3 y minus y equal to minus 3. Upon 700 point, so it is further- is equals to l 130 minus 3 y over 130 equal to minus 3. Over 200 point now is equal 130 minus 3 y over 130, equal to u to the power minus 3 over 200 point or further we have is equals to minus 3 y equal to 130 ud power minus 3 over 200 minus 130 or 3 y equals to 130, minus 130 e, the power minus 3 to 100 point or y equal 130 upon 31 minus 8 power, minus 3 t upon 200 point. It is the answer for part of the problem. Now we have part b were in part b. We have they y equals 130 upon 31 minus e power. Minus 3 t upon 200 point now y o 6. This is equal to 130 upon 31 minus. U to the power it over 200 point for the population. We have y 60 equals to 25.715. That is equal to 25.72 k, which is the answer to the given problem.
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