Problem #6: A large tank contains 100 litres of water in which 25 grams of salt is dissolved. Brine containing 10 grams of saltper litre is pumped into the tank at a rate of 8 litres per minute. The well mixed solution is pumped out of the tankat a rate of 5 litres per minute.Problem #6(a):(a) Find an expression for the amount of water in the tank after t minutes.(b) Let x(t) be the amount of salt in the tank after t minutes. Which of the following is a differential equation forx(t)?(A)(E)== 8-x(1) (B)= 80-x(t) (F)dx(1) = 80Problem 6(b): Select5x(t)100+ 3tdrEnter your answer as a symbolicfunction of t, as in theseexamples=8-dx= 85 x(t)100+ St8x(t)100+ 3t(C)dx=40-x(1) (D)dt(G) = 808x(1)100+ 3rdx 40(H)dxdt45.x(t)100 + St= 40 -8.x(1)100 + St

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Answered: Problem #6: A large tank contains 100…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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A 100-L tank was initially empty, but is to be filled with salt-and-water solution. The way the solution is being poured into the tank is through a device the controls the rate at which it is being poured. The controller ensures the solution is being poured at time t (minutes) with the expression (1 + cos(2t)) in liters per minute. It follows that the amount of salt for every liter of the solution being poured at timet (minutes) is described by the expression 0.2(1 + cos(2t)) in kilograms. This implies the rate the salt-and-water solution being poured into the tank alternately gets faster and slower. a) Show how the following differential equation describe the salt concentration s in the solution at any time t. Show full solution ds 0.2(1 + cos 2t)? dt
A 100-L tank was initially empty, but is to be filled with salt-and-water solution. The way the solution is being poured into the tank is through a device the controls the rate at which it is being poured. The controller ensures the solution is being poured at time t (minutes) with the expression (1 + cos(2t)) in liters per minute. It follows that the amount of salt for every liter of the solution being poured at time t (minutes) is described by the expression 0.2(1 + cos(2t)) in kilograms. This implies the rate the salt- and-water solution being poured into the tank alternately gets faster and slower. Determine which of the following differential equation describing the salt concentration S in the solution at any time t. O A. dS dt OB. dS dt OC. dS dt O D. dS dt = 0.2(1 + cos 21) ² = = = -0.2(1 + cos 21) ² 0.2 (1 + cos 21) ² 0.2 2 (1 + cos 21) ²
5) The population of a town is represented by the formula P(t) = A(1.2)'where A is the current population and t is time. Determine, to the nearest tenth of a year, how long it would take the #6 population to reach 3000 people if the current population is 1250.
number 2
(a) Suppose we create a model of the height of a shrub (Y) based on the amount of bacteria in the soil (X1) and whether the plant is located in partial or full sun (X2). Height is measured in cm, bacteria is measured in thousand per ml of soil (so 1000 is 1 unit), and type of sun = sun and type of sun = between shrubs with a 5500 bacteria and those with a 4000 bacteria count. Assuming the sun is the 0 if the plant is in partial 1 if the plant is in full sun. What will be the height difference on average same for both. (b) Let's say after a few training steps your model is in this state: Y = 33 + 4.6 * X1+8* X2. Explain what the effect of the plant being in sun vs. partial shade will have on the average height of the shrub if there are no bacteria in the soil, based on the model.
A tank contains 1,000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L/min. Let y(t) be the amount of salt (in kg) in the tank after t minutes. (a) Find the rate of change of amount of salt in the tank after t minutes. dy dt kg min How much salt is in the tank when t = 0? y(0) How much salt (in kg) is in the tank after t minutes? y = = kg (b) How much salt (in kg) is in the tank after 50 minutes? (Round the answer to one decimal place.) y(50) = kg
Example: A tank initially contains 800 gallons of fresh water. Water containing 1/2 lb/gal of salt flows into the tank with the rate 6 gal/min and the well stirred mixture flows out at the rate of 2 gal/min. Write the initial value problem for Q(t), the amount of salt in pounds in the tank.

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